# Amplitude of an AM wave

A message signal is modulated on a carrier wave of amplitude $50V$ using amplitude modulation. The modulation index is given as $50\%$. Find the amplitude of the AM wave.

I was able to calculate the amplitude of the message signal as $25V$. However I'm stuck at the amplitude of AM wave.

The maximum displacement of the AM wave will be $50+25=75V$ and the minimum displacement will be $50-25=25V$.

So is the amplitude of the AM wave $75V$ or is it $75-25=50V$ i.e. is it the maximum displacement or the difference between the maximum and minimum displacements?

I know amplitude is the maximum displacement from the mean position however I can't seem to figure the mean position of this AM wave.

The amplitude of the AM wave depends on the modulating signal, that's why it is called amplitude modulation. Taking your values, we have:

• If the modulating signal is at its mininum (negative peak), the AM wave amplitude is 25 V (swings between 25 and -25 V).
• If the modulating signal is at its maximum (positive peak), the AM wave amplitude is 75 V (swings between 75 and -75 V).
• If the modulating signal is 0, the AM wave amplitude is 50 V (swings between 50 and -50 V).

One thing to note is that the modulating signal is a very slow time-varying signal compared to the carrier frequency.

• So we can't assign an overall amplitude to the AM wave?. In other words, can the amplitude of an AM wave be specified only if one knows the amplitude of the modulating signal at a specific point of time? Mar 9 '15 at 18:15
• Correct, the amplitude will be changing, as it is how the information is transmitted in AM. But you can always specify an interval, for example you could say the amplitude will be between 25 V and 75 V. Mar 9 '15 at 19:04
• Andy aka is giving a good point. You can indeed assign an average amplitude (if you know the modulating signal waveform). Assuming the modulating signal is a sine, the answer is 53 V as Andy aka point out. Mar 9 '15 at 19:18

If the AM modulation index is 0.5 then the carrier amplitude varies by 50% above (and below) its unmodulated level: -

Here's where different answers may differ and it surrounds what is meant by carrier wave of amplitude 50V. Without any mention of peak values (or peak/peak), one has to assume an RMS value of 50V and a peak amplitude of 70.71 volts. This means that at a modulation index of 0.5, the peak of the modulated carrier will be 150% of 70.71 volts i.e. 106 volts.

It's also worth mentioning that the power of an AM signal is: -

$$\P_T = P_C(1 + \dfrac{M^2}{2})\$$ where M is the index and Pc is unmodulated carrier power

This basically means that a 100 watt carrier modulated by an index of 0.5 results in a power of 112.5 watts. If the power has increased by 12.5 watts to 112.5 watts, in decibels this is 0.512 dB and therefore the voltage will be 6.07% higher than what it previously was so....

An unmodulated carrier of 50V RMS becomes a modulated carrier of 53V RMS when modulated at an index of 0.5.

• It should also be noted that this formula of the power of an AM signal is only valid if the modulating signal is a sine. Mar 9 '15 at 19:36
• Uh, I'm a bit confused. (I'm still in high school and we're taught just the very basics, so please bear with me.). In the first paragraph you say that the peak of the modulated carrier is 106V and in the last paragraph, you say that the RMS of the modulated carrier is 53V. This seems like a conflict to me. Can you please explain this? Mar 10 '15 at 9:55
• The modulated carrier isn't a sinewave any more - it has an RMS of 53V but has a peak of 106V - look at the picture - at its peak it has a cyclic RMS value of about 75V - at its trough it has a cyclic RMS value of about 25V. Mar 10 '15 at 11:20