Discrete fourier transform problem

Question

We have taken $1000$ observations from signal $s(t),t \in\Bbb{R}$ $$h(k)=s(k\Delta t+t_0),k=0,1,\dots,999,$$ where $\Delta t=1/200 $ and $t_0=-2$ (in seconds). When we calculate discrete fourier transform we see that $|\hat{h}(m)|$ gets its biggest values in $m=193$ (and also when $m=807$). What's the corresponding (positive) value of frequency in signal $s$ (in hertz rounded up to $\Bbb{Z}$)?

Answer 1

The zeroth bin of the DFT corresponds to the average. The first bin $m=1$ corresponds to the frequency, where one period covers the full data length, i.e. a period length of $1000 \Delta t$ in your case.

Hence bin $m$ for $m=0 \ldots N/2$ (with $N=1000$ in your case) corresponds to a frequency of

$f_m = \frac{m}{N \Delta t}$,
which is for $m=193$, $f_{193}=38.6Hz$.

The upper bins $m=N/2+1..N$ correspond to the negative frequencies (as you are using the DFT, which has the implicit periodicity assumption in time and frequency domain)

$f_m = -\frac{m}{N \Delta t}$

Please note, that in order to get correct amplitudes, you have to divide the DFT values by $N$ (e.g. the average value is $\widehat{h}(0)/N$).

Second note, the negative frequencies components are for real values signals the conjugate complex of the positive components (hence your double peak observation). Also the amplitude is in the negative and positive frequency is only half of the true amplitude (because total power is equally shared among positive and negative frequencies).