Analysis of a LTI system using DFT

Question

Consider an LTI system $$H(z)=1-\frac{1}{2}z^{-1}+\frac{3}{4}z^{-2}$$ Let $x[n]=(\frac{1}{3})^n\cdot u[n]$ be the input signal. It is desired to determine the output for $n=0,1...,N_a$. To achieve that, a relevant part of the DFT will be used together with an uniform sampling of $H(z)$ in the unity circle with a separation of $\frac{2\pi}{N_b}$. a) Explain the process to achieve this. b) Are there any restrictions for $N_a$ and $N_b$? c) For the same $x[n]$, would the method described in a) work if $H(z)=\frac{1}{1-\frac{1}{2}z^{-1}}$? If not, correct the previous method to find the correct result.

$H(z)$ is sampled in $N_b$ points. $\hat{h}[n]$ (the sequence whose DFT is the sampled version of $H(z)$) has only 3 non-zero values, so... the sampled version of it would have $N_b -3$ zeros (I'm not sure about this). About the size of $X[k]$, I thought that, if I want to know the first $N_a +1$ values of $y[n]$, maybe it is enough to grab the first $N_a +1$ points of $x[n]$ to do the DFT and add the necessary amount of zeros to avoid aliasing? This may be false but I just don't know how to determine the relevant part of the input signal that has to be used. If everything I said above is correct (I highly doubt it), then I thought that the only restriction on $N_a$ and $N_b$ is that $N_a +3 = N_b$.

About c), I have no idea of how to do that.

Any suggestion? Thanks for your time!

Answer 1

Since the system described by $H(z)$ is causal, it's sufficient to consider the input sequence in the range $0\le n\le N_a$ in order to compute the output sequence in the same range. Each output sample can only depend on the current input sample and on past input samples. Since the length of the truncated input sequence is $N_a+1$, and the length of the system's impulse response is $3$, the length of the convolution of these two sequences is $N_a+3$. If you want to compute this linear convolution using circular convolution (i.e., using the DFT), the DFT length $N_b$ must satisfy $N_b\ge N_a+3$. So you must zero-pad the truncated input sequence as well as the impulse response to that length, then compute the DFTs of both zero-padded sequences, multiply them, and compute the IDFT of the resulting sequence, the first $N_a+1$ elements of which equal the desired output sequence. This answers a) and b).

For part c) it will not suffice to sample $H(z)$ on the unit circle, because the corresponding impulse response is infinitely long. However, for computing the first $N_a+1$ output samples it is sufficient to consider the first $N_a+1$ samples of the impulse response. Just like before, this works because the system's impulse response as well as the input signal are both zero for $n<0$. So you need to consider a system $\hat{H}(z)$ the impulse response of which is a truncated version of the infinite impulse response corresponding to $H(z)$. The length of the convolution of the truncated input signal and the truncated impulse response is $2N_a+1$, so the DFT length $N_b$ needs to satisfy $N_b\ge 2N_a+1$. The rest of the procedure is the same as before.