What's the difference in the impulse response between a filter which has the frequency-domain response |H(w)| = 1, |f| < Fs/2, 0 otherwise, angle(H(w)) = 0 and |H(w)| = 1, |f| <= Fs/2, 0 otherwise, angle(H(w)) = 0 ??? -- % Randy Yates % "...the answer lies within your soul %% Fuquay-Varina, NC % 'cause no one knows which side %%% 919-577-9882 % the coin will fall." %%%% <yates@ieee.org> % 'Big Wheels', *Out of the Blue*, ELO http://home.earthlink.net/~yatescr
sinc() question
Started by ●June 24, 2003
Reply by ●June 24, 20032003-06-24
Randy Yates wrote:> > What's the difference in the impulse response between a filter which > has the frequency-domain response > > |H(w)| = 1, |f| < Fs/2, 0 otherwise, angle(H(w)) = 0 > > and > > |H(w)| = 1, |f| <= Fs/2, 0 otherwise, angle(H(w)) = 0 > > ??? > -- > % Randy Yates % "...the answer lies within your soul > %% Fuquay-Varina, NC % 'cause no one knows which side > %%% 919-577-9882 % the coin will fall." > %%%% <yates@ieee.org> % 'Big Wheels', *Out of the Blue*, ELO > http://home.earthlink.net/~yatescrIsn't it Inline to respond to a question with another question? If you had such a pair of filters, what real-world test could tell them apart? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●June 24, 20032003-06-24
Randy Yates wrote:> What's the difference in the impulse response between a filter which > has the frequency-domain response > > |H(w)| = 1, |f| < Fs/2, 0 otherwise, angle(H(w)) = 0 > > and > > |H(w)| = 1, |f| <= Fs/2, 0 otherwise, angle(H(w)) = 0 > > ???My two cents. Since they differ by a set of dimension zero, and since the frequency domain response is calculated by mean of integrals (which have same results, in case of a null difference), the frequency response is the same. bye, -- Piergiorgio Sartor
Reply by ●June 24, 20032003-06-24
Randy Yates wrote:> What's the difference in the impulse response between a filter which > has the frequency-domain response > > |H(w)| = 1, |f| < Fs/2, 0 otherwise, angle(H(w)) = 0 > > and > > |H(w)| = 1, |f| <= Fs/2, 0 otherwise, angle(H(w)) = 0 > > ???I forgot, only if there is nothing "strange", i.e. infinite, for |f| = Fs/2. bye, -- Piergiorgio Sartor
Reply by ●June 24, 20032003-06-24
Randy Yates wrote:> > What's the difference in the impulse response between a filter which > has the frequency-domain response > > |H(w)| = 1, |f| < Fs/2, 0 otherwise, angle(H(w)) = 0 > > and > > |H(w)| = 1, |f| <= Fs/2, 0 otherwise, angle(H(w)) = 0 > > ??? > -- > % Randy Yates % "...the answer lies within your soul > %% Fuquay-Varina, NC % 'cause no one knows which side > %%% 919-577-9882 % the coin willfall."> %%%% <yates@ieee.org> % 'Big Wheels', *Out of the Blue*, ELO > http://home.earthlink.net/~yatescrIsn't it Zen like to respond to a question with another question? If you had such a pair of filters, what real-world test could tell them apart? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●June 24, 20032003-06-24
Isn't there a theorem in Fourier theory that says that the value of a spectrum or time domain waveform at a jump discontinuity is equal to the average of the left and right hand limits approaching the discontinuity? If so, the value at the discontinuity would be 0.5 in both cases and the spectra would be identical. From another approach, if the IFT is calculated, the finite value at 1 point will not change the value of the integral, so I would say they are the same. Dirk Dirk A. Bell DSP Consultant "Jerry Avins" <jya@ieee.org> wrote in message news:3EF870A4.8082D911@ieee.org...> Randy Yates wrote: > > > > What's the difference in the impulse response between a filter which > > has the frequency-domain response > > > > |H(w)| = 1, |f| < Fs/2, 0 otherwise, angle(H(w)) = 0 > > > > and > > > > |H(w)| = 1, |f| <= Fs/2, 0 otherwise, angle(H(w)) = 0 > > > > ??? > > -- > > % Randy Yates % "...the answer lies within your soul > > %% Fuquay-Varina, NC % 'cause no one knows which side > > %%% 919-577-9882 % the coin will > fall." > > %%%% <yates@ieee.org> % 'Big Wheels', *Out of the Blue*, ELO > > http://home.earthlink.net/~yatescr > > Isn't it Zen like to respond to a question with another question? If you > had such a pair of filters, what real-world test could tell them apart? > > Jerry > -- > Engineering is the art of making what you want from things you can get. > �����������������������������������������������������������������������
Reply by ●June 24, 20032003-06-24
In article 3EF83ECD.72947E50@ieee.org, Randy Yates at yates@ieee.org wrote on 06/24/2003 08:03:> What's the difference in the impulse response between a filter which > has the frequency-domain response > > |H(w)| = 1, |f| < Fs/2, 0 otherwise, angle(H(w)) = 0 > > and > > |H(w)| = 1, |f| <= Fs/2, 0 otherwise, angle(H(w)) = 0 > > ???i think there would be no difference since the inverse FT integral of the difference function of the two would be zero. r b-j
Reply by ●June 24, 20032003-06-24
Randy Yates <yates@ieee.org> wrote in message news:<3EF83ECD.72947E50@ieee.org>...> What's the difference in the impulse response between a filter which > has the frequency-domain response > > |H(w)| = 1, |f| < Fs/2, 0 otherwise, angle(H(w)) = 0 > > and > > |H(w)| = 1, |f| <= Fs/2, 0 otherwise, angle(H(w)) = 0 > > ???Not a lot. For real-valued signals, a frequency component at Fs/2 is aliased. I'll try to play with the numbers when the signal consists on one single sinusoidal at frequency Fs/2 and see what pops out: s(t)=A*cos(2*pi*Fs/2*t)+B*sin(2*pi*Fs/2*t) =A(exp(i*2*pi*Fs/2*t)+exp(i*2*pi*(-Fs/2)*t))/2+ B(exp(i*2*pi*Fs/2*t)-exp(i*2*pi*(-Fs/2)*t))/(i*2) [1] =A(exp(i*2*pi*Fs/2*t)+exp(i*2*pi*(Fs-Fs/2)*t))/2+ B(exp(i*2*pi*Fs/2*t)-exp(i*2*pi*(Fs-Fs/2)*t))/(i*2) [2] =A(exp(i*pi*Fs*t)+exp(i*pi*Fs*t))/2+ B(exp(i*pi*Fs*t)-exp(i*pi*Fs*t))/(i*2) [3] =A(cos(pi*Fs*t)+i*sin(pi*Fs*t)) [4] Since t=n*T=n/Fs, [4] simplifies to s(t)=s(n*T)=A(cos(pi*n)+i*sin(pi*n))=A*cos(pi*n)=A(-1)^(n) [5] This corresponds to my intuitive understanding, i.e. that only the even part of the signal can be recovered. With the strict inequality, the whole signal can be recovered. If f=Fs/2 is included, the whole signal can *not* be recovered. My 2c... Rune
Reply by ●June 24, 20032003-06-24
Dirk Bell wrote:> > Isn't there a theorem in Fourier theory that says that the value of a > spectrum or time domain waveform at a jump discontinuity is equal to the > average of the left and right hand limits approaching the discontinuity? If > so, the value at the discontinuity would be 0.5 in both cases and the > spectra would be identical. > > From another approach, if the IFT is calculated, the finite value at 1 point > will not change the value of the integral, so I would say they are the same. > > Dirk > > Dirk A. Bell > DSP Consultant > > "Jerry Avins" <jya@ieee.org> wrote in message > news:3EF870A4.8082D911@ieee.org... > > Randy Yates wrote: > > > > > > What's the difference in the impulse response between a filter which > > > has the frequency-domain response > > > > > > |H(w)| = 1, |f| < Fs/2, 0 otherwise, angle(H(w)) = 0 > > > > > > and > > > > > > |H(w)| = 1, |f| <= Fs/2, 0 otherwise, angle(H(w)) = 0 > > > > > > ??? > > > -- > > > % Randy Yates % "...the answer lies within your soul > > > %% Fuquay-Varina, NC % 'cause no one knows which side > > > %%% 919-577-9882 % the coin will > > fall." > > > %%%% <yates@ieee.org> % 'Big Wheels', *Out of the Blue*, ELO > > > http://home.earthlink.net/~yatescr > > > > Isn't it Zen like to respond to a question with another question? If you > > had such a pair of filters, what real-world test could tell them apart?I had in mind distinguishing 4.999999999999 KHz from 5.000000000001 in a practical way. But of course, there's a more fundamental theoretical limitation. I'm still smarting. I didn't want to go out on that limb. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●June 24, 20032003-06-24
robert bristow-johnson wrote:> > In article 3EF83ECD.72947E50@ieee.org, Randy Yates at yates@ieee.org wrote > on 06/24/2003 08:03: > > > What's the difference in the impulse response between a filter which > > has the frequency-domain response > > > > |H(w)| = 1, |f| < Fs/2, 0 otherwise, angle(H(w)) = 0 > > > > and > > > > |H(w)| = 1, |f| <= Fs/2, 0 otherwise, angle(H(w)) = 0 > > > > ??? > > i think there would be no difference since the inverse FT integral of the > difference function of the two would be zero. > > r b-jSo does it pass Fs/2 or not??? -- % Randy Yates % "...the answer lies within your soul %% Fuquay-Varina, NC % 'cause no one knows which side %%% 919-577-9882 % the coin will fall." %%%% <yates@ieee.org> % 'Big Wheels', *Out of the Blue*, ELO http://home.earthlink.net/~yatescr
