DSP Exam Time Answers

A few days ago, I posted a question from my recent DSP exam. I show it again below.


Now for the answers (under the fold).

a) To get the FTDE, one can simply read off the multipliers:

y[n] = x[n] - \frac{3}{2}x[n-1] - x[n-2] + \frac{5}{2}y[n-1] - y[n-2].

b) Again, just by reading off the multipliers:

H(z) = \frac{1 - \frac{3}{2}z^{-1} - z^{-2} }{1 - \frac{5}{2}z^{-1} + z^{-2}}

This is incomplete since we need to specify the region of convergence (ROC). Let’s put this rational function in terms of positive powers of z, get rid of the fractions, and factor:

H(z) = \frac{2z^2 - 3z - 2 }{2z^2 - 5z + 2} = \frac{(2z+1)(z-2)}{(2z-1)(z-2)}.

From this we see that this¬†causal system’s ROC is |z|>2.

c) Using polynomial long division, we find the impulse response is simply:

h[n] = \delta[n] + \frac{2}{2^n}\mu[n-1].

d) Hmm. What a strange question to ask. This system is just made of linear elements, and so should be linear. It has an IR that decays to zero fast enough that its Fourier transform exists, and so the system is stable. What is one difference between theory and practice? The number of bits for computing and storing values! Each branch of this system will thus have only a finite number of bits to store and compute values. Since this system has a pole outside the unit circle, the values in the feedback branch  grow without bound. Hence, the values in the feedback branch will saturate, and the system will no longer behave linearly.

e) Looking at the transfer function of the system, we can just “cancel” the pole and zero at z=2 and create a linear system with the same impulse response:

H(z) = \frac{2z+1}{2z-1} = \frac{1+\frac{1}{2}z^{-1}}{1-\frac{1}{2}z^{-1}}, |z|>\frac{1}{2}.

This corresponding digital filter implements the following FTDE:

$latex y[n] = x[n] + \frac{1}{2}x[n-1] + \frac{1}{2}y[n-1].


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