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HW2sol

# HW2sol - Homework 2 Solution Solutions by Jiachen Xue and...

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Homework 2 Solution Solutions by Jiachen Xue and Ramamurthy Mani Problem 5 (a) H ( S ) can be expressed as H ( s ) = K Q m j =1 ( s - z j ) Q n i =1 ( s - p i ) = n X i =1 K i s - p i (1) where z j is the j th zero and p i is the i th pole of the transfer function. K i is the coefficient. Take the inverse Laplace transformation of H ( s ) to get the impulse response. h ( t ) = L - 1 [ H ( s )] = L - 1 " n X i =1 K i s - p i # = n X i =1 K i e p i t (2) If p i < 0, that is to say there is a pole existing on the right half plane, then lim t →∞ | K i e p i t | → ∞ . (3) Thus Z -∞ | h ( t ) | dt → ∞ . (4) which makes the system unstable. (b) Assume H ( s ) has zeros z 1 , z 2 on the right plan: z 1 = σ j + j z 2 = σ j - j in which σ j > 0 and ω j 6 = 0. H ( s ) can be further expended as H ( s ) = H min ( s ) ( s - σ j - j ) ( s - σ j + j ) = H min ( s ) £ ( s - σ 2 j ) + ω 2 j / (5) where H min ( s ) has all its zeros on the left plane. Thus, H ( s ) can be written as H ( s ) = £ H min ( s ) ( ( s + σ 2 j ) + ω 2 j )/ ( s - σ 2 j ) + ω 2 j ( s + σ 2 j ) + ω 2 j = H 1 ( s ) A ( s ) (6) in which H 1 ( s ) = H min ( s ) ( ( s + σ 2 j ) + ω 2 j )

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HW2sol - Homework 2 Solution Solutions by Jiachen Xue and...

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