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# Discussion Groups | Comp.DSP | BIBO stability

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# BIBO stability - nisky - 2007-05-07 07:52:00

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 ?

# Re: BIBO stability - Tim Wescott - 2007-05-07 10:28:00

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

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

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

# Re: BIBO stability - Ikaro - 2007-05-07 10:30:00

> I tryed to prove absolute integrability (or non-integrability), but so far
> without success...

Have you tried using Cauchy-Schwarz inequality ?

# Re: BIBO stability - nisky - 2007-05-23 16:02:00

>
>> I tryed to prove absolute integrability (or non-integrability), but so
far
>> 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 ?

# Re: BIBO stability - nisky - 2007-05-23 16:08:00

>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
>
>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
>
>
>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
>

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 ?

# Re: BIBO stability - somdeb@gmail.com - 2007-05-23 23:15:00

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?

# Re: BIBO stability - nisky - 2007-06-09 05:46:00

>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

# Re: BIBO stability - Tim Wescott - 2007-06-09 14:56:00

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

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

# Re: BIBO stability - 2007-06-09 20:11:00

Tim Wescott <t...@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."

# Re: BIBO stability - Randy Yates - 2007-06-09 20:59:00

p...@remove.ieee.org (Peter K.) writes:

> Tim Wescott <t...@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\$y...@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)',
%%%% <y...@ieee.org>           % *A New World Record*, ELO
http://home.earthlink.net/~yatescr

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