1. Discrete Convolution of Two Signals
In this animation, the discrete time convolution of two signals is discussed. Convolution is the operation to obtain response of a linear system to input x[n]. Considering the input x[n] as the sum of shifted and scaled impulses, the output will be the superposition of the scaled responses of the system to each of the shifted impulses.
Course Name: Signals and Systems Level: UG
Authors
Phani Swathi
Mentor
Prof. Saravanan Vijayakumaran

2. Learning Objectives
After interacting with this Learning Object, the learner will be able to:
Explain the convolution of two discrete time signals

3. 1 2 3 4 5 Definitions of the components/Keywords:
Convolution of two signals:
The x[n] and h[n] are the two discrete signals to be convolved.
The convolution of two signals is denoted by
which means
where k is a dummy variable. 2 3 4 5

4. 1 Master Layout (Part 1) 2 3 4 5 This animation consists of 2 parts:
Part 1 – First method of convolution – Method 1
Part 2 – Alternate method of convolution – Method 2 1
Lines and dots have to appear at the same time. This is applicable to all figures
Signals taken to convolve
Shifted version of h[n]
Scaled version of h[n] 2 3 4 5

5. 3 Step 1: 1 2 4 5 Instruction for the animator
Text to be displayed in the working area (DT)
The first two points in DT has to appear before the figures.
Then the blue figure has to appear.
After that the red figure has to appear.
Then last two points in DT has to appear after the blue and red figures.
x[n] and h[n] are the two discrete signals to be convolved.
The convolution of the signals is denoted by
which means
where k is a dummy variable.
x[n] has non-zero discrete impulses at x[1], x[2] and x[3].
Therefore, y [n] can now be computed as
y[n]= x[1]h[n-1]+x[2]h[n-2]+x[3]h[n-3] ; n ≥ 1 4 5

6. 3 Step 2: 1 2 4 5 X[1]h[n-1] 3 2 1 0 1 2 3 n X[2]h[n-2] h[n-2] 6 4 2 3
Overall calculation of y[n] 1
X[1]h[n-1] 3 2 1 2 n
X[2]h[n-2]
h[n-2] 6 4 2 3 2 1 3 n n
h[n-3] 3 2 1
X[3]h[n-3] 3 2 1 4 n n
Instruction for the animator
Text to be displayed in the working area (DT)
The figures has to appear row wise. First the blue fig. then red fig. and then the green figure has to appear.
After the 9 figures appear row wise, all the fig. should appear at a time in the fashion shown above.
x[n] has non-zero discrete impulses at x[1], x[2] and x[3].
The summation of all the products of x[k]h[n-k] gives y[n].
From definition, y[n] is the superposition of the scaled responses of the system to each of the shifted impulses. 5

7. Step 3: 1
X[1]h[n-1]
x[1]h[n-1] 3 2 1 2 3 n
Instruction for the animator
Text to be displayed in the working area (DT)
Show the fig in red and then
X[1]h[n-1] has to appear
Then the green figure has to appear
After that the sentence “the scaled signal x[1]h[n-1]” should appear
The response due to the input x[k] applied at time k
The time shift of h[n] is taken
Then scaling of h[n-1] with x[1] is done.
The signal x[1]h[n-1] is the same as h[n-1] since x[1]=1. 4 5

8. Step
Step 4: 1 1
X[2]h[n-2] 6 5 4 3 2 1
h[n-2]
X[2] h[n-2] 3 2 1 2 2 3 n n
Instruction for the animator
Text to be displayed in the working area (DT)
First show the figure in red and then labeling h[n-2] should appear
Then the green figure has to appear
After that the sentence “the scaled signal x[2]h[n-2]” should appear
The next time shift of h[n-1] is taken
Then scaling of h[n-2] with x[2] is done.
The signal x[2]h[n-2] is the doubled version of h[n-2] since x[2]=2. 4 4 5 5

9. Step 5: 1
X[3]h[n-3]
h[n-3]
x[3]h[n-3] 3 2 1 3 2 1 2 3 n n
Instruction for the animator
Text to be displayed in the working area (DT)
First show the figure in red and then labeling h[n-3] should appear
Then the green figure has to appear
After that the sentence “the scaled signal x[3]h[n-3]” should appear
The next time shift of h[n-3] is taken
Then scaling of h[n-3] with x[3] is done.
The signal x[3]h[n-3] is the same as h[n-3] since x[3]=1. 4 5

10. Step 6:
Y[n] 1 1 8 7 6 5 4 3 2 1
Y[n] 2 2 3 n
Instruction for the animator
Text to be displayed in the working area (DT)
First the symbol y[n] must appear and then the figure in green must appear.
The output of the system y[n] =x[1]h[n-1]+x[2]h[n-2]+x[3]h[n-3] 4 4 5 5

11. Test your understanding Instructions/ Working area
Electrical Engineering
Use STAM template
Slide 1
Slide 3
Slide 26
Slide 28
Slide 27
Introduction
Definitions
Analogy
Test your understanding
(questionnaire)‏
Lets Sum up (summary)‏
Want to know more…
(Further Reading)‏
Interactivity:
The correct answer is shown in red
Try it yourself
Fig. A
Fig. B
Fig. C
Fig. 1
Fig. 2
Demo
Activity
Fig. A
Fig. C
Fig. B 11
Instructions/ Working area
Credits

12. 3 1 2 4 5 Interactivity option 1: Step No 1: y[n] h[n] x[n] 3 2 1 3 2
Fig. 1
Fig. 2 n n
Fig.1
Fig.2
Interactivit y type (IO 1/IO 2)
Instruction to learners
Boundar y limits & options
Instruction to animators
Results and output
Choose from fig. A,B & C.
Find the convolution of x[n] and h[n]
Show fig 1, fig.2 in the question part.
show fig. B as the output. 4 5

13. 3 1 2 4 5 Interactivity option 1: Step No 2: y[n] Y[n] Y[n] 12 10 8 6
Fig. 1
Fig. 2 n n
a) Fig. A
b) Fig. B
Interactivit y type (IO 1/IO 2)
Instruction to learners
Boundar y limits & options
Instruction to animators
Results and output
Find the convolution of x[n] and h[n]
Show fig. A, fig. B and fig. C as the options
show fig. B as the output. 4 5

14. 3 1 2 4 5 Interactivity option 1: Step No 3: y[n] Y[n] 12 10 8 6 4 2
Fig. 1
Fig. 2 n
c) Fig. C
Interactivit y type (IO 1/IO 2)
Instruction to learners
Boundar y limits & options
Instruction to animators
Results and output
show option B as the output.
If the user chooses ‘B’ remark “correct answer” should appear if not “wrong answer” should appear. 4 5

15. 1 Master Layout (Part 2)‏ 2 3 4 5 h[n] X[n] 1 2 3
This animation consists of 2 parts:
Part 1 – First method of Convolution – Method 1
Part 2 – Alternate method of Convolution – Method 2
Signals taken to convolve
h[n]
X[n] 2
…… ……. 3
Result signal of convolution
The dotted lines represent y[n]
The thicker line represents only y[4] 4 5

16. Step 1: 1
X[n]
h[n] 2 2 n 3
… …..
Instruction for the animator
Text to be displayed in the working area (DT)
First sentence in DT is to appear before the figures.
Then the blue fig. has to appear and then the red fig.
The text in last two sentences in DT has to appear after the blue figure.
x[n] and h[n] are the two signals taken to convolve.
x[k] has 7 non-zero impulses from -3 to +3 with an amplitude of 2. so, it is difficult to use method 1 .
Then Y[n] is calculated using formula. 4 5

17. 3 Step 2: 1 2 4 5 h[n-k] X[k] 3 2 2 1 k=n-2 k=n-1 k= n k
Instruction for the animator
Text to be displayed in the working area (DT)
First the blue fig. has to appear then the red fig.
The text in first two sentences in DT has to appear after the blue figure.
The sentences from 3 onwards in DT should appear after the red fig.
h[n-k] is the time reversal and shifted version of h[n] as shown in the figure.
h[n-k] is h[0] when k= n
Similarly, h[n-k] is h[1] when k=n-1
h[n-k] is h[2] when k= n-2 and so on. 4 5

18. 3 Step 3: 1 2 4 5 Instruction for the animator
Text to be displayed in the working area (DT)
The text in DT has to appear after the blue and red figures in slide 16.
If x[n] or h[n] have large number of impulses then method 1 is probably inconvenient to use.
Here y[n] can be calculated easily if the boundary conditions are known.
If x[k] is non-zero between and
h[k] is non-zero between
and h[k] has N impulses, then
From this , it is understood that y[n] is zero
if i.e,
and i.e, 3 4 5

19. Step 4: 1
X[k]
h[4-k] 2 k 3
Instruction for the animator
Text to be displayed in the working area (DT)
First the blue fig. and then the red fig. has to appear.
The text in DT has to appear after the red fig.
h[4-k] is the time reversal and shifted version of h[n] as shown in the figure. 4 5

20. 3 Step 5: 1 2 4 5 X[k]h[4-k] 6 4 2 2 3 4 Instruction for the animator 3
Instruction for the animator
Text to be displayed in the working area (DT)
The text in DT has to appear after the red fig.
Y[4] can be calculated by summing all the values of product of x[k]h[4-k].
i.e, 4 5

21. Step 6:
Y[4] 1 1 12 10 8 6 4 2
Y[4] 2 2 3 n
Instruction for the animator
Text to be displayed in the working area (DT)
First the symbol y[4] must appear and then the figure in green must appear.
Both the lines and the balls should appear at a time.
After the figure, the text in DT has to appear.
For example, to find y[4]
The output of the system y[4] is the summation of x[k]h[n-k]
That is given as
y[4] = x[0]h[4-0]+x[1]h[4-1]+x[2]h[4-2]+x[3]h[4-3]+x[4]h[4-4]
The dotted lines represent the solution for y[n]. 4 4 5 5

22. Test your understanding Instructions/ Working area
Electrical Engineering
Use STAM template
Slide 1
Slide 3
Slide 26
Slide 28
Slide 27
Introduction
Definitions
Analogy
Test your understanding
(questionnaire)‏
Lets Sum up (summary)‏
Want to know more…
(Further Reading)‏
Interactivity:
Try it yourself
Fig. a
Fig. b
Fig. c
Fig. 3
Fig. 4
Demo
Activity
Fig. a
Fig. b
Fig. c 22
Instructions/ Working area
Credits

23. 3 1 2 4 5 Interactivity option 1: Step No 1: y[n] x[n] h[5-k] 3 2 1 4
Fig. 1
Fig. 2 n n
Fig.3
Fig.4
Interactivit y type (IO 1/IO 2)
Instruction to learners
Boundar y limits & options
Instruction to animators
Results and output
Choose from fig. a, b & c
Find the value of y[5]
Show fig 3, fig.4 in the question part.
show fig. c as the output. 4 5

24. 3 1 2 4 5 Interactivity option 1: Step No 2: y[n] Y[5] Y[5] 4 3 2 1 3
Fig. 1
Fig. 2 n n
a) Fig. a
b) Fig. b
Interactivit y type (IO 1/IO 2)
Instruction to learners
Boundar y limits & options
Instruction to animators
Results and output
Hint:
h[5-k] = h[1]
When k=4
Show fig. a, fig. b and fig. c as the options
show fig. c as the output. 4 5

25. 3 1 2 4 5 Interactivity option 1: Step No 3: y[n] Y[5] 4 3 2 1 Fig. 1
Fig. 1
Fig. 2 n
c) Fig. c
Interactivit y type (IO 1/IO 2)
Instruction to learners
Boundar y limits & options
Instruction to animators
Results and output
show option c as the output.
If the user chooses ‘c’ remark “correct answer” should appear if not “wrong answer” should appear. 4 5

26. 4 Questionnaire 1 2 3 5 1. Find the value of y[6]
Answers: a) b)
2.The Convolution sum is given as ___________
Answers: ‏
The correct answers are given in red.
h[n]
x[n] 2 10 8 6 4 2 3 8 6 4 2 4 5

27. Links for further reading
Reference websites:
Books:
Signals & Systems – Alan V. Oppenheim, Alan S. Willsky, S. Hamid Nawab, PHI learning, Second edition.
Research papers:

28. Summary
In discrete time, the representation of signals is taken to be the weighted sums of shifted unit impulses.
This representation is important, as it allows to compute the response of an LTI(Linear Time Invariant) system to an arbitrary input in terms of the system’s response to a unit impulse.
The convolution sum of two discrete signals is represented as
where
The convolution sum provides a concise, mathematical way to express the output of an LTI system based on an arbitrary discrete-time input signal and the system‘s response.