# Why does the carrier wave amplitude have to be largest in amplitude modulation

I don't understand the reason for the following statements in amplitude modulation

Here's equation 3-7 described in the text

• You're storing information in the amplitude of your carrier signal. What would happen if the amplitude of the information signal exceeded the amplitude of the carrier? Commented May 6, 2019 at 18:47
• Try sketching an AM signal with $m < 1$ (try 0.5 or so), and again with $m > 1$. What's the one with $m>1$ going to sound like if you demodulate it with a simple envelope detector? Commented May 6, 2019 at 18:57
• It doesn't have to be, it just is, because of the way AM works. Commented May 6, 2019 at 19:35

Strictly speaking it doesn't. However if it isn't then demodulation gets much trickier and some would argue that the modulation is no longer purely amplitude modulation.

Lets consider an equation for an AM carrier modulated with a single sinewave. Lets consider that we perform the modulation by multiplication.

$$Y = (a\sin(\omega_st)+b)\sin(\omega_ct)$$

Where $$\\omega_s\$$ is the angular frequency of our signal and $$\\omega_c\$$ is the angular frequency of our carrier.

Now lets bash that equation around a bit to find the frequency components of our AM signal.

$$Y = a\sin(\omega_st)\sin(\omega_ct)+b\sin(\omega_ct)$$

$$Y = \frac{1}{2}a(\cos(\omega_st-\omega_ct)-\cos(\omega_st+\omega_ct))+b\sin(\omega_ct)$$

So our modulated signal consists of three frequencies, a carrier with an amplitude of $$\b\$$ and two sidebands eavh with an amplitude of $$\\frac{1}{2}a\$$.

So our attention turns to what the values of a and b should be. If $$\a \leqslant b\$$ then we are modulating the amplitude of the carrier. At the high peak of the signal the amplitude of the carrier will be b+a while at the low peak of the signal the amplitude of the carrier will be b-a.

But if $$\a \gt b\$$ then $$\a\sin(\omega_st)+b\$$ will drop below zero for part of the signal cycle. The amplitude of a signal cannot drop below zero, so with our multiplier based modulator what instead happens is the phase reverses. We will call this an over-modulated signal.

It takes quite an advanced modulator to actually implement this reversal properly, simpler modulators may instead saturate destroying information.

Ok so now we want to demodulate the signal, how do we do that? Turns out there are basically two methods.

The first is envelope detection. Basically we rectify the signal and look for the peaks. The key observation with this type of demodulation is it doesn't care about the frequency or phase of the signal only it's amplitude. In other words the result of demodulation is roughly.

$$D = |a\sin(\omega_st)+b|$$

If $$\a \leqslant b\$$, we have our signal with a DC offset, but if $$\a > b\$$ then parts of our signal will be inverted. So we cannot use envelope detection to correctly demodulate an over-modulated signal.

The other method is coherent demodulation. In this demodulation technique we multiply the signal by the carrier again. Then filter off the high frequency components.

$$D_\mathrm{uf} = (a\sin(\omega_st)+b)\sin(\omega_ct)\sin(\omega_ct)$$

$$D_\mathrm{uf} = (a\sin(\omega_st)+b)\frac{1 - \cos(2\omega_ct) }{2}$$

$$D = (a\sin(\omega_st)+b)\frac{1}{2}$$

So coherent demodulation can demodulate an over-modulated signal.

However in practice coherent demodulation is much harder to implement than envelope detection, because the frequency and phase of the carrier used for demodulation must match up with that used for modulation. Since the demodulator doesn't generally have an exact copy of the carrier used for modulation it must implement a system to estimate it.

• why does $a\sin(\omega_st)+b$ drop below zero when $a > b$ Commented May 7, 2019 at 14:04
• Because $\sin(\omega_st)$ varies between -1 and 1 Commented May 7, 2019 at 17:56
• So here's what I understand, so when $a > b$, $a\sin(\omega_st)+b$ becomes less than 1, that means $Y = -C\sin(\omega_ct)$ which can also be interpreted as $Y = C\sin(-\omega_ct)$ which is a 180 degree phase shift. Now this is hard to properly employ in modulators. And also it causes other difficulties during demodulation. So, it is better we keep modulation index less than 1. Did I get all of it right, Thank you. Commented May 8, 2019 at 19:04
• Yeah thays right. Commented May 8, 2019 at 23:29