TimeResponseLecture

# TimeResponseLecture - ME 279 Automatic Control of Dynamic...

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Time Response ME 279 Automatic Control of Dynamic Systems Dr. Robert G. Landers

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Definition 2 Time Response – response of system’s output(s) as a function of time given the input(s) Quantitative (exact) Solution – use inverse Laplace transform or other technique Qualitative Understanding – examine poles and zeros of Transfer Function (TF) Poles – values of s that cause the TF to be infinite. Roots of the TF denominator that are roots of the TF numerator Zeros – values of s that cause the TF to be zero. Roots of the TF numerator that are roots of the TF denominator Time Response Dr. Robert G. Landers
Example 3 A dynamic system is given to the right where U(s) is the input, Y(s) is the output, and G(s) is the transfer function. Determine y(t) for a unit step input and plot the pole–zero map of Y(s). Time Response Dr. Robert G. Landers When the input is a unit step input U(s) = 1/s and Y(s) is ( 29 ( 29 ( 29 ( 29 ( 29 1 8 1 9 1 1 1 3 4 12 3 12 4 12 s Y s G s U s s s s s s s +   = = = - + ÷ ÷  ÷ + + + +   Taking the inverse Laplace transform ( 29 3 4 8 9 1 12 12 12 t t y t e e - - = - + G(s) U(s) Y(s) ( 29 ( 29 ( 29 1 3 4 s G s s s + = + +

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Example 4 Time Response Dr. Robert G. Landers Imaginary -1 -2 -3 -4 Real The pole–zero map is given to the right The poles are denoted by crosses The zero is denoted by a circle The pole located at –3 contributes the (8/12)e –3t term The pole located at –4 contributes the –(9/12)e –4t term The pole located at 0 contributes the 1/12 term The zero located at –1 contributes to the magnitude of all three terms
Example 1 5 Time Response Dr. Robert G. Landers A dynamic system is given below where U(s) is the input, Y(s) is the output, and G(s) is the transfer function. Determine y(t) for a unit impulse input and plot the pole–zero map of Y(s). G(s) U(s) Y(s) ( 29 2 5 8 25 s G s s s + = + +

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Example 1 6 Time Response Dr. Robert G. Landers When the input is a unit impulse input U(s) = 1 and Y(s) is ( 29 ( 29 ( 29 ( 29 2 2 2 2 2 2 2 5 5 4 1 3 8 25 3 4 3 4 3 4 3 s s s Y s s s s s s + + + = = = + + + + + + + + + Taking the inverse Laplace transform ( 29 ( 29 ( 29 4 4 1 cos 3 cos 3 3 t t y t e t e t - - = +
Example 1 7 Time Response Dr. Robert G. Landers The pole–zero map is given to the right The poles are denoted by crosses The zero is denoted by a circle The poles located at –4±3j contribute the solution e –4t [cos(3t)+(1/3)sin(3t)] The zero located at –5 contributes to the magnitude of both terms Imaginary -4 -5 -3 Real 3

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First Order Systems 8 Time Response Dr. Robert G. Landers ( 29 ( 29 ( 29 y t y t Ku t τ + = General model of first–order system with output y(t) and input u(t) ( 29 ( 29 ( 29 0 A sY s y Y s K s - + = Let u(t) = A for all t > 0. Taking Laplace transform of both sides ( 29 ( 29 ( 29 0 1 1 y KA Y s s s s = + + + Solving for Y(s) ( 29 ( 29 / / 0 t t y t y e KA KAe - - = + - Taking inverse Laplace transform
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## This note was uploaded on 02/01/2012 for the course MECH ENG 279 taught by Professor Landers during the Spring '11 term at Missouri S&T.

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TimeResponseLecture - ME 279 Automatic Control of Dynamic...

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