Taking the fft of a signal sampled above the Nyquist rate but getting strange results

I have been trying to reproduce a frequency domain function from a time domain signal, but getting nothing like what I expect.

I am sampling the signal above the Nyquist rate. The dt below is 1e-4. When I reconstruct the signal, it looks like what I expect:

So I think I reconstructed it correctly. Notice that it is on a log-log plot.

I take the FFT by doing the following:

nfft = 2^nextpow2(length(xt));
df = 1/dt;
freq = (df/2)*linspace(0,1,nfft/2+1);
xf = fft(xt,nfft)/length(xt);

And then I get this for the single-sided fft when I type plot(freq,2*xf(1:numel(freq))):

Notice that this also is on a log-log scale. Unfortunately, this is nothing like I expect. It looks like it's been flipped about a horizontal line from what I would expect more or less. I certainly would think that it wouldn't be monotonically decreasing like this.

Have I made any error in my implementation of the fft? More specifically, does any of this make sense conceptually from the trend I see in log-log space for the time domain signal? Is there a good sanity check I can do?

(I realize this is part Matlab question, part fft concepts, but I would appreciate some insight. I'm really stuck.)

Answer 1

[Matt J]

nfft = 2^nextpow2(length(xt));

df = 1/nfft/dt;

freq = df*(0:nfft-1);

xf = abs(fft(xt,nfft))*dt;

half=1:nfft/2;

plot(freq(half),xf(half))

[L'O.G.]

That gives the same thing, and that's why I'm stuck.

[Matt J]

Well, we can't know the basis for your expectations. It's not apparent why the signal you've shown couldn't have a monotonic amplitude spectrum just by looking at the signal. Is the signal you showed us the complete signal, or is it just a half-lobe?

[L'O.G.]

Thanks for the response. It's the complete signal in the time domain: x(t) = ∑t/tau. I cover the range of all possible constants tau, and I sample above the Nyquist rate which I base on the smallest tau. I thought that should be sufficient. I'm really stumped.

[Matt J]

Well, x(t) = ∑t/tau is an infinite linear ramp. For one thing, this looks nothing like the signal you have posted. For another, it goes to infinity both at t=inf and t=-inf, so it is not Fourier transformable signal. It therefore has no bandwidth from which you could compute a Nyquist rate.