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 E_c cos (2π f_ct ) : Carrier Signal

Term 3 : b x² (t) : Squared modulating Signal

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

Term 5 : b E_c² cos² (2π f_ct ) : 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  f_c and its bandwidth is equal to 2f_m . This bandpass filter eliminates the unuseful terms from the equation of v₂(t) .

Fig 2

Hence the output voltage v_o(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 V_o(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_c .

We can express v₂(t) mathematically as under :

…………………………(4)

where,  g_p(t) is a periodic pulse train of duty cycle equal to one half cycle period i.e. T₀ /2 (where T₀ = 1/f_c) .

Fig.4

Let us express g_p(t) with the help of Fourier series as under :

…………………………….(5)

……………………………(6)

Substituting g_p(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/πE_c] . The unwanted terms can be eliminated using a band-pass filter (BPF) .