I am trying to solve these two questions, but I am stuck with the filter stability and ROC. Hope somebody could help me with it.

q1 According to what I understand, the third option is correct. Fourth option is also incorrect as it only contains one pole and on zero. I think the first and second options are incorrect; the system is unstable and ROC doesn't include unit circle because the pole lies at z = -2 and the system is casual. Not sure though. I'd appreciate some help in this regard or if any of the other options are incorrect.

q5 For this question, the pole is outside the unit circle, so I thought the system must be unstable, but then I realize the output depends on future input (x[n+1]), which makes it non-casual, also number of zeros>poles. Since for a non-casual, the ROC moves inward, this system includes unit circle and is stable? But doubtful about the second part of the option regarding being dependent on input.

I wrote the following code to find y[n],

n = 100
w = np.linspace(-np.pi,np.pi,n)
z = np.exp(1j * w)
H = z* (z - 0.5)/(z + 2)
t = np.arange(-n/2, n/2)   
x  = sp.unit_impulse(n) + 0.5 * sp.unit_impulse(n,1)
plt.plot(t,ifftshift(ifft( H * fft(x))))

Got the following result, here

I don't think, 1st, 2nd and 5th options are correct, looking at figure. As I increased the value of n, the peak became narrower, so for the 4th option, I don't know if it is showing diverging oscillation. The only peak in this question is at zero.

Any help is appreciated.

Edit: Question 1 seems to work with the above logic. Stuck with no. 5 now.

  • $\begingroup$ Why don't you show the Z Transform of the 1st question and analyze the roots and poles? $\endgroup$
    – David
    Feb 19, 2021 at 6:15
  • $\begingroup$ @David Thanks. I have solved the first question. Just needed help with the second one. I don't know why I was stuck with no. 1 for so long. It was pretty straight forward, $\endgroup$
    – Rima
    Feb 20, 2021 at 3:37

1 Answer 1


For any filter with AR(1) component the coefficient must be smaller than in order to be stable. Otherwise any input is infinitely multiplied by a non decreasing factor.


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