# Single-sided Z transform with difference equations and the system function

I am working on this problem: Given an impulse response,

1. find the system function,
2. find the difference equation representation,
3. find pole-zero plot,
4. find output $$y[n]$$ if the input is $$x[n] = 0.25^n * u[n]$$

Here is what I have so far (MATLAB code)

1. Find H(z) $$H(z) = \frac{5}{1 - 0.25z^{-1}}\quad,\qquad|z| > 0.25$$

2. For part 2 I used the fact that $$H(z)=\frac{Y(z)}{X(z)}$$ I cross multiplied, then took the inverse $$\mathcal Z$$ transform and got: $$y[n] - 0.25y[n] = 5x[n]$$

3. For part 3 I used the zplane function:

 b = [5 0];
a = [1 -0.25];
figure; zplane(b,a);
4. Part 4 is where I am getting confused. I took the one-sided $$\mathcal Z$$-transform of the difference equation to get $$Y^+(z)-0.25[y[-1]+z^{-1}Y^+(z)] = 5X^+(z)$$ Assuming the system is casual (due to unit-step function in impulse response and input $$x[n]$$, also no initial conditions given in the problem), $$y[-1] = 0$$ So, after rearranging, $$Y^+(z)=\frac{5}{1-0.5z^{-1}+0.0625z^{-2}}\quad, \qquad |z| > 0.25$$ Using the residuez function for partial fraction decomp

 b1 = [5 0 0];
a1 = [1 -0.5 0.0625];
[R p C] = residuez(b1,a1)
 R =

0
5

p =

0.2500
0.2500

C =

0


Then taking the inverse $$\mathcal Z$$-transform of the terms generated by residuez,

$$y[n] = 5(1/4)^nu[n]$$

However, when I check, this is not the case. For the check, I first create a MATLAB generated $$y[n]$$ sequence using filter


%MATLAB check for part 4
n = 0:50;  % compare first 50 samples
x = (1/4).^n;
y = filter(b,a,x);

Then, create my check sequence calculated from the inverse $$\mathcal Z$$-transform and the residuez function

ycheck =  5*(0.25).^n;
error = max(abs(y-ycheck))

The sequences are not the same, and error nowhere close to nominal. Can anyone tell me where I went wrong please?

• Residue is different for single pole and multiple poles – Ch.Siva Ram Kishore Oct 27 '20 at 8:43

You misinterpreted the result of residuez. Note that there is also a fourth output argument, namely the multiplicity of the poles (which is $$2$$ in your example). In your case, the residuez function doesn't help much, because you know the poles already, and the result is already in a partial fractions form:
$$Y(z)=X(z)H(z)=\frac{5}{\left(1-\frac14 z^{-1}\right)^2}\tag{1}$$
You can either look up the inverse $$\mathcal{Z}$$-transform of $$(1)$$ in a table, or, maybe even more straightforward, compute the result by convolution in the time domain:
$$y[n]=\sum_{k=-\infty}^{\infty}x[k]h[n-k]\tag{2}$$
Solving $$(2)$$ for the given sequences is very straightforward.