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Chapter 3Design of Discrete-Time control

systemsRoot Locus

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Outline

1 Complex variables2 Characteristic equation3 example

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Complex variables

s=+j

M=22

s=M∡

=tan−1

When

s=M e j

e j =cos j sin

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Vector representation of complex numbers : s = + j

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Vector representation of complex numbers : (s + a)

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Alternate representation of : (s + a)

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Vector representation of complex numbers : (s + 7)|s5 + j2

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Characteristic equation

T s =KG s

1KG s H s

R(s) E(s) C(s)KG(s)

H(s)

+-

1KG s H s=0Char. Eq.

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1+KG(s)H(s) = 0

KG(s)H(s) = -1

Characteristic Equation

G(s)H(s) is a complex function. We have :

KG s H s =M∡=12k1

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-a-b-c123

s1

js-plane

123=180°

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∡KG s H s =2k1

∣KG sH s∣=1

K=1

∣G s∣∣H s ∣

Value of K :

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G s =K∏i=1

m

sz i

∏j=1

n

sp i

=K∣sz1∣∣sz2∣...∣szm∣

∣sp1∣∣sp2∣...∣spn∣e j 1...m

e j 1...n

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Example (Nise)

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Vector representationof G(s) at -2+ j3

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12−3−4=56.31°71.57°−90°−108.43°

=−70.55°

Point -2+j3 is not on root locus.

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Point -2+j2/2 is on the Root Locus. We have 180 degree of angle of Ch. Eq.

For the gain

K=1

∣G sH s∣=

1M= pole lengths zero lengths

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G s=K s3s4s1s2

Example

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Complete root locus for the system example

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Example (Nise)

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Root locus and asymptotes for the system example

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Example (Nise) Sketch the root locus for the system below.

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T s =K s3

s 47s314s28K s3K

Closed-loop transfer function is

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Complex number

s47s314s28K s3K=0

Characteristic Equation

Let s= j

4− j73−142 j 8K 3K=0

4−1423K j −738K =0

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4−1423K=0

−738K =0

K=9.65 = ±1.59

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s4 1 14 3Ks3 7 8K

s2 90−K 21K

s1 −K 2−65K72090−K

s0 21K

Routh table

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From s1 row

−K 2−65K720=0

K = 9.65

Form s2 row with K=9.65

90−K s221K=80.35s2202.7=0

S = ±j1.59

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Example (Nise)

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Root locus for example showing angle of departure

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Plotting and Calibrating the Root Locus

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Transient Response Design via Gain Adjustment

The conditions justifying a second-order approximation are :

1 High order poles are much farther into the left half of the s-plane than the dominant second- order pair of poles. The response that results from a higher-order pole does not appreciably change the transient response expected from the dominant second-order poles.

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2 Closed-loop zeros near the closed-loop second-order pole pair are nearly canceled by the close proximity of higher-order closed- loop poles.3 Closed-loop zeros not canceled by the close proximity of higher-order closed-loop poles are far removed from the closed-loop second- order pole pair

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Making second-order approximations

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Design procedure for higher-order system

1 Sketch the root locus for the given system2 Assume the system is a second-order system without any zeros and then find the gain to meet the meet the transient response specification.3 Justify the second-order assumption by finding the location of all higher-order poles and evaluating the fact that they are much farther from the j-axis than the dominant second-order pair.

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4 If the assumptions cannot be justified, you solution will have to be simulated in order to be sure it meets the transient response specification.

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Example (Nise) Design the value of gain K, to yield 1.52% overshoot for the system

below. Also estimate the settling time, peak time, and steady-state error.

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Root locus of the system in slide 54

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Second- and third-order responses for case 2

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Second- and third-order responses for case 3