# Fourier components of $\cos(2\pi f_1t)$

I have the signal $s(t) = \cos(2\pi f_1t)$ and I am looking for its components vs the Fourier basis, over the interval $[0, T]$. The formula for computing the coefficients is $$s_n = \int_{t_0}^{t_1} s(t) \frac{e^{-j\frac{2\pi}{T} nt}}{\sqrt{T}} dt$$ and in this case $t_0 = 0$ and $t_1 = T$. The solution seems to be $$\frac{\sqrt{T}}{2}(-1)^n\bigg[\mathrm{sinc}\big(f_1T+n\big)e^{-j\pi f_1T} + \mathrm{sinc}\big(f_1T-n\big)e^{j\pi f_1T}\bigg]$$ but i can get no further than this $$\frac{\sqrt{T}}{2}\bigg(\frac{e^{j2\pi (f_1T-n)}-1}{j2\pi (Tf_1-n)} + \frac{e^{-j2\pi (f_1T+n)}-1}{-j2\pi (Tf_1+n)}\bigg)$$

## 1 Answer

HINT:

Going from your last equation, $$\frac{\sqrt{T}}{2}\bigg(\frac{e^{j2\pi (f_1T-n)}-1}{j2\pi (Tf_1-n)} + \frac{e^{-j2\pi (f_1T+n)}-1}{-j2\pi (Tf_1+n)}\bigg)$$ This can be simplified further down by considering the following:

\begin{align} e^{j2\pi (f_1T-n)} &= e^{j\pi (f_1T-n)}\cdot e^{j\pi (f_1T-n)}\\ 1 &=e^{j\pi (f_1T-n)}\cdot e^{-j\pi (f_1T-n)}\\ e^{-j\pi n} &= (-1)^n\\ \end{align} And that \begin{align} \mathrm{sinc}(x) &= \frac{\sin(x)}{x}\\ \sin(x) &= \frac{e^{jx} - e^{-jx}}{2j} \end{align}