# Modulation Index of Conventional AM Signal

An AM signal has the form

$$u(t)=100\left[1+\sin(2000 \pi t)+5\cos(4000\pi t)\right]\cos(2\pi f_ct)$$

Where $f_c$ is 800 kHz.

What is the modulation index?

(We can define modulation index as the ratio between the amplitude of message wave and the amplitude of carrier wave)

• Hi! What is the definition of modulation index ? – Fat32 Nov 2 '17 at 18:43
• Hi @Fat32 , I edited the question – Aldrich Taylor Nov 2 '17 at 18:55
• so your definition is there, then what point exactly causes the confusion ? – Fat32 Nov 2 '17 at 19:21

Modulation index for linear (continuous wave) amplitude modulation may be defined in different ways. But the one I found relevant to your definition is:

$$\mu = \frac{ | \min \{ m(t) \} | } {A_c}$$

where $A_c$ is the carrier amplitude, $m(t)$ is the message signal and $\mu$ is the modulation index.

For simple demodulation of a conventional AM modulated signal $s(t)$ we look for $\mu < 1$ so that the envelope of $s(t)$ never becomes less than zero.

$$s(t) = [ A_c + m(t) ] \cos(2 \pi f_0 t)$$

Your signal is: $$s(t) = [ 100 + 100 \left( ~~\sin(2000 \pi t)+5\cos(4000 \pi t)~~ \right) ] \cos(2 \pi f_c t)$$

Your message signal is: $$m(t) = 100 \left( ~~\sin(2000 \pi t)+5\cos(4000 \pi t)~~ \right)$$

whose min value is $-600$ and from which you deduce the modulation index as:

$$\mu = \frac{ | \min \{ m(t) \} | } {A_c} = \frac{ | -600 | } {100} = 6$$

I'm sorry to say but this is severely overmodulated...