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ECE2260F09_HW6p1soln

# ECE2260F09_HW6p1soln - 2260 F 09 HOMEWORK#6 prob 1 solution...

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2260 F 09 HOMEWORK #6 prob 1 solution E X : a) Find L e 5 t ( t 2) u ( t 2) { } . b) Find v ( t ) if V ( s ) = 1 s + 31 s + 39 s 2 + 2 s + 17 . c) Find lim t 0 + v ( t ) if V ( s ) = 23 s 2 4 29 s 3 + 5 s 2 + 37 . d) Plot the poles and zeros of V ( s ) in the s plane. V ( s ) = s 2 + 14 s + 45 ( s 2 + 49)( s + 19) S OL ' N : a) We may transform the delayed signals first and, later, add the exponential. We use the delay identity: L { f ( t a ) u ( t a )} = e as L { f ( t )} Here, this applies to the portion obtained by omitting the exponential: L e 5 t ( t 2) u ( t 2) { } = e 2 s L tu ( t ) { } = e 2 s L t { } = e 2 s s 2 Now, we include the exponential and apply the corresponding identity: L { e at f ( t )} = F ( s + a ) This means we replace s with s + 5 in our previous result to obtain our final answer: L e 5 t ( t 2) u ( t 2) { } = e 2( s + 5) ( s + 5) 2 An alternative approach is to write the exponential as a delayed function and use only the delay identity: L e 5 t ( t 2) u ( t 2) { } = L e 10 e 5( t 2) ( t 2) u ( t 2) { } After taking the constant term, e 10

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