The GATE Grind

GATE 2018 EC – Question 37

Communications · Analog Communications · 2 marks · Multiple choice

Let $c(t)=A_c\cos(2\pi f_ct)$ and $m(t)=\cos(2\pi f_mt)$. It is given that $f_c\gg5f_m$. The signal $c(t)+m(t)$ is applied to the input of a non-linear system, whose output $v_o(t)$ is related to the input $v_i(t)$ as $v_o(t)=av_i(t)+bv_i^2(t)$, where $a$ and $b$ are positive constants. The output of the non-linear system is passed through an ideal band-pass filter with center frequency $f_c$ and bandwidth $3f_m$, to produce an amplitude modulated (AM) wave. If it is desired to have the sideband power of the AM wave to be half of the carrier power, then $a/b$ is

  1. 0.25
  2. 0.5
  3. 1
  4. 2

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: (D) 2

Explanation

The band-pass filter keeps $aA_c\cos2\pi f_ct$ (carrier) and the cross term $2bA_c\cos2\pi f_ct\cos2\pi f_mt=bA_c[\cos2\pi(f_c+f_m)t+\cos2\pi(f_c-f_m)t]$. The carrier power is $\frac{(aA_c)^2}{2}$ and the sideband power is $2\times\frac{(bA_c)^2}{2}=(bA_c)^2$. Setting sidebands equal to half the carrier, $b^2=\frac{a^2}{4}$, so $\frac ab=2$.