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practice quiz 2 solutions

practice quiz 2 solutions - EE 341 Quiz 2 Solutions 1(a i(t...

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EE 341 Quiz 2 Solutions 1. (a) ω i ( t ) = ω c + 8000 π cos 2000 πt , B = 2000 π/ 2 π = 1000 Hz, Δ f = 8000 π/ 2 π = 4000 Hz. B EM = 2(Δ f + B ) = 2(4000 + 1000) = 10 kHz. (b) This signal can be either an FM or a PM signal. If it is a PM signal, then k p m ( t ) = 4 sin 2000 πt . For example, k p = 4 and m ( t ) = sin 2000 πt . If it is an FM signal, then k f m ( t ) = 8000 π cos 2000 πt . For example, k f = 8000 π and m ( t ) = cos 2000 πt . (c) The average transmit power is P T = 100 2 / 2 = 5000 W. 2. (a) Using the trigonometric relationship, 2 cos 2000 πt cos 300 πt = cos 1700 πt + cos 2300 πt , x ( t ) = 20 cos 2000 πt + 10 cos 2000 πt cos 300 πt = [20 + 10 cos 300 πt ] cos 2000 πt (1) so that m ( t ) = 10 cos 300 πt . (b) Using the expression for x ( t ) in the problem statement, the transmit power is P T = [5 2 +20 2 +5 2 ] / 2 = 225 W. Alternatively, using the form in (1), we showed in class that P T = [ A 2 + S m ] / 2 where A = 20 and S m = 10 2 / 2 = 50 is the average message power. P T = [20 2 + 50] / 2 = 225 W. 3. (a) m ( t ) = 4 cos 400 πt , M ( f ) = 2[ δ ( f - 200)+ δ ( f +200)], X dsb ( f ) = 2[ δ ( f - 1200)+ δ ( f +1200)+ δ ( f - 800)+ δ ( f - 800)]. From the plot of X usb ( f ) = 2[ δ ( f - 1200)+ δ ( f +1200)] in Fig P3, x ( t ) = 4 cos 2400 πt . (b) The time domain expression for the USB is
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