Control Systems Test 4

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Question 1/10
1 / -0

Consider the system shown in figure 1. To improve the performance of the system a feedback is added to this system, which results in figure 2. The value of k so that the damping ratio of the new system is 0.4, is ______ (rounded off to two decimal places)
Figure 1:
Figure 2:

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A
0.12-0.13

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B
0.124-0.131

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C
0.127-0.136

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D
0.129-0.132

Solutions

From the figure 2,
By reducing the inner loop, we get $(s) = \frac{20}{s (s + 1 + 20 K)}$
$\frac{C (s)}{R (s)} = \frac{G (s)}{1 + G (s)} = \frac{\frac{20}{s (s + 1 + 20 K)}}{1 + \frac{20}{s (s + 1 + 20 K)}} = \frac{20}{s^{2} + (1 + 20 K) s + 20}$
By comparing the above transfer function with the standard second order system
$ω_{n}^{2} = 20 \Rightarrow ω_{n} = \sqrt{20}$
$2 ξ ω_{n} = (1 + 20 K)$
$\Rightarrow 2 (0.4) (\sqrt{20}) = 1 + 20 K$
⇒ K = 0.128
Question 2/10
1 / -0

A plant has an open-loop transfer function
$G_{P} (s) = \frac{20}{(s + 0.1) (s + 2) (s + 100)}$
The approximate model obtained by retaining only one of the above poles, which is closest to the frequency response of the original transfer function at low frequency is

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A
$\frac{0.1}{s + 0.1}$

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B
$\frac{2}{s + 2}$

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C
$\frac{100}{s + 100}$

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D
$\frac{20}{s + 0.1}$

Solutions

According to dominant pole concept, we can neglect the poles which are far from origin but the DC gain should be constant.
Approximate transfer function $= \frac{20}{(s + 0.1) (2) (100)} = \frac{0.1}{(s + 0.1)}$
Question 3/10
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Consider the system shown in the following figure.

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A
The type of the system is 1

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B
For a step input r(t) = u(t), the system gives zero error

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C
For a ramp input r(t) = 5t u(t), the steady state error is 0.01

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D
For a parabolic input, r(t) = 10t² u(t), the steady state error is infinite.

Solutions

Concept:
Ka = position error constant = $lim_{s \to 0} G (s) H (s)$
Kv = velocity error constant = $lim_{s \to 0} s G (s) H (s)$
Ka = acceleration error constant = $lim_{s \to 0} s^{2} G (s) H (s)$
Steady state error for different inputs is given by
Input
Type -0
Type - 1
Type -2
Unit step
$\frac{1}{1 + K_{p}}$
0
0
Unit ramp
∞
$\frac{1}{K_{v}}$
0
Unit parabolic
∞
∞
$\frac{1}{K_{a}}$

From the above table, it is clear that for type – 1 system, a system shows zero steady-state error for step-input, finite steady-state error for Ramp-input and $\infty$ steady-state error for parabolic-input.
Calculation:
The type of the system represents the number of poles at origin.
As there is one pole at origin, the type of the system is 1.
For type 1 system, the error is zero for step input and the error is infinite for parabolic input.
$G (s) = \frac{100 (s + 2) (s + 6)}{s (s + 3) (s + 4)}$
$K_{v} = \underset{s \to 0}{l t} s G (s) = \underset{s \to 0}{l t} \frac{s .100 (s + 2) (s + 6)}{s (s + 3) (s + 4)}$
$= \frac{100 \times 2 \times 6}{8 \times 4} = 100$
Steady state error $= \frac{A}{K_{v}} = \frac{5}{100} = 0.05$
Question 4/10
1 / -0

A second order system has the following properties.
The damping ratio, ξ = 0.8 damped frequency, ω_d = 6 rad/sec
The DC gain of the system is 2.
The transfer function of the system is

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A
$\frac{200}{s^{2} + 16 s + 100}$

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B
$\frac{100}{s^{2} + 16 s + 50}$

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C
$\frac{50}{s^{2} + 12.8 s + 64}$

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D
$\frac{128}{s^{2} + 12.8 s + 64}$

Solutions

Damping ratio (ξ) = 0.8
Damped frequency (ω_d) = 6 rad/sec
$\Rightarrow ω_{n} \sqrt{1 - ξ^{2}} = 6$
$\Rightarrow ω_{n} \sqrt{1 - {0.8}^{2}} = 6$
⇒ ω_n = 10 rad/sec
Standard second order system transfer function
$T (s) = \frac{k ω_{n}^{2}}{s^{2} + 2 ξ ω_{n} s + ω_{n}^{2}}$
$= \frac{100 k}{s^{2} + 16 s + 100}$
DC gain = 2
$\Rightarrow \frac{100 k}{100} = 2$
⇒ k = 2
Now, $T (s) = \frac{200}{s^{2} + 16 s + 100}$

Question 5/10

1 / -0

Match the following:

Root locations of the characteristic equations of second order systems.

Unit step responses of second order systems

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A – T, B – S, C – P, D – R, E – Q

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A – P, B – S, C – T, D – Q, E – R

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A – S, B – T, C – R, D – P, E – Q

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A – T, B – P, C – Q, D – R, E – S

Solutions

If the poles are real and left side of s-plane, the step response approaches a steady state value without any oscillations.
If the poles are complex and left side of s-plane, the step response approaches a steady state value with the damped oscillations.
If poles are non-repeated on the jω axis, the step response will have fixed amplitude oscillations.
If the poles are real and right side of s-plane, the step response reaches infinity without any oscillations.
If the poles are real and right side of s-plane, the step response reaches infinity without any oscillations.
If the poles are complex and right side of s-plane, the step response reaches infinity with damped oscillations.

Important Points:

System	Damping ratio	Roots of the Characteristic equine.	Root in the ‘S’ plane
Undamped	ξ =0	ξ = 0 Imaginary s = ±jω_n
Underdamped (Practical system)	0 ≤ ξ ≤ 1	$- ξ ω_{n} \pm j ω_{n} \sqrt{1 - ξ^{2}}$ Complex Conjugate
Critically damped	ξ = 1	-ω_nReal and equal
Overdamped	ξ > 1	$- ξ ω_{n} \mp j ω_{n} \sqrt{ξ^{2} - 1}$ Real and unequal

Question 6/10
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The open loop transfer function of a negative feedback closed loop system is $\frac{K}{(s^{2} + 4 s)}$ where K is a non-negative real valued parameter.

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The closed loop system is stable for only K > 2

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B
The closed loop system is an underdamped second order system for K > 4

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C
For K > 4, the system has a settling time less than 4s

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D
The damping ratio is 0.8 for K = 2.

Solutions

Concept:
The transfer function of the standard second-order system is:
$T F = \frac{C (s)}{R (s)} = \frac{ω_{n}^{2}}{s^{2} + 2 ζ ω_{n} s + ω_{n}^{2}}$
ζ is the damping ratio
ωn is the natural frequency
Characteristic equation: $s^{2} + 2 ζ ω_{n} + ω_{n}^{2}$
Roots of the characteristic equation are: $- ζ ω_{n} + j ω_{n} \sqrt{1 - ζ^{2}} = - α \pm j ω_{d}$
α is the damping factor
Settling time (Ts): It is the time taken by the response to reach ± 2% tolerance band as shown in the fig above.
$e^{- ξ ω_{n} t_{s}} = \pm 5 % (o r) \pm 2 %$
$t_{s} ≃ \frac{3}{ξ ω_{n}}$ for a 5% tolerance band.
$t_{s} ≃ \frac{4}{ξ ω_{n}}$ for 2% tolerance band
Calculation:
Open loop transfer function: $\frac{K}{s^{2} + 4 s}$
Characteristic equation: 1 + G(s) H(s) = 0
⇒ s² + 4s + K = 0
The system is stable for K > 0
By comparing with standard second order system,
ω²_n= K ⇒ ω_n = √K
$2 ξ ω_{n} = 4 \Rightarrow ξ = \frac{2}{\sqrt{K}}$
For underdamped system, ξ < 1
$\Rightarrow \frac{2}{\sqrt{K}} < 1$
⇒ K > 4
For K < 4, the damping ratio, ξ > 1
For 2% tolerance, setting time, $= \frac{4}{ξ ω_{n}}$
$= \frac{4}{2} = 2 s$
For 5% tolerance, settling time $= \frac{3}{ξ ω_{n}} = \frac{3}{2} = 1.5 s$
Settling time is less than 4s.
Question 7/10
1 / -0

Poles of a second order system are shown.
The settling time for (± 2%) tolerance is _____ seconds.

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A
0.50

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B
0.66

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C
0.82

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D
0.40

Solutions

Concept:
Comparing with given diagram
ξω_n = 6
For ± 2% Tolerance:
$t_{s} = 4 τ = \frac{4}{ξ ω_{n}}$
$\Rightarrow \frac{4}{6}$
= 0.66 sec.
Question 8/10
1 / -0

For a system having transfer function $G (s) = \frac{2 - s}{2 + s}$ , a step input is applied at time t = 0. The value of the system at t = 2 sec is ______ (rounded off to three decimal places)

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A
0.96-0.97

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B
0.964-0.971

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C
0.967-0.976

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D
0.969-0.972

Solutions

$G (s) = \frac{2 - s}{2 + s}$
$r (t) = u (t)$
$\Rightarrow R (s) = \frac{1}{s}$
$\Rightarrow \frac{C (s)}{R (s)} = G (s) = \frac{2 - s}{2 + s}$
$\Rightarrow C (s) = (\frac{2 - s}{2 + s}) (\frac{1}{s})$
$\Rightarrow C (s) = \frac{1}{s} = \frac{2}{2 + s}$
⇒ C(t) = u(t) – 2e^-2t u(t)
At t = 2 sec,
c(t = 2s) = 1 – 2e^-4 = 0.9633
Question 9/10
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When the system shown in the figure (1) is subjected to a unit-step input, the system output responds as shown in figure (2).
The values of K and T from the response curve are _____
Figure 1:
Figure 2:

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A
K = 1.42, T = 1.09

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B
K = 1.65, T = 1.23

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C
K = 1.65, T = 1.42

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D
K = 1.09, T = 1.23

Solutions

From the given response curve,
Peak over shoot, (M_P) = 0.254
Peak time, t_p = 3 s
$e^{- \frac{ξ π}{\sqrt{1 - ξ^{2}}}} = 0.254$
⇒ ξ = 0.4
$t_{p} = \frac{π}{ω_{d}} = \frac{π}{ω_{n} \sqrt{1 - ξ^{2}}} = 3$
$\Rightarrow \frac{π}{ω_{n} \sqrt{1 - {(0.4)}^{2}}} = 3$
⇒ ω_n = 1.14
From the block diagram,
$\frac{C (s)}{R (s)} = \frac{K}{T s^{2} + s + K}$
$ω_{n} = \sqrt{\frac{K}{T}}, 2 ξ ω_{n} = \frac{1}{T}$
$T = \frac{1}{2 ξ ω_{n}} = \frac{1}{2 \times 0.4 \times 1.14} = 1.09$
$K = ω_{n}^{2} T = {(1.14)}^{2} \times 1.09 = 1.42$
Question 10/10
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Consider the simplified model of a DC motor in unity feedback as in the figure below, where W(s) is the disturbance torque. With B = 10, A = 2 with a proportional controller of the form D(s) = k_p. What should be the value of the magnitude of k_p such that the steady-state error with respect to disturbance inputs (unit step torque), e_ss = 0.1.

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A
49-51

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B
494-511

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C
497-516

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D
499-512

Solutions

To find the error due to the disturbance input D(s), you have to use the disturbance transfer function.
From the given block diagram, the transfer function for the error is
$\frac{E (s)}{W (s)} = - \frac{\frac{B}{A} \cdot \frac{A}{s (τ s + 1)}}{1 + D (s) \cdot \frac{A}{s (τ s + 1)}} = - \frac{10}{s (τ s + 1) + 2 k_{p}}$
$E (s) = - \frac{10}{s (τ s + 1) + 2 k_{p}} W (s)$
$W (s) = \frac{1}{s}$
Steady-state error, $e_{s s} = \begin{matrix} l t \\ s \to 0 \end{matrix} s E (s)$
$= \begin{matrix} l t \\ s \to 0 \end{matrix} s [\frac{- 10}{s (τ s + 1) + 2 k_{p}}] \cdot \frac{1}{s}$
$= \frac{- 10}{2 k_{p}} = \frac{- 5}{k_{p}}$
Given that, e_ss = 0.1
$\Rightarrow \frac{- 5}{k_{p}} = 0.1 \Rightarrow k_{p} = - 50$
The magnitude of the k_p = 50

Input	Type -0	Type - 1	Type -2
Unit step	$\frac{1}{1 + K_{p}}$	0	0
Unit ramp	∞	$\frac{1}{K_{v}}$	0
Unit parabolic	∞	∞	$\frac{1}{K_{a}}$

Question 1/10

0.12-0.13

0.124-0.131

0.127-0.136

0.129-0.132

Question 2/10

0.1s+0.1

2s+2

100s+100

20s+0.1

Question 3/10

The type of the system is 1

For a step input r(t) = u(t), the system gives zero error

For a ramp input r(t) = 5t u(t), the steady state error is 0.01

For a parabolic input, r(t) = 10t2 u(t), the steady state error is infinite.

Question 4/10

200s2+16s+100

100s2+16s+50

50s2+12.8s+64

128s2+12.8s+64

Question 5/10

A – T, B – S, C – P, D – R, E – Q

A – P, B – S, C – T, D – Q, E – R

A – S, B – T, C – R, D – P, E – Q

A – T, B – P, C – Q, D – R, E – S

Question 6/10

The closed loop system is stable for only K > 2

The closed loop system is an underdamped second order system for K > 4

For K > 4, the system has a settling time less than 4s

The damping ratio is 0.8 for K = 2.

Question 7/10

0.50

0.66

0.82

0.40

Question 8/10

0.96-0.97

0.964-0.971

0.967-0.976

0.969-0.972

Question 9/10

K = 1.42, T = 1.09

K = 1.65, T = 1.23

K = 1.65, T = 1.42

K = 1.09, T = 1.23

Question 10/10

49-51

494-511

497-516

499-512

$\frac{0.1}{s + 0.1}$

$\frac{2}{s + 2}$

$\frac{100}{s + 100}$

$\frac{20}{s + 0.1}$

For a parabolic input, r(t) = 10t² u(t), the steady state error is infinite.

$\frac{200}{s^{2} + 16 s + 100}$

$\frac{100}{s^{2} + 16 s + 50}$

$\frac{50}{s^{2} + 12.8 s + 64}$

$\frac{128}{s^{2} + 12.8 s + 64}$