Sample and reconstruct a real exponential (just one period)

Question

I have a function with an equation:

$$C = 1.6925\left( e^{-0.136t}-e^{-1.192t}\right) $$

Where $C$ is real and $t$ represents time in hours. Beneath is the representation of my function.

I am trying to find the best sampling step (based on Nyquist-Shannon theorem) to sample and reconstruct this function using a Fourier Transform.

After using a Fast Fourier Transform algorithm in MATLAB (fft), by setting:

x= fft(c);
xmag=abs(x); (amplitude estimation)

I find the below spectrum for the function for a $t$ between $0$ and $100$ sampled at $1000 \text{Hz}$ , with the highest amplitude at:

bin(1) = 1.1025e+04

and the second highest at bin(100001) with an amplitude of 9.9946e+03. the bin number 100001 represents also the length of xmag.

So my question is how would you interpret the spectrum? Specifically I would like to know how to determine the highest frequency at which I can sample this function and reconstruct it without aliasing?

Answer 1

First, since $t>0$ in your case, you can write your function as

\begin{equation} C(t)=1.6925\left(\exp^{-0.136t}- \exp^{-1.192t} \right) u(t) \end{equation} where $u(t)$ is a unit-step function. Then, denote the FT of $C(t)$ as $C(F)$, which is given by \begin{equation} C(f)=1.6925\left(\frac{1}{0.136+2\pi jf}- \frac{1}{1.192+ 2\pi jf} \right) \end{equation}

The spectrum $C(f)$ is plotted below. We can see that $C(f)$ is almost zero after 5 Hz. You can choose the sampling frequency according to the Nyquist criterion as $f_{\max} \geq 10$ Hz. You have taken a very high sampling rate (1000 Hz) that is 100 times more than the required rate. When I took 10 Hz to sample the signal above and took the FFT I got the results as follows (in order time domain $C(t)$, frequency domain signal $C(f)$, frequency domain domain signal obtained via FFT, that is, FFT(C(t)). Note the inherent periodicity of the FFT spectrum)

And here is the MATLAB/Octave code for your reference..

clear all; clc;close all;

%% Time-domain signal
Ts=0.1; % Sampling time (obtained from the frequency domain plot)
t=0:Ts:60; % in sec
C=1.6925* (exp(-0.136*t)-exp(-1.192*t));
figure,plot(t,C);
title('Time domain signal');
xlabel('Time in sec')

%Frequency domain signal
f=0:.1:10;
F1=1.6925*((1./(0.136+(j*2*pi*f)))-(1./(1.192+(j*2*pi*f))));
figure,plot(f,abs(F1));
title('Frequency domain signal');
xlabel('Frequency in Hz')
% Via FFT
F2=fft(C);
L=length(F2);
figure,plot((0:L-1)*(1/(L*Ts)),abs(F2));
title('Frequency domain signal via FFT of C(t)');
xlabel('Frequency in Hz')

Answer 2

How did you arrive at sampling frequency of 1000 Hz? Ideally speaking, you should find the Fourier transform of the given function analytically. This can be done using the FT tables for exponentials. Then you should find the highest frequency beyond which the FT is almost zero i.e less than the smallest number represented by finite precision you are using. Twice or more of this frequency is your sampling rate. How many samples should you take? The signal is almost zero beyond 5 x 10$^4$ as can be seen from the figure. So you take samples upto this time point. Now you take the FFT of these samples and it would be equal to the FT you analytically calculated i.e it would be sampled version of the FT. If you used sampling rate less than this procedure explained, the resulting FFT would be aliased version.