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1-4 - EE160 Fall 2009 San Jos State University e Solution...

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EE160 — Fall 2009 San Jos´ e State University Solution of Homework # 5 1. The modulated signal is u ( t ) = m ( t ) · c ( t ) = 100 [2 cos(2 π 2000 t ) + 5 cos(2 π 3000 t )] cos(2 πf c t ) . Therefore, U ( f ) = 50 [ δ ( f + 52000) + δ ( f + 48000) + 2 . 5 δ ( f + 53000) + 2 . 5 δ ( f + 47000) δ ( f - 52000) + δ ( f - 48000) + 2 . 5 δ ( f - 53000) + 2 . 5 δ ( f - 47000)] , which is sketched in the figure below: f (kHz) U(f) 0 47 48 52 53 -53 -52 -48 -47 50 50 50 50 125 125 125 125 2. A DSB-SC modulated signal is a bandpass signal. As discussed in class, it can be written as u ( t ) = m ( t ) cos(2 πf c t + φ ) = Re m ( t ) e j (2 πft + φ ) = m ( t ) cos(2 πf c t ) cos( φ ) - m ( t ) sin(2 πf c t ) sin( φ ) = u c ( t ) cos(2 πf c t ) - u s ( t ) sin(2 πf c t ) , where u c ( t ) = m ( t ) cos( φ ) , u s ( t ) = m ( t ) sin( φ ) . The envelope of u ( t ) is, therefore, V u ( t ) = u 2 c ( t ) + u 2 s ( t ) = m ( t ) 2 cos( φ ) 2 + m ( t ) 2 sin 2 ( φ ) = m 2 ( t ) = | m ( t ) | . 3. Write the modulated signal as u ( t ) = 20 1 + 1 2 cos(200 πt ) cos(2000 πt ) . (a) The modulating and carrier signals are, respectively m ( t ) = cos(2 π 100 t ) , c ( t ) = 20 cos(2 π 1000 t ) . (b) α = 1 2 . (c) The ratio of the power in the sidebands P s to the power in the carrier P c is P s P c = 400 α 2 / 2 400 / 2 = 50 200 = 1 4 .
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4. With a modulating signal m ( t ) = cos(2 πf
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