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Confusion with magnitude plot of a freg response of a filter, plz help

Started by Unknown November 4, 2005
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.

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.
<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