written 7.9 years ago by | modified 2.8 years ago by |
Mumbai University > Computer Engineering > Sem 3 > Electronic Circuits and Communication Fundamentals
Marks: 10 Marksr
Year: May 2014
written 7.9 years ago by | modified 2.8 years ago by |
Mumbai University > Computer Engineering > Sem 3 > Electronic Circuits and Communication Fundamentals
Marks: 10 Marksr
Year: May 2014
written 7.9 years ago by | • modified 7.9 years ago |
1.Double Sideband Suppressed Carrier (DSBSC) is an amplitude modulation technique in which the modulated wave contains both the sidebands along with the suppressed carrier.
2.Conventional AM consists of the two sidebands and a carrier where the major transmitted power is concentrated in the carrier which contains no information. Thus to increase the efficiency and to save power, the carrier is suppressed in DSBSC system.
3.The DSBSC generation using balanced modulator based on nonlinear resistance characteristics of diode is given below in Fig1.
4.The diode in the balanced modulator use the nonlinear resistance property for producing modulated signals.
5.Carrier voltage is applied in phase at both the diodes, while modulating voltage appears 180° out of phase at the diode inputs as they are at opposite ends of a center tapped transformer.
6.The modulated output currents of the two diodes are combined in the center tapped primary of the output transformer, which then gets subtracted.
7.The output of the balanced modulator contains two sidebands and sum of the harmonic components.
8.As indicated in the Fig7, the input voltage at diode D1D1 is (vc+vm)(vc+vm) and input voltage at diode D2D2 is (vc−vm)(vc−vm).
9.The primary current of the output transformer is i1=id1−id2i1=id1−id2.
Where,
id1=a+b(vc+vm)+c(vc+vm)2 andid2=a+b(vc−vm)+c(vc−vm)2id1=a+b(vc+vm)+c(vc+vm)2 andid2=a+b(vc−vm)+c(vc−vm)2
Thus, we get,
i1=id1−id2=2bvm+4cvmvci1=id1−id2=2bvm+4cvmvc
10.The modulating and carrier voltage are represented as,
vm=Vmsinωmt andvc=Vcsinωctvm=Vmsinωmt andvc=Vcsinωct
11.Substituting for vmvm and vcvc and simplifying, we get,
i1=2bVmsinωmt+4cmVc2cos(ωc−ωm)t−4cmVc2cos(ωc+ωm)ti1=2bVmsinωmt+4cmVc2cos(ωc−ωm)t−4cmVc2cos(ωc+ωm)t
12.The output voltage v0v0 is proportional to primary current i1i1 and assume constant of proportionality as α, which can be expressed as,
v0=αi1=2αbVmsinωmt+4αcmVc2cos(ωc−ωm)t−4αcmVc2cos(ωc+ωm)tv0=αi1=2αbVmsinωmt+4αcmVc2cos(ωc−ωm)t−4αcmVc2cos(ωc+ωm)t
Let P=2αbVmP=2αbVm and Q=2αcmVc2.Q=2αcmVc2.
Thus we have,
v0=Psinωmt+2Qcos(ωc−ωm)t−2Qcos(ωc+ωm)tv0=Psinωmt+2Qcos(ωc−ωm)t−2Qcos(ωc+ωm)t
13.The above equation shows that carrier has been cancelled out , leaving only two sidebands and the modulating frequencies.
14.The modulating frequencies from the output is eliminated by the tuning of the output transformer, which results in the below equation of the generated DSBSC wave.
v0=2Qcos(ωc−ωm)t−2Qcos(ωc+ωm)t