# Frequency response from Z transform

Started by June 28, 2010
```I am trying to plot the frequency response of a system given its transfer
function. I know about the freqz function in Matlab. The problem is that my
Z transfer function has fractional order term in it.

H(z) = (z^-0.5)*(1-z^-1)*(1-z^-1)

How do I get the frequency response in Matlab?
```
```On 06/28/2010 11:06 AM, analog_fever wrote:
> I am trying to plot the frequency response of a system given its transfer
> function. I know about the freqz function in Matlab. The problem is that my
> Z transfer function has fractional order term in it.
>
> H(z) = (z^-0.5)*(1-z^-1)*(1-z^-1)
>
> How do I get the frequency response in Matlab?

Calculate w = exp(2 * %pi * %i * your_frequency / your_sample_rate) for
a number of frequency points.  (Note that I'm using Scilab notation for
pi and the square root of negative 1)

Substitute w for z in your equation above.

Enjoy.

AFAIK there isn't a built-in function to do this -- you have to actually
understand the math, and know how to use Matlab.  Given a vector of w
values, you should be able to compute

h = (w .^ (-0.5)) .* (1 - w .^ (-1)) .* (1 - w .^ (-1));

and get a vector of frequency response values.  Note the '.*' and '.^'
operators -- those tell Matlab that you want to do the operation
element-by element rather than as a matrix operation (which would fail
as the vector dimensions are wrong).
```
```On Jun 28, 2:30&#2013266080;pm, Tim Wescott <t...@seemywebsite.com> wrote:
> On 06/28/2010 11:06 AM, analog_fever wrote:
>
> > I am trying to plot the frequency response of a system given its transfer
> > function. I know about the freqz function in Matlab. The problem is that my
> > Z transfer function has fractional order term in it.
>
> > H(z) = (z^-0.5)*(1-z^-1)*(1-z^-1)
>
> > How do I get the frequency response in Matlab?
>
> Calculate w = exp(2 * %pi * %i * your_frequency / your_sample_rate) for
> a number of frequency points. &#2013266080;(Note that I'm using Scilab notation for
> pi and the square root of negative 1)
>
> Substitute w for z in your equation above.
>
> Enjoy.
>
> AFAIK there isn't a built-in function to do this -- you have to actually
> understand the math, and know how to use Matlab. &#2013266080;Given a vector of w
> values, you should be able to compute
>
> h = (w .^ (-0.5)) .* (1 - w .^ (-1)) .* (1 - w .^ (-1));
>
> and get a vector of frequency response values. &#2013266080;Note the '.*' and '.^'
> operators -- those tell Matlab that you want to do the operation
> element-by element rather than as a matrix operation (which would fail
> as the vector dimensions are wrong).

If the OP is only interested in the magnitude response, the magnitude
response will be the same as for

G(z) = (1-z^-1)*(1-z^-1)

and he can use the tools he is familiar with. The phase response for G
would only be off by a linear term corresponding to a delay of 1/2
sample.

Dirk
```
```On Jun 29, 6:06&#2013266080;am, "analog_fever" <usu_vlsi@n_o_s_p_a_m.yahoo.com>
wrote:
> I am trying to plot the frequency response of a system given its transfer
> function. I know about the freqz function in Matlab. The problem is that my
> Z transfer function has fractional order term in it.
>
> H(z) = (z^-0.5)*(1-z^-1)*(1-z^-1)
>
> How do I get the frequency response in Matlab?

Remember z=exp(sT) so that z^-0.5 => exp(-0.5jwT)

=cos(wtT2)-jsin(wtT2)

This has magnitude unity (so it doesn't effect the magnitude plot) and
a phase shift of -wT/2 which you can subtract. Now the other two zeros
are easy to sketch. when z=1 the mag is zero of course (this is
dc).when z=-1 (half sampling) the gain is 4.

Hardy

Hardy
```
```>On Jun 29, 6:06=A0am, "analog_fever" <usu_vlsi@n_o_s_p_a_m.yahoo.com>
>wrote:
>> I am trying to plot the frequency response of a system given its
transfer
>> function. I know about the freqz function in Matlab. The problem is that
=
>my
>> Z transfer function has fractional order term in it.
>>
>> H(z) =3D (z^-0.5)*(1-z^-1)*(1-z^-1)
>>
>> How do I get the frequency response in Matlab?
>
>Remember z=3Dexp(sT) so that z^-0.5 =3D> exp(-0.5jwT)
>
>=3Dcos(wtT2)-jsin(wtT2)
>
>This has magnitude unity (so it doesn't effect the magnitude plot) and
>a phase shift of -wT/2 which you can subtract. Now the other two zeros
>are easy to sketch. when z=3D1 the mag is zero of course (this is
>dc).when z=3D-1 (half sampling) the gain is 4.
>
>
>Hardy
>
>
>Hardy
>

Thank you all.
```