## Example of Computation of Fourier transform of a CT SIGNAL

Compute the Fourier Transform of the signal

$\ x(t)= \int_{-\infty}^{t} \tau \sin(2 \pi \tau+ \pi/4) d\tau$

By definition the Fourier Transform of a signal is defined as:

$F[x(t)] = X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt$

First expressing the signal in as a Fourier series:

However before finding the transform we note that integration in the time domain is just division in the frequency domain. So the game plan is to find the Fourier series of x'(t) then divide by the frequency in the frequency space.

$\ x'(t)=\sin(2\pi t+ \pi/4) = \frac{e^{2 \pi jt + \pi/4}}{2j} - \frac{e^{-2 \pi jt -j \pi/4}}{2j}$

$X'(\omega)=\int_{-\infty}^{\infty} \frac{e^{j \pi/4}}{2j} e^{j2 \pi} e^{-j\omega t}dt - \int_{-\infty}^{\infty} \frac{e^{-j \pi/4}}{2j} e^{-j2 \pi} e^{-j\omega t}dt$

Using some foresight we see that a straight up integration of the expression above will yield something infinite or indeterminate, we take advantage of the known Fourier transform of a complex exponential.

$\int_{-\infty}^{\infty} x(t) dt = \frac{X(\omega)}{\omega} - X(0) \pi \delta(\omega)$

$X'(\omega)= \frac{e^{j \pi/4}}{2j} F[e^{j2 \pi}] - \frac{e^{-j \pi/4}}{2j} F[e^{-j2 \pi}]$

Noting that $\ F[e^{j\omega_0}] = 2 \pi \delta(\omega - \omega_0)$

$\ X'(\omega) = j \pi \delta(\omega + 2\pi) e^{-j \pi /4}- j \pi \delta(\omega + 2\pi) e^{j \pi /4}$

Since $\ X(\omega) = 0$

$X(\omega) =\frac{j \pi}{\omega} \delta(\omega + 2\pi) e^{-j \pi /4}- \frac{j \pi}{w} \delta(\omega + 2\pi) e^{j \pi /4} ---- [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]]$

## Alumni Liaison

BSEE 2004, current Ph.D. student researching signal and image processing. Landis Huffman