```Steve,

Let me answer your question with a question.  It looks like you're doing
okay so far.  Here's the question:

What is the magnitude when wT=pi?

"Steve" <steve@x2.org.uk> wrote in message
news:pan.2004.10.12.21.50.18.472868@x2.org.uk...
> Hi,
>
> I'm having some problems with phase angles, in the context of frequency
> response of a digital filter from its Z transform. I'm just not "getting
> it". I've been looking around the 'web for relevant material, but all the
> stuff I've been finding is beyond this and just assumes that the reader
> can do this stuff.
>
> Say you've got a filter with transfer function:
>
> H(z) =   z
>      --------
>       z + 1
>
> So on a pole/zero diagram, there's a zero at the origin of the unit circle
> and a pole at -1+j0, ie, -1 on the real axis.
>
> So using z=exp(jwT) (where T means 'delta t' and w is omega), the
> frequency response is:
>
> H(w) = exp(jwT)
>      ---------
>     exp(jwT) + 1
>
> So the amplitude response of this filter is:
>
> |H(w)| =               1
>        -------------------------------------
>        ((cos(wT) + 1)^2 + (sin(wT))^2)^(1/2)
>
> and the amplitude response is ('<' means 'the angle of'):
>
> <H(w) = wT - <(exp(jwT) + 1)
>
> and as (exp(jwT) + 1) == ((cos(wT) + 1) + j sin(wT)):
>
> theta = tan^(-1)(sin(wT)/(cos(wT) + 1))
>
> so <H(w) = wT - tan^(-1)(sin(wT)/(cos(wT) + 1))
>
> This is where things start getting confusing for me.
>
> If I try and find the phase at pi (= 3.14159...), then it all breaks down:
>
> <H(w)|wT=pi  =  pi - tan^(-1)(sin(pi)/(cos(pi) + 1))
>             =  pi - tan^(-1)(0 / (-1 + 1))
>             =  pi - tan^(-1)(undefined!!!)
>
> How do I sketch this on a phase response diagram (phase angle on the y
> axis, frequency on the x axis)?? How do I represent undefined phase??
>
> I tried the same thing with a slightly different transfer function:
>
> H(z) =  z
>      -------
>      z+(1/2)
>
> With this I get, as that second last line:
>
> <H(w)|wT=pi  =  pi - tan^(-1)(0 / -1+(1/2))
>             =  pi - tan^(-1)(0)
>             =  pi
>
> What does this mean??? That something anywhere on the real axis between -1
> and 0 will have a phase angle on pi? Looking at it again (now that I've
> written all this), it kind of makes sense, a 180degree phase shift when
> you're 180degrees around the circle. Is this correct??
>
> If anyone can tell me anything about phase frequency response of digital
> filiters, that would be a big help, I've confused myself over these last
> few days reading about it and just not "getting" these phase angles...
>
> Thanks,
>
> Steve

```
```Hi,

I'm having some problems with phase angles, in the context of frequency
response of a digital filter from its Z transform. I'm just not "getting
it". I've been looking around the 'web for relevant material, but all the
stuff I've been finding is beyond this and just assumes that the reader
can do this stuff.

Say you've got a filter with transfer function:

H(z) =   z
--------
z + 1

So on a pole/zero diagram, there's a zero at the origin of the unit circle
and a pole at -1+j0, ie, -1 on the real axis.

So using z=exp(jwT) (where T means 'delta t' and w is omega), the
frequency response is:

H(w) = exp(jwT)
---------
exp(jwT) + 1

So the amplitude response of this filter is:

|H(w)| =               1
-------------------------------------
((cos(wT) + 1)^2 + (sin(wT))^2)^(1/2)

and the amplitude response is ('<' means 'the angle of'):

<H(w) = wT - <(exp(jwT) + 1)

and as (exp(jwT) + 1) == ((cos(wT) + 1) + j sin(wT)):

theta = tan^(-1)(sin(wT)/(cos(wT) + 1))

so <H(w) = wT - tan^(-1)(sin(wT)/(cos(wT) + 1))

This is where things start getting confusing for me.

If I try and find the phase at pi (= 3.14159...), then it all breaks down:

<H(w)|wT=pi  =  pi - tan^(-1)(sin(pi)/(cos(pi) + 1))
=  pi - tan^(-1)(0 / (-1 + 1))
=  pi - tan^(-1)(undefined!!!)

How do I sketch this on a phase response diagram (phase angle on the y
axis, frequency on the x axis)?? How do I represent undefined phase??

I tried the same thing with a slightly different transfer function:

H(z) =  z
-------
z+(1/2)

With this I get, as that second last line:

<H(w)|wT=pi  =  pi - tan^(-1)(0 / -1+(1/2))
=  pi - tan^(-1)(0)
=  pi

What does this mean??? That something anywhere on the real axis between -1
and 0 will have a phase angle on pi? Looking at it again (now that I've
written all this), it kind of makes sense, a 180degree phase shift when
you're 180degrees around the circle. Is this correct??

If anyone can tell me anything about phase frequency response of digital
filiters, that would be a big help, I've confused myself over these last
few days reading about it and just not "getting" these phase angles...

Thanks,

Steve
```
```Hi,

I'm having some problems with phase angles, in the context of frequency
response of a digital filter from its Z transform. I'm just not "getting
it". I've been looking around the 'web for relevant material, but all the
stuff I've been finding is beyond this and just assumes that the reader
can do this stuff.

Say you've got a filter with transfer function:

H(z) =   z
--------
z + 1

So on a pole/zero diagram, there's a zero at the origin of the unit circle
and a pole at -1+j0, ie, -1 on the real axis.

So using z=exp(jwT) (where T means 'delta t' and w is omega), the
frequency response is:

H(w) = exp(jwT)
---------
exp(jwT) + 1

So the amplitude response of this filter is:

|H(w)| =               1
-------------------------------------
((cos(wT) + 1)^2 + (sin(wT))^2)^(1/2)

and the amplitude response is ('<' means 'the angle of'):

<H(w) = wT - <(exp(jwT) + 1)

and as (exp(jwT) + 1) == ((cos(wT) + 1) + j sin(wT)):

theta = tan^(-1)(sin(wT)/(cos(wT) + 1))

so <H(w) = wT - tan^(-1)(sin(wT)/(cos(wT) + 1))

This is where things start getting confusing for me.

If I try and find the phase at pi (= 3.14159...), then it all breaks down:

<H(w)|wT=pi  =  pi - tan^(-1)(sin(pi)/(cos(pi) + 1))
=  pi - tan^(-1)(0 / (-1 + 1))
=  pi - tan^(-1)(undefined!!!)

How do I sketch this on a phase response diagram (phase angle on the y
axis, frequency on the x axis)?? How do I represent undefined phase??

I tried the same thing with a slightly different transfer function:

H(z) =  z
-------
z+(1/2)

With this I get, as that second last line:

<H(w)|wT=pi  =  pi - tan^(-1)(0 / -1+(1/2))
=  pi - tan^(-1)(0)
=  pi

What does this mean??? That something anywhere on the real axis between -1
and 0 will have a phase angle on pi? Looking at it again (now that I've
written all this), it kind of makes sense, a 180degree phase shift when
you're 180degrees around the circle. Is this correct??

If anyone can tell me anything about phase frequency response of digital
filiters, that would be a big help, I've confused myself over these last
few days reading about it and just not "getting" these phase angles...

Thanks,

Steve
```