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ECE2260F10_HW6p1soln

C we use the initial value theorem lim v t lim sv

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Unformatted text preview: st that nothing is known about the signal before time zero. c) We use the initial value theorem: lim v ( t ) = lim sV ( s) t →∞ s→ 0 or 10 s2 + 4 lim v ( t ) = lim s 3 2 s→ 0 s + s + s t →∞ We cancel a factor of s from the top and bottom and substitute s = 0. 10 s2 + 4 lim v ( t ) = lim 2 =4 t →∞ s→ 0 s + s + 1 d) The poles are roots of the denominator, and the zeros are the roots of the numerator. s2 − s − 6 ( s − 3)( s + 2) V ( s) = 3 = s + 6 s2 + 34 s s( s + 3 + j 5)( s + 3 − j 5) The zeros are 3 and –2. The poles are 0, –3–j5, –3+j5. We denote the poles with x 's and the zeros with o 's in the complex s-plane: Im –3+j5 Re –2 –3–j5 0 3...
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