Hello, I want to solve this problem. F(z)=H(z)/sqrt(z); How can i get frequency response of F(w) ? If F(z)=H(z)/z, FFT(H(z-1)) should return frequency response. Please help me This message was sent using the Comp.DSP web interface on www.DSPRelated.com
basic question ? frequency response of non-integer shift
Started by ●May 3, 2005
Reply by ●May 3, 20052005-05-03
"lismov" <ssi@athena.kaist.ac.kr> wrote in message news:Q7qdnW_oGNeSherfRVn-jQ@giganews.com...> > Hello, > I want to solve this problem. > > F(z)=H(z)/sqrt(z); > > How can i get frequency response of F(w) ? > > If F(z)=H(z)/z, > FFT(H(z-1)) should return frequency response. > > Please help me > > This message was sent using the Comp.DSP web interface on > www.DSPRelated.comTo use F(z) and F(w) is mixing approaches. F(w) is normally about continuous, infinite-extent frequency. F(w) is evaluated on the infinte-extent jw axis in a complex r+jw plane. F(z) is about time-sampled data and periodic frequency. F(z) is evaluated on the unit circle in a complex z plane .. where the circle is e^jwT, repeating at wT=2*pi or at w=2*pi/T where T is the sample interval in time. You get frequency response F(e^jwT) by substituting e^jwT for z in the expression and evaluating over 2*pi where the circle is completed and the response repeats. T is the reciprocal of the sampling frequency. A delay of T is represented by z^-1. The sampling frequency fs=1/T with radian frequency ws=2*pi/T as above. Fred
Reply by ●May 3, 20052005-05-03
"lismov" <ssi@athena.kaist.ac.kr> writes:> Hello, > I want to solve this problem. > > F(z)=H(z)/sqrt(z); > > How can i get frequency response of F(w) ? > > If F(z)=H(z)/z, > FFT(H(z-1)) should return frequency response. > > Please help meHi, Divorce yourself for a moment from DSP and think of F(z) purely as a mathematical object. From the theory of complex variables, we know that any complex quantity can be expressed in polar form, z = r * e^{i*w}, where r is the magnitude of z and w is the angle of z in radians, r = sqrt(x^2 + y^2) w = arctan(y/x), when z = x + i*y. Then in your function, sqrt(z) = (r * e^{i*w})^(1/2) = r^(1/2) * e^{i*w/2}. Now also note that if C(z) = A(z) / B(z), then |C(z)| = |A(z)| / |B(z)|. Since r = |z|, then in your function F(z), |F(z)| = |H(z)| / |z|^(1/2). Now to get the frequency response of a system function in the z-domain, let z = e^(i*w). Since |e^(i*w)| = |z| = 1 and 1^(1/2) = 1, then |F(e^(i*w))| = |H(w)| / 1^(1/2) = |H(w)|. So you see that the sqrt(z) denominator doesn't change the magnitude response. -- Randy Yates Sony Ericsson Mobile Communications Research Triangle Park, NC, USA randy.yates@sonyericsson.com, 919-472-1124
Reply by ●May 3, 20052005-05-03
Randy Yates wrote:> "lismov" <ssi@athena.kaist.ac.kr> writes: > > >>Hello, >>I want to solve this problem. >> >> F(z)=H(z)/sqrt(z); >> >>How can i get frequency response of F(w) ?...> Hi, > > Divorce yourself for a moment from DSP and think of F(z) purely as a > mathematical object. > > From the theory of complex variables, we know that any complex > quantity can be expressed in polar form,... A fine example of how pure mathematical insight is sometimes more straightforward than physical insight. Thanks for the demonstration. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●May 3, 20052005-05-03
Jerry Avins <jya@ieee.org> writes:> Randy Yates wrote: > > "lismov" <ssi@athena.kaist.ac.kr> writes: > > > > >> Hello, I want to solve this problem. F(z)=H(z)/sqrt(z); How can i > >> get frequency response of F(w) ? > > > ... > > > Hi, > > Divorce yourself for a moment from DSP and think of F(z) purely as a > > > mathematical object. > > From the theory of complex variables, we know that any complex > > > quantity can be expressed in polar form, > > ... > > A fine example of how pure mathematical insight is sometimes more > straightforward than physical insight. Thanks for the demonstration.Ahh, shucks (hyuck hyuck!)... Thanks Jerry! -- Randy Yates Sony Ericsson Mobile Communications Research Triangle Park, NC, USA randy.yates@sonyericsson.com, 919-472-1124
Reply by ●May 3, 20052005-05-03
Randy Yates <randy.yates@sonyericsson.com> writes:> [...] > Now to get the frequency response of a system function in the z-domain, > let z = e^(i*w).I should have written "W" instead of "w", where W is normalized radian frequency, W = w*Ts, Ts the sample period. -- Randy Yates Sony Ericsson Mobile Communications Research Triangle Park, NC, USA randy.yates@sonyericsson.com, 919-472-1124
Reply by ●May 3, 20052005-05-03
Randy Yates wrote: ...> Now to get the frequency response of a system function in thez-domain,> let z = e^(i*w). Since |e^(i*w)| = |z| = 1 and 1^(1/2) = 1, > then > > |F(e^(i*w))| = |H(w)| / 1^(1/2) > = |H(w)|.Just for correctness' sake, you should have F(e^(i w)) = e^(-i w/2) H(e^(i w)) (and not H(w) ...) Regards, Andor
Reply by ●May 3, 20052005-05-03
"Andor" <an2or@mailcircuit.com> writes:> Randy Yates wrote: > ... > > Now to get the frequency response of a system function in the > z-domain, > > let z = e^(i*w). Since |e^(i*w)| = |z| = 1 and 1^(1/2) = 1, > > then > > > > |F(e^(i*w))| = |H(w)| / 1^(1/2) > > = |H(w)|. > > Just for correctness' sake, you should have > > F(e^(i w)) = e^(-i w/2) H(e^(i w)) (and not H(w) ...)That's a correct equation. The exposition would flow better if instead you write |F(e^(i*w))| = |H(e^(i*w))| / 1^(1/2) = |H(e^(i*w))|? I agree there was an error - thanks for the correction, Andor. -- Randy Yates Sony Ericsson Mobile Communications Research Triangle Park, NC, USA randy.yates@sonyericsson.com, 919-472-1124
Reply by ●May 4, 20052005-05-04
>"Andor" <an2or@mailcircuit.com> writes: > >> Randy Yates wrote: >> ... >> > Now to get the frequency response of a system function in the >> z-domain, >> > let z = e^(i*w). Since |e^(i*w)| = |z| = 1 and 1^(1/2) = 1, >> > then >> > >> > |F(e^(i*w))| = |H(w)| / 1^(1/2) >> > = |H(w)|. >> >> Just for correctness' sake, you should have >> >> F(e^(i w)) = e^(-i w/2) H(e^(i w)) (and not H(w) ...) > >That's a correct equation. The exposition would flow better >if instead you write > > |F(e^(i*w))| = |H(e^(i*w))| / 1^(1/2) > = |H(e^(i*w))|? > >I agree there was an error - thanks for the correction, Andor. >-- >Randy Yates >Sony Ericsson Mobile Communications >Research Triangle Park, NC, USA >randy.yates@sonyericsson.com, 919-472-1124 >This message was sent using the Comp.DSP web interface on www.DSPRelated.com
Reply by ●May 4, 20052005-05-04
>"Andor" <an2or@mailcircuit.com> writes: > >> Randy Yates wrote: >> ... >> > Now to get the frequency response of a system function in the >> z-domain, >> > let z = e^(i*w). Since |e^(i*w)| = |z| = 1 and 1^(1/2) = 1, >> > then >> > >> > |F(e^(i*w))| = |H(w)| / 1^(1/2) >> > = |H(w)|. >> >> Just for correctness' sake, you should have >> >> F(e^(i w)) = e^(-i w/2) H(e^(i w)) (and not H(w) ...) > >That's a correct equation. The exposition would flow better >if instead you write > > |F(e^(i*w))| = |H(e^(i*w))| / 1^(1/2) > = |H(e^(i*w))|? > >I agree there was an error - thanks for the correction, Andor. >-- >Randy Yates >Sony Ericsson Mobile Communications >Research Triangle Park, NC, USA >randy.yates@sonyericsson.com, 919-472-1124 >Thanks for many replies. I have another question ? let's assume F(z)=H1(z)+H2(z)*z^(-1/2) Then How can i find frequency response ? Looking forward fast resply .... -_-.... This message was sent using the Comp.DSP web interface on www.DSPRelated.com
