# Fourier transform of a signal

I have the following signal and need to derive it's Fourier transform: I just started learning Fourier transform and don't know how to solve this kind of question. I know though how to derive the Fourier transform for: $$cos(2\pi f_0t)$$

$$\mathcal{F}\{cos(2\pi ft)\}=\int_{-\infty}^{\infty}cos2\pi f_0*e^{-j2\pi ft}dt$$ $$=\int_{-\infty}^{\infty}\frac{1}{2}(e^{-j2\pi(f-f_0)t}dt +\int_{-\infty}^{\infty}\frac{1}{2}(e^{-j2\pi(f+f_0)t}dt$$ $$=\frac{1}{2}(\delta(f-f_0)+\delta(f+f_0))$$

• Hint: the integrand is zero outside $[-T/2, T/2]$, so you only consider the integrand on that interval. – Sean Roberson Feb 14 '18 at 0:46

## 1 Answer

We can write the function as $$g(t)=\underbrace{A\cos(2\pi f_c t)}_{x(t)}\times\underbrace{\mathrm{rect}(t/T)}_{y(t)}$$ where $T=\frac{2}{f_c}$ and the rectangular function is defined as $$\mathrm{rect}(t) =\begin{cases} 0 & \mbox{if } |t| > \frac{1}{2} \\ 1 & \mbox{if } |t| < \frac{1}{2}. \\ \end{cases}$$ The fourier transform of a product of signals is the convolution of the Fourier transforms of the signals $$\mathcal{F}\left\{g(t)\right\}=G(f)=X(f)*Y(f)$$ with \begin{align} \mathcal{F}\left\{ A\cos(2\pi f_c t) \right\}&=X(f)=\frac12\left[\delta(f-f_c)+\delta(f+f_c)\right]\\ \mathcal{F}\left\{\mathrm{rect}(t/T) \right\}&=Y(f)=T\frac{\sin(\pi Tf)}{\pi T f}=\frac{\sin(2\pi f/f_c)}{\pi f} \end{align} Observing that $z(t)*\delta(t-T)=z(t-T)$ we have \begin{align} G(f)&=\frac12\left[Y(f-f_c)+Y(f+f_c)\right]\\ &=\frac12 \left[\frac{\sin(2\pi (f-f_c)/f_c)}{\pi (f-f_c)}+\frac{\sin(2\pi (f+f_c)/f_c)}{\pi (f+f_c)}\right]\\ &=\frac{f}{\pi}\frac{\sin\left(2\pi {f}/{f_c}\right)}{f^2-f_c^2} \end{align}