Hi all, well, i struggled with this one for a while: is h(t)=2*sin(wt)*cos(Wt)/(pi*t) an impulse responce of a BIBO stable system? How can I prove it? I had several directions, but all of them lead me close but not close enough One direction is: h(t)=2(sinc((w+W)t)+sinc((w-W)t)) so I can think of it as a connection in parralel of two ideal low-pass filters, which are both not BIBO stable, but this does not insure that the whole system is unstable For the case where w=W it is a simple lawpass filter, which is not Bibo stable... I tryed to prove absolute integrability (or non-integrability), but so far without success... Thanks for the help Ilana _____________________________________ Do you know a company who employs DSP engineers? Is it already listed at http://dsprelated.com/employers.php ?
BIBO stability
Started by ●May 7, 2007
Reply by ●May 7, 20072007-05-07
nisky wrote:> Hi all, > > well, i struggled with this one for a while: > > is h(t)=2*sin(wt)*cos(Wt)/(pi*t) an impulse responce of a BIBO stable > system? How can I prove it? > > I had several directions, but all of them lead me close but not close > enough > > One direction is: > h(t)=2(sinc((w+W)t)+sinc((w-W)t)) so I can think of it as a connection in > parralel of two ideal low-pass filters, which are both not BIBO stable, > but this does not insure that the whole system is unstable > > For the case where w=W it is a simple lawpass filter, which is not Bibo > stable... > > I tryed to prove absolute integrability (or non-integrability), but so far > without success... > > Thanks for the help > > IlanaWhether or not this is homework (and it looks like something I'd assign, heh heh heh), consider that you should be able to reduce the sin(wt) * cos(Wt) to a sum -- look in a trig book. I can see by glancing at this what the answer is, but if you have to _prove_ it for your homework to be valid you'll have to revisit trig. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Posting from Google? See http://cfaj.freeshell.org/google/ Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" gives you just what it says. See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by ●May 7, 20072007-05-07
> I tryed to prove absolute integrability (or non-integrability), but so far > without success...Have you tried using Cauchy-Schwarz inequality ?
Reply by ●May 23, 20072007-05-23
> >> I tryed to prove absolute integrability (or non-integrability), but sofar>> without success... > >Have you tried using Cauchy-Schwarz inequality ? > >I did, The problem is that when you break the expression to a sum of 2 sincs, and you use the Cauchy-Shwartz inequality you can prove that the original integral over the absolute value is less then something that is infinite, which is sord of a meaningless conclusion. So I am still stuck... Bthw, it is something I gave as a homework, thought it would be a cool question, and eventually I got to the point where I cannot solve it by myself... This is the impulse responce of an ideal bandpass filter - My gutts feeling is that it cannot be BIBO stable, but I just cannot find a pretty proof for it _____________________________________ Do you know a company who employs DSP engineers? Is it already listed at http://dsprelated.com/employers.php ?
Reply by ●May 23, 20072007-05-23
>nisky wrote: >> Hi all, >> >> well, i struggled with this one for a while: >> >> is h(t)=2*sin(wt)*cos(Wt)/(pi*t) an impulse responce of a BIBO stable >> system? How can I prove it? >> >> I had several directions, but all of them lead me close but not close >> enough >> >> One direction is: >> h(t)=2(sinc((w+W)t)+sinc((w-W)t)) so I can think of it as a connectionin>> parralel of two ideal low-pass filters, which are both not BIBOstable,>> but this does not insure that the whole system is unstable >> >> For the case where w=W it is a simple lawpass filter, which is notBibo>> stable... >> >> I tryed to prove absolute integrability (or non-integrability), but sofar>> without success... >> >> Thanks for the help >> >> Ilana > >Whether or not this is homework (and it looks like something I'd assign,>heh heh heh), consider that you should be able to reduce the sin(wt) * >cos(Wt) to a sum -- look in a trig book. > >I can see by glancing at this what the answer is, but if you have to >_prove_ it for your homework to be valid you'll have to revisit trig. > >-- > >Tim Wescott >Wescott Design Services >http://www.wescottdesign.com > >Posting from Google? See http://cfaj.freeshell.org/google/ > >Do you need to implement control loops in software? >"Applied Control Theory for Embedded Systems" gives you just what itsays.>See details at http://www.wescottdesign.com/actfes/actfes.html >I have broken it into a sum, it is written in the original question: h(t)=2(sinc((w+W)t)+sinc((w-W)t)) (more or less, pardon me for disregarding the constant gains, since they are irrelevant here ) But I am stuck at this point... _____________________________________ Do you know a company who employs DSP engineers? Is it already listed at http://dsprelated.com/employers.php ?
Reply by ●May 24, 20072007-05-24
Without having tried this out myself, wouldn't it be easier to do this in Laplace domain? Get the Laplace transform and predict BIBO stability by looking at poles? Also, have you tried expanding the sinc into a Taylor series type summation form to prove absolute integrability? Will try this problem when I get home. On May 7, 4:52 am, "nisky" <ilana_ni...@yahoo.com> wrote:> Hi all, > > well, i struggled with this one for a while: > > is h(t)=2*sin(wt)*cos(Wt)/(pi*t) an impulse responce of a BIBO stable > system? How can I prove it? > > I had several directions, but all of them lead me close but not close > enough > > One direction is: > h(t)=2(sinc((w+W)t)+sinc((w-W)t)) so I can think of it as a connection in > parralel of two ideal low-pass filters, which are both not BIBO stable, > but this does not insure that the whole system is unstable > > For the case where w=W it is a simple lawpass filter, which is not Bibo > stable... > > I tryed to prove absolute integrability (or non-integrability), but so far > without success... > > Thanks for the help > > Ilana > > _____________________________________ > Do you know a company who employs DSP engineers? > Is it already listed athttp://dsprelated.com/employers.php?
Reply by ●June 9, 20072007-06-09
>Without having tried this out myself, wouldn't it be easier to do this >in Laplace domain? Get the Laplace transform and predict BIBO >stability by looking at poles? Also, have you tried expanding the sinc >into a Taylor series type summation form to prove absolute >integrability?The first idea does not work since the transform of the sinc function is not a ratinal function... I will try the second idea - I will let you know here if I succeed
Reply by ●June 9, 20072007-06-09
nisky wrote:> Hi all, > > well, i struggled with this one for a while: > > is h(t)=2*sin(wt)*cos(Wt)/(pi*t) an impulse responce of a BIBO stable > system? How can I prove it? > > I had several directions, but all of them lead me close but not close > enough > > One direction is: > h(t)=2(sinc((w+W)t)+sinc((w-W)t)) so I can think of it as a connection in > parralel of two ideal low-pass filters, which are both not BIBO stable, > but this does not insure that the whole system is unstable > > For the case where w=W it is a simple lawpass filter, which is not Bibo > stable... > > I tryed to prove absolute integrability (or non-integrability), but so far > without success... > > Thanks for the help > > Ilana >Why do you say that the ideal low-pass filters are not BIBO stable? The impulse response you give has a finite amount of energy in it, and it goes to zero over time -- that says "BIBO stable" to me. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Posting from Google? See http://cfaj.freeshell.org/google/ Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" gives you just what it says. See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by ●June 9, 20072007-06-09
Tim Wescott <tim@seemywebsite.com> writes:> Why do you say that the ideal low-pass filters are not BIBO stable? > The impulse response you give has a finite amount of energy in it, and > it goes to zero over time -- that says "BIBO stable" to me.For BIBO stability, the impulse response has to be absolutely summable (http://en.wikipedia.org/wiki/BIBO_stability). The sinc function is not absolutely summable (or integrable in continuous time http://en.wikipedia.org/wiki/Sinc_function), so the ideal low pass filter is not BIBO stable. :-) Ciao, Peter K. -- "And he sees the vision splendid of the sunlit plains extended And at night the wondrous glory of the everlasting stars."
Reply by ●June 9, 20072007-06-09
p.kootsookos@remove.ieee.org (Peter K.) writes:> Tim Wescott <tim@seemywebsite.com> writes: > >> Why do you say that the ideal low-pass filters are not BIBO stable? >> The impulse response you give has a finite amount of energy in it, and >> it goes to zero over time -- that says "BIBO stable" to me. > > For BIBO stability, the impulse response has to be absolutely summable > (http://en.wikipedia.org/wiki/BIBO_stability). > > The sinc function is not absolutely summable (or integrable in > continuous time http://en.wikipedia.org/wiki/Sinc_function), so the > ideal low pass filter is not BIBO stable. :-)Ayup. Here's a copy of the post (from Google Groups) on a proof I made a few years back: ---------- copied message ---------- From: y...@ieee.org (Randy Yates) Date: Jan 6 2003, 6:14 pm Subject: Reconstruction of a signal from discrete-time samples. To: comp.dsp "John Leonard" <JLeona...@si.rr.com> wrote in message <news:Mi6S9.4396$yi6.1300707@twister.nyc.rr.com>... > Basically, convolving the Sinc function with the original sequence seems > to interpolate the original sequence by acting as a weighting function. The > greatest weight is thrown to the points closest to the interpolated point. > The Sinc function is proportional to the <sine function divide by x>. The > interpolated value is the infinite sum of terms which are the product of the > Sinc function and the given sequence. In order for this to work the sum must > converge. It has been some time since I studied Calculus but I do not > understand how a Harmonic series (if that is what this is) converges? I am > sure you can direct me to a text where this property is demonstrated. > John Hello John, You are correct - the series does not, in general, converge. Here's a proof: The ideal interpolation filter for upsampling at an integer ratio L is given by h[n] = sin(pi*n/L) / (pi*n/L) = sinc(n/L). This interpolation filter is convolved with the input signal to get the output, y[n] = h[n] # x[n] = sum_{m = -\infty}^{+\infty} x[m] * h[n-m], where "#" denotes convolution and x[n] is the upsampled input signal. Since the upsampled input signal has L-1 zeros for every sample in the input signal, then only every Lth sample in the the interpolation filter is used to generate an output. Therefore y[n] = sum_{m = -\infty}^{+\infty} x[L*m] * h[n-L*m]. Now assume x[L*m] = sgn(h[1-L*m]). Then y[1] = sum_{m = -\infty}^{+\infty} x[L*m] * h[1-L*m] = sum_{m = -\infty}^{+\infty} x[L*m] * sinc((1-L*m)/L) = sum_{m = -\infty}^{+\infty} sgn(sinc((1-L*m)/L)) * sinc((1-L*m)/L) = sum_{m = -\infty}^{+\infty} !sinc((1-L*m)/L)!. For a specific case, let L = 2. Then y[1] = sum_{m = -\infty}^{+\infty} !sinc((1-2*m)/2)! = sum_{m = -\infty}^{+\infty} !sinc(1/2-m)! = sum_{m = -\infty}^{+\infty} |sin(pi*(1/2-m))/(pi*(1/2-m))| = sum_{m = -\infty}^{+\infty} |sin(pi*(1/2-m))|/|(pi*(1/2-m))|. Since sin(pi*(1/2 - m)) = +/- 1, |sin(pi*(1/2 - m))| = 1. Thus y[1] = sum_{m = -\infty}^{+\infty} 1/|(pi*(1/2-m))| = (1/pi) * sum_{m = -\infty}^{+\infty} 1/|(1/2-m)| = (1/pi) * [sum_{m = -\infty}^{0} 1/|(1/2-m)| + sum_{m = 1}^{+\infty} 1/|(1/2-m)|] = (1/pi) * [sum_{m = -\infty}^{0} 1/(1/2-m) + sum_{m = 1}^{+\infty} 1/|(1/2-m)|]. Now since both terms are positive, if either term diverges, the entire sum diverges. Also note that the 1/pi factor up front is irrelevent in terms of determining convergence. Therefore consider this series: y = sum_{m = -\infty}^{0} 1/(1/2-m) = sum_{m = 0}^{+\infty} 1/(1/2+m) >= sum_{m = 0}^{+\infty} 1/(1+m) = 1 + sum_{m = 1}^{+\infty} 1/m. Therefore since the second term of this expression (the so-called harmonic series) diverges, this entire expression diverges. Since this expression diverges, the entire expression for y[1] also diverges. -- Randy Yates DSP Engineer, Sony Ericsson Mobile Communications Research Triangle Park, NC, USA randy.ya...@sonyericsson.com, 919-472-1124 -- % Randy Yates % "How's life on earth? %% Fuquay-Varina, NC % ... What is it worth?" %%% 919-577-9882 % 'Mission (A World Record)', %%%% <yates@ieee.org> % *A New World Record*, ELO http://home.earthlink.net/~yatescr
