Examples

Example 1

When an arbitrary signal $x(t)$ is multiplied by a sinusoid

$$y(t)=x(t)\;cos(\omega_0 t)=x(t)\;cos(2\pi f_0 t)$$

its spectrum $X(j\omega)$ will be convolved with that of the sinusoid:

$$Y(j\omega)=X(j\omega)*\frac{1}{2}[\delta(\omega-\omega_0)+\de... ...0)] =\frac{1}{2}[X(j(\omega-\omega_0))+X(j(\omega+\omega_0))]$$

We see that the shape of the spectrum $X(j\omega)$is reserved while it is moved to some higher frequency band for transmission. This is the amplitude modulation in communication.

At the receiving end, the demodulation is done by multiplying the modulated signal $y(t)$ by the same sinusoid again:

$$z(t) = y(t)\;cos(\omega_0 t)=x(t)\;cos^2(\omega_0t) =x(t)\frac{1+cos(2\omega_0t)}{2}$$

$$= \frac{x(t)}{2} +\frac{x(t)cos(2\omega_0t)}{2}$$

The spectrum of $z(t)$ is

$$Z(j\omega) = Y(j\omega)*\frac{1}{2}[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)] =\frac{1}{2}[Y(j(\omega-\omega_0))+Y(j(\omega+\omega_0))]$$

$$= \frac{1}{4}[X(j(\omega-2\omega_0))+2X(j\omega)+X(j(\omega+2\omega_0))]$$

We see that now the spectrum $X(j\omega)$ of the signal is moved back to its original location (as well as to an even higher frequency band). The signal $x(t)$ can be recovered by suppressing the high frequency part of the signal using low-pass filtering .

Example 2:

The response of an LTI system with impulse response

$$h(t)=e^{-at} u(t)\;\;\;\;\;(a>0)$$

to an input

$$x(t)=e^{-bt} u(t)\;\;\;\;\;(b>0)$$

can be found in time domain by convolution

$$y(t) = h(t)*x(t)=\int_{-\infty}^\infty h(\tau) x(t-\tau) d\tau =\int_0^t e^{-a\tau}e^{-b(t-\tau)} d\tau$$

$$= e^{-bt} \int_0^t e^{(b-a)\tau} d\tau$$

If $a=b$, we have

$$y(t)=t e^{-at} u(t)$$

else

$$y(t)=\frac{e^{-bt}}{b-a} e^{(b-a)\tau} \vert _0^t =\frac{1}{b-a}(e^{-at}-e^{-bt}) u(t)$$

In frequency domain, we have

$$Y(j\omega)=H(j\omega)X(j\omega)=\frac{1}{(a+j\omega)(b+j\omega)}$$

By partial fraction expansion, this can be written as

$$Y(j\omega)=\frac{1}{(a+j\omega)(b+j\omega)} =\frac{A}{a+j\om... ...mega} =\frac{Ab+Aj\omega+Ba+Bj\omega}{(a+j\omega)(b+j\omega)}$$

Comparing the coefficients of $j\omega$ and the constants, we get

$$\left\{ \begin{array}{l} A+B=0 Ab+Ba=1 \end{array} \right.$$

with solution $A=-B=1/(b-a)$. Now we have

$$Y(j\omega)=\frac{1}{b-a}[\frac{1}{a+j\omega}-\frac{1}{b+j\omega}]$$

The output in time domain is obtained by inverse transform

$$y(t)={\cal F}^{-1}[Y(j\omega)] =\frac{1}{b-a}[e^{-at}-e^{-bt}]u(t)$$

When $a=b$, we have

$$Y(j\omega)=\frac{1}{(a+j\omega)^2} =j\frac{d}{d\omega}[\frac{1}{a+j\omega}]$$

But since

$${\cal F}^{-1}[\frac{1}{a+j\omega}]=e^{-at}u(t)$$

by the frequency differentiation property, the output in time domain is obtained as:

$$y(t)={\cal F}^{-1}[Y(j\omega)] ={\cal F}^{-1}[j\frac{d}{d\omega}(\frac{1}{a+j\omega})] =t\; e^{-at}u(t)$$