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What would be the best way to take the Fourier transform of $$ f(t)\cdot \cos\big(\pi(t-1)\big) $$ I'm aware that when you take the Fourier Transform of $\cos(kt)$ you get two impulse at the location of $k$. And that the $(t-1)$ would create a complex exponential. But you can only apply the time-shift when both functions are time shifted (at least to my knowledge at least). Really the $\pi$ and the non-time shifted $f(t)$ are really throwing me off. Thanks!

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    $\begingroup$ It could be useful to rewrite $f(t)\cos(\pi(t-1))$ using Euler's formula; $\cos(x)=(e^{ix}+e^{-ix})/2$. What happens to the Fourier transform then? $\endgroup$ Feb 8, 2021 at 9:57

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It is easy to see that using the trigonometric identity in Equation $(1)$ below $$ \cos(\theta_1 \pm \theta_2) = \cos(\theta_1)\cos(\theta_2) \mp \sin(\theta_1)\sin(\theta_2) \tag{1} $$ We have \begin{align} f(t)\cos\big(\pi(t - 1)\big) &= f(t)\overbrace{\cos(\pi t - \pi)}^{\text{use Equation}\ (1)}\\ &\equiv -f(t)\cos(\pi t)\tag{2} \end{align} Which can be rewritten as $$ -f(t)\cos(\pi t) = -f(t)\cos\left(2\pi \frac t2\right) = -f(t)\cos(2\pi f_c t)\quad\text{where}\quad f_c = \frac 12 $$ And we know the Fourier cosine modulation frequency-shift property in equation $(3)$: $$ \mathcal F\big\{x(t)\cos(2\pi f_c t)\big\} = \frac 12\big[X\left(f - f_c\right) + X\left(f + f_c\right)\big]\tag{3} $$ You can now easily use Equation $(2)$ and $(3)$ to find the solution.

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  • $\begingroup$ i most def forgot about that trig identity. i will try that. thank you! $\endgroup$ Feb 8, 2021 at 16:56
  • $\begingroup$ You’re welcome. If this answers your question, remember to accept the answer. $\endgroup$
    – Gilles
    Feb 8, 2021 at 18:11
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We have signal $y(t) = f(t)cos(\pi t-\pi)$. You are modulating amplitude where bias is $\pi$. By using Euler's formula you can rewrite singal as: $$ y(t) = f(t)\frac{1}{2}\left(e^{j(\pi t - \pi)} + e^{-j(\pi t - \pi)}\right) $$ $$ y(t) = \frac{1}{2}f(t)e^{j(\pi t - \pi)} + \frac{1}{2}f(t)e^{-j(\pi t - \pi)} $$ $$ y(t) = \frac{1}{2}f(t)e^{j\pi t}e^{-j\pi} + \frac{1}{2}f(t)e^{-j\pi t}e^{j\pi} $$ And then by Fourier transform we get : $$ Y(j\omega) = \frac{1}{2}F(j(\omega -\pi))e^{-j\pi} + \frac{1}{2}F(j(\omega+\pi))e^{j\pi} $$ $$ Y(j\omega) = -\frac{1}{2}F(j(\omega -\pi)) -\frac{1}{2}F(j(\omega+\pi)) $$ And we get this becuase $e^{-j\pi}=-1$ and $e^{j\pi}=-1$

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