Sample and reconstruct a real exponential (just one period)
First, since $t&gt;0$ 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')