1. CT-321 Digital Signal Processing
Yash Vasavada
Autumn 2016
DA-IICT
Lecture 14
Frequency Response of LSI Systems
13th September 2016

2. Review and Preview Review of past lecture: Preview of this lecture:
Frequency Response of LSI Systems
Preview of this lecture:
Reading Assignment
OS: Chapter 4
PM: Chapter 4 section 4.4

3. Linear Constant-Coefficient Difference Equations
An important class of LSI systems are those for which the output 𝑦(𝑛) and the input 𝑥(𝑛) satisfy the 𝑁 𝑡ℎ order linear constant coefficient difference equation (LCCDE):
What are such types of LSI systems? Exactly those for which Z Transform of the impulse response is a rational function (i.e., a ratio of polynomials) in 𝑧:
Take Z Transform of both sides of LCCDE:

4. Accumulator Expressed as LCCDE
Representation of the accumulator in time domain:
…and in Z domain:

5. Frequency Response of LSI Systems
Consider the frequency response 𝐻 𝑓 = 𝐻 𝑓 exp 𝑗∠𝐻(𝑓) of LTI systems in the polar coordinates:
Here, 𝐻 𝑓 is called the magnitude response of the filter and it is the gain that the filter applies to a complex exponential at frequency 𝑓
∠𝐻(𝑓) is the phase response and it is the phase offset that the filter applies to a complex exponential at frequency 𝑓
As we have seen, for LTI systems, the following holds:
Therefore,

6. Frequency Response of LSI Systems
Consider the magnitude response of an LTI system whose impulse response and Z Transform are given as follows:
For 𝑧 >|𝑎|
Magnitude response is obtained by substituting 𝑧= 𝑒 𝑗2𝜋𝑓 = 𝑒 𝑗𝜔 in the above:
ℎ 𝑛 𝐻(𝑧)

7. Frequency Response of LSI Systems
Vectorial interpretation of the frequency response
𝑣 3
𝑣 2
𝑣 1

8. Magnitude Response
𝑓 0 =0.125

9. Magnitude Response Varying 𝑓 0
Click on the animation from slideshow view

10. Frequency Response of LTI Systems
In general, the presence of
a pole of 𝐻(𝑧) causes the LTI system to apply a high gain at and near frequencies that are closer to the location of the pole
A zero of 𝐻(𝑧) results in the LTI system attenuating the frequencies near the location of the zero

11. Phase Response
For phase response, we start with a zero of 𝐻(𝑧) at 𝑧= 𝑧 0 =𝑎 𝑒 𝑗2𝜋 𝑓 0
For a pole of 𝐻(𝑧) at 𝑝 0 = 𝑧 0 , the phase response is the negative of the above. Why?

12. Phase Response
𝑓 0 =0.125
Radians
Radians

13. Phase Response
Varying 𝑓 0
Click on the animation from slideshow view

14. Group Delay Response
Group delay is the negative of the derivative of phase response of 𝐻(𝑧)
For a single zero of 𝐻(𝑧) at 𝑧= 𝑧 0 =𝑎 𝑒 𝑗2𝜋 𝑓 0 , the negative of the derivative of the phase response, i.e., τ H z (f)=− 𝑑∠𝐻(𝑓) 𝑑𝑓 takes the following form:
For a pole of 𝐻(𝑧) at 𝑝 0 = 𝑧 0 , the group delay is the negative of the above. Why?

15. Samples
Samples

16. An Example of the Effect of Magnitude, Phase and Group Delay Responses
See Section 1.2 of the OS text

17. 𝒇=𝟎.𝟒
𝒇=𝟎.𝟏
𝒇=𝟎.𝟐
Let us consider an input 𝑥(𝑛) which is comprised of three sinusoidal pulses of different frequencies
These are three truncated sinusoids; each of length 60 samples
They have been multiplied by a tapering function in time domain (time domain Hamming windowing)

18. 𝑋(𝑓): DTFT of 𝑥(𝑛)
Frequency domain representation of the input 𝑥(𝑛): shows six “impulses” because the input 𝑥(𝑛) is real-valued (i.e., for each frequency, there is a mirror image present)

19. Pole-Zero Diagram of an LSI system
Input 𝑥(𝑛) is passed through an LSI system whose Pole-Zero diagram is as shown here

20. Magnitude Response of the LSI system
Location of tones in the input 𝑥(𝑛)

21. Phase Response of the LSI system

22. Phase Response (Unwrapped) of the LSI system
Location of tones in the input 𝑥(𝑛)

23. Group Delay of the LSI system
Location of tones in the input 𝑥(𝑛)

24. Input 𝑥(𝑛)
𝒇=𝟎.𝟒
𝒇=𝟎.𝟏
𝒇=𝟎.𝟐

25. 𝒇=𝟎.𝟒 not visible
𝒇=𝟎.𝟐
𝒇=𝟎.𝟏
Output: y(𝑛)
Observations:
Frequency 𝑓=0.4 is no longer present
Since the group delay of the LSI system at 𝑓=0.1 is far greater than that at 𝑓=0.2, the position of the pulses is switched at the output of the filter