RootLocusLecture - ME 279 Automatic Control of Dynamic...

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Root Locus ME 279 Automatic Control of Dynamic Systems Dr. Robert G. Landers
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Definition 2 Root Locus Dr. Robert G. Landers Root Locus: graphical method to analyze the transient performance and stability of a closed–loop system as a parameters varies. E(s) + - R(s) Y(s) ( 29 C G s K = ( 29 ( 29 ( 29 G G N s G s D s = The forward loop transfer function is ( 29 ( 29 ( 29 ( 29 G G Y s N s K E s D s = The open–loop poles are not affected by K. The closed–loop transfer function is ( 29 ( 29 ( 29 ( 29 1 Y s KG s R s KG s = + The Root Locus will graphically show how the closed–loop poles change as K varies from 0 to ∞.
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Example: Motor Angular Position Control 3 Root Locus Dr. Robert G. Landers The block diagram of a closed–loop motor angular position control system is shown below. The reference motor angular position signal is R(s) and the motor angular position signal is Y(s). R(s) + - Y(s) Controller System K ( 29 1 10 s s + The closed–loop transfer function is ( 29 ( 29 2 10 Y s K R s s s K = + + The closed–loop transfer equation is s 2 + 10s + K = 0. K pole locations 0 -10 0 -2.76 -7.24 20 -5 -5 25 -5+3.87j -5-3.87j 40 -5+5j -5-5j 50
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Example: Motor Angular Position Control 4 Root Locus Dr. Robert G. Landers Im Re 0 0 20 20 50 50 40 40 25 5 -5 -5 -10 0 0 0.5 1 1.5 2 2.5 3 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 time (s) output y 1 (t): K = 15 y 2 (t): K = 25 y 3 (t): K = 100 y 4 (t): K = 1000 Output plots for four values of K and r(t) = 1 are given below. The real parts versus the imaginary parts of five closed–loop poles are plotted below.
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Vector Representation of Complex Numbers 5 Root Locus Dr. Robert G. Landers Consider a complex function F(s) = s + a. Let s = σ + j ϖ , where j is the imaginary number. In this case F(s) = ( σ +a) + j ϖ . The function s + a can be represented as a vector drawn from the zero of the function to the point s. The vector has a magnitude M and angle θ . ( 29 ( 29 2 2 1 tan M a a ϖ σ θ - = + + = ÷ + Im Re ϖ M -a σ θ σ +j ϖ
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Vector Representation of Complex Numbers 6 Root Locus Dr. Robert G. Landers A general representation of a transfer function is ( 29 ( 29 ( 29 1 1 m i i n j j s z F s s p = = + = + ( 29 ( 29 ( 29 1 1 m i i n j j s z M s s p = = + = + ( 29 ( 29 ( 29 1 1 m n i j i j s s z s p θ = = = + - + The magnitude is The angle is
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Example: TF Magnitude and Angle 7 Root Locus Dr. Robert G. Landers Im Re 1 -1 4 Calculate F(s) for s = –1 + 4j. ( 29 ( 29 1 1 s F s s s - = + 2 2 1 2 2 4 20 90 tan 116.6 4 M -   = + = ∠ = ° + = °  ÷   The magnitude and angle, respectively, of the zero at s = 1 are 2 2 1 1 1 4 17 90 tan 104.0 4 M -   = + = ∠ = °+ = °  ÷   The magnitude and angle, respectively, of the pole at s = 0 are
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Example: TF Magnitude and Angle 8 Root Locus Dr. Robert G. Landers 2 2 0 4 4 90 M = + = ∠ = ° The magnitude and angle, respectively, of the pole at s = –1 are 20 0.2712 4 17 M = = The magnitude is 116.6 104.0 90 77.4 θ= °- ° - ° = - ° The angle is
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RootLocusLecture - ME 279 Automatic Control of Dynamic...

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