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.

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

Started by ●November 4, 2005

Reply by ●November 4, 20052005-11-04

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.

Reply by ●November 4, 20052005-11-04

<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