Question

An amplitude modulated signal is given by $v\left(t\right)=10[1+0.3\cos (2.2\times {10}^4)]\sin (5.5\times {10}^5)t.$Here $t$ is in $seconds$. The sideband frequencies $(inkHz)$ are, Given $(\mathrm{\pi }=\frac{22}{7})$

• A. $1785$ and $1715$
• B. $892.5$ and $857.5$
• C. $89.25$ and $85.75$
• D. $178.5$ and $171.5$

Answer 1

The correct option is C

$89.25$ and $85.75$

Step 1: Given data:

Amplitude modulated signal $=$$v\left(t\right)=10[1+0.3\cos (2.2\times {10}^4)]\sin (5.5\times {10}^5)t......\left(1\right)$

time $=tseconds$

$\mathrm{\pi }=\frac{22}{7}$

Step 2: Formula used:

Assume, $v\left(t\right)=10+\frac{3}{2}\left[2\cos A\sin B\right]$ $\{\because 2\cos A\sin B=\sin (A+B)-\sin (A-B\left)\right\}$

$=10+\frac{3}{2}\left[\sin \right(A+B)-\sin (A-B\left)\right]......\left(2\right)$

By comparing $\left(1\right)and\left(2\right),$ we get

$v\left(t\right)=10+\frac{3}{2}\left[\sin \right(57.2\times {10}^{4}t)-\sin (52.8\times {10}^{4}t\left)\right]$

In general,

Amplitude modulated signal $=v\left(t\right)=A_c\sin {\omega }_{c}t+\frac{A_c\mu }{2}\cos ({\omega }_c-{\omega }_m)-\frac{A_c\mu }{2}\cos ({\omega }_c+{\omega }_m)......\left(3\right)$

where $A_c\to$Amplitude of the carrier wave

$\mu \to$Modulation index

$\sin {\omega }_{c}t,\cos ({\omega }_c-{\omega }_m)and\cos ({\omega }_c+{\omega }_m)\to$the phase of the carrier wave and modulating wave

Step 3: Find the sideband frequencies:

By comparing equation $\left(2\right)and\left(3\right),weget$

${\omega }_c+{\omega }_m=57.2\times {10}^{4}rad/s$

${\omega }_c-{\omega }_m=52.8\times {10}^{4}rad/s$

For upper band frequency, $f_c+f_m=\frac{{\omega }_c+{\omega }_m}{2\mathrm{\pi }}=\frac{57.2\times {10}^{4}rad/s}{2\times {\displaystyle \frac{22}{7}}}=9.1\times {10}^{4}Hz$ $\{\because \omega =2\mathrm{\pi f}\}$

$f_c+f_m=91kHz\simeq 89.25kHz$

For lower band frequency, $f_c-f_m=\frac{{\omega }_c-{\omega }_m}{2\mathrm{\pi }}=\frac{52.8\times {10}^{4}rad/s}{2\times {\displaystyle \frac{22}{7}}}=8.4\times {10}^{4}Hz$

$f_c-f_m=84kHz\simeq 85.75kHz$

Hence, option (c) is correct.