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Unformatted text preview: EE 308: Communication Systems IIT Bombay, Dept. of Electrical Engineering Semester: Autumn 2009 Instructor: B. K. Dey Solution to Assignment 2: Amplitude Modulation ( DSBFC,DSB SC,SSB,VSB ) Date: 01 September 2009 Solution (1) : m ( t ) = 2 cos 1000 t + cos 2000 t ϕ DSB SC ( t ) = m ( t ) cos 10 , 000 t = [2 cos 1000 t + cos 2000 t ] cos 10 , 000 t = [ cos 9000 t + 1 2 cos 8000 t ] + [ cos 11 , 000 t + 1 2 cos 12 , 000 t ] Baseband frequency(rad/sec) DSB frequency(rad/sec) USB frequency LSB frequency 1000 9000&11,000 11,000 9,000 2000 8,000&12,000 12,000 8,000 Solution (2) : Let us label the points on the gure : The point after the analog multiplier be ' a ', the point after low pass lter be ' b ' and after the dc blocker be ' c '. The output of the analog multiplier is given by : A(t) = [ A + m ( t )] cos 2 ω c t = . 5[ A + m ( t )] + . 5[ A + m ( t )] cos 2 ω c t The rst term in the above equation is a lowpass signal because its spectrum is located at ω =0. The lowpass lter suppresses the second term and allows the low pass signal to pass through it. The output of the lowpass lter is given by : B(t) = . 5[ A + m ( t )] Now this signal is passed through the dc blocker. Therefore the dc term 'A' is supprerssed yielding the output given by : 1 2 C(t) = . 5 m ( t ) This shows that the system can demodulate AM signal regardless of the value of A....
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This note was uploaded on 02/05/2012 for the course EE EE308 taught by Professor B.k.dey during the Spring '09 term at IIT Bombay.
 Spring '09
 B.K.Dey
 Electrical Engineering

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