In conventional AM generation using square
law modulator, the olp from the non linoor
device is expressed here. Specify the terms
which aro bolioved to be useful to get the
modulated output
1 and 2
2&3
36.4
284
Answers
Square Law Modulator
Following is the block diagram of the square law modulator
Square Modulator
Let the modulating and carrier signals be denoted as m(t)
and Acos(2πfct)
respectively. These two signals are applied as inputs to the summer (adder) block. This summer block produces an output, which is the addition of the modulating and the carrier signal. Mathematically, we can write it as
V1t=m(t)+Accos(2πfct)
This signal V1t
is applied as an input to a nonlinear device like diode. The characteristics of the diode are closely related to square law.
V2t=k1V1(t)+k2V21(t)
(Equation 1)
Where, k1
and k2
are constants.
Substitute V1(t)
in Equation 1
V2(t)=k1[m(t)+Accos(2πfct)]+k2[m(t)+Accos(2πfct)]2
⇒V2(t)=k1m(t)+k1Accos(2πfct)+k2m2(t)+
k2A2ccos2(2πfct)+2k2m(t)Accos(2πfct)
⇒V2(t)=k1m(t)+k2m2(t)+k2A2ccos2(2πfct)+
k1Ac[1+(2k2k1)m(t)]cos(2πfct)
The last term of the above equation represents the desired AM wave and the first three terms of the above equation are unwanted. So, with the help of band pass filter, we can pass only AM wave and eliminate the first three terms.
Therefore, the output of square law modulator is
s(t)=k1Ac[1+(2k2k1)m(t)]cos(2πfct)
The standard equation of AM wave is
s(t)=Ac[1+kam(t)]cos(2πfct)
Where, Ka
is the amplitude sensitivity
By comparing the output of the square law modulator with the standard equation of AM wave, we will get the scaling factor as k1
and the amplitude sensitivity ka as 2k2k1
.
Switching Modulator
Following is the block diagram of switching modulator.
Switching Modulator
Switching modulator is similar to the square law modulator. The only difference is that in the square law modulator, the diode is operated in a non-linear mode, whereas, in the switching modulator, the diode has to operate as an ideal switch.
Let the modulating and carrier signals be denoted as m(t)
and c(t)=Accos(2πfct)
respectively. These two signals are applied as inputs to the summer (adder) block. Summer block produces an output, which is the addition of modulating and carrier signals. Mathematically, we can write it as
V1(t)=m(t)+c(t)=m(t)+Accos(2πfct)
This signal V1(t)
is applied as an input of diode. Assume, the magnitude of the modulating signal is very small when compared to the amplitude of carrier signal Ac. So, the diode’s ON and OFF action is controlled by carrier signal c(t). This means, the diode will be forward biased when c(t)>0 and it will be reverse biased when c(t)<0
.
Therefore, the output of the diode is
V2(t)={V1(t)0ififc(t)>0c(t)<0
We can approximate this as
V2(t)=V1(t)x(t)
(Equation 2)
Where, x(t)
is a periodic pulse train with time period T=1fc
Fourier Series
The Fourier series representation of this periodic pulse train is
x(t)=12+2π∑n=1∞(−1)n−12n−1cos(2π(2n−1)fct)
⇒x(t)=12+2πcos(2πfct)−23πcos(6πfct)+....
Substitute, V1(t)
and x(t)
values in Equation 2.
V2(t)=[m(t)+Accos(2πfct)][12+2πcos(2πfct)−23πcos(6πfct)+.....]
V2(t)=m(t)2+Ac2cos(2πfct)+2m(t)πcos(2πfct)+2Acπcos2(2πfct)−
2m(t)3πcos(6πfct)−2Ac3πcos(2πfct)cos(6πfct)+.....
V2(t)=Ac2(1+(4πAc)m(t))cos(2πfct)+m(t)2+2Acπcos2(2πfct)−
2m(t)3πcos(6πfct)−2Ac3πcos(2πfct)cos(6πfct)+.....
The 1st term of the above equation represents the desired AM wave and the remaining terms are unwanted terms. Thus, with the help of band pass filter, we can pass only AM wave and eliminate the remaining terms.
Therefore, the output of switching modulator is
s(t)=Ac2(1+(4πAc)m(t))cos(2πfct)
We know the standard equation of AM wave is
s(t)=Ac[1+kam(t)]cos(2πfct)
Where, ka
is the amplitude sensitivity.
By comparing the output of the switching modulator with the standard equation of AM wave, we will get the scaling factor as 0.5 and amplitude sensitivity ka
as 4πAc .