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![**Section 6**
Difference between notation of discrete and continuous signals
In signal processing systems, continuous-time signals usually use the symbol 𝑡 for their
independent variables and use parentheses but discrete-time signals use the symbol n for their
independent variables and use square brackets.
Examples of continuous-time signals: 𝑦(𝑡), 𝑥(𝑡)
Examples of discrete-time signals: f[n], x[n]
HINT
In the course reference, there are no problems for just the operations on dependent variables
but what is found is a combination of operations on both dependent and independent variables.
So we can put some examples for just operations on dependent variables on both continuous-
time and discrete-time signals.
There is also a hand-created problem for each of these operations. So there are a total of 3
problems for each operation.
Amplitude scaling:
1. Fig. 1.52 – ℎ1: 2𝑥(𝑡)
2. Fig. 1.52 – ℎ2: 3𝑦(𝑡)
3. Fig. 1.56 – 𝑙1: . 3𝑥[𝑛]
4. Fig. 1.56 – 𝑙2: 1.5𝑦[𝑛]
Addition:
1. Fig. 1.52 – ℎ3: 𝑥(𝑡) + 𝑦(𝑡)
2. Fig. 1.56 – 𝑙3: 𝑥[𝑛] + 𝑦[𝑛]
Multiplication:
1. Fig. 1.52 – ℎ4: 𝑥(𝑡)𝑦(𝑡)
2. Fig. 1.56 – 𝑙4: 𝑥[𝑛]𝑦[𝑛]
Basic Operations on Signals
Before diving into the operations that can be performed over the signals it is good to know the
difference between independent and dependent variables.
For example, list the independent and dependent variables in the following signal:
z = f(x, y) = x + y + a
Here x and y are the independent variables because their values not depend on any factor but z
is the dependent variable because its value depends on both x and y.
Operations on dependent variables
Operations performed on dependent variables not affect just a single independent variable but
affects all independent variables.
So all of these operations just change the dependent variable so they don`t change the
dependent variables. The independent variable is graphed over the x-axis and the deponent
variable is graphed over the y-axis. So no change in the x-axis values before and after scaling but
the change will be in the y-axis values.
Amplitude scaling
Suppose there is a continuous-time signal 𝑥(𝑡) scaled by a scaling factor c, the resulting signal
𝑦(𝑡) is as follows:](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-12-2048.jpg)
![𝑦(𝑡) = 𝑐𝑥(𝑡)
𝑥(𝑡) the original signal, c scaling factor, 𝑦(𝑡) the scaled signal
Scaling for discrete-time signals is performed in a similar manner to continuous-time signals:
y[n] = cx[n]
Example 6.1:
𝒙(𝒕) = 𝒕 + 2 for -3 ≤ 𝑡 ≤ 3
Perform amplitude scaling for x(t) to get y(t) by a scaling factor 2 and graph the two signals.
𝑦(𝑡) = 𝑐𝑥(𝑡)
For c = 2
𝑦(𝑡) = 2𝑥(𝑡)
𝑦(𝑡) = 2(𝑡 +2) = 2𝑡 +4
Amplitude scaling:
1. Fig. 1.52 – ℎ1: 2𝑥(𝑡)
2. Fig. 1.52 – ℎ2: 3𝑦(𝑡)
3. Fig. 1.56 – 𝑙1: . 3𝑥[𝑛]
4. Fig. 1.56 – 𝑙2: 1.5𝑦[𝑛]
Addition
For the continuous-time signals 𝑥1(𝑡) and 𝑥2(𝑡) the output signal 𝑦(𝑡) is obtained by adding
these two signals as follows:
𝑦(𝑡) = 𝑥1(𝑡) + 𝑥2(𝑡)
This is similar to adding discrete-time signals:
y[n] = x1[n] + x2[n]
Example 6.2:
𝑥1(𝑡) = 𝑡
𝑥2(𝑡) = 𝑡 + 2
Using graph find 𝒚(𝒕) = 𝑥1(𝑡) + 𝑥2(𝑡) and graph the result signal where -3 ≤ 𝑡 ≤ 3.
Addition:
1. Fig. 1.52 – ℎ3: 𝑥(𝑡) + 𝑦(𝑡)
2. Fig. 1.56 – 𝑙3: 𝑥[𝑛] + 𝑦[𝑛]
Multiplication
For the continuous-time signals 𝑥1(𝑡) and 𝑥2(𝑡) the output signal 𝑦(𝑡)) is obtained by
multiplying these two signals as follows:
𝑦(𝑡) = 𝑥1(𝑡)𝑥2(𝑡)
This is similar to multiplying discrete-time signals:
y[n] = x1[n]x2[n]
Example 6.3:
Multiply the previous two signals and graph the three signals (two inputs + one output).
𝑥1(𝑡) = 𝑡
𝑥2(𝑡) = 𝑡 + 2](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-13-2048.jpg)
![Multiplication:
1. Fig. 1.52 – ℎ4: 𝑥(𝑡)𝑦(𝑡)
2. Fig. 1.56 – 𝑙4: 𝑥[𝑛]𝑦[𝑛]
Question:
𝒙(𝒕) = 𝒕 +3 where -5 ≤ 𝑡 ≤ 5
From x(t) how to return a new signal of the form using the basic operations on signals. Explain.
𝑦(𝑡) = {
𝑥(𝑡), − 3 ≤ 𝑡 ≤ 3
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
How to return a new signal that is identical to 𝑥(𝑡) for -3 ≤ 𝑡 ≤ 3 and have zero otherwise.
Operations on independent variables
These operations will of course affect the independent variables and the independent variables
and graphed over the x-axis so they affect the x-axis.
Time Scaling
For the continuous-time signal 𝑥(𝑡), the output signal 𝑦(𝑡) is obtained by scaling the
independent variable t by a scaling factor c as follows:
𝑦(𝑡) = 𝑥(𝑎𝑡)
If a > 1 then the signal get compressed and if 0 < a < 1 the signal get expanded (stretched).
This also holds for discrete-time signals:
𝑦[𝑛] = 𝑥[𝑘𝑛]
Example 6.4:
Draw the following continuous-time signal 𝒙(𝒕) based on its inputs and outputs:
X -2 -1 0 1 2
Y 0 1 2 1 0
Find the time-scaled version of that signal 𝒚(𝒕) for the following values for the scaling factor:
3
.6=3/5
Book Problems:
1.56 (a)
Question:
Why when multiplying by a value > 1 the signal gets compressed and when multiplying by a
value < 1 the signal gets expanded?
This can be explained using the unit function.
The following unit function is defined in the range -1 ≤ 𝑡 ≤ 1:](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-14-2048.jpg)
![It can be rewritten in a more representative form as follows:
𝑥(𝑡) = {
1, − 1 ≤ 𝑡 ≤ 1
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
When scaling that function by a scaling factor > 0 like 2 it will be as follows by just replacing each
𝑡 by 2𝑡:
𝑥(2𝑡) = {
1, − 1 ≤ 2𝑡 ≤ 1
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
But it is more convenient to separate the independent variables in an equation from other
factors. Here 2 and the independent variable 𝑡 are in the same side.
Divide by 2 to make the equation has this form:
𝑥(2𝑡) = { 1, −
1
2
≤ 𝑡 ≤
1
2
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
From this equation it is apparent that the range got compressed from -1:1 to -
1
2
:
1
2
.
From the last equation it is shown that when multiplying by a scaling factor greater than 1 like 2
the signal got compressed.
We can try multiplying by a factor less than 1 like
1
2
.
𝑥(𝑡) = {
1, − 1 ≤ 𝑡 ≤ 1
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
𝑥 (
𝑡
2
) = {
1, − 1 ≤
𝑡
2
≤ 1
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
𝑥 (
𝑡
2
) = {
1, − 2 ≤ 𝑡 ≤ 2
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
From the last equation it is shown that when multiplying by a scaling factor less than 1 like
1
2
the
signal gets expanded.
So we can have a rule that says if the scaling factor > 1 then the signal get compressed and if the
factor is < 1 the signal get expanded.
Reflection:
Reflection of a continuous-time signal 𝑥(𝑡) is obtained by replacing the time t by –t to generate
the reflected version of 𝑥(𝑡) which is 𝑦(𝑡).
𝑦(𝑡) = 𝑥(-t)
This is similar to discrete-time signals:
𝑦[𝑛] = 𝑥[-n]
Book Problems:](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-15-2048.jpg)
![1. Fig. 1.56 – 𝑙5: 𝑥[−𝑛]
2. Fig. 1.56 – 𝑙6: 𝑦[−𝑛]
Time Shifting:
For the continuous-time signal 𝑥(𝑡) the time-shifted version 𝑦(𝑡) is defined as:
𝑦(𝑡) = 𝑥(𝑡 + 𝑡 𝑜)
Where 𝑡 𝑜 is the time shift.
If 𝒕 𝒐> 0 then the signal goes to the right and if 𝒕 𝒐< 0 the signal goes to the left.
In case of a discrete-time signal 𝑥[𝑛], the time-shifted version is:
𝑦[𝑛] = 𝑥[𝑛 − 𝑚]
Where m is a positive or negative integer.
Book Problems:
1. Fig. 1.56 – 𝑙5: 𝑥[𝑛 + 2]
2. Fig. 1.56 – 𝑙6: 𝑦[𝑛 − 3]
Question:
When subtracting a value from the independent variable 𝒕, how the signal go to the right not
the left?
Suppose that there are two devices (𝐷1 and 𝐷2) generating signals at the same time 𝑡 which is
for example 12PM. But we need 𝐷2 to generate its signal after 𝐷1 signal by a delay of 2 minutes.
𝐷1-> S1 -> t -> 12 PM
𝐷2-> S2 -> t -> DELAY -> 𝒕 𝒐 -> 2 min -> 12PM + 2 MIN
Finally we have:
𝑡 = 12𝑃𝑀
𝑡 − 𝑡 𝑜 = 12𝑃𝑀
𝑡 − 2𝑀𝐼𝑁 = 12𝑃𝑀
𝑡 = 12𝑃𝑀 + 2𝑀𝐼𝑁
Thus time for 𝐷2 signal got delayed by 2 minutes when subtracting 2 from the time variable 𝑡.
Another better explanation:
For the following unit step function:
It has the following equation:
𝑥(𝑡) = {
1, − 1 ≤ 𝑡 ≤ 1
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
When the signal is to be shifted in time by value +2 the new equation will be as follows:
𝑥(𝑡 − 2) = {
1, − 1 ≤ 𝑡 − 2 ≤ 1
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-16-2048.jpg)
![**Section 7**
Elementary Signals in Digital Signal Processing
There are some basic signals in the field of digital signal processing that has dominant features.
These signals are important by themselves because they reflect some already existing physical
signals but also they are useful to:
Construct signals based on these basic signals
Process and apply operations over signals based on such basic signals
Because signals can be either discrete or continuous we should know:
1. Algebraic equation of continuous and discrete forms of each basic signal
2. Graph of the continuous and discrete forms of the signals
Exponential Signal
One important factor about such exponential signal is whether it is increasing or decreasing
(growing or decaying).
Continuous
𝑥(𝑡) = 𝐵𝑒 𝑎𝑡
𝑎 < 0 𝐷𝑒𝑐𝑎𝑦𝑖𝑛𝑔 𝐸𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙 𝑆𝑖𝑔𝑛𝑎𝑙
𝑎 > 0 𝐺𝑟𝑜𝑤𝑖𝑛𝑔 𝐸𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙 𝑆𝑖𝑔𝑛𝑎𝑙
a=-6, B=5 a=5, B=1
Note that e=2.7 is a number greater than 1.
When a number > 1 is raised to a power < 0 it will decrease its value by increasing the
magnitude of the power. When the power is > 0 it will increase the resultant value by increasing
the power magnitude.
In our example e is greater than 1 and when a > 0 the result will be an increasing signal and
when a < 0 result will have the form of a decreasing signal.
Discrete
𝑥[𝑛] = 𝐵𝑒 𝑎𝑛
𝑎 < 0 𝐷𝑒𝑐𝑎𝑦𝑖𝑛𝑔 𝐸𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙 𝑆𝑖𝑔𝑛𝑎𝑙
𝑎 > 0 𝐺𝑟𝑜𝑤𝑖𝑛𝑔 𝐸𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙 𝑆𝑖𝑔𝑛𝑎𝑙
a=-6, B=5 a=5, B=1](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-18-2048.jpg)
![The same explanation why x[n] is sometimes increasing and others decreasing is the same as the
continuous form.
Sinusoidal Signal
Continuous
𝑥(𝑡) = 𝐴𝑐𝑜𝑠(𝜔𝑡 + 𝛷)
𝑥(𝑡) = 𝐴𝑠𝑖𝑛(𝜔𝑡 + 𝛷)
Where 𝜔 = 2𝜋𝑓
Remember that for a signal to be periodic this condition must be valid:
𝑥(𝑡) = 𝑥(𝑡 + 𝑇)
Discrete
𝑥[𝑛] = 𝐴𝑐𝑜𝑠(𝛺𝑛 + 𝛷)
𝑥[𝑛] = 𝐴𝑠𝑖𝑛(𝛺𝑛 + 𝛷)
Where 𝛺𝑁 = 2𝜋𝑚
Discrete-Time Signal Periodicity
For a discrete-time signal like:
𝑥[𝑛] = 𝐴𝑐𝑜𝑠(𝛺𝑛 + 𝛷)
𝑥[𝑛] = 𝐴𝑠𝑖𝑛(𝛺𝑛 + 𝛷)
Where 𝛺𝑁 = 2𝜋𝑚
It will be periodic only if both m and N are integers.
Examples:
o 1.16 – p56-39
o 1.17-a, c – p56-39
o 1.18-a, d – p56-39
Exponentially Damped Sinusoidal
Unit-Step Signal
Continuous](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-19-2048.jpg)
![Discrete
Unit-Impulse Signal
Discrete
The unit-impulse signal is also called delta and δ is called delta.
The unit-impulse signal can be calculated from the unit-step signal.
𝛅[ 𝑛] = 𝑢[ 𝑛] − 𝑢[ 𝑛 − 1]
Ramp Signal
Continuous
Discrete
Note:
The ramp signal can be calculated based on the unit-step signal.
The ramp signal output is always equal to its input. So we can represent the ramp signal with
this equation:](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-20-2048.jpg)
![𝑟(𝑡) = 𝑡
It will have t this graph:
But it is just defined at 𝑡 ≥ 0.
So how to make the negative part of the ramp signal equal to zero?
We apply some of the basic operations to remove the negative part of that result.
The simplest is by using multiplication: because 𝑢(𝑡) signal is only defined at the positive part
and zero we can multiply it by the signal 𝑟(𝑡) = 𝑡 to return the ramp signal.
So it will be as follows for continuous and discrete forms:
𝑟(𝑡) = 𝑡𝑢(𝑡)
𝑟[𝑛] = 𝑛𝑢[𝑛]
Basic Operations on the elementary signals
Examples:
o Show step by step how to create a unit-impulse signal using just unit-step signals.
o Show step by step how to create a ramp signal using unit-step signals.
o How to create the signal shown in the following graphs using just the elementary
signals:
A B
A B
Figure. P1
Solutions:
1. 𝑢(𝑡 + 2) − 𝑢(𝑡 − 2) + 𝑢(𝑡 + 1) − 𝑢(𝑡 − 1)
2. 2δ[ 𝑛] + 4δ[ 𝑛 + 1] − δ[ 𝑛 + 2] − 5δ[ 𝑛 − 1] − 2δ[ 𝑛 − 2]
Advanced problems:](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-21-2048.jpg)
![ How to represent the signal shown in P1-B using just unit-step signals.
At first the signal is directly represented using unit-impulse signals. Unit impulse signals can be
represented using just the unit-step signals.
So solution steps are:
1. Represent the signal using just unit-impulse signals.
2. Replace each unit-impulse signals by its equivalent unit-step signals.
Solution is:
δ[ 𝑛] = 𝑢[ 𝑛] − 𝑢[𝑛 − 1]
δ[ 𝑛 + 1] = 𝑢[ 𝑛 + 1] − 𝑢[𝑛]
δ[ 𝑛 + 2] = 𝑢[ 𝑛 + 2] − 𝑢[𝑛 + 1]
δ[ 𝑛 − 1] = 𝑢[ 𝑛 − 1] − 𝑢[𝑛 − 2]
δ[ 𝑛 − 2] = 𝑢[ 𝑛 − 2] − 𝑢[𝑛 − 3]
The final equation is:
2( 𝑢[ 𝑛] − 𝑢[ 𝑛 − 1]) + 4( 𝑢[ 𝑛 + 1] − 𝑢[ 𝑛]) − ( 𝑢[ 𝑛 + 2] − 𝑢[ 𝑛 + 1])
− 5( 𝑢[ 𝑛 − 1] − 𝑢[ 𝑛 − 2]) − 2( 𝑢[ 𝑛 − 2] − 𝑢[𝑛 − 3])
Book Problems:
o 1.54-a, d – p106-89
Some operations will involve just a single type of the above basic signals and others may contain
more than one type.
Interconnection of operations
o Book examples
o 1.12 – p70-53
o Book problems
o 1.25-p71-54
It is simpler to represent the system as a graph rather than an equation.
System operations can be connected using two ways:
1. Parallel: Each operation takes a separate path and implemented by a separate circuit.
This is simple but adds more costs to create a separate circuit for each operation.
2. Cascade: The single operation is used multiple times. This saves costs but adds more
complexity to reuse the same circuit.](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-22-2048.jpg)
![there is another input 𝒙 𝟐(𝒕) generating output 𝒚 𝟐(𝒕) then 𝒙 𝟏( 𝒕) + 𝒙 𝒚(𝒕) will
generate 𝒚 𝟏( 𝒕) + 𝒚 𝟐(𝒕).
But what is the goal of determining whether a system is linear or not?
As we previously discussed that a linear signal changes by a fixed amount. This will help us
predict future values of the signal. If a linear signal accepts inputs 1, 2, and 3 to generate
outputs 10, 20, and 30 then we can predict future values like input 4 that will generate output
40. We can predict output because the system is linear and changes by a fixed amount each
time.
To prove that a signal is linear or not we can follow two ways:
1. Algebraic
2. Graph
For each signal to be linear or non-linear it must be both homogenous and additive.
At first we need to prove that it is homogenous.
So for an input 𝒙(𝒕) the output is 𝒙(𝒕) 𝟐
.
For that system to be homogenous then scaling the input by a will also scale the output by the
same scale.
So the input will be 𝒂𝒙(𝒕) and the output is 𝒂 𝟐
𝒙(𝒕) 𝟐
Because the output was not scaled by the same scale as the input the signal is not homogenous.
Because at least one condition was not met then the signal is not linear.
But we can try to find whether the signal is additive or not.
For these two input signals:
𝒙 𝟏(𝒕) => 𝒚 𝟏(𝒕) 𝟐
𝒙 𝟐(𝒕) => 𝒚 𝟐(𝒕) 𝟐
The signal will be additive if we 𝒙 𝟏(𝒕) + 𝒙 𝟐(𝒕) will generate the sum of their outputs which is
𝒚 𝟏(𝒕) 𝟐
+ 𝒚 𝟐(𝒕) 𝟐
The output for the input 𝒙 𝟏(𝒕) + 𝒙 𝟐(𝒕) is:
[𝒙 𝟏(𝒕) + 𝒙 𝟐(𝒕)] 𝟐
𝒙 𝟏(𝒕) 𝟐
+ 2𝒙 𝟏(𝒕)𝒙 𝟐(𝒕) + 𝒙 𝟐(𝒕) 𝟐
Because the output is not equal to the sum of the individual outputs we conclude that the signal
is not additive.
Examples:
8.1.1. 𝑦(t) = x(t)
8.1.2. y(t) = x(t − 5)
8.1.3. y(t) = x(t + 2) + x(3 − t)
8.1.4. y(t) = x(2t − 1)
8.1.5. y(t) = x(t2
)
8.1.6. y(t) = x2(t) = [x(t)]2
8.1.7. y(t) = 2x(t) + 4
8.1.8. y(t) = 2x(t) + 4t
8.1.9. y(t) = x(t)cos(t)
Answers:
8.1.1. Linear, Time invariant, causal, memoryless, stable](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-25-2048.jpg)
![8.1.2. Linear, Time invariant, causal, memory, stable
8.1.3. Linear, Time variant, non-causal, memory, stable
8.1.4. Linear, Time variant, non-causal, memory, stable
8.1.5. Linear, Time variant, non-causal, memory, stable
8.1.6. Non-linear, Time invariant, causal, memoryless, stable
8.1.7. Non-linear, Time invariant, causal, memoryless, stable
8.1.8. Non-linear, Time variant, causal, memoryless, stable
8.1.9. Linear, Time variant, causal, memoryless, stable
8.1.7. Solution
Homogenous:
Multiply x(t) by a factor say a to be ax(t). Then the result will be 2ax(t) + 4
The output y(t) when get multiplied by factor a is a[2x(t) + 4] = 2ax(t) + 4𝑎
Because scaling the input by a factor a does not scale the output by the same scale then the
system is not homogenous. Because at least of the two properties (homogenous & additivity) is
not met then the system is non-linear.
Additivity:
Say there are two signals 𝑥1(t) and 𝑥2(t) generating outputs 𝑦1(t) and 𝑦2(t).
The sum of the two input signals which is 𝑥1(t) + 𝑥2(t) will be the new input to the system. So
the output will be:
𝑦(𝑥1(t) + 𝑥2(t)) = 2(𝑥1(t) + 𝑥2(t)) + 4. Eq1
The sum of the outputs is:
𝑦1(t) = 2𝑥1(t) + 4
𝑦2(t) = 2𝑥2(t) + 4
𝑦1(t) + 𝑦2(t) = (2𝑥1(t) + 4) + (2𝑥2(t) + 4). Eq2
Because Eq1 and Eq2 are not equal then the system is not additive.
Because the system is neither homogenous nor additive then it is non-linear.
8.1.9. Solution
Homogenous:
Multiply x(t) by a factor say a to be ax(t). Then the result will be ax(t)cos(𝑡).
The output y(t) when get multiplied by factor a is a[x(t)cos(𝑡)].
Because scaling the input by a factor a does scale the output by the same scale then the system
is homogenous.
Additivity:
Say there are two signals 𝑥1(t) and 𝑥2(t) generating outputs 𝑦1(t) and 𝑦2(t).
The sum of the two input signals which is 𝑥1(t) + 𝑥2(t) will be the new input to the system. So
the output will be:
𝑦(𝑥1(t) + 𝑥2(t)) = (𝑥1(t) + 𝑥2(t))cos(𝑡). Eq1
The sum of the outputs is:
𝑦1(t) = 𝑥1(t)cos(𝑡)
𝑦2(t) = 𝑥2(t)cos(𝑡)
𝑦1(t) + 𝑦2(t) = 𝑥1(t) cos(𝑡) + 𝑥2(t)cos(𝑡) Eq2](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-26-2048.jpg)
![Because Eq1 and Eq2 are equal then the system is additive.
Because the system is homogenous and additive then it is linear.
Time Invariance
A system is called time invariant if it does not change over time. But a time variant system does
change.
So the time invariant system responds identically regardless of when the input was applied. Soa
time invariant system does not change by shift (time delay or advance).
If the input was applied at 5PM and generated an output y, so if it was delayed to 4PM or
moved forward in time to 6PM the results will be the same.
If the system was time invariant so we can know the signal behavior in the past and future
because it will get the same output whatever the time is.
To prove that the system is time invariant make sure that if it generates an output y at time 𝒕 𝟏
then it will generate the same output at different times 𝒕 𝟐. That is if:
𝒚(𝒕) = 𝒙(𝒕)
And the input was shifted by amount d to be 𝒙(𝒕 − 𝒅) then the same output will be generated
but shifted by the same amount d to be 𝒚(𝒕 − 𝒅).
Steps to prove that a system is time invariant or not:
1. Shifting input be replacing each 𝒙(𝒕) by 𝒙(𝒕 − 𝒅) to generate an output 𝒚 𝒅(𝒕).
2. Shifting the output by replacing each 𝒕 by 𝒕 − 𝒅 to generate an output 𝒚(𝒕 − 𝒅).
3. If 𝒚 𝒅(𝒕) = 𝒚(𝒕 − 𝒅) then the system is time invariant.
Examples:
Which of the following systems are time invariant?
Examples:
8.2.1. 𝑦(t) = x(t)
8.2.2. y(t) = x(t − 5)
8.2.3. y(t) = x(t + 2) + x(3 − t)
8.2.4. y(t) = x(2t − 1)
8.2.5. y(t) = x(t2
)
8.2.6. y(t) = x2(t) = [x(t)]2
8.2.7. y(t) = 2x(t) + 4
8.2.8. y(t) = 2x(t) + 4t
8.2.9. y(t) = x(t)cos(t)
Answers:
8.2.1. Linear, Time invariant, causal, memoryless, stable
8.2.2. Linear, Time invariant, causal, memory, stable
8.2.3. Linear, Time variant, non-causal, memory, stable
8.2.4. Linear, Time variant, non-causal, memory, stable
8.2.5. Linear, Time variant, non-causal, memory, stable
8.2.6. Non-linear, Time invariant, causal, memoryless, stable
8.2.7. Non-linear, Time invariant, causal, memoryless, stable](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-27-2048.jpg)
![A system is said to be BIBO stable if for every set of bounded inputs there is a set of bounded
outputs. That is the system is called BIBO (Bounded-Input, Bounded-Output) and BIBO means
that if the number of inputs to the system was less than infinity then the number of outputs
must be less than infinity.
The following conditions must hold:
Where 𝑀 𝑦 and 𝑀 𝑥 are finite positive integers.
Why stable system is important?
A stable system is better than unstable system because stable systems are predictable and we
can know how it can behave compared to unstable systems.
So if we applied a set of inputs we can predict the system behavior and know its outputs.
This is similar to a programmer that writes code and knows the behavior of the code. For an
input x there will be output y compared to another programmer that writes code and doesn`t
know how it can behave responding to an input x.
If there is another input z, the first programmer is stable and can predict the output responding
to the applied input but the second one is unstable because we can`t predict the output.
Another example is about a car. A stable car system will increase the car speed if the input was
pressing the gas pedal.
Unstable car will not just increase the car speed responding to pressing the gas pedal but can
make unpredicted behavior like opening the windows, closing the air conditioner, firing the
horn.
Examples:
8.3.1. 𝑦(t) = x(t)
8.3.2. y(t) = x(t − 5)
8.3.3. y(t) = x(t + 2) + x(3 − t)
8.3.4. y(t) = x(2t − 1)
8.3.5. y(t) = x(t2
)
8.3.6. y(t) = x2(t) = [x(t)]2
8.3.7. y(t) = 2x(t) + 4
8.3.8. y(t) = 2x(t) + 4t
8.3.9. y(t) = x(t)cos(t)
Answers:
8.3.1. Linear, Time invariant, causal, memoryless, stable
8.3.2. Linear, Time invariant, causal, memory, stable
8.3.3. Linear, Time variant, non-causal, memory, stable
8.3.4. Linear, Time variant, non-causal, memory, stable
8.3.5. Linear, Time variant, non-causal, memory, stable
8.3.6. Non-linear, Time invariant, causal, memoryless, stable
8.3.7. Non-linear, Time invariant, causal, memoryless, stable](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-29-2048.jpg)
![8.3.8. Non-linear, Time variant, causal, memoryless, stable
8.3.9. Linear, Time variant, causal, memoryless, stable
Causality
A system is called to be causal if it depends on the present and past values of the input signal
but not on the future values.
Real-time systems are causal because they depend on the present and the past but not on the
future.
A system is called non-causal if it depends on at least one value from the future.
Examples:
8.4.1. 𝑦(t) = x(t)
8.4.2. y(t) = x(t − 5)
8.4.3. y(t) = x(t + 2) + x(3 − t)
8.4.4. y(t) = x(2t − 1)
8.4.5. y(t) = x(t2
)
8.4.6. y(t) = x2(t) = [x(t)]2
8.4.7. y(t) = 2x(t) + 4
8.4.8. y(t) = 2x(t) + 4t
8.4.9. y(t) = x(t)cos(t)
Answers:
8.4.1. Linear, Time invariant, causal, memoryless, stable
8.4.2. Linear, Time invariant, causal, memory, stable
8.4.3. Linear, Time variant, non-causal, memory, stable
8.4.4. Linear, Time variant, non-causal, memory, stable
8.4.5. Linear, Time variant, non-causal, memory, stable
8.4.6. Non-linear, Time invariant, causal, memoryless, stable
8.4.7. Non-linear, Time invariant, causal, memoryless, stable
8.4.8. Non-linear, Time variant, causal, memoryless, stable
8.4.9. Linear, Time variant, causal, memoryless, stable
Memory
A system is said to have memory if it depends on past or future values of the input signal.
A system is said to be memoryless if it depends only on the present values.
Examples:
8.5.1. 𝑦(t) = x(t)
8.5.2. y(t) = x(t − 5)
8.5.3. y(t) = x(t + 2) + x(3 − t)
8.5.4. y(t) = x(2t − 1)
8.5.5. y(t) = x(t2
)
8.5.6. y(t) = x2(t) = [x(t)]2
8.5.7. y(t) = 2x(t) + 4
8.5.8. y(t) = 2x(t) + 4t
8.5.9. y(t) = x(t)cos(t)
Answers:](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-30-2048.jpg)
![**Section 10**
Summation Rules
∑ 𝛼 𝑘
𝑁
𝑘=𝑚
=
𝛼 𝑚
− 𝛼 𝑁+1
1 − 𝛼
∑ 𝛼 𝑘
∞
𝑘=𝑚
=
𝛼 𝑚
1 − 𝛼
∑ 𝑐
𝑁
𝑘=𝑚
= 𝑐(𝑁 − 𝑚 + 1)
Convolution Sum
𝑦[ 𝑛] = ∑ 𝑥( 𝑘)ℎ(𝑛 − 𝑘)
∞
−∞
Note:
For ℎ[n], n is replaced by 𝑛 − 𝑘. So ℎ[−n] will be ℎ[−(n − k)].
Find the discrete-time convolution sum given the input signals and system impulse response in
the following examples:
Sheet
2.33-a (p200-183)
2.33-b (p200-183)
2.34-a (p201-184)
Sheet-2.33-a (p200-183)
𝑦[𝑛] = 𝑢[𝑛 + 3) ∗ 𝑢[𝑛 − 3]
𝑥[𝑛] = 𝑢[𝑛 + 3), ℎ[𝑛] = 𝑢[𝑛 − 3]
The first step to solve the convolution problems is to write the correct formula you are to use:
𝑦[𝑛] = ∑ 𝑥(𝑘)ℎ(𝑛 − 𝑘)
∞
−∞
𝑦[𝑛] = ∑ 𝑢(𝑘 + 3)𝑢(𝑛 − 𝑘 − 3)
∞
−∞](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-32-2048.jpg)
![At this step, the summation is defined from −∞ to ∞. So we have to substitute in 𝑢(𝑘 +
3)𝑢(𝑛 − 𝑘 − 3) by values like −∞, …, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, … , ∞.
Remember that 𝑢(𝑘 + 3) is only defined at (𝑘 + 3) > 0 and this requires k to be in this range -
3, -2, -1, 0, 1, 2, 3, … , ∞.
𝑢(𝑛 − 𝑘 − 3) is only defined at 𝑛 − 𝑘 − 3 ≥ 0 == 𝑘 ≤ 𝑛 − 3 and this requires k to be in this
range 3, 4, 5, 6, … , ∞.
So why we substitute by values like -5 where nothing of these signals are defined! Ata value like
-5 both of these equations gives 0 and we don`t have to add 0 to the summation and loses time.
The good solution is what saves time by not adding zeros to the summation.
To overcome this problem, the range of the summation should change to just start and end at
values defined by the signals.
The first part 𝑢(𝑘 + 3) is just defined at 𝑘 ≥ −3 and the second part 𝑢(𝑛 − 𝑘 − 3) is just
defined at 𝑘 ≤ 𝑛 − 3. The final range at which at least one of these equations are defined is
from −3 to 𝑛 − 3.
So the summation should change to reflect the new range. AT first change the lower end of the
summation to -3 as shown in the next equation:
𝑦[𝑛] = ∑ 1. 𝑢(𝑛 − 𝑘 − 3)
∞
−3
When the range starts at -3 the signal 𝑢(𝑘 + 3) will always be 1. So we can replace it by the
constant 1.
Similarly, when the higher end of the summation is replace to be 𝑛 − 3 the summation will have
tis form:
𝑦[𝑛] = ∑ 1.1
𝑛−3
−3
When the range ends at 𝑛 − 3 the signal 𝑢(𝑛 − 𝑘 − 3) will always be 1. So we can replace it by
the constant 1.
Finally the summation will be as follows:
𝑦[𝑛] = ∑ 1
𝑛−3
−3
Using this summation rule:
∑ 𝑐
𝑁
𝑘=𝑚
= 𝑐(𝑁 − 𝑚 + 1)
So
∑ 1
𝑛−3
𝑘=−3
= 1. ((𝑛 − 3) − (−3) + 1) = 𝑛 + 1
The final equation of the solution is as follows:
𝑦[𝑛] = {
𝑛 + 1, 𝑛 ≥ 0
0, 𝑛 < 0
Sheet-2.33-b (p200-183)
𝑦[𝑛] = 3 𝑛
𝑢[−𝑛 + 3] ∗ 𝑢[𝑛 − 2]
Solution:](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-33-2048.jpg)
![𝑦[𝑛] = ∑ 𝑥[𝑘]ℎ[𝑛 − 𝑘]
∞
−∞
𝑦[𝑛] = ∑ 3 𝑘
𝑢[−𝑘 + 3] ∗ 𝑢[𝑛 − 𝑘 − 2]
∞
−∞
Next is to find the range of the summation.
𝑢[−𝑘 + 3] => 𝑘 ≤ 3
𝑢[𝑛 − 𝑘 − 2] => 𝑘 ≤ 𝑛 − 2
The problem is having two upper limits 𝑘 ≤ 3 & 𝑘 ≤ 𝑛 − 2.
The solution to this ambiguous situation is as follows:
For 𝑛 − 2 ≤ 3 == 𝑛 ≤ 5, the upper limit is 𝑛 − 2 and the summation is:
𝑦[𝑛] = ∑ 3 𝑘
𝑛−2
−∞
=
3 𝑛−2+1
1 − 3
=
3 𝑛−1
2
=
3 𝑛
. 3−1
2
=
3 𝑛
3
2
=
3 𝑛
6
For 𝑛 − 2 ≥ 4 == 𝑛 ≥ 6, the upper limit is 3 and the summation is:
𝑦[𝑛] = ∑ 3 𝑘
3
−∞
=
33+1
1 − 3
=
34
2
=
81
2
Final result is:
𝑦[𝑛] = {
3 𝑛
6
, 𝑛 ≤ 5
81
2
, 𝑛 ≥ 6
Sheet-2.34-a (p201-184)
𝑦[n] = x[n] ∗ 𝑧[𝑛]
In this problem, you are given a graph of two signals and want to find the convolution between
them.
The first step in the solution is to find the algebraic equation that represents each of these
graphs.
𝑥[n] = u[n + 5] − 𝑢[𝑛 − 1]
𝑧[n] = u[n] − 𝑢[𝑛 − 8] + 𝑢[𝑛 − 4] − 𝑢[𝑛 − 8] = u[n] − 2𝑢[𝑛 − 8] + 𝑢[𝑛 − 4]
The impulse response here is ℎ[n] = 𝑧[𝑛].](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-34-2048.jpg)
![Then find the convolution between these equations. Because these signals are discrete so we
will find the convolution sum using this equation:
𝑦[𝑛] = ∑ 𝑥(𝑘)ℎ(𝑛 − 𝑘)
∞
−∞
By substituting in the summation:
𝑦[𝑛] = ∑(u[k + 5] − 𝑢[𝑘 − 1]) ∗ (u[n − 𝑘] − 2𝑢[𝑛 − 𝑘 − 8] + 𝑢[𝑛 − 𝑘 − 4])
∞
−∞
The equation can be represented as follows:
𝑦[𝑛] = ∑ u[k + 5]u[n − 𝑘] − 2 ∑ u[k + 5]𝑢[𝑛 − 𝑘 − 8]
∞
−∞
+ ∑ u[k + 5]𝑢[𝑛 − 𝑘 − 4]
∞
−∞
∞
−∞
− ∑ 𝑢[𝑘 − 1]u[n − 𝑘]
∞
−∞
+ 2 ∑ 𝑢[𝑘 − 1]𝑢[𝑛 − 𝑘 − 8]
∞
−∞
− ∑ 𝑢[𝑘 − 1]𝑢[𝑛 − 𝑘 − 4]
∞
−∞
Next is to determine the range of each summation:
𝑦[𝑛] = ∑ u[k + 5]u[n − 𝑘] − 2 ∑ u[k + 5]𝑢[𝑛 − 𝑘 − 8]
𝑛−8
−5
+ ∑ u[k + 5]𝑢[𝑛 − 𝑘 − 4]
𝑛−4
−5
𝑛
−5
− ∑ 𝑢[𝑘 − 1]u[n − 𝑘]
𝑛
1
+ 2 ∑ 𝑢[𝑘 − 1]𝑢[𝑛 − 𝑘 − 8]
𝑛−8
1
− ∑ 𝑢[𝑘 − 1]𝑢[𝑛 − 𝑘 − 4]
𝑛−4
1
After that we are not in need of the unit-step signals because in these ranges they will be 1.
𝑦[𝑛] = ∑ 1 − 2 ∑ 1
𝑛−8
−5
+ ∑ 1
𝑛−4
−5
𝑛
−5
− ∑ 1
𝑛
1
+ 2 ∑ 1
𝑛−8
1
− ∑ 1
𝑛−4
1
But how to find the correct ranges at which the output 𝑦[𝑛] is defined?
Follow that procedure:
1. Give a number to each summation for indexing.
2. Find the least value among all the ranges and set it as the beginning of the first range.
3. Next is to substitute by this value in the equation to find 𝑦[𝑣] where 𝑣 is set to that
value. Reject all the summations that makes the summation invalid. The summation is
valid when it have the higher end value greater than or equal to the lower end value.
For example 0 and 0 or -3 and 1.
4. The remaining valid summations will be the summations used to calculate the input
value.
5. Increment the last value used by 1 and try to find which summations are used to
calculate the new output.
6. Whenever a new valid summation is used to calculate the output value other than the
summations used to calculate the previous value stop. This is an indication that we
reached the end of the previous range started before and reached also the beginning of](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-35-2048.jpg)
![a new range. This recent value is the end of the first range and the start value of the
new range.
7. Go back to step 3.
Applying that procedure to our example:
There are 6 summations and will be indexed by numbers from 1 to 6 where each summation
takes an index equal to its position in the equation.
The least value across all ranges is -5.
We need to find 𝑦[−5]. But make sure that each summation is valid.
𝑦[−5] = ∑ 1 − 2 ∑ 1
−5−8
−5
+ ∑ 1
−5−4
−5
−5
−5
− ∑ 1
−5
1
+ 2 ∑ 1
−5−8
1
− ∑ 1
−5−4
1
𝑦[−5] = ∑ 1 − 2 ∑ 1
−13
−5
+ ∑ 1
−9
−5
−5
−5
− ∑ 1
−5
1
+ 2 ∑ 1
−13
1
− ∑ 1
−9
1
The last 5 summations are invalid because the higher end is lower than the lower end. Finally to
calculate 𝑦[−5] we will just use the first summation with index 1.
Next is to increment the last value -5 by 1 to be -4 and substitute to find the summations used
to calculate the output.
𝑦[−4] = ∑ 1 − 2 ∑ 1
−12
−5
+ ∑ 1
−8
−5
−4
−5
− ∑ 1
−4
1
+ 2 ∑ 1
−12
1
− ∑ 1
−8
1
The valid summation is also the first one.
Next is to increment the last value -4 by 1 to be -3 and substitute to find the summations used
to calculate the output.
𝑦[−3] = ∑ 1 − 2 ∑ 1
−11
−5
+ ∑ 1
−7
−5
−3
−5
− ∑ 1
−3
1
+ 2 ∑ 1
−11
1
− ∑ 1
−7
1
The valid summation is also the first one.
The process will be repeated until the value of n used is -1.
When the value used is -1, the output will be as follows:
𝑦[−1] = ∑ 1 − 2 ∑ 1
−9
−5
+ ∑ 1
−5
−5
−1
−5
− ∑ 1
−1
1
+ 2 ∑ 1
−9
1
− ∑ 1
−5
1
Here there is a new valid summation used to calculate the output value. It is the third one. Then
we reached the end of the previous range started before and got the start of a new range.
The previous range is −5 ≤ 𝑛 ≤ −2 𝑜𝑟 − 5 ≤ 𝑛 < −1.
The next range starts at -1.
Increment -1 by 1 to be 0 and substitute to find the output.
After following that procedure the final ranges are:
−5 ≤ 𝑛 < −1, 1
−1 ≤ 𝑛 < 1, 1, 3
1 ≤ 𝑛 < 3, 1, 3, 4
3 ≤ 𝑛 < 5, 1, 2, 3, 4
5 ≤ 𝑛 < 9, 1, 2, 3, 4, 6](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-36-2048.jpg)
![9 ≤ 𝑛, 1, 2, 3, 4, 5, 6
Next is to formulate the output of that equation in a better form:
𝑦[𝑛] = ∑ 1 − 2 ∑ 1
𝑛−8
−5
+ ∑ 1
𝑛−4
−5
𝑛
−5
− ∑ 1
𝑛
1
+ 2 ∑ 1
𝑛−8
1
− ∑ 1
𝑛−4
1
𝑦[𝑛] =
{
0
𝑛 + 6
2𝑛 + 8
𝑛 + 8
−𝑛 + 12
−2𝑛 − 16
0
,
{
𝑛 < 0
−5 ≤ 𝑛 < −1
−1 ≤ 𝑛 < 1
1 ≤ 𝑛 < 3
3 ≤ 𝑛 < 5
5 ≤ 𝑛 < 9
9 ≤ 𝑛 {
𝑆1
𝑆1, 𝑆3
𝑆1, 𝑆3, 𝑆4
𝑆1, 𝑆3, 𝑆4, 𝑆2
𝑆1, 𝑆3, 𝑆4, 𝑆2, 𝑆6
𝐴𝐿𝐿
Lecture Example
𝑥[n] = u[n] − 𝑢[𝑛 − 10]
ℎ[n] =
1
4
(u[n] − 𝑢[𝑛 − 4])
Book Problems
Problem 2.2 –b (p130-113)
Problem 2.2 –b (p130-113)
𝑦[𝑛] = (
1
2
) 𝑛
𝑢[𝑛 − 2] ∗ u[n]
Using this equation:
𝑦[𝑛] = ∑ 𝑥(𝑘)ℎ(𝑛 − 𝑘)
∞
−∞
𝑦[𝑛] = ∑(
1
2
) 𝑘
𝑢[𝑘 − 2]𝑢[𝑛 − 𝑘]
∞
−∞
𝑦[𝑛] = ∑(
1
2
) 𝑘
𝑢[𝑘 − 2]𝑢[𝑛 − 𝑘]
𝑛
2
𝑦[𝑛] = ∑(
1
2
) 𝑘
𝑛
2
Using this summation rule:
∑ 𝛼 𝑘
𝑁
𝑘=𝑚
=
𝛼 𝑚
− 𝛼 𝑁+1
1 − 𝛼
𝑦[𝑛] = ∑(
1
2
) 𝑘
𝑛
𝑘=2
=
(
1
2)2
− (
1
2) 𝑛+1
1 −
1
2](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-37-2048.jpg)
![𝑦[𝑛] =
1
4 − (
1
2)(
1
2) 𝑛
1
2
𝑦[𝑛] =
1
2
− (
1
2
) 𝑛
The ranges are as follows:
𝑦[𝑛] = {
0, 𝑛 < 2
1
2
− (
1
2
)
𝑛
, 𝑛 ≥ 2
Convolution Integral
𝑦( 𝑡) = ∫ 𝑥(Τ)h(t − Τ)dΤ
∞
−∞
Note:
For ℎ(𝑡), n is replaced by 𝑡 − 𝑇. So ℎ[−𝑡] will be ℎ[−(t − T)].
Sheet
2.39-a (p202-185)
2.40-a (p202-185)
2.39-a (p202-185)
𝒚(𝒕) = (𝒖(𝒕) − 𝒖(𝒕 − 𝟐)) ∗ 𝒖(𝒕)
Following the convolution integral equation:
𝑦(𝑡) = ∫ 𝑥(Τ)h(t − Τ)dΤ
∞
−∞
𝑦(𝑡) = ∫ 𝑥(𝑢(𝑡) − 𝑢(𝑡 − 2))𝑢(𝑡)dΤ
∞
−∞
𝑦(𝑡) = ∫ 𝑢(Τ)𝑢(𝑡 − Τ) − ∫ 𝑢(Τ − 2)𝑢(𝑡 − Τ)dΤ
∞
−∞
∞
−∞
𝑦(𝑡) = ∫ 𝑢(Τ)𝑢(𝑡 − Τ)dΤ − ∫ 𝑢(Τ − 2)𝑢(𝑡 − Τ)dΤ
𝑡
2
𝑡
0
𝑦(𝑡) = ∫ 𝑑Τ − ∫ dΤ
𝑡
2
𝑡
0
Ranges are as follows:
𝑦(𝑡) = {
0
𝑡
2
{
𝑡 < 0
0 ≤ 𝑡 < 2
2 ≤ 𝑡
2.40-a (p202-185)
𝑚(𝑡) = 𝑥(𝑡) ∗ 𝑦(𝑡)](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-38-2048.jpg)
![𝑥(𝑡) = 𝑢(𝑡 + 1) − 𝑢(𝑡 − 1)
𝑦(𝑡) = 𝑢(𝑡) − 𝑢(𝑡 − 4) + 𝑢(𝑡 − 2) + 𝑢(𝑡 − 4)
𝑦(𝑡) = 𝑢(𝑡) − 2𝑢(𝑡 − 4) + 𝑢(𝑡 − 2)
𝑦(𝑡) = ∫ 𝑥(Τ)h(t − Τ)dΤ
∞
−∞
𝑦(𝑡) = ∫ (𝑢(𝑡 + 1) − 𝑢(𝑡 − 1))(𝑢(𝑡) − 2𝑢(𝑡 − 4) + 𝑢(𝑡 − 2))dΤ
∞
−∞
𝑦(𝑡) = ∫ 𝑢(Τ + 1)𝑢(𝑡 − Τ)dΤ − 2 ∫ 𝑢(Τ + 1)𝑢(𝑡 − Τ − 4)dΤ
∞
−∞
∞
−∞
+ ∫ 𝑢(Τ + 1)𝑢(𝑡 − Τ − 2)dΤ
∞
−∞
− ∫ 𝑢(Τ − 1)𝑢(𝑡 − Τ)dΤ
∞
−∞
+ 2 ∫ 𝑢(Τ − 1)𝑢(𝑡 − Τ − 4)dΤ
∞
−∞
− ∫ 𝑢(Τ − 1)𝑢(𝑡 − Τ − 2)dΤ
∞
−∞
𝑦(𝑡) = ∫ 𝑢(Τ + 1)𝑢(𝑡 − Τ)dΤ − 2 ∫ 𝑢(Τ + 1)𝑢(𝑡 − Τ − 4)dΤ
𝑡−4
−1
𝑡
−1
+ ∫ 𝑢(Τ + 1)𝑢(𝑡 − Τ − 2)dΤ
𝑡−2
−1
− ∫ 𝑢(Τ − 1)𝑢(𝑡 − Τ)dΤ
𝑡
1
+ 2 ∫ 𝑢(Τ − 1)𝑢(𝑡 − Τ − 4)dΤ
𝑡−4
1
− ∫ 𝑢(Τ − 1)𝑢(𝑡 − Τ − 2)dΤ
𝑡−2
1
𝑦(𝑡) = ∫ dΤ − 2 ∫ dΤ
𝑡−4
−1
+ ∫ dΤ
𝑡−2
−1
− ∫ dΤ
𝑡
1
+ 2 ∫ dΤ
𝑡−4
1
− ∫ dΤ
𝑡−2
1
𝑡
−1
Ranges are:
𝑦[𝑛] =
{
0
𝑡 + 1
𝑡 + 1
10 − 2𝑡
0
,
{
𝑡 < −1
−1 ≤ 𝑡 < 1
1 ≤ 𝑡 < 3
3 ≤ 𝑡 < 5
5 ≤ 𝑡 {
𝐼1
𝐼1, 𝐼3, 𝐼4
𝐼1, 𝐼3, 𝐼4, 𝐼2, 𝐼6
𝐼1, 𝐼3, 𝐼4, 𝐼2, 𝐼6, 𝐼5
Lecture Example
Book Problems
Problem 2.6 (p141-124)
Sheet 5
Book Problems](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-39-2048.jpg)
![***Section 11***
Unit Impulse as System Impulse Response
The effect of using the unit impulse signal over an input signal is to time shift the input signal by
the time shift value in the unit impulse.
𝑥(𝑡) ∗ δ(t − τ) = 𝑥(𝑡 − τ)
Example 11.1:
Find the convolution integral 𝑦(𝑡) when the input signal is 𝑥(𝑡) = 𝑢(𝑡 − 3) and the system
impulse response is ℎ(𝑡) = δ(t + 2) + δ(t − 4).
Solution:
𝑦( 𝑡) = 𝑥( 𝑡) ∗ ℎ(𝑡)
𝑦( 𝑡) = 𝑢( 𝑡 − 3) ∗ [δ(t + 2) + δ(t − 4)]
Using the distributive property of LTI systems:
𝑦( 𝑡) = 𝑢( 𝑡 − 3) ∗ δ(t + 2) + u(t) ∗ δ(t − 4)
Using the convolution by unit impulse property:
𝑦( 𝑡) = 𝑢( 𝑡 − 3 + 2) + u(t − 4)
𝑦( 𝑡) = 𝑢( 𝑡 − 1) + u(t − 4)
System Properties
Each property of the LTI systems will apply for both discrete-time and continuous-time signals.
Distributive
𝑥[𝑛] ∗ [ℎ1[𝑛] ± ℎ2[𝑛]] = 𝑥[𝑛] ∗ ℎ1[𝑛] ± 𝑥[𝑛] ∗ ℎ2[𝑛]
𝑥( 𝑡) ∗ [ℎ1( 𝑡) ± ℎ2( 𝑡)] = 𝑥( 𝑡) ∗ ℎ1( 𝑡) ± 𝑥( 𝑡) ∗ ℎ2(𝑡)
Example 11.2:
Using the distributive property how to construct the same output illustrated in the following
diagram using only one circuit? Just draw the block diagram.
Answer:
Using the distributive property the system can be designed using a cascaded diagram rather
than parallel to have just a single circuit as shown in the following diagram.](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-41-2048.jpg)
![Associative
(𝑥[ 𝑛] ∗ ℎ1[ 𝑛]) ∗ ℎ2[ 𝑛] = 𝑥[ 𝑛] ∗ (ℎ1[ 𝑛] ∗ ℎ2[ 𝑛])
[𝑥( 𝑡) ∗ ℎ1( 𝑡)] ∗ ℎ2( 𝑡) = 𝑥[ 𝑛] ∗ [ℎ1( 𝑡) ∗ ℎ2( 𝑡)]
Example 11.3:
Using the associative property how to construct the same output illustrated in the following
diagram using only one circuit? Just draw the block diagram.
Answer:
Using the associative property the system can be designed as shown in the following diagram to
have just a single circuit.
Commutative
ℎ1[ 𝑛] ∗ ℎ2[ 𝑛] = ℎ2[ 𝑛] ∗ ℎ1[ 𝑛]
ℎ1(𝑡) ∗ ℎ2(𝑡) = ℎ2(𝑡) ∗ ℎ1(𝑡)
Example 11.4:
Using the following block diagram find the overall system impulse response.
Steps to get the impulse response is as follows:
1. Divide each parallel connection into two serial connections](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-42-2048.jpg)
![Linear Time Invariant (LTI) Systems Properties
Memoryless
The discrete-time LTI system is memoryless if and only if:
ℎ[𝑘] = 𝑐δ[k]
The continuous-time LTI system is memoryless if and only if:
ℎ(𝑡) = 𝑐δ(t)
Causality
The discrete-time LTI system is causal if and only if:
ℎ[𝑘] = 0, 𝑘 < 0
The continuous-time LTI system is causal if and only if:
ℎ(𝑡) = 0, 𝑡 < 0
Stability
The discrete-time LTI system is stable if and only if:
)(nh
The continuous-time LTI system is stable if and only if:
dtth )(
Note that an impulse response with this part 𝑎 𝑏
is stable only if a<1 or b<0.
Examples
)(
2
1
)(
)1()(
)(2)(
)()(
2
nunh
nuenh
nunh
nunh
n
n
n
n
First one is stable only if < 1.
Second one is unstable because 2 > 1.
Third one is unstable because e > 1.
Fourth one is stable because 1/2 < 1.
Lecture Examples:](https://image.slidesharecdn.com/introductiontodigitalsignalprocesisngdsp-170519210334/75/Introduction-to-Digital-Signal-Processing-DSP-Course-Notes-45-2048.jpg)
Uploaded byAhmed Gad
Introduction to Digital Signal Processing (DSP) - Course Notes
These course notes on Digital Signal Processing (DSP) provide a comprehensive overview of the subject, covering topics from signal classification, system properties, to Fourier transforms and MATLAB applications. The material addresses both continuous and discrete signals, outlines key operations such as convolution and signal manipulation, and offers examples and exercises for practical understanding. It is structured in sections, each focusing on distinct aspects of DSP, including definitions, mathematical foundations, and practical implementations using MATLAB.
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Introduction to Digital Signal Processing (DSP) - Course Notes
- 1. Introduction to DigitalSignal Processing (DSP) Course Notes By Ahmed Fawzy Gad Faculty of Computers and Information (FCI) Menoufia University Egypt [email protected] المنوفية جامعة والمعلومات الحاسبات كلية قسمالمعلومات تكنولوجيا معالجةالرقمية اإلشارات MENOUFIA UNIVERSITY FACULTY OF COMPUTERS AND INFORMATION INFORMATION TECHNOLOGY DEPARTMENT DIGITAL SIGNAL PROCESSING المنوفية جامعة
- 2. Course Syllabus Section 1,2 MATLAB Section 3, 4 Revision Section 5 Introduction to DSP What is signal? What is system? Signal Classification o Continuous & Discrete o Even & Odd Using graph Using algebra Even and Odd Signals Characteristics Finding even and odd components o Periodic & Non-periodic Continuous only Using graph Using algebra o Deterministic & Random Section 6 Difference between notation of discrete and continuous signals Basic Operations on Signals o Operations on dependent variables Amplitude scaling Addition Multiplication Differentiation Integration o Operations on independent variables Time scaling Reflection Time shifting Section 7 Elementary signals in digital signal processing o Exponential o Sinusoidal
- 3. Discrete-time signalperiodicity o Exponentially damped sinusoidal o Unit-step o Unit-impulse o Ramp Basic Operations on the elementary signals System o System Definition o Grouping operations to create a system o Interconnection of operations MATLAB Graphical User Interface (GUI) o Signal Basic Operations Program Section 8 System Cont. o System Properties Linear Homogenous o Algebraic o Graph Additive o Algebraic o Graph Time invariant Memoryless Causal Stable Stable systems Why stable system is important? Section 9 Midterm Exam Section 10 Summation Rules Convolution o Discrete Convolution Sum o Continuous Convolution Integral Section 11 Unit Impulse as System Impulse Response System Properties (Discrete & Continuous) o Distributive o Associative o Commutative
- 4. Linear TimeInvariant (LTI) Systems Properties (Discrete & Continuous) o Memory o Causality o Stability System Block Diagram Section 12 Introduction to Fourier Transform o What is Fourier? o Why Using Frequency Domain? o Example to show importance of the frequency domain. o Fourier Representations. Fourier Transform o Moving between Time-Domain and Frequency-Domain o Useful Rules o Some uses of the Fourier transform in signal processing Exercises o FT o IFT Section 13 Steps to find Z-transform of a signal 1. Find 𝑋(𝑧) 2. Determine Region Of Convergence (ROC) 3. Calculate Poles and Zeros Z-transform examples
- 5. ***Section 5*** Introduction toDSP Some students asked why using MATLAB and what is the relation between MATLAB & the course of digital signal processing. Just we will make a hint that to process a signal a signal processing tool is used. MATLAB is one of these tools and there are other tools like Python. This diagram helps students to understand the journey of a signal from being an input continuous signal to being an output continuous signal. Digital signal processing is responsible for the operations to convert the signal from continuous to discrete, processing it and converting it back to continuous. Even and Odd Signals Characteristics General Characteristics o 𝑥(𝑡) = 𝑥 𝑒(𝑡) + 𝑥 𝑜(𝑡) o 𝑥 𝑒(−𝑡) = 𝑥 𝑒(𝑡) o 𝑥 𝑜(−𝑡) = −𝑥 𝑜(𝑡) Addition Characteristics o 𝑥 𝑒 + 𝑦𝑒 = 𝑧 𝑒 o 𝑥 𝑜 + 𝑦𝑜 = 𝑧 𝑜 o 𝑥 𝑒 + 𝑦𝑜 = 𝑧 𝑒 + 𝑧 𝑜 Multiplication Characteristics o 𝑥 𝑒 * 𝑦𝑒 = 𝑧 𝑒 o 𝑥 𝑜 * 𝑦𝑜 = 𝑧 𝑒 o 𝑥 𝑒 * 𝑦𝑜 = 𝑧 𝑜 Numbers Classification Units of measurements
- 6. Signal Classification Even andOdd We can determine whether a function is even or odd using: 1. Graph 2. Algebra Even and Odd using Graph Some functions can be easily classified as either even or odd from its graph. If the function can be folded into two parts around the vertical axis and be identical (symmetric about vertical axis) then the function is even. If the graph is antisymmetric about the time origin it is odd. Examples: **Notebook** Even and Odd using Algebraic Operations We should know these characteristics at first: General Characteristics 𝑥(𝑡) = 𝑥 𝑒(𝑡) + 𝑥 𝑜(𝑡) 𝑥 𝑒(−𝑡) = 𝑥 𝑒(𝑡) 𝑥 𝑜(−𝑡) = −𝑥 𝑜(𝑡) Examples: Determine whether the following signals are even or odd 1. f(t) = sin(t) 2. f(t) = cos(t) 3. f(t) = t + 5 4. **Example 1.1** p18-35 Answers: 1. Odd 2. Even 3. Not even and not odd 4. Odd
- 7. Finding even andodd components Revise the previous characteristics. Find the equations to calculate 𝒙 𝒆(𝒕) and 𝒙 𝒐(𝒕). The equation 𝑥(𝑡) = 𝑥 𝑒(𝑡) + 𝑥 𝑜(𝑡) has two unknowns 𝑥 𝑒(𝑡) and 𝑥 𝑜(𝑡) so we need two equations. One equation is as follows: 𝑥(𝑡) = 𝑥 𝑒(𝑡) + 𝑥 𝑜(𝑡) The other one can be found by substituting for –t rather than t to get: 𝑥(−𝑡) = 𝑥 𝑒(−𝑡) + 𝑥 𝑜(−𝑡) Based on the following properties: 𝑥 𝑒(−𝑡) = 𝑥 𝑒(𝑡) 𝑥 𝑜(−𝑡) = −𝑥 𝑜(𝑡) We can write: 𝑥(−𝑡) = 𝑥 𝑒(𝑡) - 𝑥 𝑜(𝑡) So the two equations are: 1. 𝑥(𝑡) = 𝑥 𝑒(𝑡) + 𝑥 𝑜(𝑡) 2. 𝑥(−𝑡) = 𝑥 𝑒(𝑡) - 𝑥 𝑜(𝑡) We can add both of these equations to get rid of 𝑥 𝑜(𝑡): 𝑥(𝑡) + 𝑥(−𝑡) = 2𝑥 𝑒(𝑡) So finally: 𝑥 𝑒(𝑡) = 𝑥(𝑡) + 𝑥(−𝑡) 2 Similarly we can find 𝑥 𝑜(𝑡): 𝑥 𝑜(𝑡) = 𝑥(𝑡) − 𝑥(−𝑡) 2 Examples: Find the even and odd components of the following signals: 1. **Example 1.2** p19-36 Note that the first step to make when solving such problems is to prepare 𝑥(−𝑡) because both the even and odd components depend on it and also the given 𝑥(𝑡). Periodic & Non-periodic Using Graph
- 8. A signal issaid to be periodic using just its graph if it consists of a single part that repeats itself. Examples: 1. Sine wave 2. Square waves 3. Discrete wave Examples of non-periodic waves: Using Algebra A signal is said to be periodic if it satisfies the following condition: 𝑥(𝑡) = 𝑥(𝑡 + 𝑇) Where T is the fundamental period of the signal 𝑥(𝑡). Thus the fundamental period is required. It is calculated by mapping the existing function to its standard form. Examples: 𝑒 𝑖𝑡 => 𝑒5𝑖𝜋𝑡 cos 𝑡 => cos(2𝜋𝑡) Examples: 1. 𝑒 𝑖2𝜋𝑡/10 2. cos(𝑡 + 𝜋/2) Solutions: 1) For the signal 𝑥(𝑡) to be periodic it must satisfy the following condition: 𝑥(𝑡) = 𝑥(𝑡 + 𝑇) where T is the fundamental period First step is to find the fundamental period of the signal. The angular frequency ω = 2𝜋 𝑇 The angular frequency ω is the part multiplied at the independent variable t. In our case, 𝑥(𝑡) = 𝑒 𝑖2𝜋𝑡/10 , the angular frequency ω is the part multiplied by t which is 2𝜋 10 . So ω = 2𝜋 10 Because ω = 2𝜋 10 = 2𝜋 𝑇 so ( 𝑇 =10). As we calculated the fundamental period 𝑇 we can find 𝑥(𝑡 + 𝑇). 𝑥(𝑡 + 𝑇)= 𝑥(𝑡 + 10) = 𝑒 𝑖2𝜋(𝑡+10)/10 The equation can be simplified as follows: 𝑒 𝑖2𝜋(𝑡+10)/10 = 𝑒 𝑖2𝜋𝑡/10+𝑖2𝜋10/10 = 𝑒 𝑖2𝜋𝑡/10+𝑖2𝜋 = 𝑒 𝑖2𝜋𝑡/10 . 𝑒 𝑖2𝜋 Because 𝑒 𝑖2𝜋 = cos(2𝜋) + 𝑖sin(2𝜋)
- 9. And because sin(2𝜋)= 0 and 𝑐𝑜𝑠(2𝜋) = 1 So 𝑒 𝑖2𝜋 = 1 So 𝑒 𝑖2𝜋𝑡/10 . 𝑒 𝑖2𝜋 = 𝑒 𝑖2𝜋𝑡/10 As we reached that 𝑥(𝑡) = 𝑥(𝑡 + 𝑇) we can conclude that the signal is periodic. Why sin(-t) = sin(t) and cos(-t) = cos(t)? A function has two types of variables: 1. Independent variables 2. Dependent variables Independent variables are the inputs to the functions which not depend on any form of variables to define their values. Dependent variables are the outputs of the functions and they do depend on other variables to determine their values. The variables that they depend on are the independent variables. Suppose that there is a function y = f(x) = x + 5 The independent variable here in f(x) is just the variable x and the dependent variable is y. It is dependent here because y values can only be determined when values of x are known. So depending on x values y values can be determined. When drawing such a function on the Euclidean space the independent variables are drawn on the x-axis and the dependent variables are drawn on the y-axis. The x-axis values change in sign when moving horizontal from right to left and vice versa. The y-axis values change in sign when moving vertical from top to bottom and vice versa. We can conclude that the independent variables change in sign when moving vertically and dependent variables change in sign when moving vertically. For the sine and cosine functions we know the following: sin = 𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 cos = 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 . For the left part of the next figure, the opposite of θ is the y-axis and the adjacent of θ is the x- axis. So sine depends on the opposite and the opposite in our case is the y-axis and the y-axis change sign when moving vertically above and below the horizontal x-axis. Also cosine depends on the adjacent and the adjacent in our case is the x-axis and the x-axis change sign when moving horizontally about the vertical y-axis. Summary: sine -> opposite -> y-axis -> vertically cosine -> adjacent -> x-axis -> horizontally
- 10. When selecting apositive angle like 60 it will be drawn counter-clockwise. To draw a negative angle it will be drawn clockwise. For a negative angle –θ as in our example it will be drawn counter-clockwise. As it is apparent in the right part of the figure that drawing a negative angle will flip over the x- axis. In our example, flipping over the x-axis will make the x-axis values remain the same as if the angle still positive but it changes the sign of the y-axis. What depends on the y-axis? It is the sine. So the sine will change sign too. But because the x-axis remains the same and the cosine depends on it so no changes will occur to the cosine. At this point we should have known whey sin(-t) = sin(t) and cos(-t) = cos(t).
- 11. MATLAB Task Write aMATLAB function named check_periodicity that checks whether the following signal is periodic or non-periodic. 𝑥(𝑡) = 𝑒 𝑖𝒂𝜋𝑡/𝒃 The function will accept two positive integers as inputs from the user: 1. Coefficient in the enumerator (a in the equation) 2. Coefficient in the denominator (b in the equation) After accepting these inputs the complete equation will be displayed on the screen. Notes: Clarify your code using comments. Write your name and section commented immediately after the function header. Grades will be based on these points: Creating a function Accepting the right inputs Displaying the equation correctly on the screen Correct algorithm that can successfully classify the signal Code clarification Copy minus.
- 12. **Section 6** Difference betweennotation of discrete and continuous signals In signal processing systems, continuous-time signals usually use the symbol 𝑡 for their independent variables and use parentheses but discrete-time signals use the symbol n for their independent variables and use square brackets. Examples of continuous-time signals: 𝑦(𝑡), 𝑥(𝑡) Examples of discrete-time signals: f[n], x[n] HINT In the course reference, there are no problems for just the operations on dependent variables but what is found is a combination of operations on both dependent and independent variables. So we can put some examples for just operations on dependent variables on both continuous- time and discrete-time signals. There is also a hand-created problem for each of these operations. So there are a total of 3 problems for each operation. Amplitude scaling: 1. Fig. 1.52 – ℎ1: 2𝑥(𝑡) 2. Fig. 1.52 – ℎ2: 3𝑦(𝑡) 3. Fig. 1.56 – 𝑙1: . 3𝑥[𝑛] 4. Fig. 1.56 – 𝑙2: 1.5𝑦[𝑛] Addition: 1. Fig. 1.52 – ℎ3: 𝑥(𝑡) + 𝑦(𝑡) 2. Fig. 1.56 – 𝑙3: 𝑥[𝑛] + 𝑦[𝑛] Multiplication: 1. Fig. 1.52 – ℎ4: 𝑥(𝑡)𝑦(𝑡) 2. Fig. 1.56 – 𝑙4: 𝑥[𝑛]𝑦[𝑛] Basic Operations on Signals Before diving into the operations that can be performed over the signals it is good to know the difference between independent and dependent variables. For example, list the independent and dependent variables in the following signal: z = f(x, y) = x + y + a Here x and y are the independent variables because their values not depend on any factor but z is the dependent variable because its value depends on both x and y. Operations on dependent variables Operations performed on dependent variables not affect just a single independent variable but affects all independent variables. So all of these operations just change the dependent variable so they don`t change the dependent variables. The independent variable is graphed over the x-axis and the deponent variable is graphed over the y-axis. So no change in the x-axis values before and after scaling but the change will be in the y-axis values. Amplitude scaling Suppose there is a continuous-time signal 𝑥(𝑡) scaled by a scaling factor c, the resulting signal 𝑦(𝑡) is as follows:
- 13. 𝑦(𝑡) = 𝑐𝑥(𝑡) 𝑥(𝑡)the original signal, c scaling factor, 𝑦(𝑡) the scaled signal Scaling for discrete-time signals is performed in a similar manner to continuous-time signals: y[n] = cx[n] Example 6.1: 𝒙(𝒕) = 𝒕 + 2 for -3 ≤ 𝑡 ≤ 3 Perform amplitude scaling for x(t) to get y(t) by a scaling factor 2 and graph the two signals. 𝑦(𝑡) = 𝑐𝑥(𝑡) For c = 2 𝑦(𝑡) = 2𝑥(𝑡) 𝑦(𝑡) = 2(𝑡 +2) = 2𝑡 +4 Amplitude scaling: 1. Fig. 1.52 – ℎ1: 2𝑥(𝑡) 2. Fig. 1.52 – ℎ2: 3𝑦(𝑡) 3. Fig. 1.56 – 𝑙1: . 3𝑥[𝑛] 4. Fig. 1.56 – 𝑙2: 1.5𝑦[𝑛] Addition For the continuous-time signals 𝑥1(𝑡) and 𝑥2(𝑡) the output signal 𝑦(𝑡) is obtained by adding these two signals as follows: 𝑦(𝑡) = 𝑥1(𝑡) + 𝑥2(𝑡) This is similar to adding discrete-time signals: y[n] = x1[n] + x2[n] Example 6.2: 𝑥1(𝑡) = 𝑡 𝑥2(𝑡) = 𝑡 + 2 Using graph find 𝒚(𝒕) = 𝑥1(𝑡) + 𝑥2(𝑡) and graph the result signal where -3 ≤ 𝑡 ≤ 3. Addition: 1. Fig. 1.52 – ℎ3: 𝑥(𝑡) + 𝑦(𝑡) 2. Fig. 1.56 – 𝑙3: 𝑥[𝑛] + 𝑦[𝑛] Multiplication For the continuous-time signals 𝑥1(𝑡) and 𝑥2(𝑡) the output signal 𝑦(𝑡)) is obtained by multiplying these two signals as follows: 𝑦(𝑡) = 𝑥1(𝑡)𝑥2(𝑡) This is similar to multiplying discrete-time signals: y[n] = x1[n]x2[n] Example 6.3: Multiply the previous two signals and graph the three signals (two inputs + one output). 𝑥1(𝑡) = 𝑡 𝑥2(𝑡) = 𝑡 + 2
- 14. Multiplication: 1. Fig. 1.52– ℎ4: 𝑥(𝑡)𝑦(𝑡) 2. Fig. 1.56 – 𝑙4: 𝑥[𝑛]𝑦[𝑛] Question: 𝒙(𝒕) = 𝒕 +3 where -5 ≤ 𝑡 ≤ 5 From x(t) how to return a new signal of the form using the basic operations on signals. Explain. 𝑦(𝑡) = { 𝑥(𝑡), − 3 ≤ 𝑡 ≤ 3 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 How to return a new signal that is identical to 𝑥(𝑡) for -3 ≤ 𝑡 ≤ 3 and have zero otherwise. Operations on independent variables These operations will of course affect the independent variables and the independent variables and graphed over the x-axis so they affect the x-axis. Time Scaling For the continuous-time signal 𝑥(𝑡), the output signal 𝑦(𝑡) is obtained by scaling the independent variable t by a scaling factor c as follows: 𝑦(𝑡) = 𝑥(𝑎𝑡) If a > 1 then the signal get compressed and if 0 < a < 1 the signal get expanded (stretched). This also holds for discrete-time signals: 𝑦[𝑛] = 𝑥[𝑘𝑛] Example 6.4: Draw the following continuous-time signal 𝒙(𝒕) based on its inputs and outputs: X -2 -1 0 1 2 Y 0 1 2 1 0 Find the time-scaled version of that signal 𝒚(𝒕) for the following values for the scaling factor: 3 .6=3/5 Book Problems: 1.56 (a) Question: Why when multiplying by a value > 1 the signal gets compressed and when multiplying by a value < 1 the signal gets expanded? This can be explained using the unit function. The following unit function is defined in the range -1 ≤ 𝑡 ≤ 1:
- 15. It can berewritten in a more representative form as follows: 𝑥(𝑡) = { 1, − 1 ≤ 𝑡 ≤ 1 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 When scaling that function by a scaling factor > 0 like 2 it will be as follows by just replacing each 𝑡 by 2𝑡: 𝑥(2𝑡) = { 1, − 1 ≤ 2𝑡 ≤ 1 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 But it is more convenient to separate the independent variables in an equation from other factors. Here 2 and the independent variable 𝑡 are in the same side. Divide by 2 to make the equation has this form: 𝑥(2𝑡) = { 1, − 1 2 ≤ 𝑡 ≤ 1 2 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 From this equation it is apparent that the range got compressed from -1:1 to - 1 2 : 1 2 . From the last equation it is shown that when multiplying by a scaling factor greater than 1 like 2 the signal got compressed. We can try multiplying by a factor less than 1 like 1 2 . 𝑥(𝑡) = { 1, − 1 ≤ 𝑡 ≤ 1 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 𝑥 ( 𝑡 2 ) = { 1, − 1 ≤ 𝑡 2 ≤ 1 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 𝑥 ( 𝑡 2 ) = { 1, − 2 ≤ 𝑡 ≤ 2 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 From the last equation it is shown that when multiplying by a scaling factor less than 1 like 1 2 the signal gets expanded. So we can have a rule that says if the scaling factor > 1 then the signal get compressed and if the factor is < 1 the signal get expanded. Reflection: Reflection of a continuous-time signal 𝑥(𝑡) is obtained by replacing the time t by –t to generate the reflected version of 𝑥(𝑡) which is 𝑦(𝑡). 𝑦(𝑡) = 𝑥(-t) This is similar to discrete-time signals: 𝑦[𝑛] = 𝑥[-n] Book Problems:
- 16. 1. Fig. 1.56– 𝑙5: 𝑥[−𝑛] 2. Fig. 1.56 – 𝑙6: 𝑦[−𝑛] Time Shifting: For the continuous-time signal 𝑥(𝑡) the time-shifted version 𝑦(𝑡) is defined as: 𝑦(𝑡) = 𝑥(𝑡 + 𝑡 𝑜) Where 𝑡 𝑜 is the time shift. If 𝒕 𝒐> 0 then the signal goes to the right and if 𝒕 𝒐< 0 the signal goes to the left. In case of a discrete-time signal 𝑥[𝑛], the time-shifted version is: 𝑦[𝑛] = 𝑥[𝑛 − 𝑚] Where m is a positive or negative integer. Book Problems: 1. Fig. 1.56 – 𝑙5: 𝑥[𝑛 + 2] 2. Fig. 1.56 – 𝑙6: 𝑦[𝑛 − 3] Question: When subtracting a value from the independent variable 𝒕, how the signal go to the right not the left? Suppose that there are two devices (𝐷1 and 𝐷2) generating signals at the same time 𝑡 which is for example 12PM. But we need 𝐷2 to generate its signal after 𝐷1 signal by a delay of 2 minutes. 𝐷1-> S1 -> t -> 12 PM 𝐷2-> S2 -> t -> DELAY -> 𝒕 𝒐 -> 2 min -> 12PM + 2 MIN Finally we have: 𝑡 = 12𝑃𝑀 𝑡 − 𝑡 𝑜 = 12𝑃𝑀 𝑡 − 2𝑀𝐼𝑁 = 12𝑃𝑀 𝑡 = 12𝑃𝑀 + 2𝑀𝐼𝑁 Thus time for 𝐷2 signal got delayed by 2 minutes when subtracting 2 from the time variable 𝑡. Another better explanation: For the following unit step function: It has the following equation: 𝑥(𝑡) = { 1, − 1 ≤ 𝑡 ≤ 1 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 When the signal is to be shifted in time by value +2 the new equation will be as follows: 𝑥(𝑡 − 2) = { 1, − 1 ≤ 𝑡 − 2 ≤ 1 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
- 17. 𝑥(𝑡 − 2)= { 1, − 1 + 2 ≤ 𝑡 ≤ 1 + 2 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 𝑥(𝑡 − 2) = { 1, 1 ≤ 𝑡 ≤ 3 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 Thus the signal is shifted to the right when it shifted by a positive value like +2. The signal will also get shifted to the left when it shifted by a negative value like -2. Sheet: 1. Problems solved in the section 2. Book Problems: a. 1.50 b. 1.51 c. 1.52 d. 1.56
- 18. **Section 7** Elementary Signalsin Digital Signal Processing There are some basic signals in the field of digital signal processing that has dominant features. These signals are important by themselves because they reflect some already existing physical signals but also they are useful to: Construct signals based on these basic signals Process and apply operations over signals based on such basic signals Because signals can be either discrete or continuous we should know: 1. Algebraic equation of continuous and discrete forms of each basic signal 2. Graph of the continuous and discrete forms of the signals Exponential Signal One important factor about such exponential signal is whether it is increasing or decreasing (growing or decaying). Continuous 𝑥(𝑡) = 𝐵𝑒 𝑎𝑡 𝑎 < 0 𝐷𝑒𝑐𝑎𝑦𝑖𝑛𝑔 𝐸𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙 𝑆𝑖𝑔𝑛𝑎𝑙 𝑎 > 0 𝐺𝑟𝑜𝑤𝑖𝑛𝑔 𝐸𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙 𝑆𝑖𝑔𝑛𝑎𝑙 a=-6, B=5 a=5, B=1 Note that e=2.7 is a number greater than 1. When a number > 1 is raised to a power < 0 it will decrease its value by increasing the magnitude of the power. When the power is > 0 it will increase the resultant value by increasing the power magnitude. In our example e is greater than 1 and when a > 0 the result will be an increasing signal and when a < 0 result will have the form of a decreasing signal. Discrete 𝑥[𝑛] = 𝐵𝑒 𝑎𝑛 𝑎 < 0 𝐷𝑒𝑐𝑎𝑦𝑖𝑛𝑔 𝐸𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙 𝑆𝑖𝑔𝑛𝑎𝑙 𝑎 > 0 𝐺𝑟𝑜𝑤𝑖𝑛𝑔 𝐸𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙 𝑆𝑖𝑔𝑛𝑎𝑙 a=-6, B=5 a=5, B=1
- 19. The same explanationwhy x[n] is sometimes increasing and others decreasing is the same as the continuous form. Sinusoidal Signal Continuous 𝑥(𝑡) = 𝐴𝑐𝑜𝑠(𝜔𝑡 + 𝛷) 𝑥(𝑡) = 𝐴𝑠𝑖𝑛(𝜔𝑡 + 𝛷) Where 𝜔 = 2𝜋𝑓 Remember that for a signal to be periodic this condition must be valid: 𝑥(𝑡) = 𝑥(𝑡 + 𝑇) Discrete 𝑥[𝑛] = 𝐴𝑐𝑜𝑠(𝛺𝑛 + 𝛷) 𝑥[𝑛] = 𝐴𝑠𝑖𝑛(𝛺𝑛 + 𝛷) Where 𝛺𝑁 = 2𝜋𝑚 Discrete-Time Signal Periodicity For a discrete-time signal like: 𝑥[𝑛] = 𝐴𝑐𝑜𝑠(𝛺𝑛 + 𝛷) 𝑥[𝑛] = 𝐴𝑠𝑖𝑛(𝛺𝑛 + 𝛷) Where 𝛺𝑁 = 2𝜋𝑚 It will be periodic only if both m and N are integers. Examples: o 1.16 – p56-39 o 1.17-a, c – p56-39 o 1.18-a, d – p56-39 Exponentially Damped Sinusoidal Unit-Step Signal Continuous
- 20. Discrete Unit-Impulse Signal Discrete The unit-impulsesignal is also called delta and δ is called delta. The unit-impulse signal can be calculated from the unit-step signal. 𝛅[ 𝑛] = 𝑢[ 𝑛] − 𝑢[ 𝑛 − 1] Ramp Signal Continuous Discrete Note: The ramp signal can be calculated based on the unit-step signal. The ramp signal output is always equal to its input. So we can represent the ramp signal with this equation:
- 21. 𝑟(𝑡) = 𝑡 Itwill have t this graph: But it is just defined at 𝑡 ≥ 0. So how to make the negative part of the ramp signal equal to zero? We apply some of the basic operations to remove the negative part of that result. The simplest is by using multiplication: because 𝑢(𝑡) signal is only defined at the positive part and zero we can multiply it by the signal 𝑟(𝑡) = 𝑡 to return the ramp signal. So it will be as follows for continuous and discrete forms: 𝑟(𝑡) = 𝑡𝑢(𝑡) 𝑟[𝑛] = 𝑛𝑢[𝑛] Basic Operations on the elementary signals Examples: o Show step by step how to create a unit-impulse signal using just unit-step signals. o Show step by step how to create a ramp signal using unit-step signals. o How to create the signal shown in the following graphs using just the elementary signals: A B A B Figure. P1 Solutions: 1. 𝑢(𝑡 + 2) − 𝑢(𝑡 − 2) + 𝑢(𝑡 + 1) − 𝑢(𝑡 − 1) 2. 2δ[ 𝑛] + 4δ[ 𝑛 + 1] − δ[ 𝑛 + 2] − 5δ[ 𝑛 − 1] − 2δ[ 𝑛 − 2] Advanced problems:
- 22. How torepresent the signal shown in P1-B using just unit-step signals. At first the signal is directly represented using unit-impulse signals. Unit impulse signals can be represented using just the unit-step signals. So solution steps are: 1. Represent the signal using just unit-impulse signals. 2. Replace each unit-impulse signals by its equivalent unit-step signals. Solution is: δ[ 𝑛] = 𝑢[ 𝑛] − 𝑢[𝑛 − 1] δ[ 𝑛 + 1] = 𝑢[ 𝑛 + 1] − 𝑢[𝑛] δ[ 𝑛 + 2] = 𝑢[ 𝑛 + 2] − 𝑢[𝑛 + 1] δ[ 𝑛 − 1] = 𝑢[ 𝑛 − 1] − 𝑢[𝑛 − 2] δ[ 𝑛 − 2] = 𝑢[ 𝑛 − 2] − 𝑢[𝑛 − 3] The final equation is: 2( 𝑢[ 𝑛] − 𝑢[ 𝑛 − 1]) + 4( 𝑢[ 𝑛 + 1] − 𝑢[ 𝑛]) − ( 𝑢[ 𝑛 + 2] − 𝑢[ 𝑛 + 1]) − 5( 𝑢[ 𝑛 − 1] − 𝑢[ 𝑛 − 2]) − 2( 𝑢[ 𝑛 − 2] − 𝑢[𝑛 − 3]) Book Problems: o 1.54-a, d – p106-89 Some operations will involve just a single type of the above basic signals and others may contain more than one type. Interconnection of operations o Book examples o 1.12 – p70-53 o Book problems o 1.25-p71-54 It is simpler to represent the system as a graph rather than an equation. System operations can be connected using two ways: 1. Parallel: Each operation takes a separate path and implemented by a separate circuit. This is simple but adds more costs to create a separate circuit for each operation. 2. Cascade: The single operation is used multiple times. This saves costs but adds more complexity to reuse the same circuit.
- 23. MATLAB GUI Task Createa MATLAB GUI application to perform the basic operations on a sine wave as explained in the following diagram. Sheet 3: Problems solved in the section Book Problems o Discrete-time signal periodicity Book examples Example 1.7-a – p55-38 Book problems 1.16 – p56-39 1.17 – p56-39 1.18 1.58 o Elementary signals operations Book problems 1.54 1.55 o Interconnection of operations Book problems 1.63 1.65 1.69 - a
- 24. **Section 8** The coursetitle is digital signal processing. Before making processing over a digital signal we should know the digital signal itself. Also before studying digital signals we should know signals at all. So we should know signals then digital signals and finally processing over such digital signals. A system performs some operations (processing) over its input signals. Because a system operation (processing) will get applied over an input signal so we should study signals at first. We concentrated on signals and knew different types of signals like sinusoidal and step and also knew their properties like (even/odd, periodic/non-periodic). So we knew these points: Signal definition Signal digitization Signal classification Examples of basic signals Next is to know how operations applied to such signals. We started by practicing some simple operations like: Addition Scaling Shifting After that we will concentrated on systems. A system performs multiple operations over the signal so we should know how to perform a group of operations on the same signal. A system with multiple operations will be hard to view using just its equation but it will be easier to look at its graph. We also should study how to graph a system. After that we should know properties of a system. So we will study: System definition System operations Interconnection of operations System properties System Properties Linearity For a system to be linear it must exhibit two properties: Homogeneity: A system is called homogeneous if scaling the input signal by a factor will also scale the output by the same factor. So it the input is 𝒙(𝒕) and the output is 𝒚(𝒕) then scaling the input by factor say a to have 𝒂𝒙(𝒕) will generate the output scaled by the same factor to be 𝒂𝒚(𝒕). So if a sine input signal was having an amplitude of 1 and the output was having an amplitude of 3. Then scaling the input signal by a factor 2 to double the signal will also double the output signal. Additivity: A system is called additive if the sum of input signals generates the sum of their output signals. That is if an input signal is 𝒙 𝟏(𝒕) and its output is 𝒚 𝟏(𝒕) and
- 25. there is anotherinput 𝒙 𝟐(𝒕) generating output 𝒚 𝟐(𝒕) then 𝒙 𝟏( 𝒕) + 𝒙 𝒚(𝒕) will generate 𝒚 𝟏( 𝒕) + 𝒚 𝟐(𝒕). But what is the goal of determining whether a system is linear or not? As we previously discussed that a linear signal changes by a fixed amount. This will help us predict future values of the signal. If a linear signal accepts inputs 1, 2, and 3 to generate outputs 10, 20, and 30 then we can predict future values like input 4 that will generate output 40. We can predict output because the system is linear and changes by a fixed amount each time. To prove that a signal is linear or not we can follow two ways: 1. Algebraic 2. Graph For each signal to be linear or non-linear it must be both homogenous and additive. At first we need to prove that it is homogenous. So for an input 𝒙(𝒕) the output is 𝒙(𝒕) 𝟐 . For that system to be homogenous then scaling the input by a will also scale the output by the same scale. So the input will be 𝒂𝒙(𝒕) and the output is 𝒂 𝟐 𝒙(𝒕) 𝟐 Because the output was not scaled by the same scale as the input the signal is not homogenous. Because at least one condition was not met then the signal is not linear. But we can try to find whether the signal is additive or not. For these two input signals: 𝒙 𝟏(𝒕) => 𝒚 𝟏(𝒕) 𝟐 𝒙 𝟐(𝒕) => 𝒚 𝟐(𝒕) 𝟐 The signal will be additive if we 𝒙 𝟏(𝒕) + 𝒙 𝟐(𝒕) will generate the sum of their outputs which is 𝒚 𝟏(𝒕) 𝟐 + 𝒚 𝟐(𝒕) 𝟐 The output for the input 𝒙 𝟏(𝒕) + 𝒙 𝟐(𝒕) is: [𝒙 𝟏(𝒕) + 𝒙 𝟐(𝒕)] 𝟐 𝒙 𝟏(𝒕) 𝟐 + 2𝒙 𝟏(𝒕)𝒙 𝟐(𝒕) + 𝒙 𝟐(𝒕) 𝟐 Because the output is not equal to the sum of the individual outputs we conclude that the signal is not additive. Examples: 8.1.1. 𝑦(t) = x(t) 8.1.2. y(t) = x(t − 5) 8.1.3. y(t) = x(t + 2) + x(3 − t) 8.1.4. y(t) = x(2t − 1) 8.1.5. y(t) = x(t2 ) 8.1.6. y(t) = x2(t) = [x(t)]2 8.1.7. y(t) = 2x(t) + 4 8.1.8. y(t) = 2x(t) + 4t 8.1.9. y(t) = x(t)cos(t) Answers: 8.1.1. Linear, Time invariant, causal, memoryless, stable
- 26. 8.1.2. Linear, Timeinvariant, causal, memory, stable 8.1.3. Linear, Time variant, non-causal, memory, stable 8.1.4. Linear, Time variant, non-causal, memory, stable 8.1.5. Linear, Time variant, non-causal, memory, stable 8.1.6. Non-linear, Time invariant, causal, memoryless, stable 8.1.7. Non-linear, Time invariant, causal, memoryless, stable 8.1.8. Non-linear, Time variant, causal, memoryless, stable 8.1.9. Linear, Time variant, causal, memoryless, stable 8.1.7. Solution Homogenous: Multiply x(t) by a factor say a to be ax(t). Then the result will be 2ax(t) + 4 The output y(t) when get multiplied by factor a is a[2x(t) + 4] = 2ax(t) + 4𝑎 Because scaling the input by a factor a does not scale the output by the same scale then the system is not homogenous. Because at least of the two properties (homogenous & additivity) is not met then the system is non-linear. Additivity: Say there are two signals 𝑥1(t) and 𝑥2(t) generating outputs 𝑦1(t) and 𝑦2(t). The sum of the two input signals which is 𝑥1(t) + 𝑥2(t) will be the new input to the system. So the output will be: 𝑦(𝑥1(t) + 𝑥2(t)) = 2(𝑥1(t) + 𝑥2(t)) + 4. Eq1 The sum of the outputs is: 𝑦1(t) = 2𝑥1(t) + 4 𝑦2(t) = 2𝑥2(t) + 4 𝑦1(t) + 𝑦2(t) = (2𝑥1(t) + 4) + (2𝑥2(t) + 4). Eq2 Because Eq1 and Eq2 are not equal then the system is not additive. Because the system is neither homogenous nor additive then it is non-linear. 8.1.9. Solution Homogenous: Multiply x(t) by a factor say a to be ax(t). Then the result will be ax(t)cos(𝑡). The output y(t) when get multiplied by factor a is a[x(t)cos(𝑡)]. Because scaling the input by a factor a does scale the output by the same scale then the system is homogenous. Additivity: Say there are two signals 𝑥1(t) and 𝑥2(t) generating outputs 𝑦1(t) and 𝑦2(t). The sum of the two input signals which is 𝑥1(t) + 𝑥2(t) will be the new input to the system. So the output will be: 𝑦(𝑥1(t) + 𝑥2(t)) = (𝑥1(t) + 𝑥2(t))cos(𝑡). Eq1 The sum of the outputs is: 𝑦1(t) = 𝑥1(t)cos(𝑡) 𝑦2(t) = 𝑥2(t)cos(𝑡) 𝑦1(t) + 𝑦2(t) = 𝑥1(t) cos(𝑡) + 𝑥2(t)cos(𝑡) Eq2
- 27. Because Eq1 andEq2 are equal then the system is additive. Because the system is homogenous and additive then it is linear. Time Invariance A system is called time invariant if it does not change over time. But a time variant system does change. So the time invariant system responds identically regardless of when the input was applied. Soa time invariant system does not change by shift (time delay or advance). If the input was applied at 5PM and generated an output y, so if it was delayed to 4PM or moved forward in time to 6PM the results will be the same. If the system was time invariant so we can know the signal behavior in the past and future because it will get the same output whatever the time is. To prove that the system is time invariant make sure that if it generates an output y at time 𝒕 𝟏 then it will generate the same output at different times 𝒕 𝟐. That is if: 𝒚(𝒕) = 𝒙(𝒕) And the input was shifted by amount d to be 𝒙(𝒕 − 𝒅) then the same output will be generated but shifted by the same amount d to be 𝒚(𝒕 − 𝒅). Steps to prove that a system is time invariant or not: 1. Shifting input be replacing each 𝒙(𝒕) by 𝒙(𝒕 − 𝒅) to generate an output 𝒚 𝒅(𝒕). 2. Shifting the output by replacing each 𝒕 by 𝒕 − 𝒅 to generate an output 𝒚(𝒕 − 𝒅). 3. If 𝒚 𝒅(𝒕) = 𝒚(𝒕 − 𝒅) then the system is time invariant. Examples: Which of the following systems are time invariant? Examples: 8.2.1. 𝑦(t) = x(t) 8.2.2. y(t) = x(t − 5) 8.2.3. y(t) = x(t + 2) + x(3 − t) 8.2.4. y(t) = x(2t − 1) 8.2.5. y(t) = x(t2 ) 8.2.6. y(t) = x2(t) = [x(t)]2 8.2.7. y(t) = 2x(t) + 4 8.2.8. y(t) = 2x(t) + 4t 8.2.9. y(t) = x(t)cos(t) Answers: 8.2.1. Linear, Time invariant, causal, memoryless, stable 8.2.2. Linear, Time invariant, causal, memory, stable 8.2.3. Linear, Time variant, non-causal, memory, stable 8.2.4. Linear, Time variant, non-causal, memory, stable 8.2.5. Linear, Time variant, non-causal, memory, stable 8.2.6. Non-linear, Time invariant, causal, memoryless, stable 8.2.7. Non-linear, Time invariant, causal, memoryless, stable
- 28. 8.2.8. Non-linear, Timevariant, causal, memoryless, stable 8.2.9. Linear, Time variant, causal, memoryless, stable Please note: The system will be time invariant if and only if the output generated by applying an input shifted by amount d to be 𝒙(𝒕 − 𝒅) will be identical to the same output 𝒚(𝒕) of the original non-shifted input but shifted by the same amount d to be 𝒚(𝒕 − 𝒅). 8.2.7. a. Shifting the input 𝒙(𝒕) by amount d to be 𝒙(𝒕 − 𝒅) will generates an output 𝟐𝒙(𝒕 − 𝒅) + 𝟒. b. Shifting the output 𝒚(𝒕) by amount d to be 𝒚(𝒕 − 𝒅) by replacing each 𝒕 by 𝒕 − 𝒅 will generates 𝟐𝒙(𝒕 − 𝒅) + 𝟒. c. Because shifting the input and the output generates the same signal then the system is time invariant. 8.2.8. a. Shifting the input 𝒙(𝒕) by amount d to be 𝒙(𝒕 − 𝒅) will generates an output 𝟐𝒙(𝒕 − 𝒅) + 𝟒𝑡. b. Shifting the output 𝒚(𝒕) by amount d to be 𝒚(𝒕 − 𝒅) will generate an output 𝟐𝒙(𝒕 − 𝒅) + 𝟒(𝒕 − 𝒅). c. Because the results generated by shifting both the input and the output are different then the system is time variant. 8.2.9. a. Shifting the input 𝒙(𝒕) by amount d to be 𝒙(𝒕 − 𝒅) will generates an output 𝒙(𝒕 − 𝒅)𝒄𝒐𝒔(𝒕). b. Shifting the output 𝒚(𝒕) by amount d to be 𝒚(𝒕 − 𝒅) by replacing each 𝒕 by 𝒕 − 𝒅 will generate an output 𝒙(𝒕 − 𝒅)𝒄𝒐𝒔(𝒕 − 𝒅). c. Because the results generated by shifting both the input and the output are different then the system is time variant. A rule to be used to determine whether a system is time invariant or time variant: In most cases, if there was an independent variable outside of the input signal then the system will be time variant. 8.2.7 is time invariant because the independent variable is inside the input signal 𝒙(𝒕) and no 𝒕 outside of it. 8.2.8 & 8.2.9 are time variant because there is at least one independent variable outside the input signal 𝒙(𝒕). But this rule fails in some cases like 𝒙(𝟒 − 𝒕). In this case the independent variable 𝒕 is inside the input signal but the system is time variant. BIBO Stability There are multiple ways to define that a system is stable: 1. SISO (Single Input Single Output) 2. SIMO (Single Input Multiple Output) 3. MISO (Multiple Input Single Output) 4. MIMO (Multiple Input Multiple Output) 5. BIBO (Bounded Input Bounded Output)
- 29. A system issaid to be BIBO stable if for every set of bounded inputs there is a set of bounded outputs. That is the system is called BIBO (Bounded-Input, Bounded-Output) and BIBO means that if the number of inputs to the system was less than infinity then the number of outputs must be less than infinity. The following conditions must hold: Where 𝑀 𝑦 and 𝑀 𝑥 are finite positive integers. Why stable system is important? A stable system is better than unstable system because stable systems are predictable and we can know how it can behave compared to unstable systems. So if we applied a set of inputs we can predict the system behavior and know its outputs. This is similar to a programmer that writes code and knows the behavior of the code. For an input x there will be output y compared to another programmer that writes code and doesn`t know how it can behave responding to an input x. If there is another input z, the first programmer is stable and can predict the output responding to the applied input but the second one is unstable because we can`t predict the output. Another example is about a car. A stable car system will increase the car speed if the input was pressing the gas pedal. Unstable car will not just increase the car speed responding to pressing the gas pedal but can make unpredicted behavior like opening the windows, closing the air conditioner, firing the horn. Examples: 8.3.1. 𝑦(t) = x(t) 8.3.2. y(t) = x(t − 5) 8.3.3. y(t) = x(t + 2) + x(3 − t) 8.3.4. y(t) = x(2t − 1) 8.3.5. y(t) = x(t2 ) 8.3.6. y(t) = x2(t) = [x(t)]2 8.3.7. y(t) = 2x(t) + 4 8.3.8. y(t) = 2x(t) + 4t 8.3.9. y(t) = x(t)cos(t) Answers: 8.3.1. Linear, Time invariant, causal, memoryless, stable 8.3.2. Linear, Time invariant, causal, memory, stable 8.3.3. Linear, Time variant, non-causal, memory, stable 8.3.4. Linear, Time variant, non-causal, memory, stable 8.3.5. Linear, Time variant, non-causal, memory, stable 8.3.6. Non-linear, Time invariant, causal, memoryless, stable 8.3.7. Non-linear, Time invariant, causal, memoryless, stable
- 30. 8.3.8. Non-linear, Timevariant, causal, memoryless, stable 8.3.9. Linear, Time variant, causal, memoryless, stable Causality A system is called to be causal if it depends on the present and past values of the input signal but not on the future values. Real-time systems are causal because they depend on the present and the past but not on the future. A system is called non-causal if it depends on at least one value from the future. Examples: 8.4.1. 𝑦(t) = x(t) 8.4.2. y(t) = x(t − 5) 8.4.3. y(t) = x(t + 2) + x(3 − t) 8.4.4. y(t) = x(2t − 1) 8.4.5. y(t) = x(t2 ) 8.4.6. y(t) = x2(t) = [x(t)]2 8.4.7. y(t) = 2x(t) + 4 8.4.8. y(t) = 2x(t) + 4t 8.4.9. y(t) = x(t)cos(t) Answers: 8.4.1. Linear, Time invariant, causal, memoryless, stable 8.4.2. Linear, Time invariant, causal, memory, stable 8.4.3. Linear, Time variant, non-causal, memory, stable 8.4.4. Linear, Time variant, non-causal, memory, stable 8.4.5. Linear, Time variant, non-causal, memory, stable 8.4.6. Non-linear, Time invariant, causal, memoryless, stable 8.4.7. Non-linear, Time invariant, causal, memoryless, stable 8.4.8. Non-linear, Time variant, causal, memoryless, stable 8.4.9. Linear, Time variant, causal, memoryless, stable Memory A system is said to have memory if it depends on past or future values of the input signal. A system is said to be memoryless if it depends only on the present values. Examples: 8.5.1. 𝑦(t) = x(t) 8.5.2. y(t) = x(t − 5) 8.5.3. y(t) = x(t + 2) + x(3 − t) 8.5.4. y(t) = x(2t − 1) 8.5.5. y(t) = x(t2 ) 8.5.6. y(t) = x2(t) = [x(t)]2 8.5.7. y(t) = 2x(t) + 4 8.5.8. y(t) = 2x(t) + 4t 8.5.9. y(t) = x(t)cos(t) Answers:
- 31. 8.5.1. Linear, Timeinvariant, causal, memoryless, stable 8.5.2. Linear, Time invariant, causal, memory, stable 8.5.3. Linear, Time variant, non-causal, memory, stable 8.5.4. Linear, Time variant, non-causal, memory, stable 8.5.5. Linear, Time variant, non-causal, memory, stable 8.5.6. Non-linear, Time invariant, causal, memoryless, stable 8.5.7. Non-linear, Time invariant, causal, memoryless, stable 8.5.8. Non-linear, Time variant, causal, memoryless, stable 8.5.9. Linear, Time variant, causal, memoryless, stable Sheet 4: Problems solved in the section Book problems: o 1.64 – a, b, g, I, k, l o 1.67
- 32. **Section 10** Summation Rules ∑𝛼 𝑘 𝑁 𝑘=𝑚 = 𝛼 𝑚 − 𝛼 𝑁+1 1 − 𝛼 ∑ 𝛼 𝑘 ∞ 𝑘=𝑚 = 𝛼 𝑚 1 − 𝛼 ∑ 𝑐 𝑁 𝑘=𝑚 = 𝑐(𝑁 − 𝑚 + 1) Convolution Sum 𝑦[ 𝑛] = ∑ 𝑥( 𝑘)ℎ(𝑛 − 𝑘) ∞ −∞ Note: For ℎ[n], n is replaced by 𝑛 − 𝑘. So ℎ[−n] will be ℎ[−(n − k)]. Find the discrete-time convolution sum given the input signals and system impulse response in the following examples: Sheet 2.33-a (p200-183) 2.33-b (p200-183) 2.34-a (p201-184) Sheet-2.33-a (p200-183) 𝑦[𝑛] = 𝑢[𝑛 + 3) ∗ 𝑢[𝑛 − 3] 𝑥[𝑛] = 𝑢[𝑛 + 3), ℎ[𝑛] = 𝑢[𝑛 − 3] The first step to solve the convolution problems is to write the correct formula you are to use: 𝑦[𝑛] = ∑ 𝑥(𝑘)ℎ(𝑛 − 𝑘) ∞ −∞ 𝑦[𝑛] = ∑ 𝑢(𝑘 + 3)𝑢(𝑛 − 𝑘 − 3) ∞ −∞
- 33. At this step,the summation is defined from −∞ to ∞. So we have to substitute in 𝑢(𝑘 + 3)𝑢(𝑛 − 𝑘 − 3) by values like −∞, …, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, … , ∞. Remember that 𝑢(𝑘 + 3) is only defined at (𝑘 + 3) > 0 and this requires k to be in this range - 3, -2, -1, 0, 1, 2, 3, … , ∞. 𝑢(𝑛 − 𝑘 − 3) is only defined at 𝑛 − 𝑘 − 3 ≥ 0 == 𝑘 ≤ 𝑛 − 3 and this requires k to be in this range 3, 4, 5, 6, … , ∞. So why we substitute by values like -5 where nothing of these signals are defined! Ata value like -5 both of these equations gives 0 and we don`t have to add 0 to the summation and loses time. The good solution is what saves time by not adding zeros to the summation. To overcome this problem, the range of the summation should change to just start and end at values defined by the signals. The first part 𝑢(𝑘 + 3) is just defined at 𝑘 ≥ −3 and the second part 𝑢(𝑛 − 𝑘 − 3) is just defined at 𝑘 ≤ 𝑛 − 3. The final range at which at least one of these equations are defined is from −3 to 𝑛 − 3. So the summation should change to reflect the new range. AT first change the lower end of the summation to -3 as shown in the next equation: 𝑦[𝑛] = ∑ 1. 𝑢(𝑛 − 𝑘 − 3) ∞ −3 When the range starts at -3 the signal 𝑢(𝑘 + 3) will always be 1. So we can replace it by the constant 1. Similarly, when the higher end of the summation is replace to be 𝑛 − 3 the summation will have tis form: 𝑦[𝑛] = ∑ 1.1 𝑛−3 −3 When the range ends at 𝑛 − 3 the signal 𝑢(𝑛 − 𝑘 − 3) will always be 1. So we can replace it by the constant 1. Finally the summation will be as follows: 𝑦[𝑛] = ∑ 1 𝑛−3 −3 Using this summation rule: ∑ 𝑐 𝑁 𝑘=𝑚 = 𝑐(𝑁 − 𝑚 + 1) So ∑ 1 𝑛−3 𝑘=−3 = 1. ((𝑛 − 3) − (−3) + 1) = 𝑛 + 1 The final equation of the solution is as follows: 𝑦[𝑛] = { 𝑛 + 1, 𝑛 ≥ 0 0, 𝑛 < 0 Sheet-2.33-b (p200-183) 𝑦[𝑛] = 3 𝑛 𝑢[−𝑛 + 3] ∗ 𝑢[𝑛 − 2] Solution:
- 34. 𝑦[𝑛] = ∑𝑥[𝑘]ℎ[𝑛 − 𝑘] ∞ −∞ 𝑦[𝑛] = ∑ 3 𝑘 𝑢[−𝑘 + 3] ∗ 𝑢[𝑛 − 𝑘 − 2] ∞ −∞ Next is to find the range of the summation. 𝑢[−𝑘 + 3] => 𝑘 ≤ 3 𝑢[𝑛 − 𝑘 − 2] => 𝑘 ≤ 𝑛 − 2 The problem is having two upper limits 𝑘 ≤ 3 & 𝑘 ≤ 𝑛 − 2. The solution to this ambiguous situation is as follows: For 𝑛 − 2 ≤ 3 == 𝑛 ≤ 5, the upper limit is 𝑛 − 2 and the summation is: 𝑦[𝑛] = ∑ 3 𝑘 𝑛−2 −∞ = 3 𝑛−2+1 1 − 3 = 3 𝑛−1 2 = 3 𝑛 . 3−1 2 = 3 𝑛 3 2 = 3 𝑛 6 For 𝑛 − 2 ≥ 4 == 𝑛 ≥ 6, the upper limit is 3 and the summation is: 𝑦[𝑛] = ∑ 3 𝑘 3 −∞ = 33+1 1 − 3 = 34 2 = 81 2 Final result is: 𝑦[𝑛] = { 3 𝑛 6 , 𝑛 ≤ 5 81 2 , 𝑛 ≥ 6 Sheet-2.34-a (p201-184) 𝑦[n] = x[n] ∗ 𝑧[𝑛] In this problem, you are given a graph of two signals and want to find the convolution between them. The first step in the solution is to find the algebraic equation that represents each of these graphs. 𝑥[n] = u[n + 5] − 𝑢[𝑛 − 1] 𝑧[n] = u[n] − 𝑢[𝑛 − 8] + 𝑢[𝑛 − 4] − 𝑢[𝑛 − 8] = u[n] − 2𝑢[𝑛 − 8] + 𝑢[𝑛 − 4] The impulse response here is ℎ[n] = 𝑧[𝑛].
- 35. Then find theconvolution between these equations. Because these signals are discrete so we will find the convolution sum using this equation: 𝑦[𝑛] = ∑ 𝑥(𝑘)ℎ(𝑛 − 𝑘) ∞ −∞ By substituting in the summation: 𝑦[𝑛] = ∑(u[k + 5] − 𝑢[𝑘 − 1]) ∗ (u[n − 𝑘] − 2𝑢[𝑛 − 𝑘 − 8] + 𝑢[𝑛 − 𝑘 − 4]) ∞ −∞ The equation can be represented as follows: 𝑦[𝑛] = ∑ u[k + 5]u[n − 𝑘] − 2 ∑ u[k + 5]𝑢[𝑛 − 𝑘 − 8] ∞ −∞ + ∑ u[k + 5]𝑢[𝑛 − 𝑘 − 4] ∞ −∞ ∞ −∞ − ∑ 𝑢[𝑘 − 1]u[n − 𝑘] ∞ −∞ + 2 ∑ 𝑢[𝑘 − 1]𝑢[𝑛 − 𝑘 − 8] ∞ −∞ − ∑ 𝑢[𝑘 − 1]𝑢[𝑛 − 𝑘 − 4] ∞ −∞ Next is to determine the range of each summation: 𝑦[𝑛] = ∑ u[k + 5]u[n − 𝑘] − 2 ∑ u[k + 5]𝑢[𝑛 − 𝑘 − 8] 𝑛−8 −5 + ∑ u[k + 5]𝑢[𝑛 − 𝑘 − 4] 𝑛−4 −5 𝑛 −5 − ∑ 𝑢[𝑘 − 1]u[n − 𝑘] 𝑛 1 + 2 ∑ 𝑢[𝑘 − 1]𝑢[𝑛 − 𝑘 − 8] 𝑛−8 1 − ∑ 𝑢[𝑘 − 1]𝑢[𝑛 − 𝑘 − 4] 𝑛−4 1 After that we are not in need of the unit-step signals because in these ranges they will be 1. 𝑦[𝑛] = ∑ 1 − 2 ∑ 1 𝑛−8 −5 + ∑ 1 𝑛−4 −5 𝑛 −5 − ∑ 1 𝑛 1 + 2 ∑ 1 𝑛−8 1 − ∑ 1 𝑛−4 1 But how to find the correct ranges at which the output 𝑦[𝑛] is defined? Follow that procedure: 1. Give a number to each summation for indexing. 2. Find the least value among all the ranges and set it as the beginning of the first range. 3. Next is to substitute by this value in the equation to find 𝑦[𝑣] where 𝑣 is set to that value. Reject all the summations that makes the summation invalid. The summation is valid when it have the higher end value greater than or equal to the lower end value. For example 0 and 0 or -3 and 1. 4. The remaining valid summations will be the summations used to calculate the input value. 5. Increment the last value used by 1 and try to find which summations are used to calculate the new output. 6. Whenever a new valid summation is used to calculate the output value other than the summations used to calculate the previous value stop. This is an indication that we reached the end of the previous range started before and reached also the beginning of
- 36. a new range.This recent value is the end of the first range and the start value of the new range. 7. Go back to step 3. Applying that procedure to our example: There are 6 summations and will be indexed by numbers from 1 to 6 where each summation takes an index equal to its position in the equation. The least value across all ranges is -5. We need to find 𝑦[−5]. But make sure that each summation is valid. 𝑦[−5] = ∑ 1 − 2 ∑ 1 −5−8 −5 + ∑ 1 −5−4 −5 −5 −5 − ∑ 1 −5 1 + 2 ∑ 1 −5−8 1 − ∑ 1 −5−4 1 𝑦[−5] = ∑ 1 − 2 ∑ 1 −13 −5 + ∑ 1 −9 −5 −5 −5 − ∑ 1 −5 1 + 2 ∑ 1 −13 1 − ∑ 1 −9 1 The last 5 summations are invalid because the higher end is lower than the lower end. Finally to calculate 𝑦[−5] we will just use the first summation with index 1. Next is to increment the last value -5 by 1 to be -4 and substitute to find the summations used to calculate the output. 𝑦[−4] = ∑ 1 − 2 ∑ 1 −12 −5 + ∑ 1 −8 −5 −4 −5 − ∑ 1 −4 1 + 2 ∑ 1 −12 1 − ∑ 1 −8 1 The valid summation is also the first one. Next is to increment the last value -4 by 1 to be -3 and substitute to find the summations used to calculate the output. 𝑦[−3] = ∑ 1 − 2 ∑ 1 −11 −5 + ∑ 1 −7 −5 −3 −5 − ∑ 1 −3 1 + 2 ∑ 1 −11 1 − ∑ 1 −7 1 The valid summation is also the first one. The process will be repeated until the value of n used is -1. When the value used is -1, the output will be as follows: 𝑦[−1] = ∑ 1 − 2 ∑ 1 −9 −5 + ∑ 1 −5 −5 −1 −5 − ∑ 1 −1 1 + 2 ∑ 1 −9 1 − ∑ 1 −5 1 Here there is a new valid summation used to calculate the output value. It is the third one. Then we reached the end of the previous range started before and got the start of a new range. The previous range is −5 ≤ 𝑛 ≤ −2 𝑜𝑟 − 5 ≤ 𝑛 < −1. The next range starts at -1. Increment -1 by 1 to be 0 and substitute to find the output. After following that procedure the final ranges are: −5 ≤ 𝑛 < −1, 1 −1 ≤ 𝑛 < 1, 1, 3 1 ≤ 𝑛 < 3, 1, 3, 4 3 ≤ 𝑛 < 5, 1, 2, 3, 4 5 ≤ 𝑛 < 9, 1, 2, 3, 4, 6
- 37. 9 ≤ 𝑛,1, 2, 3, 4, 5, 6 Next is to formulate the output of that equation in a better form: 𝑦[𝑛] = ∑ 1 − 2 ∑ 1 𝑛−8 −5 + ∑ 1 𝑛−4 −5 𝑛 −5 − ∑ 1 𝑛 1 + 2 ∑ 1 𝑛−8 1 − ∑ 1 𝑛−4 1 𝑦[𝑛] = { 0 𝑛 + 6 2𝑛 + 8 𝑛 + 8 −𝑛 + 12 −2𝑛 − 16 0 , { 𝑛 < 0 −5 ≤ 𝑛 < −1 −1 ≤ 𝑛 < 1 1 ≤ 𝑛 < 3 3 ≤ 𝑛 < 5 5 ≤ 𝑛 < 9 9 ≤ 𝑛 { 𝑆1 𝑆1, 𝑆3 𝑆1, 𝑆3, 𝑆4 𝑆1, 𝑆3, 𝑆4, 𝑆2 𝑆1, 𝑆3, 𝑆4, 𝑆2, 𝑆6 𝐴𝐿𝐿 Lecture Example 𝑥[n] = u[n] − 𝑢[𝑛 − 10] ℎ[n] = 1 4 (u[n] − 𝑢[𝑛 − 4]) Book Problems Problem 2.2 –b (p130-113) Problem 2.2 –b (p130-113) 𝑦[𝑛] = ( 1 2 ) 𝑛 𝑢[𝑛 − 2] ∗ u[n] Using this equation: 𝑦[𝑛] = ∑ 𝑥(𝑘)ℎ(𝑛 − 𝑘) ∞ −∞ 𝑦[𝑛] = ∑( 1 2 ) 𝑘 𝑢[𝑘 − 2]𝑢[𝑛 − 𝑘] ∞ −∞ 𝑦[𝑛] = ∑( 1 2 ) 𝑘 𝑢[𝑘 − 2]𝑢[𝑛 − 𝑘] 𝑛 2 𝑦[𝑛] = ∑( 1 2 ) 𝑘 𝑛 2 Using this summation rule: ∑ 𝛼 𝑘 𝑁 𝑘=𝑚 = 𝛼 𝑚 − 𝛼 𝑁+1 1 − 𝛼 𝑦[𝑛] = ∑( 1 2 ) 𝑘 𝑛 𝑘=2 = ( 1 2)2 − ( 1 2) 𝑛+1 1 − 1 2
- 38. 𝑦[𝑛] = 1 4 −( 1 2)( 1 2) 𝑛 1 2 𝑦[𝑛] = 1 2 − ( 1 2 ) 𝑛 The ranges are as follows: 𝑦[𝑛] = { 0, 𝑛 < 2 1 2 − ( 1 2 ) 𝑛 , 𝑛 ≥ 2 Convolution Integral 𝑦( 𝑡) = ∫ 𝑥(Τ)h(t − Τ)dΤ ∞ −∞ Note: For ℎ(𝑡), n is replaced by 𝑡 − 𝑇. So ℎ[−𝑡] will be ℎ[−(t − T)]. Sheet 2.39-a (p202-185) 2.40-a (p202-185) 2.39-a (p202-185) 𝒚(𝒕) = (𝒖(𝒕) − 𝒖(𝒕 − 𝟐)) ∗ 𝒖(𝒕) Following the convolution integral equation: 𝑦(𝑡) = ∫ 𝑥(Τ)h(t − Τ)dΤ ∞ −∞ 𝑦(𝑡) = ∫ 𝑥(𝑢(𝑡) − 𝑢(𝑡 − 2))𝑢(𝑡)dΤ ∞ −∞ 𝑦(𝑡) = ∫ 𝑢(Τ)𝑢(𝑡 − Τ) − ∫ 𝑢(Τ − 2)𝑢(𝑡 − Τ)dΤ ∞ −∞ ∞ −∞ 𝑦(𝑡) = ∫ 𝑢(Τ)𝑢(𝑡 − Τ)dΤ − ∫ 𝑢(Τ − 2)𝑢(𝑡 − Τ)dΤ 𝑡 2 𝑡 0 𝑦(𝑡) = ∫ 𝑑Τ − ∫ dΤ 𝑡 2 𝑡 0 Ranges are as follows: 𝑦(𝑡) = { 0 𝑡 2 { 𝑡 < 0 0 ≤ 𝑡 < 2 2 ≤ 𝑡 2.40-a (p202-185) 𝑚(𝑡) = 𝑥(𝑡) ∗ 𝑦(𝑡)
- 39. 𝑥(𝑡) = 𝑢(𝑡+ 1) − 𝑢(𝑡 − 1) 𝑦(𝑡) = 𝑢(𝑡) − 𝑢(𝑡 − 4) + 𝑢(𝑡 − 2) + 𝑢(𝑡 − 4) 𝑦(𝑡) = 𝑢(𝑡) − 2𝑢(𝑡 − 4) + 𝑢(𝑡 − 2) 𝑦(𝑡) = ∫ 𝑥(Τ)h(t − Τ)dΤ ∞ −∞ 𝑦(𝑡) = ∫ (𝑢(𝑡 + 1) − 𝑢(𝑡 − 1))(𝑢(𝑡) − 2𝑢(𝑡 − 4) + 𝑢(𝑡 − 2))dΤ ∞ −∞ 𝑦(𝑡) = ∫ 𝑢(Τ + 1)𝑢(𝑡 − Τ)dΤ − 2 ∫ 𝑢(Τ + 1)𝑢(𝑡 − Τ − 4)dΤ ∞ −∞ ∞ −∞ + ∫ 𝑢(Τ + 1)𝑢(𝑡 − Τ − 2)dΤ ∞ −∞ − ∫ 𝑢(Τ − 1)𝑢(𝑡 − Τ)dΤ ∞ −∞ + 2 ∫ 𝑢(Τ − 1)𝑢(𝑡 − Τ − 4)dΤ ∞ −∞ − ∫ 𝑢(Τ − 1)𝑢(𝑡 − Τ − 2)dΤ ∞ −∞ 𝑦(𝑡) = ∫ 𝑢(Τ + 1)𝑢(𝑡 − Τ)dΤ − 2 ∫ 𝑢(Τ + 1)𝑢(𝑡 − Τ − 4)dΤ 𝑡−4 −1 𝑡 −1 + ∫ 𝑢(Τ + 1)𝑢(𝑡 − Τ − 2)dΤ 𝑡−2 −1 − ∫ 𝑢(Τ − 1)𝑢(𝑡 − Τ)dΤ 𝑡 1 + 2 ∫ 𝑢(Τ − 1)𝑢(𝑡 − Τ − 4)dΤ 𝑡−4 1 − ∫ 𝑢(Τ − 1)𝑢(𝑡 − Τ − 2)dΤ 𝑡−2 1 𝑦(𝑡) = ∫ dΤ − 2 ∫ dΤ 𝑡−4 −1 + ∫ dΤ 𝑡−2 −1 − ∫ dΤ 𝑡 1 + 2 ∫ dΤ 𝑡−4 1 − ∫ dΤ 𝑡−2 1 𝑡 −1 Ranges are: 𝑦[𝑛] = { 0 𝑡 + 1 𝑡 + 1 10 − 2𝑡 0 , { 𝑡 < −1 −1 ≤ 𝑡 < 1 1 ≤ 𝑡 < 3 3 ≤ 𝑡 < 5 5 ≤ 𝑡 { 𝐼1 𝐼1, 𝐼3, 𝐼4 𝐼1, 𝐼3, 𝐼4, 𝐼2, 𝐼6 𝐼1, 𝐼3, 𝐼4, 𝐼2, 𝐼6, 𝐼5 Lecture Example Book Problems Problem 2.6 (p141-124) Sheet 5 Book Problems
- 40. o Problem 2.2– a, b, c, d, f (p130-113) o Problem 2.5 (p140-123) o Problem 2.6 (p141-124) Sheet Problems o 2.33-a, b, c, e, i o 2.34-all o 2.39-a, b, d, e, f, i, j, k o 2.40-all except (o)
- 41. ***Section 11*** Unit Impulseas System Impulse Response The effect of using the unit impulse signal over an input signal is to time shift the input signal by the time shift value in the unit impulse. 𝑥(𝑡) ∗ δ(t − τ) = 𝑥(𝑡 − τ) Example 11.1: Find the convolution integral 𝑦(𝑡) when the input signal is 𝑥(𝑡) = 𝑢(𝑡 − 3) and the system impulse response is ℎ(𝑡) = δ(t + 2) + δ(t − 4). Solution: 𝑦( 𝑡) = 𝑥( 𝑡) ∗ ℎ(𝑡) 𝑦( 𝑡) = 𝑢( 𝑡 − 3) ∗ [δ(t + 2) + δ(t − 4)] Using the distributive property of LTI systems: 𝑦( 𝑡) = 𝑢( 𝑡 − 3) ∗ δ(t + 2) + u(t) ∗ δ(t − 4) Using the convolution by unit impulse property: 𝑦( 𝑡) = 𝑢( 𝑡 − 3 + 2) + u(t − 4) 𝑦( 𝑡) = 𝑢( 𝑡 − 1) + u(t − 4) System Properties Each property of the LTI systems will apply for both discrete-time and continuous-time signals. Distributive 𝑥[𝑛] ∗ [ℎ1[𝑛] ± ℎ2[𝑛]] = 𝑥[𝑛] ∗ ℎ1[𝑛] ± 𝑥[𝑛] ∗ ℎ2[𝑛] 𝑥( 𝑡) ∗ [ℎ1( 𝑡) ± ℎ2( 𝑡)] = 𝑥( 𝑡) ∗ ℎ1( 𝑡) ± 𝑥( 𝑡) ∗ ℎ2(𝑡) Example 11.2: Using the distributive property how to construct the same output illustrated in the following diagram using only one circuit? Just draw the block diagram. Answer: Using the distributive property the system can be designed using a cascaded diagram rather than parallel to have just a single circuit as shown in the following diagram.
- 42. Associative (𝑥[ 𝑛] ∗ℎ1[ 𝑛]) ∗ ℎ2[ 𝑛] = 𝑥[ 𝑛] ∗ (ℎ1[ 𝑛] ∗ ℎ2[ 𝑛]) [𝑥( 𝑡) ∗ ℎ1( 𝑡)] ∗ ℎ2( 𝑡) = 𝑥[ 𝑛] ∗ [ℎ1( 𝑡) ∗ ℎ2( 𝑡)] Example 11.3: Using the associative property how to construct the same output illustrated in the following diagram using only one circuit? Just draw the block diagram. Answer: Using the associative property the system can be designed as shown in the following diagram to have just a single circuit. Commutative ℎ1[ 𝑛] ∗ ℎ2[ 𝑛] = ℎ2[ 𝑛] ∗ ℎ1[ 𝑛] ℎ1(𝑡) ∗ ℎ2(𝑡) = ℎ2(𝑡) ∗ ℎ1(𝑡) Example 11.4: Using the following block diagram find the overall system impulse response. Steps to get the impulse response is as follows: 1. Divide each parallel connection into two serial connections
- 43. 2. If anyof these serial connections have a parallel connection divide it until reaching the most inner serial connection. 3. Combine all of the impulse responses in each serial connection to create a single response using convolution. 4. When the individual serial connections in the parallel connection have single response combine the parallel connection serial connections into a single response using summation. 5. Go to the higher parallel connection and repeat steps 1 to 4 until having a single response for the overall system. Solution:
- 44. Sheet Problems: 2.46-a(p205-188) 2.47-a(p205-188)
- 45. Linear Time Invariant(LTI) Systems Properties Memoryless The discrete-time LTI system is memoryless if and only if: ℎ[𝑘] = 𝑐δ[k] The continuous-time LTI system is memoryless if and only if: ℎ(𝑡) = 𝑐δ(t) Causality The discrete-time LTI system is causal if and only if: ℎ[𝑘] = 0, 𝑘 < 0 The continuous-time LTI system is causal if and only if: ℎ(𝑡) = 0, 𝑡 < 0 Stability The discrete-time LTI system is stable if and only if: )(nh The continuous-time LTI system is stable if and only if: dtth )( Note that an impulse response with this part 𝑎 𝑏 is stable only if a<1 or b<0. Examples )( 2 1 )( )1()( )(2)( )()( 2 nunh nuenh nunh nunh n n n n First one is stable only if < 1. Second one is unstable because 2 > 1. Third one is unstable because e > 1. Fourth one is stable because 1/2 < 1. Lecture Examples:
- 46. )( 2 1 )( )1()( )(2)( )()( )( )1(2)()( )1()1()( )()( 2 2 nunh nuenh nunh tueth eth tututh tututh nunh n n n at t n System Block Diagram BookProblems: 2.9-(p149-132) Sheet Problems: 2.54-b Section 11 Sheet Problems Solved in the Section Book Problems o 2.8-(p149-132) o 2.9-(p149-132) o 2.10-(p153-136) Book Sheet o 2.46 o 2.47 o 2.48 o 2.49 o 2.57: Just draw the block diagram of the system.
- 47. ***Section 12*** Introduction toFourier Analysis We have previously discussed the elementary signals in digital signal processing like unit step, unit impulse, ramp, sinusoidal, and exponential. We also said that the basic functions of these signals are to create new signals or to process an existing signal. In the previous two sections we knew how a signal (DT/CT) can be represented using the unit impulse (delta) signals. In this section we will know how to represent a signal (DT/CT) using the sinusoidal signal. Fourier theory is what represents the signals using only sine signals. When the signal is composed of sine waves we move from the time domain to the frequency domain. The following figure illustrates how useful is the frequency domain. Example to show importance of the frequency domain The following example illustrates the benefits of using the frequency domain. Assume we need to delete the green and red signals from the following signal and leave only the blue signal. How to do that task in both time and frequency domain? It seems impossible to do such task in time domain because there is overlab among all the three signals in time domain. The solution is to use the frequency domain. The previous figure is not what appears on the final result but a one like this appears. We use the previous figure just for simplification.
- 48. The individual signalsare shown as follows:
- 49. The general solutionusing the frequency domain is as follows: Investigate the signal and find the different frequencies constituting the signal. Assume there are three signals 𝐴, 𝐵, 𝐶 with the frequencies 𝐹𝐴, 𝐹𝐵, 𝑎𝑛𝑑 𝐹𝐶 respectively and you need the 𝐴 signal and remove the other signals. Because the 𝐴 signal has the 𝐹𝑎 frequency then the solution is to create a filter that passes the 𝐹𝑎 frequency and stop the other two frequencies.
- 50. For our example,the solution is to know the frequencies of both the red and green signals and cut-off these frequencies to leave only the blue signal. The red signal has a frequency equal to .3Hz and the green signal has a frequency equal to 1Hz. Using a filter we can remove these frequencies to leave only the blue signal. Fourier Representations There are four representations to Fourier each targets a class of signals. The classes are based on whether the signal is DT or CT and whether it is periodic or non-periodic. We will use the FT class. Now there is a fast version of the FT called Fast FT (FFT). Fourier Transform Moving between Time-Domain and Frequency-Domain For a discrete-time signal 𝑥(𝑡), the Fourier transform (FT) of the signal 𝑋(𝑗𝜔) is computed by the following equation: 𝑋(𝑗𝜔) = ∫ 𝑥(𝑡)𝑒−𝑗𝜔𝑡 ∞ −∞ For a given frequency-domain representation 𝑋(𝑗𝜔), the time-domain version 𝑋(𝑡) is computed by the inverse Fourier transform (IFT) is as follows: 𝑥(𝑡) = 1 2𝜋 ∫ 𝑋(𝑗𝜔)𝑒−𝑗𝜔𝑡 ∞ −∞ Note: 𝜔 = 2𝜋𝑓 ∫ 𝑒 𝑡 𝑏 𝑎 𝑑𝑡 = 𝑒 𝑘 𝑘 A special signal to find the FT of is delta. 𝐹𝑇{δ( 𝑡)} = 1 Useful Rules 𝑒 𝑗𝜃 = cos( 𝜃) + 𝑗𝑠𝑖𝑛(𝜃) cos(𝑡) = ( 𝑒𝑗𝑡 + 𝑒−𝑗𝑡) 2 sin(𝑡) = ( 𝑒𝑗𝑡 − 𝑒−𝑗𝑡) 2𝑗
- 51. cosh(𝑡) = ( 𝑒𝑡+ 𝑒−𝑡) 2 sinh(𝑡) = ( 𝑒𝑡 − 𝑒−𝑡) 2 Some uses of the Fourier transform in signal processing Able to find the frequency components of the signal. Convolution for time-domain signals is equivalent to multiplication in frequency- domain: 𝑧(𝑡) = 𝑥(𝑡) ∗ ℎ(𝑡) 𝑍(𝑡) = 𝑋(𝑡). 𝐻(𝑡) Using FFT, it is faster to implement the convolution in frequency-domain rather than time-domain. Signal Energy. Signal Power. Exercises FT 3.14-e – (p262-245) IFT 𝑋(𝑗𝜔) = 𝑒−2𝜔 𝑢(𝜔)
- 52. ***Section 13*** 𝑋( 𝑧)= ∑ 𝑥(𝑘)𝑧−𝑘 ∞ −∞ Examples Example 1 Find Z-transform of this signal: 𝑥(𝑛) = { 𝑎 𝑛 , 𝑛 ≥ 0 0, 𝑂. 𝑊 𝑥(𝑛) = 𝑎 𝑛 𝑢(𝑛) Using this equation to find the Z-transform: 𝑋(𝑧) = ∑ 𝑥(𝑘)𝑧−𝑘 ∞ −∞ 𝑋(𝑧) = ∑ 𝑎 𝑘 𝑢(𝑘)𝑧−𝑘 ∞ −∞ 𝑋(𝑧) = ∑ 𝑎 𝑘 𝑧−𝑘 ∞ 0 𝑋(𝑧) = ∑(𝑎𝑧−1 ) 𝑘 ∞ 0 𝑋(𝑧) = ∑( 𝑎 𝑧 ) 𝑘 ∞ 0 Using this equation to find the summation result: ∑ 𝛼 𝑘 = ∞ 𝑘=𝑚 𝛼 𝑚 1 − 𝛼 𝑋(𝑧) = ∑ ( 𝑎 𝑧 ) 𝑘 = ( 𝑎 𝑧) 0 1 − 𝑎 𝑧 ∞ 0 = 1 1 − 𝑎 𝑧 By multiplying the nominator and denominator by 𝑧: 𝑋(𝑧) = 𝑧 𝑧 − 𝑎 ROC is the region at which 𝑋(𝑧) is defined. So it is required to avoid values that make it undefined like ∞. From this equation to be defined: 𝑋(𝑧) = ∑( 𝑎 𝑧 ) 𝑘 ∞ 0 The | 𝑎 𝑧 | value must be less than 1. So ROC is | 𝑎 𝑧 | < 1 𝑂𝑅 |𝑧| > |𝑎|.
- 53. Next is tofind poles and zeros. Zeros are the values that makes 𝑋(𝑧) equal to zero. In other words the zeros are the values that makes the nominator equal to 0. Zeros in this case are just 𝑧 = 0. Poles are the values that makes the denominator equal to 0. Poles in this case are just 𝑧 = 𝑎. Example 2 Find poles and zeros of the following example: 𝑋(𝑧) = 𝑧(𝑧 − 1 3) (𝑧 − 2)(𝑧 + 1 2) To find zeros: 𝑧 (𝑧 − 1 3 ) = 0 𝑧 = 0 𝑂𝑅 (𝑧 − 1 3 ) = 0 Zeros are 0 and 1 3 . To find poles: (𝑧 − 2) (𝑧 + 1 2 ) = 0 (𝑧 − 2) = 0 𝑂𝑅 (𝑧 + 1 2 ) = 0 So poles are 2 and − 1 2 . Example 3 Find Z-transform of this signal: 𝑥(𝑛) = −𝑎 𝑛 𝑢(−𝑛 − 1) Using this equation to find the Z-transform: 𝑋(𝑧) = ∑ 𝑥(𝑘)𝑧−𝑘 ∞ −∞ 𝑋(𝑧) = − ∑ 𝑎 𝑘 𝑢(−𝑘 − 1)𝑧−𝑘 ∞ −∞ 𝑋(𝑧) = − ∑ 𝑎 𝑘 𝑧−𝑘 1 −∞ 𝑋(𝑧) = − ∑(𝑎𝑧−1 ) 𝑘 1 −∞ = − ∑(𝑎𝑧−1 )−𝑘 ∞ 1 = − ∑(𝑎−1 𝑧) 𝑘 ∞ 1 𝑋(𝑧) = − ∑( 𝑧 𝑎 ) 𝑘 ∞ 1 Using this equation to find the summation result:
- 54. ∑ 𝛼 𝑘 = ∞ 𝑘=𝑚 𝛼𝑚 1 − 𝛼 𝑋(𝑧) = − ∑ ( 𝑎 𝑧 ) 𝑘 = − ( 𝑧 𝑎) 1 1 − 𝑧 𝑎 ∞ 0 = − 𝑧 𝑎 1 − 𝑧 𝑎 By multiplying the nominator and denominator by 𝑎: 𝑋(𝑧) = − 𝑧 𝑎 − 𝑧 = 𝑧 𝑧 − 𝑎 ROC is the region at which 𝑋(𝑧) is defined. So it is required to avoid values that make it undefined like ∞. From this equation to be defined: 𝑋(𝑧) = − ∑( 𝑧 𝑎 ) 𝑘 ∞ 1 The | 𝑧 𝑎 | value must be less than 1. So ROC is | 𝑧 𝑎 | < 1 𝑂𝑅 |𝑧| < |𝑎|. Next is to find poles and zeros. Zeros are the values that makes 𝑋(𝑧) equal to zero. In other words the zeros are the values that makes the nominator equal to 0. Zeros in this case are just 𝑧 = 0. Poles are the values that makes the denominator equal to 0. Poles in this case are just 𝑧 = 𝑎. Example 4 Find Z-transform of each of this signal: 𝑥(𝑛) = −( 1 3 ) 𝑛 𝑢(−𝑛 − 1) Using this equation to find the Z-transform: 𝑋(𝑧) = ∑ 𝑥(𝑘)𝑧−𝑘 ∞ −∞ 𝑋(𝑧) = − ∑( 1 3 ) 𝑘 𝑢(−𝑘 − 1)𝑧−𝑘 ∞ −∞ 𝑋(𝑧) = − ∑( 1 3 ) 𝑘 𝑧−𝑘 1 −∞ 𝑋(𝑧) = − ∑( 1 3 𝑧−1 ) 𝑘 1 −∞ = − ∑( 1 3 𝑧−1 )−𝑘 ∞ 1 = − ∑(3𝑧) 𝑘 ∞ 1 𝑋(𝑧) = − ∑(3𝑧) 𝑘 ∞ 1 Using this equation to find the summation result: ∑ 𝛼 𝑘 = ∞ 𝑘=𝑚 𝛼 𝑚 1 − 𝛼
- 55. 𝑋(𝑧) = −∑ ( 𝑧 3 ) 𝑘 = − (3𝑧)1 1 − 𝑧 3 ∞ 0 = − 3𝑧 1 − 3𝑧 By dividing the nominator and denominator by 3: 𝑋(𝑧) = − 𝑧 1 3 − 𝑧 = 𝑧 𝑧 − 1 3 ROC is the region at which 𝑋(𝑧) is defined. So it is required to avoid values that make it undefined like ∞. From this equation to be defined: 𝑋(𝑧) = − ∑(3𝑧) 𝑘 ∞ 1 The |3𝑧| value must be less than 1. So ROC is |3𝑧| < 1 𝑂𝑅 |𝑧| < | 1 3 |. Next is to find poles and zeros. Zeros are the values that makes 𝑋(𝑧) equal to zero. In other words the zeros are the values that makes the nominator equal to 0. Zeros in this case are just 𝑧 = 0. Poles are the values that makes the denominator equal to 0. Poles in this case are just 𝑧 = 1 3 .
- 56. GUI Task Create aMATLAB GUI application as shown in the following diagram: The application have 8 buttons divided as 2 buttons per group. Each group performs an operation and its opposite. Each button shows both the discrete and continuous versions of the signal except for the Continuous & Discrete group it will show only Continuous or Discrete signal. For example clicking on APeriodic button will show a non-periodic signal in both continuous and discrete forms as shown in the following diagram: