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Topic: BIBO Instability of Ideal LPF (Read 4322 times) |
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william wu
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BIBO Instability of Ideal LPF
« on: Oct 11th, 2003, 8:45am » |
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What's an easy way to show that an Ideal Lowpass Filter is BIBO (Bounded Input Bounded Output) unstable? That is, I want to show that for the system with impulse response h(t) = sinc(t) = sin([pi]t)/([pi]t), there exists a Bounded Input
x(t) : [forall]t, |x(t)| < r, r[in][bbr]
such that the output signal
y(t) = (sinc [smiley=star.gif] x)(t) = [int]-inf to inf sinc([lambda])x(t - [lambda]) d[lambda]
diverges in magnitude for some t.
Note 1:
Some facts that may be useful: sinc(t) = sin([pi]t)/([pi]t). This wiggly curve shows up in a LOT of places ... see Note 3 for more ...
Let [calf]{} denote the Fourier Transform operator. Then
[calf]{sinc(t)} = [int]-inf,infsinc(t)e-j2[pi]ftdt = [prod](f)
where [prod](f) = 1 for |f|[le]0.5, and 0 elsewhere.
Since convolution ([smiley=star.gif]) is multiplcation in the frequency domain, the output signal y(t) can be given by
y(t) = [calf]-1{X(f) [cdot] [prod](f)}
where [calf]-1{} is the inverse Fourier Transform operator, and X(f) is the Fourier Transform of x(t).
Note 2:
I know of three ways to show BIBO instability for a system with impulse response h(t):
1) Simply come up with some bounded input that results in an unbounded output.
2) Take Laplace transform of the impulse response h(t) and show that there exist poles not strictly in the left-hand plane of the complex plane.
3) Show that [int]-inf to inf || h(t) || dt [to] [infty].
I would really prefer to use method 1), but oddly I couldn't think of an input that would work ... it's been a while since I've done these stability problems.
Regarding method 2), I don't know how to do LT of sinc(t) ... well, maybe I'm too lazy / lookup-table-brainwashed to figure out how to integrate it. (Mathematica returns Arctan[1/s] though, and the poles are on the jw axis which is not strictly in the left hand plane.)
Regarding method 3), this is the one I ended up using just for the sake of solving the problem. I showed the integral diverges by breaking up the integral from -inf to inf into integrals over the limits [k,k+1], where k[in][bbz]. I then did a comparison test with the harmonic sequence. I feel that this solution is unsatisfyingly complicated though, and there should be a very simple way to solve this problem.
Note 3:
"It's fair to say that many EE's see the sinc function in their dreams."
- quote and accompanying pics by Dr. Brad Osgood
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| « Last Edit: Oct 14th, 2003, 2:52am by william wu » |
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James Fingas
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Re: BIBO Instability of Ideal LPF
« Reply #1 on: Oct 14th, 2003, 12:58pm » |
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This is not too hard: Take an input that is a series of pulses (not delta functions, but short rectangular pulses, so they remain bounded) that occur at the times {0, 2, 4, 6, ...}.
The output is the convolution of the input and h(t), which you can see is a sinusoid with amplitude that goes roughly as the sum of 1/([pi]t), which goes to infinity.
Basically, any bounded signal that is periodic at a frequency in the LPF's pass band will work, because it will have delta functions in the frequency domain that get passed untouched through the filter. But note that the signal has to have infinite duration (or else it can't possibly have infinite energy).
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william wu
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Re: BIBO Instability of Ideal LPF
« Reply #2 on: Oct 16th, 2003, 12:37am » |
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I'm not sure what your h(t) is. Do you mean a train of rectangular pulses of unit width, where each pulse is centered at even integers? If so, I don't find it obvious that the sum of all these shifted pulse responses to the sinc produces a sinusoid, much less a sinusoid whose amplitude decays like that.
on Oct 14th, 2003, 12:58pm, James Fingas wrote:
Basically, any bounded signal that is periodic at a frequency in the LPF's pass band will work, because it will have delta functions in the frequency domain that get passed untouched through the filter. But note that the signal has to have infinite duration (or else it can't possibly have infinite energy). |
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By "will work" I assume you mean those are the conditions for the bounded inputs which will produce bounded outputs, not conditions for the bounded inputs I was looking for to demonstrate unbounded outputs.
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william wu
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Re: BIBO Instability of Ideal LPF
« Reply #3 on: Oct 16th, 2003, 12:56am » |
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Here are the dirty details of my [int]|sinc(t)|dt [to] [infty] proof if anyone is curious:
|sinc(t)| = (1/[pi]) * (|sin([pi]t)|/|t|)
Note |sin([pi]t)| > 1/2 over all intervals In = [1/6 + n, 5/6 + n], where n is an integer. Note that the interval has length 5/6 - 1/6 = 2/3.
Also, over the interval In, |t|[le]5/6 + n. So 1/|t| [ge] (5/6 + n)-1, over the interval In.
Let's evaluate |sinc(t)| over one of these intervals:
[int]I_n |sin[pi]t|/([pi]|t|) dt [ge] (2[pi])-1 (5/6 + n)-1 [int]I_ndt = (2[pi])-1 (2/3) (5/6 + n)-1
Now apply this result to the integral of |sinc(t)| over all time (finding a lower bound):
[center] [int]-inf to inf |sinc(t)| [ge] [sum]n [int]I_n |sin[pi]t|/([pi]|t|) dt = (3[pi])-1 [sum]n=1 to inf (5/6 + n)-1 [to] [infty]
where the last step follows from the divergence of the harmonic series.
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| « Last Edit: Oct 16th, 2003, 12:57am by william wu » |
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James Fingas
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Here are some diagrams showing my pulse train and how it convolves with the sinc function.
You're right. It's not enough to have a delta function in the frequency domain.
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