hw8_solution.pdf

# Because two systems are both causal the overall

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Because two systems are both causal, the overall system is still causal (intersection of right- sided ROCs is still right-sided ROC). So the step response is S ( s ) = 4 ( s + 2) - 2 s + 3 s ( t ) = 4 e - 2 t u ( t ) - 2 e - 3 t u ( t ) . 4. You are given that a causal, stable system has a transfer function of H ( s ) = ( s 2 + 1) ( s + 1)( s + 3) Find the inverse of the system expressed in terms of a transfer function and the differential equation that describes the inverse system. Assuming the inverse is causal, would it also be stable? Solution: The inverse system transfer function H I ( s ) = 1 H ( s ) = ( s + 1)( s + 3) s 2 + 1 = s 2 + 4 s + 3 s 2 + 1 . The corresponding differential equation is d 2 y ( t ) dt + y ( t ) = d 2 x ( t ) dt + 4 dx ( t ) dt + x ( t ) . Given the inverse system has poles s = ± j on the axis, the system is not stable . 5. Consider an LTI system with DC gain of 1, a zero at s = 1, and poles at s = - 1 , - 4 , 2. (a) Find H ( s ). Solution: Based on the poles and zeros, the transfer function should be in the form of H ( s ) = A ( s - 1) ( s + 1)( s + 4)( s - 2) . 4

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Because it has DC gain of 1, H (0) = - A - 8 = 1 A = 8 . So the transfer function H ( s ) = 8( s - 1) ( s + 1)( s + 4)( s - 2) . (b) Find the ROC that would correspond to a stable system. Solution: For stable system, the ROC must contain the axis. The ROC is shown in Figure 5. Figure 5: The ROC of stable H ( s ). (c) Find the ROC that would correspond to a causal system. Solution: For causal system, the ROC must be right-sided, as shown in Figure 6. Figure 6: The ROC of causal H ( s ).
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• Fall '09
• LTI system theory, Impulse response

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