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hw2sol

# hw2sol - Instructor Christopher Wayne Walker Ph.D TA...

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Instructor: Christopher Wayne Walker, Ph.D. TA: Federico Cattivelli EE132A Winter 2008 EE 132A Homework 2 Solutions Problem 1. Proakis & Salehi 3.1. Solution: The modulated signal is u ( t ) = m ( t ) c ( t ) = Am ( t ) cos(2 π 4 × 10 3 t ) = A bracketleftbigg 2 cos(2 π 200 π t ) + 4 sin(2 π 250 π t + π 3 ) bracketrightbigg cos(2 π 4 × 10 3 t ) = A cos(2 π (4 × 10 3 + 200 π ) t ) + A cos(2 π (4 × 10 3 - 200 π ) t ) +2 A sin(2 π (4 × 10 3 + 250 π ) t + π 3 ) - 2 A sin(2 π (4 × 10 3 - 250 π ) t - π 3 ) Taking the Fourier transform of the previous relation, we obtain U ( f ) = A bracketleftbigg δ ( f - 200 π ) + δ ( f + 200 π ) + 2 j e j π 3 δ ( f - 250 π ) - 2 j e - j π 3 δ ( f + 250 π ) bracketrightbigg * 1 2 [ δ ( f - 4 × 10 3 ) + δ ( f + 4 × 10 3 )] = A 2 bracketleftbigg δ ( f - 4 × 10 3 - 200 π ) + δ ( f - 4 × 10 3 + 200 π ) +2 e - j π 6 δ ( f - 4 × 10 3 - 250 π ) + 2 e j π 6 δ ( f - 4 × 10 3 + 250 π ) + δ ( f + 4 × 10 3 - 200 π ) + δ ( f + 4 × 10 3 + 200 π ) +2 e - j π 6 δ ( f + 4 × 10 3 - 250 π ) + 2 e j π 6 δ ( f + 4 × 10 3 + 250 π ) bracketrightbigg Figure 1 depicts the magnitude and the phase of the spectrum U(f). Figure 1: Figure of problem 1 1

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Instructor: Christopher Wayne Walker, Ph.D. TA: Federico Cattivelli EE132A Winter 2008 Problem 2. Proakis & Salehi 3.5. Solution: u ( t ) = m ( t ) · c ( t ) = 100[2 cos(2 π 2000 t ) + 5 cos(2 π 3000 t )] cos(2 πf c t ) U ( f ) = 100 2 bracketleftbigg δ ( f - 2000) + δ ( f + 2000) + 5 2 ( δ ( f - 3000) + δ ( f + 3000)) bracketrightbigg * [ δ ( f - 50000) + δ ( f + 50000)] = 50 bracketleftbigg δ ( f - 52000) + δ ( f + 48000) + 5 2 δ ( f - 53000) + 5 2 δ ( f + 47000) + δ ( f + 52000) + δ ( f + 48000) + 5 2 δ ( f + 53000) + 5 2 δ ( f + 47000) bracketrightbigg A plot of the spectrum of the modulated signal is given in Figure 2. Figure 2: Figure of problem 2 Problem 3. Proakis & Salehi 3.6. Solution: The mixed signal y(t) is given by y ( t ) = u ( t ) · x L ( t ) = Am ( t ) cos(2 πf c t ) cos(2 πf c t + θ ) = A 2 m ( t )[cos(2 π 2 f c t + θ ) + cos( θ )] The lowpass filter will cut-off the frequencies above W, where W is the bandwidth of the message signal m(t). Thus, the output of the lowpass filter is z ( t ) = A 2 m ( t ) cos( θ ) If the power of m(t) is P M , then the power of the output signal z(t) is P o ut = P M A 2 4 cos 2 ( θ ). The power of the modulated signal u ( t ) = Am ( t ) cos(2 πf c t ) is P u = A 2 4 cos 2 ( θ ) . Hence, P out P u = 1 2 cos 2 ( θ ) A plot of P out P u for 0 θ π is given in Figure 3. Problem 4. Proakis & Salehi 3.15. Solution: 1) The modulated signal is written as u ( t ) = 100(2 cos(2 π 10 3 t ) + cos(2 π 3 × 10 3 t )) cos(2 πf c t ) = 200 cos(2 π 10 3 t ) cos(2 πf c t ) + 100 cos(2 π 3 × 10 3 t ) cos(2 πf c t ) = 100[cos(2 π ( f c + 10 3 ) t ) + cos(2 π ( f c - 10 3 ) t )] +50[cos(2 π ( f c + 3 × 10 3 ) t ) + cos(2 π ( f c - 3 × 10 3 ) t )] 2
Instructor: Christopher Wayne Walker, Ph.D. TA: Federico Cattivelli EE132A Winter 2008 0 0.5 1 1.5 2 2.5 3 3.5 0 0.1 0.2 0.3 0.4 0.5 Theta (rad) Pout/Pu Figure 3: Figure of problem 3 Taking the Fourier transform of the previous expression, we obtain U ( f ) = 50 bracketleftbigg δ ( f - ( f c + 10 3 )) + δ ( f + f c + 10 3 ) + δ ( f - ( f c - 10 3 )) + δ ( f + f c - 10 3 ) bracketrightbigg +25 bracketleftbigg δ ( f - f c - 3 · 10 3 ) + δ ( f + f c + 3 · 10 3 ) + δ ( f - f c + 3 · 10 3 ) + δ ( f + f c - 3 · 10 3 ) bracketrightbigg The spectrum of the signal is depicted in Figure 4.

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