minimum sampling rate for very short duration signals

I understand that sampling rate should be twice the signal bandwidth for successful reconstruction. But what I don't get is how this translates in the time domain.

For example, I have a sine wave pulse with frequency 100MHz and duration of say 50ns. This signal has a bandwidth of 20MHz (1/50ns = 20MHz). Theoretically, I can sample this signal at any frequency above 40Mhz. Let's say I sample at 50MHz, then I only get 2 samples per pulse!! Now what can you possibly do with those 2 samples is my question. In other words, although Nyquist criterion is satisfied in this example, there's not enough data points to proceed.

This kind of signals are common in radar and generally you want to run some sort of matched filtering/correlation once you have digitized the pulse. But you can't run any descent signal processing on just two samples? am I missing something here?

• Just a thought, perhaps the 2 samples is enough if the signal is perfect but if there is noise then the bandwidth is larger and you have to sample more often. – geometrikal Aug 15 '13 at 23:00
• It's alluded to in the below answers, but one fallacious assumption is that the pulsed sine wave has a bandwidth of 20 MHz. The baseband one-sided null-to-null bandwidth of a 50 nanosecond rectangular pulse is 20 MHz, but a rectangular pulse actually has infinite frequency support. – Jason R Aug 16 '13 at 13:16
• Is there a specific question here? Or are you just looking for general discussion of processing short-duration signals? – nispio Aug 16 '13 at 17:03

In many pulsed radar applications, one pulse is not enough to be informative, mainly for the reason you mention. However, coherent processing of multiple pulses can be used to extract useful information.

As an example, take a time-domain signal of sufficient length to accurately extract frequency information. Now multiply (not convolve) that signal with a rectangular pulse train in the time domain. In effect, you are taking a large number of your samples, and setting them to zero. Now consider the result in the frequency-domain. It will look like the convolution of the original spectrum with the spectrum of the pulse train.

Here is a MATLAB example:

K = 3;                              % Samples per pulse
N = 50;                             % Number of pulses
duty = 0.1;                         % Duty Cycle
Fs = 48e3;                          % Sampling Frequency
f0 = 13e3;                          % Center frequency
fd = f0*(1+0.03);                   % Doppler shifted frequency
SNR = 10;                           % Signal/Noise Ratio (dB)

% Helper function for plotting FFTs
plotFFT=@(x,Fs) plot(Fs*(0:length(x)-1)/length(x)-Fs/2, abs(fftshift(fft(x))));

% Create a train of rectangular pulses
rect_pulse_train = repmat([ones(1,K), zeros(1,K*(1/duty-1))],1,N);
figure(1); plotFFT(rect_pulse_train,Fs); title('Spectrum of pulse train');

% Create a carrier signal plus a doppler shifted version
t = (0:length(rect_pulse_train)-1)/Fs;
x = cos(2*pi*f0*t) + 0.7*cos(2*pi*fd*t);

nstd = norm(x) / sqrt(2 * length(x) * 10^(SNR/10));
x_noisy = x + nstd*(randn(size(x)) + 1j*randn(size(x)));
x_pulse = [x_noisy(1:K), zeros(1,length(x)-K)];
figure(2); plotFFT(x_pulse,Fs); title('Spectrum of single pulse');

% Capture a coherently-pulsed version of the signal
y = rect_pulse_train.*x_noisy;
figure(3); plotFFT(y,Fs); title('Spectrum of coherently pulsed signal');


As you try various values of K, you can see how it affects your frequency resolution. Even though the peaks are spectrally smeared, it is still possible to identify small frequency shifted copies of the carrier when using coherent doppler processing.

• are you suggesting that the radar algorithms actually operate on the full coherent integration interval? I thought that the target detection has to happen every pulse. what is the fast-time processing all about then? – user4673 Aug 16 '13 at 8:19
• @user4673: Those are very application specific questions. Do you have a specific algorithm in mind? The above is intended as one example of when short pulses can be useful. In reality, my example is only half-baked, because there is not much gained versus just taking a zero-padded FFT of the original short pulse. Coherent processing is most useful in the identification of doppler shifts. Perhaps I should update the example to reflect this. – nispio Aug 16 '13 at 15:44
• let's say I am interested in estimating the frequency (and nothing else) of an incoming radar pulse of 50ns with IF carrier frequency of 200MHz. Algorithms that can do this naturally operate in time-domain (by introducing a known delay to incoming signal and then digitally mixing the delayed/non-delayed versions to derive frequency-dependant phase information. Because delay is known, and phase difference has been measured, frequency can be estimated (phase = frequency x time delay). The trouble is that you need a lot of samples for this and sampling frequency can easily go into GHz range!!! – user4673 Aug 16 '13 at 16:37
• It is easy to find the phase based on frequency and time delay, but not nearly so straight forward to solve the problem in reverse due to modulo arithmetic. This is one motivation for using LFM chirped pulses rather than CF carriers. – nispio Aug 16 '13 at 16:59
• can I see the matlab script you wrote earlier, please? – user4673 Aug 17 '13 at 0:08

Any signal of finite extent (duration) in one domain (time) has infinite extent (bandwidth) in the other (frequency). The bandwidth of a short pulse above some noise floor can be extremely high, due to the pulse window's transform convolved with the carrier frequency.

Sharp edges or cutoffs imply very high bandwidths. Thus in practice, most designed pulses are shaped closer to Gaussian, with much wider tails than the main pulse lobe (in both domains).

Even so, if you know the exact frequency of an unmodulated fragment of a pure sine wave, then in zero noise, 2 or 3 non-aliased points are theoretically enough to solve the equations for the sinewave's amplitude and phase.