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how can determine the region of convergence?

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sara
sara on 1 Apr 2016
Commented: John BG on 2 Apr 2016
hi all. I am a student, I should solve this question in matlab. can you help me please?

Answers (1)

John BG
John BG on 2 Apr 2016
Edited: John BG on 2 Apr 2016
From Digital Signal Processing using MATLAB,
by Vinay K Ingle, John G Proakis
pg 104
Region of Convergence of z domain functions is defined
as the abs(z) where H(z) exists, z: complex frequency.
Since you have defined in the same question that X is the input signal,
H is the system, both have finite energy,
X(z) ROC is abs(z)>.25 and H(z) ROC is abs(z)>.25
X(z)=1/(1+.5*z^-1) % ROC abs(z)>.5
x(n)=(-.5)^n*u(n)
H(z)=1/(1-.25*z^-1) % ROC abs(z)>.25
h(n)=(.25)^n*u(n)
if X(z) and H(z) had been defined with ROCs abs(z)<.5 and abs(z)<.25 respectively then
x(n)=.5^n*u(-n-1)
h(n)=-.25^n*u(-n-1)
but it's not the case.
Since
Y(z)=H(z)*X(z)
the overlapping ROCs define Y(z) ROC.
Y(z) has same ROC as input signal
abs(z)>.5
If you find this answer of any help solving your question, please click on the thumbs-up vote link,
thanks in advance John
  2 Comments
sara
sara on 2 Apr 2016
Edited: sara on 2 Apr 2016
thanks dear John. it is so helpful. but I am confused. z is variable and changes. for x: abs(z)>.5 and for h: abs(z)>1/4... how can I define this? and is there any way to show this in a plot?
John BG
John BG on 2 Apr 2016
it's defined by the exercise, not you: signal X(z) gets in H(z), Y(z)=H(z)*X(z) out it goes.
z is complex, you usually plot the magnitude with abs(z) and phase with angle(z).

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