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The signal $$x(t)=\cos(10\pi t+\phi)+\cos(20\pi t)$$ is sampled with a sampling frequency $F_s$ as $25 \mathrm{Hz}$ where the phase $\phi$ is unknown.

Sampling the continuous time signal $$y[n]=x(nT)$$ where $T=1/F_s$.

After calculating the DFT of $y[n]$ with the length of DFT $N=100$, which of $k$ will give us the maximum of the magnitude of $Y[k]$?

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  • $\begingroup$ Any ideas that you might have to start with? $\endgroup$
    – GKH
    Commented Feb 3, 2020 at 9:32
  • $\begingroup$ May I know why you have introduced one more unknown $\phi$ in seeking an answer? $\endgroup$
    – jomegaA
    Commented Feb 3, 2020 at 9:42
  • $\begingroup$ I thought of converting to Euler and then converting to DFT but the answer I get is incorrect. @GKH $\endgroup$
    – ori
    Commented Feb 3, 2020 at 9:48
  • $\begingroup$ This is the qeustion I got... @jomegaA $\endgroup$
    – ori
    Commented Feb 3, 2020 at 9:49

2 Answers 2

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Note that the frequencies of the FFT grid are given by

$$f_k=\frac{kf_s}{N},\qquad k=0,1,.\ldots,N-1\tag{1}$$

where $N$ is the FFT length.

Now observe that the two frequencies of the given signal fall exactly on the grid. So you only have to determine these two frequencies, and then use $(1)$ to figure out the corresponding indices.


EDIT:

It's always good to check your results using some simulation software. The following Octave/Matlab script confirms what you came up with.

phi = 0;    % exact value irrelevant
n = 0:99;
T = 1/25;
y = cos( 10*pi*n*T+phi ) + cos( 20*pi*n*T );
Y = fft(y);
subplot(2,1,1), plot(n,abs(Y),'x'), grid on

enter image description here

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  • $\begingroup$ I already did it and I get k=20,40,60,80. its incorrect. The answer should be k=10,20,80,90 $\endgroup$
    – ori
    Commented Feb 3, 2020 at 10:56
  • $\begingroup$ @ori: What you got is correct. Why don't you just use some software and check the result? The given answer is wrong. $\endgroup$
    – Matt L.
    Commented Feb 3, 2020 at 11:03
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    $\begingroup$ @ori: I've added a simulation to my answer to convince you that we're right and the given solution is wrong. $\endgroup$
    – Matt L.
    Commented Feb 3, 2020 at 11:17
  • $\begingroup$ I understand, thank you very much $\endgroup$
    – ori
    Commented Feb 3, 2020 at 11:20
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    $\begingroup$ @jomegaA: If you want the zeroth bin in the center you can use fftshift. $\endgroup$
    – Matt L.
    Commented Feb 3, 2020 at 11:35
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My humble attempt and I could be totally wrong in trivial things. Phase $\phi=0$

enter image description here

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