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I am trying to write a program to compute the magnitude and phase of a specific, non-integer frequency component (i.e. given a sampled finite signal of length $N$, I want to know the magnitude and phase of the spectrum at frequency 40.678 Hz).

I tried using the equation for the Discrete Time Fourier Transform:

$$ X(\omega) = \sum_{n=-\infty}^{\infty} x[n] e^{-j\omega n} = \sum_{n=0}^{N-1} x[n] e^{-j\omega n} $$

such that

$$ X(2\pi\cdot40.678) = \sum_{n=0}^{N-1} x[n] e^{-j 2\pi\cdot40.678\cdot n} $$

I did this as follows:

t = np.linspace(0,1,100)
x = np.sin(2*np.pi*40.678*t)

Xf = 0
for n in range(len(x)):       
    Xf += (x[n]*np.exp((-np.complex(0,1))*2*np.pi*40.678*n))
print(abs(Xf))

Which printed $0.6149015687259044$.

I know that this is wrong because if I do an FFT as follows:

X = np.fft.fft(x)
plt.figure();
plt.plot(abs(X))

I can see peaks of magnitude $50$ at around (what I assume after correcting the axis is) 40.678 Hz.

enter image description here

What is the reason for this difference? I also tried the equation of the DFT just like I did with the DTFT code above this time to measure magnitude at 50 Hz of a sinusoid of frequency 50 Hz and I got a completely wrong value again as compared to what the plot of the FFT shows.

What causes this discrepancy?

Thank you.

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1 Answer 1

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HINT:

Try introducing the sampling frequency $F_s$. Then $f$ in the relation in Equation $(1)$ is relative to the sampling frequency $$ \omega = 2\pi f\quad\text{where}\quad f = \frac FF_s\tag{1} $$ Where the units are as follows:

  • $\omega$ in [radians/sample]
  • $f$ in [cycles/sample]
  • $F_s$ in [samples/second]
  • $F$ in [cycles/second] or [Hz]

Your discrete time vector for your signal should be as shown in equation $(2)$ $$ t = nT_s\quad\text{where}\quad T_s = \frac 1F_s\quad\text{and}\quad n = 0, \ldots, N - 1\tag{2} $$ You seem to use the frequency $F$ in [Hz] instead of the relative frequency $f$ in [cycles/sample].

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  • $\begingroup$ Thank you very much! $\endgroup$ Commented Jan 14, 2021 at 12:12

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