I have a question as follows. This is not a homework question I just need to clarify my doubt on how this modulation index is defined.
Suppose a 2kHz audio tone having 2V amplitude is to be amplitude modulated on a carrier \$x_c(t) = 5\cos(6\pi 10^5t)\$ with a modulation index of 0.8. For the resulting AM signal
- Derive the mathematical expression
$$x(t) = A_c[1+\mu x_m(t)]\cos(\omega_c t)$$
$$x(t) = 5[1 + \frac{0.8 * 2}{5}x_m(t)]\cos(2\pi*3 *10^{5}t)$$
is this correct? Isn't modulation index made from \$\frac{2}{5}\$ so do I have to use it like this?
$$x(t) = 5[1 + (0.8 * 2)x_m(t)]\cos(2\pi*3 *10^{5}t)$$
I carried on with the first formula
- Sketch the frequency spectrum
I calculated the amplitudes as follows
Carrier Amplitude = \$\frac{A_c}{2}\$
Side band amplitude = \$\frac{\mu A_c a}{4}\$ where a = 2
-
Find the bandwidth
\$2 \times f_m\$ = 4 kHz
-
Find the power of the carrier frequency component
\$\frac{A_c^2}{2} = \frac{5^2}{2}\$ = 12.5 W
- Express the total sideband power as a ratio to the carrier power
\$(A_c[1+\mu x_m(t)]\cos(\omega_c t))^2\$ simplifies into \$\frac{A_c^2[1+\mu ^2 x_m^2(t)]}{2}\$
so the carrier power is \$\frac{A_c^2}{2}\$ and total sideband power is \$\frac{A_c^2\mu^2x_m^2(t)}{2}\$
so as a ratio to the carrier power, it is \$\mu^2x_m^2(t)\$ which simplifies as \$\frac{2^2*0.8^2}{2}\$ (Because amplitude of modulating signal is 2V)
Is this assumption correct?
Best Answer
This formula seems to be misinterpreted by you:
$$x(t) = A_c[1+\mu x_m(t)]\cos(\omega_c t)$$ \$x_m(t)\$ is any message signal, not necessarily a sine wave. Therefore,
$$x(t)=[A_c + A_m x_m(t)]cos(\omega_c t)$$ \$A_c\$ : carrier amplitude
\$A_m\$ : message amplitude
Then the carrier amplitude \$A_c\$ term is taken common to yield:
$$x(t)=A_c[1+\mu x_m(t)]cos(\omega _ct)$$ Where \$\mu\$ = modulation index \$\dfrac{A_m}{A_c}\$ and \$0 \leq\mu \leq 1\$.
So according to this link you cannot give both the carrier and message amplitude and expect a signal with a modulation index within reality.