GATE 2018 EC Q.27: solution and answer

Official answer: DSource: Official answer key, GATE 2018 EC (IIT Guwahati)

1. Let ${\omega }_c=2\pi f_c$ and ${\omega }_m=2\pi f_m$. The input to the device is $v_i(t)=A_c\cos {\omega }_{c}t+\cos {\omega }_{m}t$, and $v_o=a{v}_i+b{v}_i^2$.

2. Expand the square: $v_i^2=A_c^2{\cos }^2{\omega }_{c}t+{\cos }^2{\omega }_{m}t+2A_c\cos {\omega }_{c}t\cos {\omega }_{m}t$.

3. Frequencies present in each term. The linear term $a{v}_i$ has $f_c$ and $f_m$. The term $A_c^2{\cos }^2{\omega }_{c}t$ has DC and $2f_c$. The term ${\cos }^2{\omega }_{m}t$ has DC and $2f_m$. The cross term has $f_c\pm f_m$.

4. The ideal band-pass filter is centred at $f_c$ with bandwidth $3f_m$, so it passes $f_c-1.5f_m\le f\le f_c+1.5f_m$. Because $f_c\gg 5f_m$, the components at DC, $f_m$ and $2f_m$ lie far below the pass-band and $2f_c$ lies far above it. They are all rejected. The carrier at $f_c$ and the sidebands at $f_c\pm f_m$ (inside $\pm 1.5f_m$) are passed.

5. Filter output: $$y(t)=a{A}_c\cos {\omega }_{c}t+2b{A}_c\cos {\omega }_{m}t\cos {\omega }_{c}t=a{A}_c\left[1+\frac{2b}{a}\cos {\omega }_{m}t\right]\cos {\omega }_{c}t.$$

6. This has the standard single-tone AM form $A[1+\mu \cos {\omega }_{m}t]\cos {\omega }_{c}t$ with modulation index $\mu =\frac{2b}{a}$.

7. For single-tone AM, the carrier power is $P_c=\frac{A^2}{2}$ and the total sideband power is $P_{sb}=\frac{A^2{\mu }^2}{4}$. Hence $\frac{P_{sb}}{P_c}=\frac{{\mu }^2}{2}$.

8. Setting $\frac{{\mu }^2}{2}=\frac{1}{2}$ gives $\mu =1$ (taking the positive value, since $a,b>0$). Then $\frac{2b}{a}=1$, so $\frac{a}{b}=2$, which is option (D).