Square law demodulator. How can we eliminate the term containing $m^2(t)$?

We are trying to extract the message signal from the AM signal using square law demodulator.

In the final expression, $V_2(t)$, the term encircled in blue has to be retained, and the others have to be eliminated. The terms encircled in green can be eliminated by low pass filter(the career frequency being high). The constant term can be eliminated with the help of coupling capacitor as mentioned below the derivation. But the problem is that the spectrum of the term $\frac{k_2 {A_{c}}^{2}{k_{a}}^{2}m^2\left ( t \right )}{2}$ will lie around the origin. It is just that the width of the spectrum of this signal component will be twice of that of $m(t)$[Multiplication in time domain is convolution in frequency domain]. The spectrum of the the component $k_2{A_{c}}^{2}k_am\left ( t \right )$ will also lie around the origin. Will the spectrum of $k_2{A_{c}}^{2}k_am\left ( t \right )$ and $\frac{k_2 {A_{c}}^{2}{k_{a}}^{2}m^2\left ( t \right )}{2}$ not interfare? How can then we extract the message signal?

$|m(t)| \leq 1.0$, so for analog signals, ${m^2(t)}$ is going to be small most of the time. Also $k_a < 1.0$, so ${k_a}^2 \ll 1.0$. You can ignore the term usually. If you can't, then use something other than a square law demodulator.
• Sir, I understand that $k_a < 1.0$ so as to avoid overmodulation. However, I am unable to understand why $|m(t)| \leq 1.0$. – Soumee Apr 9 '18 at 11:59
• Generally the transmitted message, $m(t)$, is normalized to not exceed +/- 1.0. If the message wasn't bounded, the modulation index, $k_a$, would be somewhat pointless. :) – Andy Walls Apr 9 '18 at 12:56
• Sir, for a single toned frequency signal, the modulation index is given by $k_a=\frac{A_m}{A_c}$. Since signals are transmitted over large distances, $A_c$ must be large. So whenever $A_m<A_c$, $k_a < 1.0$. So, why should $|m(t)| \leq 1.0$. Thankyou. – Soumee Apr 9 '18 at 13:50
• The product $k_a m(t)$ is a design parameter of the transmitter to avoid overmodulation. The terms never appear separately, so you have a degree of freedom on distributing amplitude between $k_a$ and $m(t)$ (or $A_m$). There is generally no point in having $m(t)$ exceed $\pm 1.0$. For the bigger picture look at the ratio of the two terms you care about $20 \log\left(\dfrac{\dfrac{k_2 {A_c}^2{k_a}^2 m^2(t)}{2}}{k_2 {A_c}^2{k_a}m(t)}\right) = 20 \log\left(\dfrac{k_a m(t)}{2}\right)$. As long as the product $k_a m(t)$ is small thing are fine. You can make $m(t)$ large, but what's the point. – Andy Walls Apr 9 '18 at 14:29