timeresponse10

# timeresponse10 - 29 29 29 cos 4 sin 4 y t A B t C t = Since...

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Time Response Q UESTION 10 For each of the transfer functions below, find the location of the poles and zeros and write an expression for the general form of the step response without solving for the inverse Laplace transform. Determine if the time response is stable, marginally stable, or unstable and determine if the time response is overdamped, underdamped, critically damped, or undamped. a. ( 29 2 5 3 18 T s s s = + - b. ( 29 2 2 16 T s s = + c. ( 29 2 5 2 100 T s s s = + + d. ( 29 ( 29 2 5 1 s T s s + = - For the transfer function ( 29 2 5 3 18 T s s s = + - , the poles are at 3 and –6, and there are no finite zeros. Since the poles are purely real, the time response is overdamped and the general time response to a step input is ( 29 3 6 t t y t A Be Ce - = + + . Since the real part of one of the poles is positive, the time response is unstable. For the transfer function ( 29 2 2 16 T s s = + , the poles are at –4 j and 4 j , and there are no finite zeros. Since the poles are purely imaginary, the time response is undamped and the general time response to a step input is

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Unformatted text preview: ( 29 ( 29 ( 29 cos 4 sin 4 y t A B t C t = + + . Since the poles are purely imaginary, the time response is marginally stable. Time Response For the transfer function ( 29 2 5 2 100 T s s s = + + , the poles are at 1 99 j- + and 1 99 j- -, and there are no finite zeros. Since the poles are complex, the time response is underdamped and the general time response to a step input is ( 29 ( 29 ( 29 ( 29 cos 99 sin 99 cos 99 t t t y t A Be t A Ce t De t φ---= + + = + + . Since the real parts of the poles are negative, the time response is stable. For the transfer function ( 29 ( 29 2 5 1 s T s s + =-, the poles are at 1 and 1, and there is a zero at –5. Since the poles are real and equal, the time response is critically damped and the general time response to a step input is ( 29 t t y t A Be Cte = + + . Since the real parts of the poles are positive, the time response is unstable. 2...
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timeresponse10 - 29 29 29 cos 4 sin 4 y t A B t C t = Since...

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