Question

An AM wave is expressed as $e=10(1+0.6cos2000\pi t)cos2\times {10}^8\pi t$ volts. Then, the minimum and maximum values of modulated carrier wave are

• A. 10 V, 20 V
• B. 4 V, 8 V
• C. 16 V, 4 V
• D. 8 V, 20 V

Answer 1

The correct option is C 16 V, 4 V

We have the modulation index as, $m=0.6$ and $E_c=10$.
Thus the amplitude of the side frequencies is given as, $\frac{m{E}_c}{2}=\frac{0.6\times 10}{2}=3$
Thus the maximum amplitude value of the modulated wave is given as amplitude of carrier wave plus the amplitude of lower and upper side frequencies. i.e. 10+2(3) and 10-2(3) i.e 16 and 4.