Reply by rickman ●December 2, 20142014-12-02
On 12/2/2014 1:47 AM, Evgeny Filatov wrote:

> On 02.12.2014 7:27, radams2000@gmail.com wrote:
>> I believe that the OP is asking a different question. He's not looking
>> for the maximum value that could ever be produced by reconstructing
>> samples that are constrained to be between -1 and +1. Rather I think
>> he's asking the question, given a finite set of samples produced by
>> some signal of interest, what is the maximum value that can occur
>> after reconstruction. Perhaps the OP can clarify?
>>
>> Bob
>>
>
> Bob, I agree that information about either the waveform or occupied
> spectrum of the signal of interest is required for a meaningful answer.
>
> Indeed, there's an important aspect which makes the example I cited
> unrealistic -- namely that the analog signal has strong spectral
> components closely approaching the Nyquist frequency. In a practical
> system you'd have an anti-aliasing filter suppressing some frequency
> range near Nyquist -- which I believe limits the maximum peak value of
> reconstructed samples.


I believe he has given enough information.  He is looking for the 
max/min of the actual analog signal before sampling.  See his second 
post on 11/27.

-- 

Rick
Reply by Evgeny Filatov ●December 2, 20142014-12-02
On 02.12.2014 7:27, radams2000@gmail.com wrote:

> I believe that the OP is asking a different question. He's not looking for the maximum value that could ever be produced by reconstructing samples that are constrained to be between -1 and +1. Rather I think he's asking the question, given a finite set of samples produced by some signal of interest, what is the maximum value that can occur after reconstruction. Perhaps the OP can clarify?
>
> Bob
>


Bob, I agree that information about either the waveform or occupied 
spectrum of the signal of interest is required for a meaningful answer.

Indeed, there's an important aspect which makes the example I cited 
unrealistic -- namely that the analog signal has strong spectral 
components closely approaching the Nyquist frequency. In a practical 
system you'd have an anti-aliasing filter suppressing some frequency 
range near Nyquist -- which I believe limits the maximum peak value of 
reconstructed samples.

Evgeny.
Reply by ●December 2, 20142014-12-02
I believe that the OP is asking a different question. He's not looking for the maximum value that could ever be produced by reconstructing samples that are constrained to be between -1 and +1. Rather I think he's asking the question, given a finite set of samples produced by some signal of interest, what is the maximum value that can occur after reconstruction. Perhaps the OP can clarify?

Bob
Reply by Evgeny Filatov ●December 1, 20142014-12-01
On 29.11.2014 9:24, glen herrmannsfeldt wrote:

> Next, consider all samples -1, except for two consecutive +1.
> I believe this gets the largest positive peak.
>
> -- glen
>


Let's consider the digital signal you've proposed. Let the length of the 
digital signal be 20 samples. So we have:

-1,-1,-1,-1,-1,-1,-1,-1,-1, 1, 1,-1,-1,-1,-1,-1,-1,-1,-1,-1
The positive peak for its analog reconstruction is about 1.57.

Now let's replace two of -1 with 1:
-1,-1,-1,-1,-1,-1,-1, 1,-1, 1, 1,-1, 1,-1,-1,-1,-1,-1,-1,-1
Now the positive peak is around 2.08.

Keep replacing -1 with 1:
For
-1,-1,-1,-1,-1, 1,-1, 1,-1, 1, 1,-1, 1,-1, 1,-1,-1,-1,-1,-1
the positive peak is 2.37,

for
-1,-1,-1, 1,-1, 1,-1, 1,-1, 1, 1,-1, 1,-1, 1,-1, 1,-1,-1,-1
it's 2.56

for
-1, 1,-1, 1,-1, 1,-1, 1,-1, 1, 1,-1, 1,-1, 1,-1, 1,-1, 1,-1
the positive peak is 2.71.

Here you can see the plots of the considered sampled signals along with 
their analog reconstructions:

http://tinyurl.com/nwaws63

The tendency is for the value of positive peak to rise as long as you 
increase the length of the sequence of alternating -1 and 1. It's due to 
the fact that the positive peak is the sum of shifted versions of the 
sinc function multiplied by values of the respective samples. Since the 
sinc function has alternating lobes of positive and negative signs, by 
properly choosing signs of the samples you can increase the magnitude of 
the positive peak.

Also, you can always increase the value of the positive peak if you can 
add more samples. The trick is that while the integral of sinc(x)dx from 
-inf to +inf is finite, the integral of |sinc(x)|dx from -inf to +inf is 
infinite. Which means that in theory you can drive the value of the 
positive peak above any predefined positive value as long as you can 
choose the number of samples in the sequence.

Hope this helps.

Evgeny.
Reply by Evgeny Filatov ●December 1, 20142014-12-01
On 29.11.2014 9:24, glen herrmannsfeldt wrote:


>
> Next, consider all samples -1, except for two consecutive +1.
> I believe this gets the largest positive peak.
>
> -- glen
>


Glen, no. The largest positive peak is unlimited, and is achieved for 
the sequence of alternating samples 1 and -1, with a single -1 sample 
removed from the sequence, so that you have two consecutive +1.

The value of the peak depends on the length of the sequence, and each 
sample in the sequence contributes to the peak. See the details in my 
other post.

Evgeny.
Reply by Eric Jacobsen ●November 29, 20142014-11-29
On Thu, 27 Nov 2014 03:44:26 -0600, "Detlef _A" <25706@dsprelated>
wrote:


>Group,
>
>I sample a bandlimited signal satisfying Nyquist. How can I get the
>maximum/minimum of the continious time signal from the samples,  i.e.
>upper/lower bounds or the dedicated value.
>
>Any pointers welcome.
>
>THX
>Cheers
>Detlef
>


Given that computation power is often cheap these days, you can
numerically up-sample it (i.e., increase the sample rate by
interpolation), and take the numerical min/max of the result.   The
higher the upsampling rate the more accurate the results.   If you
want to get some sort of confidence indication on the results, you can
successively increase the sample rate and take the max/min of each
result.   The max/min values may stabilize or show an asymptotic trend
toward a particular value, but you would likely see when the values
stopped changing within the required amount of precision.


Eric Jacobsen
Anchor Hill Communications
http://www.anchorhill.com
Reply by glen herrmannsfeldt ●November 29, 20142014-11-29
Rick Lyons <R.Lyons@_bogus_ieee.org> wrote:

> On Thu, 27 Nov 2014 11:09:21 -0600, Tim Wescott <tim@seemywebsite.com>
> wrote:
>>On Thu, 27 Nov 2014 03:44:26 -0600, Detlef _A wrote:



>>> I sample a bandlimited signal satisfying Nyquist. How can I get the
>>> maximum/minimum of the continious time signal from the samples,  i.e.
>>> upper/lower bounds or the dedicated value.



>> I'm sure that there's some nice closed-form solution to this, 
>> but I don't know it.


(snip)


>  It seems to me the answer to Detlef's question depends 
> on the nature of the time-domain signal.  If his signal 
> is a pure tone (a single sinusoid) then computing the 
> variance of his signal will lead to the signal's max 
> amplitude value.


Some time ago, I wondered about the maximum reconstructed amplitude
for a sampled signal with fixed max/min sample values. 

First, as you note, consider sinusoids.

A sine at Fs/4, sampled at (in degrees) 45, 135, 225, 315, such
that the sample values are +1, +1, -1, -1, has amplitude sqrt(2).

Note also that the RMS value of the sine equal the RMS of 
the samples.

Or a sampled sinusoid giving samples of +1, +1, -1 with frequency Fs/3.
First, the average values is 1/3, and samples at 30, 150, 270 degrees.
The AC component has amplitude 4/3, with positve peak at 5/3.
The amplitude of the sinusoid is smaller than Fs/4, but the positive
peak is higher.

Removing the constraint for sinusoids, consider the signal whose
samples are all -1, except at n=0, where it is +1. It isn't hard
to see that 2*sinc(n)-1 fits the points. The DC component is
(in the limit) -1.

Next, consider all samples -1, except for two consecutive +1.
I believe this gets the largest positive peak.

-- glen
Reply by Rick Lyons ●November 28, 20142014-11-28
On Thu, 27 Nov 2014 11:09:21 -0600, Tim Wescott <tim@seemywebsite.com>
wrote:


>On Thu, 27 Nov 2014 03:44:26 -0600, Detlef _A wrote:
>
>> Group,
>> 
>> I sample a bandlimited signal satisfying Nyquist. How can I get the
>> maximum/minimum of the continious time signal from the samples,  i.e.
>> upper/lower bounds or the dedicated value.
>> 
>> Any pointers welcome.
>> 
>> THX Cheers Detlef
>
>I'm sure that there's some nice closed-form solution to this, but I don't 
>know it.
>
>If you were to oversample at ten times the bandwidth or more, my gut feel 
>is that (a) you'd get pretty close just by taking the maximum and minimum 
>of the sampled signal, and (b) you could get closer yet by taking one 
>sample on each side of the extremum, making a quadratic expression out of 
>it, and calculating the extremum of that.
>
>Are you looking for some practical "close enough" way to do this, or some 
>academically perfect closed form?


Hi,
  It seems to me the answer to Detlef's question depends 
on the nature of the time-domain signal.  If his signal 
is a pure tone (a single sinusoid) then computing the 
variance of his signal will lead to the signal's max 
amplitude value.

If his signal is an ampltitude modulation (AM) signal 
then generating a complex-valued (analytic) version 
of his time signal and then performing 'envelope detection' 
might be the thing to do.

But your (Tim's) suggestion of oversampling and searching 
for max and min time sample values should do the job for 
generic time signals.  If oversampling is not possible, 
then Detlef could perform time-domain interpolation and 
then search for max and min sample values.
Alternately, he could compute the FFT of his time signal, 
insert lots of zero-valued samples in the center of the 
FFT samples, and compute a larger-sized inverse FFT to 
implement time-doamin interpolation.  Follow that, as before, 
by searching for max and min time sample values.

I wonder what sort of sig processing software Detlef is using.

[-Rick-]
Reply by Evgeny Filatov ●November 28, 20142014-11-28
On 27.11.2014 12:44, Detlef _A wrote:

> Group,
>
> I sample a bandlimited signal satisfying Nyquist. How can I get the
> maximum/minimum of the continious time signal from the samples,  i.e.
> upper/lower bounds or the dedicated value.
>
> Any pointers welcome.
>
> THX
> Cheers
> Detlef
>
> 	
>
> _____________________________		
> Posted through www.DSPRelated.com
>


According to the sampling theorem, you can get a perfect analog 
reconstruction of a digital signal if you interpolate it using sin(x)/x 
function.

The problem is that the infinite series 1/x is divergent. Exploiting 
that property, we can get arbitrarily high maximums of the continuous 
time signal by increasing the length of the sampled series without 
altering its magnitude.

For example, consider a digital signal consisting of alternating 1 and 
-1, with a single sample removed from the center of the sequence. So you 
get signal of the form [1 -1 ... 1 -1 1 1 -1 ... 1 -1]. For such digital 
signal consisting of N = 2*1e6 samples, the maximum of the corresponding 
analog signal exceeds 10! Here, I have zoomed in the central part, so 
you can see a part of the digital signal and its analog reconstruction:

http://tinypic.com/r/2urxgg0/8

Hope this helps.

Evgeny.
Reply by glen herrmannsfeldt ●November 27, 20142014-11-27
Tim Wescott <tim@seemywebsite.com> wrote:

> On Thu, 27 Nov 2014 03:44:26 -0600, Detlef _A wrote:


(snip)

>> I sample a bandlimited signal satisfying Nyquist. How can I get the
>> maximum/minimum of the continious time signal from the samples,  i.e.
>> upper/lower bounds or the dedicated value.

 
(snip)


> If you were to oversample at ten times the bandwidth or more, my gut feel 
> is that (a) you'd get pretty close just by taking the maximum and minimum 
> of the sampled signal, and (b) you could get closer yet by taking one 
> sample on each side of the extremum, making a quadratic expression out of 
> it, and calculating the extremum of that.


Seems to me that you could fit an Nth order polynomial to some number
of points and find its maximum or minimum. For increasing N, you
get closer.

In theory, all the points can contribute to the peak, but
diminishing, I believe 1/x, with distance away. 

I have thought before about a sine at half nyquist with a peak
half way between sample points, or at 2/3 nyquist, again with
a peak between sample points.

How about a sinc, again with the peak half way between two points.
I think sinc is the peakiest band-limited function, so that
would seem a place to look.

-- glen