Question

The expression for an AM voltage is

$e=5(1+0.6\cos 100\pi t\cos 5\times {10}^6\pi t)volt.$. Frequencies of sideband is :

• A. $2.50005\times {10}^{6}Hz$, $2.49995\times {10}^{6}Hz$
• B. $2.505\times {10}^{6}Hz$, $2.495\times {10}^{6}Hz$
• C. $2.505\times {10}^{6}kHz$, $2.495\times {10}^{6}kHz$
• D. $2.505MHz$, $2.495kHz$

Answer 1

The correct option is A $2.50005\times {10}^{6}Hz$, $2.49995\times {10}^{6}Hz$

Amplitude modulated wave is given by :

$e=K(1+A\cos 2\pi f_{m}t\cos 2\pi f_{c}t)$

comparing with given equation $f_m=50$ and $f_c=2.5\times {10}^6$

Frequency of side bands are given by $f_1=f_c+f_m$ and $f_2=f_c-f_m$

Putting values

$f_1=2.5\times {10}^6-50=2.49995\times {10}^{6}Hz$

$f_1=2.5\times {10}^6+50=2.50005\times {10}^{6}Hz$

Hence option A is correct