ECE405HW2Su2010Sol

# ECE405HW2Su2010Sol - ECE 405 HOMEWORK #2 Solutions SU 2010...

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ECE 405 HOMEWORK #2 Solutions SU 2010 ©DR. JAMES S. KANG 1 (a) F[sinc 2 (ct)] = (1/c) tri(f/c). M(f) = F[sinc 2 (t)] = tri(f/1). -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 0 0.5 1 1.5 2 f M(f) (b) AM: -12 -11 -10 -9 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 0 0.5 1 1.5 f Sam(f) DSBSC -12 -11 -10 0 1 2 3 4 5 6 7 8 9 10 11 12 0 0.5 1 1.5 f Sdsb(f) LSSB -12 -11 -10 0 1 2 3 4 5 6 7 8 9 10 11 12 0 0.5 1 1.5 f Slssb(f) 1

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USSB -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 0 0.5 1 1.5 f Sussb(f) (c) AM -1.5 -0.5 0 0.5 1 1.5 2 0 1 2 3 t sam(t) DSBSC 0 1 2 3 4 5 -1.5 -0.5 0 0.5 1 1.5 2 t sdsbsc(t) 2 (a) () ( ) 53 1 c o s2 1 0 c o 0 c o 0 mt t t t ππ π ×= × × 5 11 cos 2 10 10 cos 2 10 10 22 tt ⎡⎤ + ⎣⎦ ( ) 55 2 cos 2 10 cos 2 4,000 cos 2 10 t t t × × cos 2 100,000 4,000 cos 2 100,000 4,000 + 2
() 11 ( ) cos 2 100,000 1,000 cos 2 100,000 4,000 22 y tt ππ ⎡⎤ + ⎣⎦ t (2 99,000 ) cos(2 104,000 ) + × (b) y(t) LPF at 100 kHz HPF at 100 kHz LPF LPF Oscillator cos(2 π × 10 5 t) m 1 (t) m 2 (t) 3 (a) 1 0 1 2 –2 –1 f (kHz) X(f) (b) 1 0 1 –1 f (kHz) Y(f) 4 0 f (kHz) Y(f) 15 20 25 75 80 85 –20 –80 0 f (kHz) Z(f) 15 20 25 –20 5 3

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(a) f (kHz) M s (f) 51 0 15 20 14 6 –5 –10 –15
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## This note was uploaded on 08/09/2010 for the course ECE ECE 405 taught by Professor Jameskang during the Summer '10 term at Cal Poly Pomona.

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ECE405HW2Su2010Sol - ECE 405 HOMEWORK #2 Solutions SU 2010...

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