This document covers Root locus Example

1. By Prof. Hitesh Dholakiya
Root Locus
Example
Control Engineering
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2. Root Locus plot
Asymptotes
Angle of Departure
Break away point
Outlines of Session
Intersection to imaginary axis
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3. Root Locus Plot
❖ Plot root locus of system given by
𝑮 𝒔 =
𝒌(𝒔+𝟐)
𝒔𝟐+𝟐𝒔+𝟏𝟎
❖ Step 1 : Find position of poles, zeros and loci
❖ Position of zeros
❖ Position of poles
𝑍1 = −2
𝑃1 = −1 + 𝑗3, 𝑃2 = −1 − 𝑗3
Imag
j4
j3
j2
j1
0
-j1
-j2
-j3
-j4
Real
-6 -5 -4 -3 -2 -1
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4. Asymptotes
❖ Find Number of Asymptotes, Centroid of Asymptotes and Angle of Asymptotes
❖ Number of Asymptotes
❖ Centroid of Asymptotes
❖ Angle of Asymptotes
Imag
j4
j3
j2
j1
0
-j1
-j2
-j3
-j4
Real
-6 -5 -4 -3 -2 -1
Number of Asymptotes = P – Z = 2 – 1 = 1
𝑮 𝒔 =
𝒌(𝒔 + 𝟐)
𝒔𝟐 + 𝟐𝒔 + 𝟏𝟎
𝑍1 = −2 𝑃1 = −1 + 𝑗3, 𝑃2 = −1 − 𝑗3
𝜎𝑐 =
σ 𝑅𝑒𝑎𝑙 𝑝𝑎𝑟𝑡𝑠 𝑜𝑓 𝑃𝑜𝑙𝑒𝑠 − σ 𝑅𝑒𝑎𝑙 𝑝𝑎𝑟𝑡𝑠 𝑜𝑓 𝑃𝑜𝑙𝑒𝑠
𝑃 − 𝑍
𝜎𝑐 =
−1 − 1 − (2)
2 − 1
= 0
𝜃 =
(2𝑘 + 1)
𝑃 − 𝑍
× 1800 = 1800
𝜎𝑐 = 0
𝜃 =180
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5. Angle of Departure
❖ Angle of Departure
Imag
j4
j3
j2
j1
0
-j1
-j2
-j3
-j4
Real
-6 -5 -4 -3 -2 -1 𝜎𝑐 = 0
𝜃 =180
𝜙𝐷 = 180 − ෍ 𝐴𝑛𝑔𝑙𝑒 𝑜𝑓 𝑝𝑜𝑙𝑒 − ෍ 𝐴𝑛𝑔𝑙𝑒 𝑜𝑓 𝑧𝑒𝑟𝑜𝑠
𝜙𝐷 = 180 − 𝜙𝑝1 − 𝜙𝑧1 = 180 − 90 − 71 = 161
𝝓𝒑𝟏= 90
𝝓𝒛𝟏= 71
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6. Break away point
❖ Characteristics equation
❖ Find root at
𝒅𝒌
𝒅𝒔
= 𝟎
Imag
j4
j3
j2
j1
0
-j1
-j2
-j3
-j4
Real
-6 -5 -4 -3 -2 -1 𝜎𝑐 = 0
𝜃 =180
𝝓𝒑𝟏= 90
𝝓𝒛𝟏= 71
∴ 1 + G S H S = 0
𝑮 𝒔 =
𝒌(𝒔 + 𝟐)
𝒔𝟐 + 𝟐𝒔 + 𝟏𝟎
𝑍1 = −2 𝑃1 = −1 + 𝑗3, 𝑃2 = −1 − 𝑗3
∴ 1 +
𝒌(𝒔 + 𝟐)
𝒔𝟐 + 𝟐𝒔 + 𝟏𝟎
= 0
∴ 𝑘 = −
(𝑠2 + 2𝑠 + 10)
(𝑠 + 2)
∴
𝑑𝑘
𝑑𝑠
= 0 = −
𝑠 + 2 2𝑠 + 2 − (𝑠2 + 2𝑠 + 10)
(𝑠 + 2)2
∴ 0 = 𝑠2 + 4𝑠 − 6
∴ 𝑠 = 1.16 ∴ 𝑠 = −5.16
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7. Inter section to imaginary axis
❖ As per Diagram You don’t need to plot Intersection to imaginary axis
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