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

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

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:

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

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,

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)

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

Magnitude Response 𝑓 0 =0.125

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

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

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?

Phase Response 𝑓 0 =0.125 Radians Radians

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

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?

Samples Samples

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

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)

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)

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

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

Phase Response of the LSI system

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

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

Input 𝑥(𝑛) 𝒇=𝟎.𝟒 𝒇=𝟎.𝟏 𝒇=𝟎.𝟐

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