State and prove the following properties of Fourier Transform with example

(i) Time Shifting.

(ii) Convolution in time domain.

Mumbai University > Information Technology > Sem 3 > Principles of analog and digital communication

Marks:- 5M

Year:- Dec 2015

Obtain the Fourier transform of the rectangular pulse as shown below:

Solution:

• A pulse which has an amplitude A and which extends from

t=-T/2 to t =+ T/2 is expressed as follows:

x(t) =A rect (t/T)………….(1)

• The rectangular pulse of fig is the delayed version of the standard rectangular pulse expressed in equation(2) , with a time delay of T/2.Hence it can be mathematically represented as,

$x_1$(t) =A rect [(t-T/2)/T]……………….(2)

• Using the property of “time shifting “we get its Fourier transform as,

F[$x_1$(t)] = $X_1$(f) = $e^{-j2πfT/2} .X(f) = $e^{-jπfT}.ATsinc (fT)

$X_1$(f) = AT$e^{-jπfT} sinc (fT)………….Ans **Convolution Theorems** The convolution theorem states that convolution in time domain corresponds to multiplication in frequency domain and vice versa: F[x(t)*y(t)] = X(jw) Y(jw)...(a) F[x(t) y(t)]= X(jw) * Y(jw)...(b) Proof of (a): F[x(t)*y(t)] : $ \int^{\infty}_{-\infty}$ [ $ \int^{\infty}_{-\infty} x(T)y(t-T) dT] {e^{-jwt}} dt$ : $ \int^{\infty}_{-\infty}$ x(T) [ $ \int^{\infty}_{-\infty} y(t-T) {e^{-jwt}} dt$}dT Proof of (b): F[x(t)y(t): $ \int^{\infty}_{-\infty}$ x(t)y(t) $e^{-jwt}$ dt : $ \int^{\infty}_{-\infty}$ [1/2π $ \int^{\infty}_{-\infty}$ X (jw')$e^{jw't}$dw'] y(t) $e^{-jwt} dt$ : 1/2π $ \int^{\infty}_{-\infty}$X(jw')[ $ \int^{\infty}_{-\infty}$ y(t) $e^{jw't}$ $e^{-jwt}$ dt] dw'

: 1/2π $ \int^{\infty}_{-\infty}$X(jw')[ $ \int^{\infty}_{-\infty}$ y(t) $e^{-j(w-w')t}$ dt] dw'

: 1/2π $ \int^{\infty}_{-\infty}$X(jw') Y(j(w-w'))dw' = X(jw)* Y(jw)

Example: Find the convolution, z(t), of the following two signals, x(t) and y(t), by using (a) the integral representation of the convolution equation and (b) multiplication in the Laplace domain.

The signal y(t) is simply the Heaviside step, u(t).

The signal x(t) is given by the following infinite sinusoid, x0(t), and windowing function, xw(t):

$ x_0$(t) =sin (t)

xw(t)=u(t)-u(t-2π)

Thus, the convolution we wish to perform is therefore:

z (t) =x(t)*y(t)

z (t) =sin(t)[u(t)- u(t-2π)]*u(t)

z(t)=[sin(t)u(t)-sin(t) u(t-2π)]*u(t)

From the distributive law:

z(t)= sin(t)u(t)u(t)-sin(t) u(t-2π)]u(t)