PROBLEM 3 Fig. 3 shows a scheme for coherent (synchronous) demodulation. Show th

Question

PROBLEM 3 Fig. 3 shows a scheme for coherent (synchronous) demodulation. Show that this scheme can demodulate the AM signal [A+mt)cos(t) regardless of the value of A. A cos Low-pass de blocker Figure 3: Block Diagram of Problem 3 PROBLEM 4 In the early days of radio, AM signals were demodulated by a crystal detector followed by a lowpass filter and a DC blocker, as shown in Fig. 4. Assume a crystal detector to be basically a squaring device. Determine the signals at points a, b, c, and d. Point out the distortion term in the output ytt). Show that if AIm)I the distortion is small xit) de block filter Figure 4: Block Diagram of Problem 4 PROBLEM 5 A modulating signal m(t) is given by (a) m)cos(I00) (b) m) cos(1001) +2 cos(300) (c) mt) cos(100r)cos(500r). In each case: i) Sketch the spectrum of m(t) (i) Find and sketch the spectrum of the DSB-SC signal 2m(t) cos(1000r). iii) From the spectrum obtained in (i) suppress the LSB spectrum to obtain the USB spectrum. (iv) Knowing the USB inii, write the expression uss(t) for the USB signal. (v) Repeat (iii) and (iv) to obtain the LSB signal ?1SB(t). PROBLEM 6* Find ?LSB (1) and Pusa(1) for the modulating signal m(t)-B sinc(2nBI) with B-1000 and carrier frequency 10,000:T (a) Sketch spectra of m(t) and the corresponding DSB-SC signal 2m(t)cos(wt) (b) To find the LSB spectrum, suppress the USB in the DSB-SC spectrum found in (a). (c) Find the LSB signal ?1Sa(1), which is the inverse Fourier transform of the LSB spectrum found in part (b) cuc Follow a similar procedure to find ?USB(1).

Explanation / Answer

you have posted multiple independent questions as chegg Question Answer (QA) policy expert have to answer only first question completely so I am providing answer for first question only.

after product modulator we will get

=((A+m(t))*(cosWc*t))*cos(Wc*t) = ((A+m(t))*(1+cos2Wc*t))/2

=(A+m(t))/2)+((A+m(t))*(cos2Wc*t))/2

when above signal passes through LPF it will block the high frequencies which is 2Wc

means it will block the second signal that is ((A+m(t)) *(cos2Wc*t)) /2

so we will get only (A+m(t))/2 = (A/2)+(m(t))/2 after LPF

NOW dc blocker will block the dc component present in the signal which A/2

so after dc blocker we will left the message signal only that is (m(t)/2

so finally we will get (m(t))/2 which our original message signal

now if you use amplifier of gain of 2 you will get m(t).

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