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So the exercise is basically a signal $f(t)$ that is going to modulate the carrier $Acosw_ct$ using a modulation index of $m=1$. I have to find $A$ and the power of the modulated signal.:


The minimum amplitude of $f(t)$ is -2. Then $A = 2$. The power of the signal is, assuming that $R = 1ohm$:

$P = P_c+P_s=\frac{A^2}{2}+\frac{\overline{f^2(t)}}{2}$

Having in mind that:


The power is:

$P =\frac{A^2}{2}+\frac{\overline{f^2(t)}}{2}=\frac{2^2}{2}+\frac{5}{4}=3.25$

In the book the author uses an effective modulation index that is defined as $m_t = \sqrt{m_1^2+m_2^2}$ where $m_1=1/2$ and $m_2 = 2/2$. So the power is:

$P = Pc(1+\frac{m_t^2}{2})=2(1+\frac{1.12^2}{2})=3.25$

My question is, why would I want to define a modulation index for each tone? What do I get from that?

Another thing that I don't understand is that according to this guy the condition that ensures that there's no overmodulation, regardless of the tone frequencies, is $m_1+m_2<=1$. Obviously in this case the condition is not met but I'm pretty sure there's no overmodulation with $A=2$.

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I'm no expert on this, but until the big guys give an answer, here's my take:

For f(t) = cos(100*t)+2*cos(2*100*t), so w_m = 100 Hz for example:

f(t) = cos(100*t)+2*cos(2*100*t)

The amplitude's peak value=3 exceeds A=2, so there is overmodulation. This answer only addresses your last point of "I'm pretty sure there's no overmodulation".

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No. What really matters is the negative peak amplitude. – user3680 Jul 9 '12 at 13:31

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