ECE405HW3Su2010Sol

ECE405HW3Su2010Sol - ECE 405 HOMEWORK#3 Solutions SU 2010 DR JAMES S KANG 1 If x(t = A rect(t then x(t = A t ln t 2 2(a m(t =(b 1 000 t 3 t mUSSB t

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ECE 405 HOMEWORK #3 Solutions SU 2010 ©DR. JAMES S. KANG 1 If () ( / ) x tA r e c t t τ = , then 2 ˆ() ln 2 t A xt t π + = (a) 13 3 t mt t + = (b) 1,000 3 1,000 cos(2 1,000 ) ln sin(2 1,000 ) 63 USSB tt r e c t t t t ππ + ⎛⎞ ⎜⎟ ⎝⎠ × + t o t 2 ( ) 4 sin(12 ) cos(7 ) cos(3 ) 2 sin(12 )[cos(4 ) cos(10 )] t t t t t t == sin(8 ) sin(16 ) sin(2 ) sin(22 ) t =+ ++ . ˆ ( ) sin(8 90 ) sin(16 90 ) sin(2 90 ) sin(22 90 ) oo o t t t t =− + +− + cos(8 ) cos(16 ) cos(2 ) cos(22 ) t 3 –10 10 t m(t) 4 (a) . ( ) 1 22 2 1 1 ()cos ()1 cos 2 2 cc c v t Am t t t ωω c + 2 1 ()cos 2 2 vt Amt t ω = 2 1 cos 2 oc A t c = (b) ( ) ( ) ( ) 1 ˆˆ ( )cos ( )sin 2 ( ) ( )sin cos c m t t t mtmt t t c + 1
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() 22 11 ˆˆ ()1 cos 2 () ()s in 2 cc m t t t mtmt t c ω ωω ⎡⎤ =+ +− + ⎣⎦ () ( ) ( ) 2 2 cos ( ) 1 cos ( ) ( )sin v t t t t c −+ (c) yes for DSB-SC, no for SSB-SC. 5 2sin( )sin( ) cos( ) cos( ) x yx =− y ( ) cos(2,000 ) cos(12,000 ) mt t t ˆ ( ) sin(2,000 ) sin(12,000 ) t t (a) 77 ˆ ()cos10 ()s in10 LSSB st m t t m t t [ ] [ ] ( ) cos(2,000 ) cos(12,000 ) cos 10 sin(2,000 ) sin(12,000 ) sin 10 tt t t t + t ( ) cos(2,000 )cos 10 sin(2,000 )sin 10 t t ( ) cos(12,000 )cos 10 sin(12,000 )sin 10 t t ) ( cos 10 2,000 cos 10 12,000 =−−− (b) magnitude response 0 10 7 –2000
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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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ECE405HW3Su2010Sol - ECE 405 HOMEWORK#3 Solutions SU 2010 DR JAMES S KANG 1 If x(t = A rect(t then x(t = A t ln t 2 2(a m(t =(b 1 000 t 3 t mUSSB t

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