Equation for square law circuit modulator

Question

For DSBFC AM (double side band full carrier amplitude modulation ) the message signal \$m(t)\$ must be multiplied by carrier maybe $$ A_c\cos(\omega_c(t)) $$ (For modulation)

This modulation is done in reality by using the non-linear characteristics of the diode, where

$$ i(t)=av + bv^2 $$

\$v =\$ applied voltage to diode

Here we apply

$$ v =V_c(t)+V_m(t)$$ $$ i(t)=a(V_c+_Vm)+b(V_c+V_m)^2 $$ $$ i(t)= aV_c +bV_m +bV_c^2+bV_m^2 +2V_cV_m $$

Here $$V_c=A_c\cos(\omega_c(t))$$

Thus $$i(t)=aA_c\cos(\omega_c(t))+bm(t)+b(A_c\cos(\omega_c(t)))^2 +bm(t)^2 +2m(t)A_c\cos(\omega_c(t))$$

After simplification

$$i(t)=aA_c\cos(\omega_c(t))+bm(t)+\frac{bA_c}{2} +\frac{bA_c\cos(2\omega_c(t))}{2} +bm(t)^2 +2m(t)A_c\cos(\omega_c(t))$$

In frequency domain I can clearly understand following components

$$ f(\text{message frequency}),f_c,f_c+f,f_c-f,2f_c $$

but my book tells me there are additional components at \$2f_m\$.

Can someone help me understand where it is?

Answer 1

2fm

This frequency comes from the message signal; specifically \$(m_{(t)})^2\$. Multiplication of a signal with another (or itself) in time domain is represented in frequency domain as a convolution. So the corresponding signal in frequency domain is

$$ M(s) \circledast M(s) $$

If a signal with frequency contents from 0 to \$f_{max}\$ is convolved with itself, the resulting spectrum will have frequency content from 0 to \$2f_{max}\$.

From Wikipedia