A modulated signal C_m(t) has the form C_m(t) | vedclass

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(B) The standard equation for an $AM$ wave is $C_m(t) = A_c \sin(\omega_c t) + \frac{\mu A_c}{2} \cos((\omega_c - \omega_m)t) - \frac{\mu A_c}{2} \cos

Vedclass · Accepted Answer

$f_c = 150 	ext{ Hz}, f_m = 50 	ext{ Hz}, \mu = 2/3$

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(B) The standard equation for an $AM$ wave is $C_m(t) = A_c \sin(\omega_c t) + \frac{\mu A_c}{2} \cos((\omega_c - \omega_m)t) - \frac{\mu A_c}{2} \cos((\omega_c + \omega_m)t)$.Comparing this with the given equation $C_m(t) = 30 \sin(300\pi t) + 10 \cos(200\pi t) - 10 \cos(400\pi t)$:$1$. Carrier frequency: $\omega_c = 300\pi \Rightarrow 2\pi f_c = 300\pi \Rightarrow f_c = 150 	ext{ Hz}$.$2$. Sideband frequencies: $\omega_c - \omega_m = 200\pi$ and $\omega_c + \omega_m = 400\pi$.Subtracting the two: $2\omega_m = 200\pi \Rightarrow \omega_m = 100\pi \Rightarrow 2\pi f_m = 100\pi \Rightarrow f_m = 50 	ext{ Hz}$.$3$. Modulation index: $\frac{\mu A_c}{2} = 10$. Since $A_c = 30$,we have $\frac{\mu(30)}{2} = 10 \Rightarrow 15\mu = 10 \Rightarrow \mu = 10/15 = 2/3$.

$A$ modulated signal $C_m(t)$ has the form $C_m(t) = 30 \sin(300\pi t) + 10 \cos(200\pi t) - 10 \cos(400\pi t)$. The carrier frequency $f_c$,the modulating frequency $f_m$,and the modulation index $\mu$ are respectively given by:

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