0
$\begingroup$

I am currently working the figure through below.

as it is an FIR Filter i have worked out using convolution that the output is 4,2,4,6,0,0.

i am trying to obtain the 'z' domain transfer function of the filter below using h(n)

using the $h(n)$ 2,-1,3

I have started to obtain the $z$ domain transfer function with

$y(n)=2x(n)-1x(n-1)+3x(n-2)$

any help?

enter image description here

Improve this question
$\endgroup$
0
$\begingroup$

You can solve this with high-school algebra and the time delay property

$$ \mathcal{Z}(x(n-k)) = z^{-k} X(z) $$

Applying this to your expression

$$y(n)=2x(n)-1x(n-1)+3x(n-2)$$

we get

$$Y(z) = 2 X(z) - X(z)z^{-1} + 3X(z)z^{-2}$$

$$\frac{Y(z)}{X(x)} = 2z^{-2} - z^{-1} + 3$$

The denominator is null because the filter is not recursive

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ Sorry to be be nit picky: the denominator is 1 not null. The transfer function has three zeros and three poles, the poles are just all at $z = 0$ $\endgroup$
    – Hilmar
    Mar 4 at 16:08
  • $\begingroup$ so would it read more like this?Y(z) = 2X(z) - X(z)z-1 + 3X(z)z-2 Pull out the common X(z) Y(z) = X(z) (2 - z-1 + 3 z-2) Y(z)/X(z) = 2 - z-1 + 3 z-2 $\endgroup$
    – user56023
    Mar 4 at 16:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy