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

# Max/Min of continuous signal

Started by ●November 27, 2014

Reply by ●November 27, 20142014-11-27

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 DetlefI'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? -- www.wescottdesign.com

Reply by ●November 27, 20142014-11-27

On Thursday, November 27, 2014 10:44:29 PM UTC+13, 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.comIf it is random and Gaussian you could work out the variance (assume zero mean) and roughly take +/- 3 sigma .

Reply by ●November 27, 20142014-11-27

>>>>>Are you looking for some practical "close enough" way to do this, or some academically perfect closed form? It must not be perfect, but I'm driven by interest in theory. The pragmatic solution could be simple 'fast enough' sampling with some margin. Cheers Detlef _____________________________ Posted through www.DSPRelated.com

Reply by ●November 27, 20142014-11-27

On 27.11.2014 10:44, 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.Provided I understood your problem definition correctly... There is a theoretical bound, given by the Young inequality. As the Fourier transformation is essentially a convolution, and the min/max is the L^\infty bound, you find an inequality that tells you that you can bound the min/max by a constant times the L^1-norm of the Fourier-coefficients, which is again bound by the size of the integral (the Nyquist-frequency) times the L^\infty bound of the coefficients. For details, I'd suggest to google for Young-Inequality. The Reed-Simon (Functional Analysis) should have details. HTHH, Thomas

Reply by ●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

Reply by ●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 ●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 ●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 ●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