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How can i calculate the Fourier transform of a delayed cosine? I haven't found anywhere how to do that.

This is my attempt in hoping for a way to find it without using the definition:

$$ x(t) = cos(2\pi f_ct -θ) = cos\bigg(2\pi f_c\bigg(t -\frac{θ}{2πf_c}\bigg)\bigg) $$ ($f_c$ stands for the fundamental frequency of the signal and $θ$ is the phase shift) Now using the Fourier time-shift property $:$ $$ x(t-θ) \longrightarrow X(f)e^{-j2πfθ} $$ and knowing the fourier transform of $$cos(2πf_ct) = \frac{1}{2}δ(f-f_c)+\frac{1}{2}δ(f+f_c)$$ i get: $$ cos\bigg(2\pi f_c\bigg(t -\frac{θ}{2πf_c}\bigg)\bigg) = \bigg[ \frac{1}{2}δ(f-f_c)+\frac{1}{2}δ(f+f_c) \bigg]e^{-j2πf\frac{θ}{2πf_c}} = $$ $$ = \frac{1}{2}δ(f-f_c)e^{-\frac{jfθ}{f_c}} +\frac{1}{2}δ(f+f_c)e^{-\frac{jfθ}{f_c}} $$

Is this a way to find it? If yes can i simplify it further? And what happens with the simplifications when you're given the $f_c$? Thanks in advance.

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  • $\begingroup$ Why not taking this approach? $$ x(t) = cos(2\pi f_ct -θ) = cos(2\pi f_ct)cos(θ)-sin(2\pi f_ct)sin(θ)$$ $\endgroup$
    – Moti
    Aug 24, 2015 at 0:49
  • $\begingroup$ This seems harder to find the Fourier transform. Why not complete that and add as the answer? $\endgroup$
    – KeyC0de
    Aug 24, 2015 at 7:59
  • $\begingroup$ What is the transform for $cos(2\pi f_c t)$? $\endgroup$
    – Moti
    Aug 25, 2015 at 4:27
  • $\begingroup$ It's right there. I have wrote it.. Do you see it? I will write it in one whole line, just so it's better to see. $\endgroup$
    – KeyC0de
    Aug 25, 2015 at 8:19
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    $\begingroup$ You just need to multiply the cos and sin transforms by the phase correction. It looks like what you got is the right result. $\endgroup$
    – Moti
    Aug 26, 2015 at 3:58

1 Answer 1

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Although the question is old, I would like to provide a solution since recently I have been asked a similar question.

Fourier transform gives the locations and the (complex) amplitudes of the exponential,i.e. $e^{jwt}$, terms. By using the Euler identity $cos(\theta)=\frac{e^{j\theta}+e^{-j\theta}}{2}$ Fourier transform of $cos(wt)$ and $sin(wt)$ can be found. This is due to the fact that $F(e^{jw_0t})=2\pi\delta(w-w_0)$.

Thus the Fourier transform of shifted cosine $x(t)=cos(w_0t-\theta)$ is $$ cos(w_0t-\theta)=\frac{e^{j(w_0t-\theta)}+e^{-j(w_0t-\theta)}}{2} \\ \rightarrow F(cos(w_0t-\theta))=F(\frac{e^{j(w_0t-\theta)}+e^{-j(w_0t-\theta)}}{2})\\ =\frac{F(e^{j(w_0t-\theta)})+F(e^{-j(w_0t-\theta)})}{2} \\ =\frac{e^{-j\theta}F(e^{jw_0t})+e^{j\theta}F(e^{-jw_0t})}{2} \\ =\frac{e^{-j\theta}2\pi\delta(w-w_0)+e^{j\theta}2\pi\delta(w+w_0)}{2} \\ =\pi(e^{-j\theta}\delta(w-w_0)+e^{j\theta}\delta(w+w_0)) $$

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