# Calculating an output of a system (Z- transform question)

I have a following question to answer:

An LTI system is described by its impulse response h[n]. For input x[n] it gives output y[n].

$$h[n] = u(n) - u(n-N)$$ $$x[n] = u(n) - u(n-M)$$

I want to calculate it's output function. I have used the Z-transform property of convolution in time being multiplication in Z-domain to calculate Y(z) which looks like this $$Y(z) = \frac{1-z^{-M}}{1-z^{-N}}$$ However I have trouble to apply inverse Z-transform to it. Is there some nice way I can transform this resulting function to time(sample) domain?

• Your $Y(z)$ is wrong. It should be $$Y(z) = H(z) X(z) = \frac{1 - z^{-N} } {1-z^{-1} } \frac{ 1 - z^{-M} }{ 1-z^{-1} } = \frac{ (1 - z^{-N})(1 - z^{-M}) } {(1-z^{-1})^2}$$ – Fat32 Nov 25 '18 at 12:17

## 1 Answer

For this specific example, using a partial fraction expansion (PFE) would be an overkill, so yo should better consider using the formula :

$$1 - z^{-N} = (1- z^{-1}) (1 + z^{-1} + z^{-2} + ... + z^{-N+1}$$

to simplify the Z-transform $$Y(z)$$ of your output as:

\begin{align} Y(z) &= H(z) X(z) = \frac{1 - z^{-N} } {1-z^{-1} } \frac{ 1 - z^{-M} }{ 1-z^{-1} } \\ \\ &= \left( 1 + z^{-1} + z^{-2} + ... + z^{1-N} \right) \left( 1 + z^{-1} + z^{-2} + ... + z^{1-M} \right)\\ Y(z) &= 1 + c_1 z^{-1} + c_2 z^{-2} + ... + c_{N+M-2} z^{2-N-M} \\ Y(z) &= \sum_{n=0}^{K} y[n] z^{-n} = 1 + c_1 z^{-1} + c_2 z^{-2} + ... + c_{N+M-2} z^{2-N-M} \\ \end{align}

Now as you can see, $$Y(z)$$ is a simple polynomial in $$z^{-1}$$. You can either explicitly multiply those two polynomials in parenthesis to find those coefficients $$c_k$$ to get th sample values $$y[n]$$ as indicated by: $$y[k] = c_k ~~~,~~~ k = 0,1,2,...,N+M-2$$

you can easily get those (coefficient $$c_k$$) values by the following convolution: $$y[n] = c[n] = a[n] \star b[n]$$ where $$a[n]$$ and $$b[n]$$ are all ones of length $$N$$ and $$M$$ respectively.

Now if you insist on a PFE method, the you could also do that as follows: \begin{align} Y(z) &= H(z) X(z) = \frac{1 - z^{-N} } {1-z^{-1} } \frac{ 1 - z^{-M} }{ 1-z^{-1} } \\ \\ &= \frac{1 - z^{-N} - z^{-M} + z^{-N-M} } { (1-z^{-1})^2 }\\ \\ &= \frac{1} { (1-z^{-1})^2 } - \frac{z^{-N}} { (1-z^{-1})^2 } - \frac{ - z^{-M} } { (1-z^{-1})^2 } + \frac{z^{-N-M} } { (1-z^{-1})^2 }\\ \\ \end{align}

Now as can be seen, I have separated the expression into four pieces each of which is a delayed version of the first one (on the left without a delay). So I will express $$y[n]$$ as a sum of those four delayed functions, denoting the first one as $$f[n]$$ we have :

$$y[n] = f[n] - f[n-N] - f[n-M] + f[n-M-N]$$ where the function $$f[n]$$ is the inverse Z- transform of $$f[n] = \mathcal{Z}^{-1} \{ \frac{1}{(1-z^{-1})^2} \}$$

looking from a table or simply finding it yourself, one can find that $$f[n] = u[n] \star u[n] = r[n]$$

where $$r[n]$$ is the following function : $$r[n] = (n+1) u[n]$$

hence the result is simplified as:

$$y[n] = r[n] - r[n-N] - r[n-M] + r[n-M-N]$$

instead of denoting the result in terms of ramp functions, you can also simplfity the shifted sum formula through its convolution operatos as: $$y[n] = (u[n] \star u[n]) -(u[n] \star u[n-N])-(u[n] \star u[n-M])+(u[n] \star u[n-N-M])$$ $$y[n] = u[n] \star \left( u[n] - u[n-N] - u[n-M] + u[n-N-M] \right)$$

denoting the finite length sequence on the right by $$g[n]$$, then the output can be found to be

$$y[n] = u[n] \star g[n] = \sum_{k=-\infty}^{n} g[k] = \sum_{k=0}^{n} g[k]$$

The rightmost sum occurs assuming $$g[n]$$ is a causal sequence.