Question

(a) How is amplitude modulation achieved ?
(b) The frequencies of two side bands in an AM wave are 640 kHz and 660 kHz respectively. Find the frequencies of carrier and modulating signal. What is the bandwidth required for amplitude modulation ?

Answer 1

(a) Process to achieve AM:

Here the modulating signal $A_{m}sin{\omega }_{m}t$ is added to the carrier signal $A_{m}sin{\omega }_{m}t$ to produce the signal x(t). This signal $x\left(t\right)=A_{m}sin{\omega }_{c}t$ is passed through a square law device which is a non- linear device which produces an output $y\left(t\right)=Bx\left(t\right)+C{x}^2\left(t\right)$ Where B and C are constant, Thus, $y\left(t\right)=B{A}_{m}sin{\omega }_{m}t+B{A}_{c}sin{\omega }_{c}t+C[A_m^{2}si{n}^2{\omega }_{m}t+A_c^{2}si{n}^2{\omega }_{c}t+2A_m{A}_{c}sin{\omega }_{m}tsin{\omega }_{c}t]...\left(ii\right)=B{A}_{m}sin{\omega }_{m}t+B{A}_{c}sin{\omega }_{c}t+\frac{C{A}_m^2}{2}+\frac{C}{2}{A}_c^2-\frac{C{A}_m^2}{2}cos2{\omega }_{m}t-\frac{C{A}_c^2}{2}cos2{\omega }_{c}t+C{A}_m{A}_{c}cos({\omega }_c-{\omega }_m)t-C{A}_m{A}_{c}cos({\omega }_c+{\omega }_m)t...\left(iii\right)$ In equation (iii) , there is a dc term $C/2(A_m^2+A_c^2)$ and sinusoids of frequencies ${\omega }_m,2{\omega }_m,{\omega }_c,2{\omega }_c,{\omega }_c-{\omega }_m$ and ${\omega }_c+{\omega }_m$. As shown in the above figure , this signal is passed through a band pass filter which rejects dc, the sinusoids of frequencies ${\omega }_m,2{\omega }_m,2{\omega }_c$ and retains the frequencies ${\omega }_c,{\omega }_c-{\omega }_m$ and ${\omega }_c+{\omega }_m$. The output of the band pass filter is an AM wave.

(b) Given : $f_c+f_m=660\left(i\right)f_c+f_m=640\left(ii\right)$
Adding (i) and (ii), $2f_c=1300f_c=650khz$
Putting in (i),
$f_m=660-650$
$f_m=10khz$
Bandwidth = $2f_m$
$=2\times 10$
= 20 kHz