# Stability of IIR Filters

Started by December 5, 2015
```Folks,

First let me ask this: Does the Fs/2-wide sinc function, interpolating
at some fractional sample offset, have a z-transform? I don't think so,
but I thought I'd verify.

Now it can be shown that a fractionally interpolating sinc function can
generate an infinite output with bounded input. This is commonly
described as an unstable filter. BI does not imply BO.

For the class of IIRs which have rational z-transforms, we can factor
the denominator and find all the poles, and it is common knowledge that
if the poles are inside the unit circle, the filter is called "stable."
Bounded input ==> bounded output.

Now here's the \$64,000 question: Is bounded input ==> bounded output
stability only valid for digital signals?

Asked another way, consider an analog (continuous-time) signal that
blows up at some point in time t_bu, but is finite otherwise. Assume
we sample that signal with a sample phase that avoids t_bu. Then the
resulting digital signal is bounded, right?

So now let's say we run this signal through an IIR with rational
z-transform. I think we can certainly say that the result is always
bounded if there are no poles on or outside the unit circle. Right? But
if we convert that signal back to continuous-time using the ideal
interpolator, can we say the resulting continuous-time signal is also
bounded?

Inquiring minds want to know...
--
Randy Yates, DSP/Embedded Firmware Developer
Digital Signal Labs
http://www.digitalsignallabs.com
```
```On Fri, 04 Dec 2015 23:07:05 -0500, Randy Yates wrote:

> Folks,
>
> First let me ask this: Does the Fs/2-wide sinc function, interpolating
> at some fractional sample offset, have a z-transform? I don't think so,
> but I thought I'd verify.

If it's a sinc in time, it would be sinc(<mumble-mumble>k)/z^k, wouldn't
it?  Presumably if you have a fractional shift there would still be some
closed-form function you could write.

> Now it can be shown that a fractionally interpolating sinc function can
> generate an infinite output with bounded input. This is commonly
> described as an unstable filter. BI does not imply BO.

Are you implying that only straight people start to smell after a while?

A sinc filter is still BIBO stable.  Having an infinite response doesn't
mean something isn't BIBO stable.

> For the class of IIRs which have rational z-transforms, we can factor
> the denominator and find all the poles, and it is common knowledge that
> if the poles are inside the unit circle, the filter is called "stable."
> Bounded input ==> bounded output.

And the response tails off infinitely in time.

> Now here's the \$64,000 question: Is bounded input ==> bounded output
> stability only valid for digital signals?
>
> Asked another way, consider an analog (continuous-time) signal that
> blows up at some point in time t_bu, but is finite otherwise. Assume we
> sample that signal with a sample phase that avoids t_bu. Then the
> resulting digital signal is bounded, right?

Ah ha -- I can see the error in your thinking from here.  If you can
straddle the point where it "blows up" then you are, ipso facto, sampling
too slowly.

> So now let's say we run this signal through an IIR with rational
> z-transform. I think we can certainly say that the result is always
> bounded if there are no poles on or outside the unit circle. Right?

Right.

> But
> if we convert that signal back to continuous-time using the ideal
> interpolator, can we say the resulting continuous-time signal is also
> bounded?

Yes.

--
www.wescottdesign.com
```
```Tim Wescott <tim@seemywebsite.com> writes:

> On Fri, 04 Dec 2015 23:07:05 -0500, Randy Yates wrote:
>
>> Folks,
>>
>> First let me ask this: Does the Fs/2-wide sinc function, interpolating
>> at some fractional sample offset, have a z-transform? I don't think so,
>> but I thought I'd verify.
>
> If it's a sinc in time, it would be sinc(<mumble-mumble>k)/z^k, wouldn't
> it?  Presumably if you have a fractional shift there would still be some
> closed-form function you could write.
>
>> Now it can be shown that a fractionally interpolating sinc function can
>> generate an infinite output with bounded input. This is commonly
>> described as an unstable filter. BI does not imply BO.
>
> Are you implying that only straight people start to smell after a
> while?

<snicker>

> A sinc filter is still BIBO stable.  Having an infinite response doesn't
> mean something isn't BIBO stable.

Er, what?!!!!?
--
Randy Yates, DSP/Embedded Firmware Developer
Digital Signal Labs
http://www.digitalsignallabs.com
```
```Does infinite response mean infinite in time or
infinite in amplitude?
```
```It's true that if you apply abs(sinc(x)) as an input to a sinc filter you will get an unbounded output but only when the alignment between input and filter taps is in a certain range. So like you say it's infinite for a short time but then comes back to earth again. My understanding of an unstable filter is that the output continues to grow after the input goes to zero, which implies a pole in the RHP or outside the unit circle. So I don't know that I would call this an unstable filter.

Bob
```
```Randy Yates <yates@digitalsignallabs.com> Wrote in message:
> Tim Wescott <tim@seemywebsite.com> writes:
>
>> On Fri, 04 Dec 2015 23:07:05 -0500, Randy Yates wrote:
>>
>>> Folks,
>>>
>>> First let me ask this: Does the Fs/2-wide sinc function, interpolating
>>> at some fractional sample offset, have a z-transform? I don't think so,
>>> but I thought I'd verify.
>>
>> If it's a sinc in time, it would be sinc(<mumble-mumble>k)/z^k, wouldn't
>> it?  Presumably if you have a fractional shift there would still be some
>> closed-form function you could write.
>>
>>> Now it can be shown that a fractionally interpolating sinc function can
>>> generate an infinite output with bounded input. This is commonly
>>> described as an unstable filter. BI does not imply BO.
>>
>> Are you implying that only straight people start to smell after a
>> while?
>
> <snicker>
>
>> A sinc filter is still BIBO stable.  Having an infinite response doesn't
>> mean something isn't BIBO stable.
>
> Er, what?!!!!?
>

Oops. Having a response that's infinite in time doesn't make
something unstable. Better?

--
www.wescottdesign.com

http://usenet.sinaapp.com/
```
```On 05.12.2015 18:03, radams2000@gmail.com wrote:
> It's true that if you apply abs(sinc(x)) as an input to a sinc filter you will get an unbounded output but only when the alignment between input and filter taps is in a certain range. So like you say it's infinite for a short time but then comes back to earth again. My understanding of an unstable filter is that the output continues to grow after the input goes to zero, which implies a pole in the RHP or outside the unit circle. So I don't know that I would call this an unstable filter.
>
>
> Bob
>

If you absolutely (pun intended) insist on applying abs(sinc(x)) as an
input to a sinc filter, you should know that will converge allright due
to the alternating series test.

Better luck in applying sinc(x) as an input to a sinc filter.

--
Evgeny.

```
```radams2000@gmail.com Wrote in message:
> It's true that if you apply abs(sinc(x)) as an input to a sinc filter you will get an unbounded output but only when the alignment between input and filter taps is in a certain range. So like you say it's infinite for a short time but then comes back to earth again. My understanding of an unstable filter is that the output continues to grow after the input goes to zero, which implies a pole in the RHP or outside the unit circle. So I don't know that I would call this an unstable filter.
>
>
> Bob
>

Nope. Sinc (p*n) has a finite sum, so its response to any
all-positive sequence will be finite.

--
www.wescottdesign.com

http://usenet.sinaapp.com/
```
```radams2000@gmail.com Wrote in message:

<snip>
> My understanding of an unstable filter is that the output continues to grow after the input goes to zero, which implies a pole in the RHP or outside the unit circle.
>
<snip>

there are many different definitions of stability; that's one.

--
www.wescottdesign.com

http://usenet.sinaapp.com/
```
```On 05.12.2015 19:31, Tim Wescott wrote:
>> It's true that if you apply abs(sinc(x)) as an input to a sinc filter you will get an unbounded output but only when the alignment between input and filter taps is in a certain range. So like you say it's infinite for a short time but then comes back to earth again. My understanding of an unstable filter is that the output continues to grow after the input goes to zero, which implies a pole in the RHP or outside the unit circle. So I don't know that I would call this an unstable filter.
>>
>>
>> Bob
>>
>
> Nope. Sinc (p*n) has a finite sum, so its response to any
>   all-positive sequence will be finite.
>

Kinda like Dirichlet's test?

--
Evgeny.

```