Discrete Convolution of Two Signals

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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 xn. Considering the input xn 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
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Learning Objectives

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

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Definitions of the components/Keywords
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• Convolution of two signals
• The xn and hn are the two discrete signals to
  be convolved.
• The convolution of two signals is denoted by
• which means
• where k is a dummy variable.

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Master Layout (Part 1)
This animation consists of 2 parts Part 1
First method of convolution Method 1 Part 2
Alternate method of convolution Method 2
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Lines and dots have to appear at the same time.
This is applicable to all figures
Signals taken to convolve
Shifted version of hn
Scaled version of hn
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Step 1
1
2
3
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. xn and hn are the two discrete signals to be convolved. The convolution of the signals is denoted by which means where k is a dummy variable. xn has non-zero discrete impulses at x1, x2 and x3. Therefore, y n can now be computed as yn x1hn-1x2hn-2x3hn-3 n 1
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Step 2
Overall calculation of yn
1
X1hn-1
3 2 1
2
0 1 2 3 n
X2hn-2
hn-2
6 4 2
3 2 1
3
0 1 2 3 4 n
0 1 2 3 4 n
hn-3
3 2 1
X3hn-3
3 2 1
4
0 1 2 3 4 5 n
0 1 2 3 4 5 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. xn has non-zero discrete impulses at x1, x2 and x3. The summation of all the products of xkhn-k gives yn. From definition, yn is the superposition of the scaled responses of the system to each of the shifted impulses.
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Step 3
1
X1hn-1
x1hn-1
3 2 1
2
0 1 2 3 n
3
Instruction for the animator Text to be displayed in the working area (DT)
Show the fig in red and then X1hn-1 has to appear Then the green figure has to appear After that the sentence the scaled signal x1hn-1 should appear The response due to the input xk applied at time k The time shift of hn is taken Then scaling of hn-1 with x1 is done. The signal x1hn-1 is the same as hn-1 since x11.
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Step 4
Step
1
1
X2hn-2
6 5 4 3 2 1
hn-2
X2 hn-2
3 2 1
2
2
3
0 1 2 3 4 n
0 1 2 3 4 n
Instruction for the animator Text to be displayed in the working area (DT)
First show the figure in red and then labeling hn-2 should appear Then the green figure has to appear After that the sentence the scaled signal x2hn-2 should appear The next time shift of hn-1 is taken Then scaling of hn-2 with x2 is done. The signal x2hn-2 is the doubled version of hn-2 since x22.
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Step 5
1
X3hn-3
hn-3
x3hn-3
3 2 1
3 2 1
2
3
0 1 2 3 4 5 n
0 1 2 3 4 5 n
Instruction for the animator Text to be displayed in the working area (DT)
First show the figure in red and then labeling hn-3 should appear Then the green figure has to appear After that the sentence the scaled signal x3hn-3 should appear The next time shift of hn-3 is taken Then scaling of hn-3 with x3 is done. The signal x3hn-3 is the same as hn-3 since x31.
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Step 6
Yn
1
1
8 7 6 5 4 3 2 1
Yn
2
2
3
0 1 2 3 4 5 6 7 8 n
Instruction for the animator Text to be displayed in the working area (DT)
First the symbol yn must appear and then the figure in green must appear. The output of the system yn x1hn-1x2hn-2x3hn-3
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Use STAM template
Electrical Engineering
Slide 1
Slide 3
Slide 26
Slide 28
Slide 27
Introduction
Definitions
Test your understanding (questionnaire)?
Lets Sum up (summary)?
Want to know more (Further Reading)?
Analogy
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
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Instructions/ Working area
Credits
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Interactivity option 1 Step No 1
1
yn
hn
xn
3 2 1
3 2 1
3 2 1
2
0 1 2 3 4 5 0 1
2 3 4 5
3
Fig. 1
Fig. 2
0 1 2 3 4 5 n
-1 0 1 2 3 4 5
n
Fig.1
Fig.2
Interactivity type (IO 1/IO 2) Instruction to learners Boundary limits options Instruction to animators Results and output
Choose from fig. A,B C. Find the convolution of xn and hn Show fig 1, fig.2 in the question part. show fig. B as the output.
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Interactivity option 1 Step No 2
1
yn
Yn
Yn
12 10 8 6 4 2
3 2 1
12 10 8 6 4 2
2
0 1 2 3 4 5 0 1
2 3 4 5
3
Fig. 1
Fig. 2
0 1 2 3 4 5 6 n
0 1 2 3 4 5 6 n
a) Fig. A
b) Fig. B
Interactivity type (IO 1/IO 2) Instruction to learners Boundary limits options Instruction to animators Results and output
Find the convolution of xn and hn Show fig. A, fig. B and fig. C as the options show fig. B as the output.
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Interactivity option 1 Step No 3
1
yn
Yn
12 10 8 6 4 2
3 2 1
2
0 1 2 3 4 5 0 1
2 3 4 5
3
Fig. 1
Fig. 2
0 1 2 3 4 5 6 n
c) Fig. C
Interactivity type (IO 1/IO 2) Instruction to learners Boundary 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.
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Master Layout (Part 2)?
1
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
hn
Xn
2
1 2 3
. -3 -2 -1 0 1 2 3 .
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Result signal of convolution The dotted lines
represent yn The thicker line represents only
y4
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Step 1
1
Xn
hn
2
2

1 2 3
n
3
.-3 -2 -1 0 1 2 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. xn and hn are the two signals taken to convolve. xk has 7 non-zero impulses from -3 to 3 with an amplitude of 2. so, it is difficult to use method 1 . Then Yn is calculated using formula.
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Step 2
1
hn-k
Xk
3
2
2
1
2
kn-2 kn-1 k n
k
3
-3 -2 -1 0 1 2 3
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. hn-k is the time reversal and shifted version of hn as shown in the figure. hn-k is h0 when k n Similarly, hn-k is h1 when kn-1 hn-k is h2 when k n-2 and so on.
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Step 3
1
2
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 xn or hn have large number of impulses then method 1 is probably inconvenient to use. Here yn can be calculated easily if the boundary conditions are known. If xk is non-zero between and hk is non-zero between and hk has N impulses, then From this , it is understood that yn is zero if i.e, and i.e,
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Step 4
1
Xk
h4-k
2
2 3 4 k
3
-3 -2 -1 0 1 2 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. h4-k is the time reversal and shifted version of hn as shown in the figure.
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Step 5
1
Xkh4-k
Xkh4-k
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4
2
2
2 3 4
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. Y4 can be calculated by summing all the values of product of xkh4-k. i.e,
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Step 6
Y4
1
1
12 10 8 6 4 2
Y4
2
2
3
0 1 2 3 4 5 6 7
8 9 n
Instruction for the animator Text to be displayed in the working area (DT)
First the symbol y4 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 y4 The output of the system y4 is the summation of xkhn-k That is given as y4 x0h4-0x1h4-1x2h4-2x3h4-3x4h4-4 The dotted lines represent the solution for yn.
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Use STAM template
Electrical Engineering
Slide 1
Slide 3
Slide 26
Slide 28
Slide 27
Introduction
Definitions
Test your understanding (questionnaire)?
Lets Sum up (summary)?
Want to know more (Further Reading)?
Analogy
Interactivity
Try it yourself

• Fig. a
• Fig. b
• Fig. c

Fig. 3
Fig. 4
Demo
Activity
Fig. a
Fig. b
Fig. c
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Instructions/ Working area
Credits
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Interactivity option 1 Step No 1
1
yn
xn
h5-k
3 2 1
3 2 1
4 3 2 1
2
0 1 2 3 4 5 0 1
2 3 4 5
3
Fig. 1
Fig. 2
0 1 2 3 4 5 6 n
0 1 2 3 4 5 6 n
Fig.3
Fig.4
Interactivity type (IO 1/IO 2) Instruction to learners Boundary limits options Instruction to animators Results and output
Choose from fig. a, b c Find the value of y5 Show fig 3, fig.4 in the question part. show fig. c as the output.
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5
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Interactivity option 1 Step No 2
1
yn
4 3 2 1
3 2 1
3 2 1
Y5
Y5
2
0 1 2 3 4 5 0 1
2 3 4 5
3
Fig. 1
Fig. 2
0 1 2 3 4 5 6 n
0 1 2 3 4 5 6 n
a) Fig. a
b) Fig. b
Interactivity type (IO 1/IO 2) Instruction to learners Boundary limits options Instruction to animators Results and output
Hint h5-k h1 When k4 Show fig. a, fig. b and fig. c as the options show fig. c as the output.
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Interactivity option 1 Step No 3
1
yn
Y5
3 2 1
4 3 2 1
2
0 1 2 3 4 5 0 1
2 3 4 5
3
Fig. 1
Fig. 2
0 1 2 3 4 5 6 n
c) Fig. c
Interactivity type (IO 1/IO 2) Instruction to learners Boundary 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.
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Questionnaire
3 2 1
1
3 2 1
1. Find the value of y6 Answers
a) b) 2.The
Convolution sum is given as ___________
Answers ? The correct answers are given in
red.
hn
xn
2
0 1 2 3 4 5 6
0 1 2 3 4 5 6
10 8 6 4 2
8 6 4 2
3
4
0 1 2 3 4 5 6
0 1 2 3 4 5 6
5
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Links for further reading

• Reference websites
• Books
• Signals Systems Alan V. Oppenheim, Alan S.
  Willsky, S. Hamid Nawab, PHI learning, Second
  edition.
• Research papers

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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 systems 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 systems response.