ECE108HW%231sol_06

ECE108HW%231sol_06 - ECE 108 HW#1 Solutions Due Text...

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ECE 108 HW #1 --- Solutions Due 1/30/06 Text: Oppenheim and Willsky, 2 nd Edition 1. Determine the values of and for each of the following signals P E (a) () xt t = 33 22 2 2 2 2 31 2 TT T T T tT Ex t d t t d t −− == = = ∫∫ Then lim T T E →∞ =∞ Also 2 lim 12 T T ET PP T →∞ = This is neither an energy or power signal (b) () ( ) () cos t xt e tut π = 2 0 2 cos t T T t d t e t d t But 2 2 2 cos 24 jt ee e e t ππ  ++ +   Substituting into the above expression for T E
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2 () 22 2 2 2 2 00 0 222 21 2 000 2 2 0 11 2 44 4 112 444 2 81 8 1 2 8 8 TT T tj t t j t t T TTT jt t T t t t jT T E e e d te e d d t ed t t e d t ee e jj e ππ −− −+ =+ + =++ =−  + + +   ∫∫ ∫∫∫ Then 2 2 2 lim 2 2 1 8 14 1 T T EE π →∞ ++ + == + = + = lim 0 T T E PP P T →∞ =⇒ = = This is an Energy signal (c) 2 4 Here we observe 1 j xt e xt  +   2 T Ex t d t d t T = Then E =∞ 1 1 T T E T ⇒= This is a power signal
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3 (d) () 10sin 2 10cos 2 10 2 xn n n π ππ  =+ = =   22 2 100 100 NN N nn Ex n N =− =− == = ∑∑ Therefore lim 100 lim 100 N N N N N N EE E P N PP →∞ This is a power signal (e) () () xt jut = 2 0 2 2 lim 1 2 1 lim 2 TT T T T T T T T T T t d t d t E P T →∞ →∞ = ∫∫ This is a power signal
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4 (f) () () () 1 2 2 n n xn un  ==   1 22 2 2 1 00 0 2 2 11 1 1 44 4 24 13 4 1 4 lim 3 41 1 1 34 4 lim 0 N N NN N N n n n N N nn n n N N N N N N N Ex n EE E P PP + −− = =− →∞ = =    ∑∑ This is an energy signal 2. Textbook problem 1.10 ( ) ( ) ( ) 2cos 10 1 sin 4 1 xt t t =+ Calculate ( ) ( ) ( ) ( ) ()( ) ( ) ( ) ( ) 2cos 10 1 sin 4 1 2cos 10 1 10 sin 4 1 4 2cos 10 cos 10 1 2sin 10 sin 10 1 -sin 4 cos 4 1 cos 4 sin 4 1 xt T t T tT t T Tt T t T t += ++
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ECE108HW%231sol_06 - ECE 108 HW#1 Solutions Due Text...

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