# Difficulty with a Fourier Transform

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!

• 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? – Dan Pollard Feb 8 at 9:57

## 2 Answers

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.

• i most def forgot about that trig identity. i will try that. thank you! – Minato Namikaze Feb 8 at 16:56
• You’re welcome. If this answers your question, remember to accept the answer. – Gilles Feb 8 at 18:11

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 $$e^{j\pi}=-1$$