# Explain the Generation of AM Waves using Square Law Modulator and Switching Modulator

## Generation of AM Waves

The circuit that generates the AM waves is called as amplitude modulator and in this post we will discuss two such modulator circuits namely :

1. Square Law Modulator
2. Switching Modulator

Both of these circuits use a non-linear elements such as a diode for their implementation . Both these modulators are low power modulator circuits .

## Square Law Modulator

Generation of AM Waves using the square law modulator could be understood in a better way by observing the square law modulator circuit shown in fig.1 . Fig 1

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It consists of the following :

1. A non-linear device
2. A bandpass filter
3. A carrier source and modulating signal

The modulating signal and carrier are connected in series with each other and their sum V1(t) is applied at the input of the non-linear device, such as diode, transistor etc.

Thus, …………………………(1)

The input output relation for non-linear device is as under : …………………………….(2)

where a and b are constants.

Now, substituting the expression (1) in (2), we get Or, Or, The five terms in the expression for V2(t) are as under :

Term 1: ax(t) : Modulating Signal

Term 2 : a Ecos (2π fct ) : Carrier Signal

Term 3 : b x(t) : Squared modulating Signal

Term 4 : 2 b x(t) cos ( 2π fct ) : AM wave with only sidebands

Term 5 : b Eccos(2π fct ) : Squared Carrier

Out of these five terms, terms 2 and 4 are useful whereas the remaining terms are not useful .

Let us club terms 2, 4 and 1, 3, 5 as follows to get , The LC tuned circuit acts as a bandpass filter . Its frequency responce is shown in fig 2 which shows that the circuit is tuned to frequency  fand its bandwidth is equal to 2f. This bandpass filter eliminates the unuseful terms from the equation of v2(t) . Fig 2

Hence the output voltage vo(t) contains only the useful terms . Or, Therefore , ………………………….(3)

Comparing this with the expression for standard AM wave i.e. ,

We find that the expression for Vo(t) of equation (3) represents an AM wave with m = (2b/a) .

Hence, the square law modulator produces an AM wave .

## Switching Modulator

Generation of AM Waves using the switching modulator could be understood in a better way by observing the switching modulator diagram. The switching modulator using a diode has been shown in fig 3(a) .  Fig 3 (a)                                                                  Fig 3(b)

This diode is assumed to be operating as a switch .

The modulating signal x(t) and the sinusoidal carrier signal c(t) are connected in series with each other. Therefore, the input voltage to the diode is given by : The amplitude of carrier is much larger than that of x(t) and c(t) decides the status of the diode (ON or OFF ) .

### Working Operation and Analysis

Let us assume that the diode acts as an ideal switch . Hence, it acts as a closed switch when it is forward biased in the positive half cycle of the carrier and offers zero impedance . Whereas it acts as an open switch when it is reverse biased in the negative half cycle of the carrier and offers an infinite impedance .

Therefore, the output voltage  v2(t) = v1(t) in the positive half cycle of c(t) and v2(t) = 0 in the negative half cycle of c(t) .

Hence ,               v2(t) = v1(t)      for c(t) > 0

v2(t) = 0           for c(t) < 0

In other words , the load voltage v2(t) varies periodically between the values v1(t) and zero at the rate equal to carrier frequency f.

We can express v2(t) mathematically as under : …………………………(4)

where,  gp(t) is a periodic pulse train of duty cycle equal to one half cycle period i.e. T/2 (where T= 1/fc) . Fig.4

Let us express gp(t) with the help of Fourier series as under : …………………………….(5) ……………………………(6)

Substituting gp(t) into equation (4), we get Therefore, ………………………..(7)

The odd harmonics in this expression are unwanted, and therefore, are assumed to be eliminated .

Hence, In this expression, the first and the fourth terms are unwanted terms whereas the second and third terms together represents the AM wave .

Clubing the second and third terms together , we obtain This is the required expression for the AM wave with m=[4/πEc] . The unwanted terms can be eliminated using a band-pass filter (BPF) .