GATE 2018 IN Q.24: solution and answer

Official answer: 0.3Source: Official answer key, GATE 2018 IN (IIT Guwahati)

1. Read the envelope from the figure. The envelope peaks (the horizontal line from the 13 label) at $A_{max}=13\text{V}$. It dips at its troughs (the horizontal line from the 7 label) to $A_{min}=7\text{V}$. The waveform is symmetric about the zero line.

2. For standard AM, $s(t)=A_c[1+\mu \cos (2\pi f_{m}t)]\cos (2\pi f_{c}t)$. The envelope is $A_c[1+\mu \cos (2\pi f_{m}t)]$, so $A_{max}=A_c(1+\mu )$ and $A_{min}=A_c(1-\mu )$.

3. Add and subtract: $A_c=\frac{A_{max}+A_{min}}{2}=\frac{13+7}{2}=10\text{V}$ and $\mu A_c=\frac{A_{max}-A_{min}}{2}=\frac{13-7}{2}=3\text{V}$.

4. Modulation index: $\mu =\frac{A_{max}-A_{min}}{A_{max}+A_{min}}=\frac{13-7}{13+7}=\frac{6}{20}=0.3$.

5. Since $\mu <1$, the envelope never reaches zero. This agrees with the figure, where the envelope bottoms out at 7 V and the signal is not over-modulated.