hw8_solution.pdf

# Solution based on the differential equation we can

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Solution: Based on the differential equation, we can find the transfer function H ( s ) = - s + 2 s 2 + 5 s + 4 . 2

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Using partial fraction method, we can rewrite H ( s ) = - s + 2 ( s + 1)( s + 4) = A s + 1 + B s + 4 = ( A + B ) s + (4 A + B ) ( s + 1)( s + 4) . Based on this, we can get A = 1 , B = - 2 and the transfer function can be written as H ( s ) = 1 s + 1 - 2 s + 4 . Alternatively, you can use: A = H ( s )( s + 1) | s = - 1 = 3 / 3 = 1 B = H ( s )( s + 4) | s = - 4 = 6 / ( - 3) = - 2 Based on that the system is causal (right-sided ROC as shown in Figure 4) and the LT table, Figure 4: The ROC of H ( s ). the impulse response of the system is h ( t ) = e - t u ( t ) - 2 e - 4 t u ( t ) . Given that the axis is contained in the ROC, the system is stable . The DC gain H (0) = 1 2 . 3. You are given two causal systems in series, where the first system has H 1 ( s ) = 2 s s 2 + 5 s + 6 and the second system is described by the differential equation y ( t ) = 4 x ( t ) + dx ( t ) /dt . Use Laplace transforms to find the step response of the system. Solution: Based on the differential equation of the 2nd system, the corresponding transfer function is H 2 ( s ) = s + 4 . Given the two system is in series, the overall system transfer function H ( s ) = H 1 ( s ) H 2 ( s ) = 2 s ( s + 4) ( s + 2)( s + 3) . 3
For the step response, the input is u ( t ), which has LT 1 /s , so the Laplace transform of the output s ( t ) is S ( s ) = 1 s H ( s ) = 2( s + 4) ( s + 2)( s + 3) . Again, using the partial fraction method, we can reformulate S ( s ) = 2( s + 4) ( s + 2)( s + 3) = A ( s + 2) + B ( s + 3) where A = H ( s )( s + 2) | s = - 2 = 2(2) (1) = 4 B = H ( s )( s + 3) | s = - 3 = 2(1) ( - 1) = - 2 Alternatively, you can solve for A and B , using 2( s + 4) ( s + 2)( s + 3) = A ( s + 2) + B ( s + 3) = ( A + B ) s + (3 A + 2 B ) ( s + 2)( s + 3) .
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• Fall '09
• LTI system theory, Impulse response

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