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I want to find a DFT of a pure cosine wave cos(θ) sampled at N equally spaced points on the interval $[0, 2\pi)$

so for our cosine wave, I put my $x$ like this

$x=cos(\phi)$

then I just put it in DFT formula

$X[k]=\sum \cos(\phi)𝑒^{−j2\pi kn/N}$

and use euler furmula

$\cos(\theta) = (e^{j\theta} + e^{-j\theta})/2$

now we have

$X[k]=\sum(e^{j\theta} + e^{-j\theta})/2.(e^{−j2\pi nk/N})$

mathematically we can move 1/2 before $\sum$ and separate the $\sum$ like down(because we have + )

$X[k]=1/2\sum(e^{j\theta})(e^{−j2\pi nk/N})+1/2\sum(e^{-j\theta})(e^{−j\pi nk/N})$

know I do not know if I doing this right or not!?

and what should I do next?

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    $\begingroup$ You probably need to try again with $cos[\omega n]$, since cos($\theta$) is a constant since it is not a function of any independent variable n. $\endgroup$
    – DSP Rookie
    Apr 5, 2020 at 22:55
  • $\begingroup$ Decide a digital frequency $\omega$ of your cosine wave and then compute it's DFT. $\endgroup$
    – DSP Rookie
    Apr 5, 2020 at 22:57
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    $\begingroup$ Spoiler alert: dsprelated.com/showarticle/771.php and dsprelated.com/showarticle/1120.php You got to (14) in the first article. The simplification of (16) has great significance. $\endgroup$ Apr 6, 2020 at 1:05
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    $\begingroup$ Side note to the downvoter I reversed. Why would you want to stifle this kind of inquiry? This is the fundamental question to understanding the behavior of a DFT: How does a single pure tone behave in it? It is what the alphabet is to spelling, what words are to writing, and just below what notes are to chords. $\endgroup$ Apr 6, 2020 at 1:24

2 Answers 2

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If you want to take DFT of a cosine wave $cos(\theta)$ sampled at N equally spaced values between $[0, 2\pi]$, then you need to consider taking N-point DFT of the sequence : $$x[n] = cos[2\pi \frac{n}{N}], n = 0,1,2,...,N-1$$

<span class=$cos(\theta)$">

And, for this $x[n]$, you dont even have to apply the DFT formula. It can be done pretty simply by using the Euler's formula $$cos(\theta) = \frac{e^{j\theta} + e^{-j\theta}}{2}$$. $$x[n] = cos[2\pi \frac{n}{N}] = \frac{e^{j2\pi\frac{n}{N}} + e^{-j2\pi\frac{n}{N}}}{2} = \frac{e^{j2\pi\frac{n}{N}} + e^{j2\pi\frac{n.(N-1)}{N}}}{2}$$ Now, see the above expression as Invere DFT and figure out that $X[k]$ will be non-zero only for 2 values of $k$, i.e. $k=1$ and $k= (N-1)$. But the magnitude of $X[1]$ and $X[N-1]$ won't be $\frac{1}{2}$.

I think you can do it now yourself to figure out the magnitude of $X[1]$ and $X[N-1]$.

Hint: The Magnitude will get scaled by the length of DFT.

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  • $\begingroup$ "figure out that X[k] will be non-zero only for 2 values of k, i.e. k=1 and k=(N−1)", what is the reason for this? I am not able to see it. $\endgroup$ Jan 24 at 14:35
  • $\begingroup$ @OuttaSpaceTime If you observe the Euler's formula expansion of $cos[2\pi\frac{n}{N}]$, there are only 2 terms. And these terms can be viewed as if Inverse DFT summation has been spread with all the terms equal to 0 except for $k=1$ and $k=N-1$. $\endgroup$
    – DSP Rookie
    Jan 25 at 9:29
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Your $\theta$ is not a function of n sample number. After sampling, any continuous function $x(t)$ becomes a function of n $x(nT_s)$ where $T_s$ is the sampling period. If $f$ if the frequency of the cosine signal in Hz, the discrete cosine signal can be expressed as $cos[2 \pi fn/N]$. Now that cosine is a function of n, you can apply the DFT formula.

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