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Unformatted text preview: 3.3 Effects of Pole Locations For a rational transfer function H ( s ), it can be written as H ( s ) = b ( s ) a ( s ) The roots of b ( s ) = 0 are called the zeros of the system. The roots of a ( s ) = 0 are called the poles of the system. The poles and zeros of the system determine the behavior of the system response. We say that the system is stable if all poles are in the open left half plane, unstable if there is any poles in the open right half plane, marginally stable (or unstable) if there is any poles on the imaginary axis. 1 1st order plant: H ( s ) = 1 s + σ , σ > 0. Its impulse response is given by h ( t ) = L 1 [ H ( s )] = e σt 1 ( t ) . If σ < 0: unstable. The impulse response for σ > 0 is shown below for σ = 1 and it decays with a time constant τ = 1 /σ . h ( t ) 1.0 0.8 0.6 0.4 0.2 1.0 2.0 3.0 4.0 Time (sec) t H11005 t e H11002 s t 1 e 2 Time (sec) 1 3 4 0.2 0.4 0.6 0.8 1 h ( t ), y ( t ) (b) y h (a) The unit step response Y ( s ) = H ( s ) /s is shown in Figure (b). 2 Example 3.22: H ( s ) = 2 s + 1 s 2 + 3 s + 2 = 2( s + 1 / 2) ( s + 1)( s + 2) z = 1 / 2, p 1 = 1, p 2 = 2. Using partial fractional expansion, H ( s ) = 1 s + 1 + 3 s + 2 . Thus h ( t ) = ( e t + 3 e 2 t ) 1 ( t ). 1 H11002 1 H11002 2 H11002 j j Im( s ) Re( s ) H11005 Zero H11005 Pole >> num=[2,1], den=[1,3,2], pzmap(num, den) Im( s ) Re( s ) LHP RHP STABLE UNSTABLE The shape of the components of h(t), e t and e 2 t are determined by the locations of the poles. 3 For small t > 0, e t ≈ 1 and e 2 t ≈ 1, hence 3 e 2 t dominates the response because of the larger coefficient. As t becomes large, 3 e 2 t decays to zero a lot faster than e t does because of the smaller time constant. Hence e t dominates the response for large t . The impulse response plot below verifies the above analysis. h ( t ) 2.0 1.5 1.0 0.5 H11002 0.5 6 5 4 3 2 1 Time (sec) 4 2nd order plant with complex roots: (prototype) H ( s ) = ω 2 n s 2 + 2 ζω n s + ω 2 n = ω 2 n ( s + σ ) 2 + ω 2 d , < ζ < 1 ω n : undamped natural frequency ζ : damping ratio Note that s 2 + 2 ζω n s + ω 2 n = ( s + σ jω d )( s + σ + jω d ) with ω d = ω n √ 1 ζ 2 ( damped frequency ), σ = ζω n . Im( s ) Re( s ) u H11005 sin H11002 1 z v n v d s The two poles lie on LHP located at σ ± jω d . The angle θ is given by θ = sin 1 ζ . The smaller the θ angle, the closer the poles to the jω axis. The value of σ measures the distance of the two poles to the jωaxis. The impulse response is h ( t ) = ω n √ 1 ζ 2 e σt sin( ω d t ) 1 ( t ) . 5 (b) 0.9 z H11005 1 2 4 6 8 10 12 1.0 0.8 0.6 0.4 0.2 0.0 H11002 0.2 H11002 0.4 H11002 0.6 H11002 0.8 H11002 1.0 y ( t ) v n t (a) 2 4 6 8 10 12 v n t 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 y ( t ) z H11005 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 z H11005 0.1 0.2 0.3 0.4 0.5 0.6 z H11005 0.7 0.8 0.9 1.0 6 3.4 TimeDomain Specifications A typical step response is shown below: t M p...
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This note was uploaded on 02/03/2012 for the course EE 3530 taught by Professor Chen during the Fall '07 term at LSU.
 Fall '07
 Chen

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