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132A_1_Midterm-W07_sol

# 132A_1_Midterm-W07_sol - UCLA — Electrical Engineering...

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Unformatted text preview: UCLA — Electrical Engineering Dept. EE132A: Communication Systems Midterm Exam — Solutions 1. An FM modulator operates with frequency sensitivity k f = 100 Hz/V, carrier amplitude A c = 10 V, and carrier frequency f c = 100 kHz. The input message is m ( t ) = A m cos(2 πf m t ) , where f m = 1 kHz. The modulator is followed by a bandpass filter with bandwidth of 3 kHz, centered around f = f c . Let x ( t ) be the filter output. (a) i. Write the time-domain expression for the signal x ( t ) when A m = 24 V. Write it as a summation of terms of the type cos(2 π ( f c + nf m ) t ) . Here, β = k f A m /f m = 2400 / 1000 = 2 . 4. The signal at the input of the filter is y ( t ) = ∑ ∞ n =-∞ A c J n ( β )cos(2 π ( f c + nf m ) t ). Only the cosines that fit in the filter bandwidth pass the filter, so only the terms corresponding to n =- 1 , , 1 survive. Therefore x ( t ) = A c J ( β )cos(2 πf c t )+ A c J- 1 ( β )cos(2 π ( f c- f m ) t )+ A c J 1 ( β )cos(2 π ( f c + f m ) t ) = A c J ( β )cos(2 πf c t )- A c J 1 ( β )cos(2 π ( f c- f m ) t ) + A c J 1 ( β )cos(2 π ( f c + f m ) t ) . In this case, J (2 . 4) = 0, so we have that x ( t ) =- A c J 1 ( β )cos(2 π ( f c- f m ) t ) + A c J 1 ( β )cos(2 π ( f c + f m ) t ) . ii. Plot the amplitude spectrum of x ( t ) . We only have two deltas at f = f c + f m and f = f c- f m , both with amplitude (1 / 2) A c J 1 ( β ) (same for the negative frequencies)....
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132A_1_Midterm-W07_sol - UCLA — Electrical Engineering...

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