Determine the expression for the frequency response H(Omega) of a causal IIR LTI

Question

Determine the expression for the frequency response H(Omega) of a causal IIR LTI discrete-time system characterized by the input-output relation:
y[n]=x[n]-ay[n-R]
where |a|<1, and y[n] and x[n] denote, respectively, the output and input sequences. Determine the maximum and minimum values of its magnitude and phase response.
Sketch the magnitude and phase response for R=5.

So I think I determined the expression for frequency response: -1/(1+ae^-Rjw)

I just need help determining the maximum and minimum values of its magnitude and phase response. And should the sketch be a unit circle?? Please help!!!

Explanation / Answer

y[n]=x[n] -ay[n-R]
taking the fourier transform on the both side
y[w]= x[w] -ae-jRw y[w]

or y[w](1+ae-jRw)= x[w]

or y[w]=x[w]/(1+ae-jRw).

magnitude = |x[w]/(1+ae-jRw)|=   |x[w]|/|(1+ae-jRw)|

       now, |(1+ae-jRw)|= | 1+acos(Rw) - Jasin(Rw)|= [(1+acos(Rw))2 + (asin(Rw))2 ]0.5

                                   = [1+ a2 + 2acos(Rw)]0.5

hence , magnitude = |x[w]|/[1+ a2 + 2acos(Rw)]0.5

               (magnitude)max= |x[w]|/[1+ a2 + 2a]0.5 when , cos(Rw)=1

                                          = |x[w]|/|1+a|

                         (magnitude)min = |x[w]|/|1-a| , when , cos(Rw)=-1

phase= arc(tan[ asin(Rw)/(1+acos(Rw)])=

(phase)min= 0, when sin(wR)=0. that is wR =n

(phase)max= a , when sin(wR)=1,

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