Ampptude Modulation

A continuous-wave goes on continuously without any intervals and it is the baseband message signal, which contains the information. This wave has to be modulated.

According to the standard definition, “The ampptude of the carrier signal varies in accordance with the instantaneous ampptude of the modulating signal.” Which means, the ampptude of the carrier signal containing no information varies as per the ampptude of the signal containing information, at each instant. This can be well explained by the following figures.

The first figure shows the modulating wave, which is the message signal. The next one is the carrier wave, which is a high frequency signal and contains no information. While, the last one is the resultant modulated wave.

It can be observed that the positive and negative peaks of the carrier wave, are interconnected with an imaginary pne. This pne helps recreating the exact shape of the modulating signal. This imaginary pne on the carrier wave is called as Envelope. It is the same as that of the message signal.

Mathematical Expressions

Following are the mathematical expressions for these waves.

Time-domain Representation of the Waves

Let the modulating signal be,

$$mleft ( t ight )=A_mcosleft ( 2pi f_mt ight )$$

and the carrier signal be,

$$cleft ( t ight )=A_ccosleft ( 2pi f_ct ight )$$

Where,

$A_m$ and $A_c$ are the ampptude of the modulating signal and the carrier signal respectively.

$f_m$ and $f_c$ are the frequency of the modulating signal and the carrier signal respectively.

Then, the equation of Ampptude Modulated wave will be

$s(t)= left [ A_c+A_mcosleft ( 2pi f_mt ight ) ight ]cos left ( 2pi f_ct ight )$ (Equation 1)

Modulation Index

A carrier wave, after being modulated, if the modulated level is calculated, then such an attempt is called as Modulation Index or Modulation Depth. It states the level of modulation that a carrier wave undergoes.

Rearrange the Equation 1 as below.

$s(t)=A_cleft [ 1+left ( frac{A_m}{A_c} ight )cos left ( 2pi f_mt ight ) ight ]cos left ( 2pi f_ct ight )$

$Rightarrow sleft ( t ight ) = A_cleft [ 1 + mu cos left ( 2 pi f_m t ight ) ight ] cosleft ( 2 pi f_ct ight )$ (Equation 2)

Where, $mu$ is Modulation index and it is equal to the ratio of $A_m$ and $A_c$. Mathematically, we can write it as

$mu = frac{A_m}{A_c}$ (Equation 3)

Hence, we can calculate the value of modulation index by using the above formula, when the ampptudes of the message and carrier signals are known.

Now, let us derive one more formula for Modulation index by considering Equation 1. We can use this formula for calculating modulation index value, when the maximum and minimum ampptudes of the modulated wave are known.

Let $A_max$ and $A_min$ be the maximum and minimum ampptudes of the modulated wave.

We will get the maximum ampptude of the modulated wave, when $cos left ( 2pi f_mt ight )$ is 1.

$Rightarrow A_max = A_c + A_m$ (Equation 4)

We will get the minimum ampptude of the modulated wave, when $cos left ( 2pi f_mt ight )$ is -1.

$Rightarrow A_min = A_c - A_m$ (Equation 5)

Add Equation 4 and Equation 5.

$$A_max + A_min = A_c+A_m+A_c-A_m = 2A_c$$

$Rightarrow A_c = frac{A_max + A_min}{2}$ (Equation 6)

Subtract Equation 5 from Equation 4.

$$A_max - A_min = A_c + A_m - left (A_c -A_m ight )=2A_m$$

$Rightarrow A_m = frac{A_max - A_min}{2}$ (Equation 7)

The ratio of Equation 7 and Equation 6 will be as follows.

$$frac{A_m}{A_c} = frac{left ( A_{max} - A_{min} ight )/2}{left ( A_{max} + A_{min} ight )/2}$$

$Rightarrow mu = frac{A_max - A_min}{A_max + A_min}$ (Equation 8)

Therefore, Equation 3 and Equation 8 are the two formulas for Modulation index. The modulation index or modulation depth is often denoted in percentage called as Percentage of Modulation. We will get the percentage of modulation, just by multiplying the modulation index value with 100.

For a perfect modulation, the value of modulation index should be 1, which imppes the percentage of modulation should be 100%.

For instance, if this value is less than 1, i.e., the modulation index is 0.5, then the modulated output would look pke the following figure. It is called as Under-modulation. Such a wave is called as an under-modulated wave.

If the value of the modulation index is greater than 1, i.e., 1.5 or so, then the wave will be an over-modulated wave. It would look pke the following figure.

As the value of the modulation index increases, the carrier experiences a 180^o phase reversal, which causes additional sidebands and hence, the wave gets distorted. Such an over-modulated wave causes interference, which cannot be epminated.

Bandwidth of AM Wave

Bandwidth (BW) is the difference between the highest and lowest frequencies of the signal. Mathematically, we can write it as

$$BW = f_{max} - f_{min}$$

Consider the following equation of ampptude modulated wave.

$$sleft ( t ight ) = A_cleft [ 1 + mu cos left ( 2 pi f_m t ight ) ight ] cosleft ( 2 pi f_ct ight )$$

$$Rightarrow sleft ( t ight ) = A_ccos left ( 2pi f_ct ight )+ A_cmu cos(2pi f_ct)cos left ( 2pi f_mt ight )$$

$Rightarrow sleft ( t ight )= A_ccos left ( 2pi f_ct ight )+frac{A_cmu }{2}cos left [ 2pi left ( f_c+f_m ight ) t ight ]+frac{A_cmu }{2}cos left [ 2pi left ( f_c-f_m ight ) t ight ]$

Hence, the ampptude modulated wave has three frequencies. Those are carrier frequency $f_c$, upper sideband frequency $f_c + f_m$ and lower sideband frequency $f_c-f_m$

Here,

$f_{max}=f_c+f_m$ and $f_{min}=f_c-f_m$

Substitute, $f_{max}$ and $f_{min}$ values in bandwidth formula.

$$BW=f_c+f_m-left ( f_c-f_m ight )$$

$$Rightarrow BW=2f_m$$

Thus, it can be said that the bandwidth required for ampptude modulated wave is twice the frequency of the modulating signal.

Power Calculations of AM Wave

Consider the following equation of ampptude modulated wave.

$ sleft ( t ight )= A_ccos left ( 2pi f_ct ight )+frac{A_cmu }{2}cos left [ 2pi left ( f_c+f_m ight ) t ight ]+frac{A_cmu }{2}cos left [ 2pi left ( f_c-f_m ight ) t ight ]$

Power of AM wave is equal to the sum of powers of carrier, upper sideband, and lower sideband frequency components.

$$P_t=P_c+P_{USB}+P_{LSB}$$

We know that the standard formula for power of cos signal is

$$P=frac{{v_{rms}}^{2}}{R}=frac{left ( v_m/ sqrt{2} ight )^2}{2}$$

Where,

$v_{rms}$ is the rms value of cos signal.

$v_m$ is the peak value of cos signal.

First, let us find the powers of the carrier, the upper and lower sideband one by one.

Carrier power

$$P_c=frac{left ( A_c/sqrt{2} ight )^2}{R}=frac{{A_{c}}^{2}}{2R}$$

Upper sideband power

$$P_{USB}=frac{left ( A_cmu /2sqrt{2} ight )^2}{R}=frac{{A_{c}}^{2}{_{mu }}^{2}}{8R}$$

Similarly, we will get the lower sideband power same as that of the upper side band power.

$$P_{LSB}=frac{{A_{c}}^{2}{_{mu }}^{2}}{8R}$$

Now, let us add these three powers in order to get the power of AM wave.

$$P_t=frac{{A_{c}}^{2}}{2R}+frac{{A_{c}}^{2}{_{mu }}^{2}}{8R}+frac{{A_{c}}^{2}{_{mu }}^{2}}{8R}$$

$$Rightarrow P_t=left ( frac{{A_{c}}^{2}}{2R} ight )left ( 1+frac{mu ^2}{4}+frac{mu ^2}{4} ight )$$

$$Rightarrow P_t=P_cleft ( 1+frac{mu ^2}{2} ight )$$

We can use the above formula to calculate the power of AM wave, when the carrier power and the modulation index are known.

If the modulation index $mu=1$ then the power of AM wave is equal to 1.5 times the carrier power. So, the power required for transmitting an AM wave is 1.5 times the carrier power for a perfect modulation.