Control systems miscellaneous
- If the open-loop transfer function of the system is
G(s) H(s) = K(s + 10) s(s + 8) (s + 16) (s + 72)
then a closed loop pole will be located at s = –14, when the value of K is—
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Here the value of K can be obtained by
|K| = All the poles of G(s) H(s) ⎪at s = – 14 All the zeros of G(s) H(s)
or|K| = s(s + 8) (s + 16) (s + 72) (s + 10)
or|K| = – 14 (– 14 + 8) (– 14 + 16) (– 14 + 72) (– 14 + 10)
or|K| = – 14 × – 6 × 2 × 58 – 4
or
K = |– 2436| = 2436Correct Option: B
Here the value of K can be obtained by
|K| = All the poles of G(s) H(s) ⎪at s = – 14 All the zeros of G(s) H(s)
or|K| = s(s + 8) (s + 16) (s + 72) (s + 10)
or|K| = – 14 (– 14 + 8) (– 14 + 16) (– 14 + 72) (– 14 + 10)
or|K| = – 14 × – 6 × 2 × 58 – 4
or
K = |– 2436| = 2436
- In the root-locus for open-loop transfer function
G (s)H (s) = K(s + 6) (s + 3) (s + 5)
the ‘break away’ and ‘break-in' points are located respectively at—
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Given G(s) H(s) = K(s + 6) (s + 3)(s + 6)
the pole-zero plot of the given function is shown below C. E. is given by
1 + G(s) H(s) = 01 + K(s + 6) = 0 (s + 3)(s + 5)
or
K = – (s + 3) (s + 5) = 0 (s + 6) or dK = - 
(s + 6) d/ds (s2 + 8s + 15) – (s2 + 8s + 15) d/ds (s + 6) 
ds (s + 6)2 = - (s + 6) (2s + 8) – (s2 + 8s + 15)1 (s + 6)2
= – 2s2 + 208 + 48 – s2 – 8s – 15 (s + 6)2
ordK = – s2 + 12s + 33 ds (s + 6)2
dK/ds = 0 gives
s2 + 12s + 33 = 0
on solving we gets = – 12 ± √10 2
= – 7.58, – 4.42
Note: Since break away point lies between two poles and the break in point lies between two zeros. Since here one zero at s = – 6 and other at infinity. Hence only option (D) is the correct choice not option (C).
Correct Option: D
Given G(s) H(s) = K(s + 6) (s + 3)(s + 6)
the pole-zero plot of the given function is shown below C. E. is given by
1 + G(s) H(s) = 01 + K(s + 6) = 0 (s + 3)(s + 5)
or
K = – (s + 3) (s + 5) = 0 (s + 6) or dK = - (s + 6) d/ds (s2 + 8s + 15) – (s2 + 8s + 15) d/ds (s + 6) ds (s + 6)2 = - (s + 6) (2s + 8) – (s2 + 8s + 15)1 (s + 6)2
= – 2s2 + 208 + 48 – s2 – 8s – 15 (s + 6)2
ordK = – s2 + 12s + 33 ds (s + 6)2
dK/ds = 0 gives
s2 + 12s + 33 = 0
on solving we gets = – 12 ± √10 2
= – 7.58, – 4.42
Note: Since break away point lies between two poles and the break in point lies between two zeros. Since here one zero at s = – 6 and other at infinity. Hence only option (D) is the correct choice not option (C).
- The transfer function of a compensating network is of the form (1 + α Ts)/ (1 + Ts). If this is a phase-lag network, the value of α should be—
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Given Vo(s) = (1 + αTs) Vi(s) (1 + sT)
The output leads to the input by an angleφ = tan–1 
αT 
- tan–1(T) 1
If α is > 1, φ is positive and if 0 < α < 1 then φ becomes negative meaning that output will lag the input by φ. Since the given function is to represent a phase lag network, so α should be between 0 and 1.Correct Option: B
Given Vo(s) = (1 + αTs) Vi(s) (1 + sT)
The output leads to the input by an angleφ = tan–1 αT - tan–1(T) 1
If α is > 1, φ is positive and if 0 < α < 1 then φ becomes negative meaning that output will lag the input by φ. Since the given function is to represent a phase lag network, so α should be between 0 and 1.
- The maximum phase shift that can be provided by a lead compensator with transfer function
is—G (s) = 1 + 6s 1 + 2s
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Given GC(s) = 1 + 6s ....…(i) 1 + 2s
In phase lead compensatorVo(s) = G(s) = α(1 + sT) .......…(ii) Vi(s) (1 + αsT)
on comparing equations (i) and (ii), we have
T = 6
αT = 2
or α = 2/6 = 1/3
Now,
sin φm = 1 – α = 1 - (1/3) 1 + α 1 + (1/3)
or
sin φm = 1/2
or
φm = 30°
Alternative methodGc(s) = 1 + 6s 1 + 2s
Phase shift is φ = tan–1 6ω – tan–1 2ωdφ = 6 - 2 dω 1 + 36ω2 1 + 4ω2 For maximum value of φ put dφ = 0 dω
∴ 3(1 + 4ω2) = 1 + 36ω2
or
12ω2 = 1
orω = 1 √12
Thus,φ = tan–1 6 - tan–1 1 √12 √3 tan–1(√3) - tan–1 1 √3
= 60° – 30° = 30°Correct Option: B
Given GC(s) = 1 + 6s ....…(i) 1 + 2s
In phase lead compensatorVo(s) = G(s) = α(1 + sT) .......…(ii) Vi(s) (1 + αsT)
on comparing equations (i) and (ii), we have
T = 6
αT = 2
or α = 2/6 = 1/3
Now,
sin φm = 1 – α = 1 - (1/3) 1 + α 1 + (1/3)
or
sin φm = 1/2
or
φm = 30°
Alternative methodGc(s) = 1 + 6s 1 + 2s
Phase shift is φ = tan–1 6ω – tan–1 2ωdφ = 6 - 2 dω 1 + 36ω2 1 + 4ω2 For maximum value of φ put dφ = 0 dω
∴ 3(1 + 4ω2) = 1 + 36ω2
or
12ω2 = 1
orω = 1 √12
Thus,φ = tan–1 6 - tan–1 1 √12 √3 tan–1(√3) - tan–1 1 √3
= 60° – 30° = 30°
- A transfer function G (s) has the pole-zero plot as shown in the given figure. Given that the steady state is 2, the transfer function G (s) will be given by—
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Here, there are 2 pole and 1 zero. Let the transfer function G(s) of the type
G(s) = K(s + 1) (s + 2 + j)(s + 2 – j)
orG(s) = K(s + 1) = K(s + 1) (s + 2)2 – j2 s2 + 4s + 5
Also given that steady state gain is 2. i.e.,
G(0) = 2 = K(0 + 1) 02 + 4·0 + 5
or
K = 10
Now, theT.F. G(s) = 10(s + 1) 02 + 4s + 5 
Correct Option: C
Here, there are 2 pole and 1 zero. Let the transfer function G(s) of the type
G(s) = K(s + 1) (s + 2 + j)(s + 2 – j)
orG(s) = K(s + 1) = K(s + 1) (s + 2)2 – j2 s2 + 4s + 5
Also given that steady state gain is 2. i.e.,
G(0) = 2 = K(0 + 1) 02 + 4·0 + 5
or
K = 10
Now, theT.F. G(s) = 10(s + 1) 02 + 4s + 5 
