Question

Question 4 (35 points). Two message signals, ?1 (?) = 200??(100??) cos(103πt) volts and ?2 (?) = 900??2 (450??) cos(103πt) volts, are required to be combined and then transmitted over a channel around 2 MHz using amplitude double-side modulation. We assume that the channel is ideal with no delay and no attenuation. Design a system consisting of mixers, filters, etc. to transmit and restore perfectly these two signals (i.e. ?1 (?) and ?2 (?)) at the receiving side after the channel. One key requirement is that the transmission uses minimal channel bandwidth. For your design, a) (8 points) Find the Fourier transform of ?1 (?) and ?2 (?) and sketch their magnitude spectrum, respectively. b) (10 points) Sketch the block diagram of the modulator and specify the parameters of its each component (e.g. mixer and filter) . c) (8 points) Sketch the output signal spectrum of each component of the modulator. d) (9 points) Sketch the block diagram of the demodulator, and specify its parameters of each component.

Question 4 (35 points). Two message signals, m (t) = 2008a(100mt) cos(10’nt) volts and m2(t) = 900Sa?(450nt) cos(103Ttt) volts, are required to be combined and then transmitted over a channel around 2 MHz using amplitude double-side modulation. We assume that the channel is ideal with no delay and no attenuation. Design a system consisting of mixers, filters, etc. to transmit and restore perfectly these two signals (i.e. m(t) and mz(t)) at the receiving side after the channel. One key requirement is that the transmission uses minimal channel bandwidth. For your design, a) (8 points) Find the Fourier transform of m (t) and m2(t) and sketch their magnitude spectrum, respectively. b) (10 points) Sketch the block diagram of the modulator and specify the parameters of its each component(e.g. mixer and filter). c) (8 points) Sketch the output signal spectrum of each component of the modulator. d) (9 points) Sketch the block diagram of the demodulator, and specify its parameters of each component.

Answer 1

4. 9) milt): 200 Sa Cloot +) cos (1037) malt)= goo sah (450lt) COS Clot) fouriel transform of miltian Milf we know that cos Clott) <F.T , 0.58 (f-500] +0.58(6+500) and f 200 Sa Cloont) <F.T, 2T ( Too) So. MI CF) = T1 (-509) + T (4+500) f Spectrum e -1 f(Hz) 450 500 550 -550 - 500 - 450

Ala Malt) tourice transform of malt) we know that cos Clotit) <f. t. , 0.58 (f-500) +0.58(4+500) and goo Sa(45011 t) esity 2 Alison t + 500 so Malt) = Alf-500) ta 450 A 50 Spectrum f (Hz 50 +500 950 - 950 -50 - 500

4.b. Block diagram of modulator: Vilt) Volt) Samit) Band- Pass sawe law device filter mit) - Mixel G milt) & malt) L.O. 2x106H2 fca mit) - Input or message signal. L.o.o local oscillator of carries frequency 2 MHZ vilt- mlt)t clt) where ct) Calrich signal. sawe law device have volt) = 9, Vilt) +9qvikt) BPF = Band pass filter of suitable center to frequency and Bandwidth pass the Amplituele double side-band signal.

4.0) Vilt) miltofhalt) + c(t) Vilt= Milt) +Malt) +(6+) sketch AYE A42 f(kHz) 0.5 0.05 o.gr 2000 -0.5 -0.05 -20000.95 now by passing through BPF Sam(t) = b {M, 14 - 2x10W) + M(f+2x106] + Ya ( melt +2x104) +Me (f - 2xvel) + /Ac 861-8X16+%4804+2016) Samlf) - 2000.5 2000.5 f(kHz)

A.d) for demodula tion of signal envelope detector Rs С. RL mit) Sainit) Vc for proper detection, RSC should be possibly minimum. and here we also keep in mind I <fic <yfm fc for our problem fez 2 MHZ and fM=.950, 5sola