<VijaKhara@gmail.com> wrote in message
news:1131134435.441548.34430@g14g2000cwa.googlegroups.com...
> Dear members:
>
> Plz tell me what is the point I am wrong. It is my exam.
>
> I have to plot the magnitude of the freq response a Low Pass filter:
>
> y[n]=1/3*(x[n]+x[n-1]+x[n-2]);
>
>
> I used Z transform and found
>
> Y(z)/X(z)=H(z)=1/3*(z^2+z+1)/z^2.
>
> So there are two zeros at z1=-1/2 +sqrt(3)/2 and z2=-1/2-sqrt(3)/2, and
> two double poles at zp=0.
>
> I plotted it as follows: the magnitude equals zero at omega1=2*pi/3 and
> omega2=-2*pi/3 and goes to infinity at omega=zero (since poles are
> zero).
>
> But in the solution, prof solves as follows:
>
> H(omega)=1/2* (sin(3*omega/2)/sin(omega/2))*exp(-j*omega) and at
> omega=zero the maginitude |H(omega)|=1, not =infinity as what I made.
>
> Plz tell me what are my wrong points?
>
> Thank you very much.
>

You need to substitute z=exp(j*theta) and then play around with teh
expression a bit. This is where most people get lost. As an example consider

H(z) = 1-z^-2

we substitute z^-1 =exp(-j*theta) and end up with

1-exp(-2jtheta).

Now do this little tric - write the above as

{exp(jtheta)-exp(-jtheta) }X exp(-jtheta)  ... so it's just the same really
except the exp(-jtheta) term is outside the brackets.

then use the fundamental expression for sin(theta) =
{exp(jtheta)-exp(-jtheta)}/2j - you will need to multiply by 2j of course to
keep the expression unchanged.
We then get

2jsin(theta) exp(-jtheta)

Which has magnitude

2mod(sin(theta))

the exp(-jtheta) term effects the phase only  - the expression has linear
phase. This is a comb filter of sorts with two zeros at z=1 and z=-1 (dc and
half sampling).

Thats the sort of tricks you use.

McC

VijaKhara@gmail.com wrote:
> Dear members:
> 
> Plz tell me what is the point I am wrong. It is my exam.
> 
> I have to plot the magnitude of the freq response a Low Pass filter:
> 
> y[n]=1/3*(x[n]+x[n-1]+x[n-2]);
> 
> 
> I used Z transform and found
> 
> Y(z)/X(z)=H(z)=1/3*(z^2+z+1)/z^2.
> 
> So there are two zeros at z1=-1/2 +sqrt(3)/2 and z2=-1/2-sqrt(3)/2, and
> two double poles at zp=0.
> 
> I plotted it as follows: the magnitude equals zero at omega1=2*pi/3 and
> omega2=-2*pi/3 and goes to infinity at omega=zero (since poles are
> zero).
> 
> But in the solution, prof solves as follows:
> 
> H(omega)=1/2* (sin(3*omega/2)/sin(omega/2))*exp(-j*omega) and at
> omega=zero the maginitude |H(omega)|=1, not =infinity as what I made.
> 
> Plz tell me what are my wrong points?
> 
> Thank you very much.

It's easy to get lost if you're just trying to go straight from z into 
frequency without any intermediate step.  Try creating the pole-zero 
plot for the z transform... then see what it looks like.

Dear members:

Plz tell me what is the point I am wrong. It is my exam.

I have to plot the magnitude of the freq response a Low Pass filter:

y[n]=1/3*(x[n]+x[n-1]+x[n-2]);


I used Z transform and found

Y(z)/X(z)=H(z)=1/3*(z^2+z+1)/z^2.

So there are two zeros at z1=-1/2 +sqrt(3)/2 and z2=-1/2-sqrt(3)/2, and
two double poles at zp=0.

I plotted it as follows: the magnitude equals zero at omega1=2*pi/3 and
omega2=-2*pi/3 and goes to infinity at omega=zero (since poles are
zero).

But in the solution, prof solves as follows:

H(omega)=1/2* (sin(3*omega/2)/sin(omega/2))*exp(-j*omega) and at
omega=zero the maginitude |H(omega)|=1, not =infinity as what I made.

Plz tell me what are my wrong points?

Thank you very much.