c = fft(a);
takes the fft of both columns independently and then when you plot() that you get both plots on the same axis. The dt will be a single spike and much larger magnitude than the spectrum of the signal so that's what is visible in the plot.
Use
Y=fft(E);
instead.
I recommend
doc fft % study example for computing/plotting a one-sided PSD
ADDENDUM
Your data are grossly over-sampled for the actual frequency content it appears...
I built a little routine from the fft example
>> type psdfft
function psd=psdfft(y,Fs)
L=length(y);
NFFT = 2^nextpow2(L);
Y = fft(y,NFFT)/L;
f = Fs/2*linspace(0,1,NFFT/2+1);
psd=2*abs(Y(1:NFFT/2+1));
plot(f,psd)
title('Single-Sided Amplitude Spectrum of y(t)')
xlabel('Frequency (Hz)')
ylabel('|Y(f)|')
>>
As other poster notes, you also have a mean that is significant in comparison to the amplitude and there's a very abrupt end to the recorded signal so I used a Hamming window and with those--
>> T=t(end); Fs=1/T;
>> figure
>> psdfft(hamming(length(e)).*detrend(e),Fs);
>> set(gca,'xscale','log','yscale','log')
>>
A couple peaks around 10^4 which given sample rate of 10^11 is many orders of magnitude less, there's mostly just noise elsewhere and it takes log scale both axes to see anything at all...
If I just decimate the input signal, downsampling by 128,
>> es=e(1:128:end);
>> Fse=Fs/128;
>> psdfft(detrend(es),Fse);
the result is
If you downspace the x axes on the above figure you'll see when you get the two axes ranges to overlay that the peaks are in the same place; simply owing to not having all the extra sampling in the signal the resolution is better for the actual significant energy content.
You can do similar thing by averaging the fft of the sampled signal over multiple shorter windows to improve the estimation of the energy actually in the signal.
