i have attached questions with answer key solve accordingly KTU ECE 2019 SCHEME CONTROL SYSTEM SUBJECT 2019 scheme---13 b) Consider a unity feedback control system with the closed loop transfer function given by
C(s) s+k
--- = ------- . Determine the open loop transfer function. Show that the steady (8)
R(s) s^2+as+b
state error in the unit ramp input response is e_ss = (a-k)/b
14 a) Starting from the generalized transfer function, derive expression for peak time of (9)
second order under-damped system subjected to unit step function.
b) The open loop transfer function of a unity feedback control system is (5)
G(s) = K / (s(s+1)(s+2))
i) Determine the type and order of the system
ii) Find the minimum value of K for which the steady state error is less than
0.2 for a unit ramp input.
Module -3
15 a) Given the characteristic equation of a system. Using R.H criterion, Find the location (5)
of roots in s-plane and hence comment whether the system is fully stable, unstable
or conditionally stable.
F(s) = s^4 + 2s^3 + 11s^2 + 18s + 18 = 0
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b) Sketch the root locus for the given open loop transfer function and find the value of (9)
K and w for marginal stability.
G(s)H(s) = K / (s(s+2)(s+3))
16 a) Given the characteristic equation of a system. Using R.H criterion, Find the range of (6)
K for the system to be stable. Also find the frequency of sustained oscillation at the
marginal stability.
F(s) = s^4 + 20s^3 + 15s^2 + 2s + K = 0
b) Sketch the root locus for the given open loop transfer function and comment on the (8)
system stability.
G(s)H(s) = K(s+2) / (s(s+1)(s+4))
Module -4
17 a) Compare lead, lag and lag-lead compensators. (4)
b) A unity feedback control system with given G(s), Draw the Bode plot. Find the gain (10)
margin and phase margin. Also check for the stability. (Use semi-log sheet)
G(s) = 5(1+2s) / ((1+4s)(1+0.25s))
18 a) Explain the design procedure of phase lead compensator using Bode plot method. (5)
b) Draw the Nyquist plot for the system whose open loop transfer function is (9)
G(s)H(s) = K / (s(s+2)(s+10))
Also comment on closed loop stability.
Module -5
19 a) Obtain the state model for the electrical network shown. (7)
Chart Description:
- Type: Electrical circuit diagram.
- Main Elements:
- Components: Voltage source (V), capacitor (C), inductor (L), resistor (R).
- Connections: V and C are in series, with voltage across C labeled Vc. This series combination is in parallel with the series combination of L and R. The current through L is iL, and the voltage across R is VR.
- Labels: V, Vc, C, i, iL, L, R, VR. Arrows indicate direction of current (i entering parallel branches, iL through L) and polarity of voltage source.
- Relative Position: V and C are on the left, L and R are on the right, connected in parallel.
b) Check the controllability and observability of the following system. (7)
.
x = [-1 0; 0 -2] x + [0; 1] u; y = [1 2] x
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20 a) Determine the transfer function of a system represented by (5)
.
x = [-2 -2; 4 -8] x + [1; 1] u; y = [1 0] x
b) An LTI system is represented by the state equation x = Ax + Bu, where (9)
A = [-3 0 0; 0 -1 1; 0 0 -1] and B = [1; 0; 1], find the state transition matrix Phi(t).
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**Module -5**
**19 a)** Let
T(s) = 1 / (s² + 20s + 100) is the transfer function of a system.
Also represent the system in the state variable form (phase variable).
[ẋ₁] = [ 0 1 ] [x₁] + [0] r(t)
[ẋ₂] [-100 -20] [x₂] [1]
y(t) = [1 0] [x₁]
[x₂]
Draw its signal flow graph of above for the phase variable form.
Marks: 3, 1.5, 2.5
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**b)** Find the state transition matrix of a system represented by two state variables and having state coefficient matrix, A = [ 0 6 ]
[-1 -5]
Find Φ(s) = [sI - A]⁻¹
Inverse transform L⁻¹{[sI - A]⁻¹}
Marks: 4, 3
**20 a)** A single-input single-output system has the matrix equations
ẋ = [ 0 1 ] x + [1] u
[-3 -4] [1]
y = [10 0] x
Draw signal flow graph and get the transfer function using the signal flow model.
Marks: 3
Transfer function expression in transform domain in terms of state equation matrices and state transition matrix.
G(s) = Y(s)/U(s) = CΦ(s)B
Marks: 4
**b)** [ẋ₁] = [ 0 1 0 ] [x₁] + [0] u(t)
[ẋ₂] = [ 0 0 1 ] [x₂] [0]
[ẋ₃] [-6 -11 -6] [x₃] [2]
...3 Marks
Controllability Analysis:
Q = [b Ab A²b]
Q = [ 0 0 2 ]
[ 0 2 -12 ]
[ 2 -12 50 ]
|Q| ≠ 0 System is controllable ......2 Marks
Observability Analysis:
Q = [Cᵀ AᵀCᵀ (Aᵀ)²Cᵀ]
Q = [ 1 0 0 ]
[ 0 1 0 ]
[ 0 0 1 ]
System is Observable ...... 2 Marks
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**Module -3**
15 a) A system has characteristic equation, $s^3 + 3s^2 + (K+1)s + 4 = 0$.
Routh table
$s^3$ | 1 | K+1
$s^2$ | 3 | 4
$s^1$ | $b = \frac{(3K+1)-4}{3}$ | 0
$s^0$ | 4 |
For stability apply the RH criterion
$3K - 1 > 0$
The range of K for a stable system is
$K > 1/3$.
b) For a system having open loop transfer function,
$G(s)H(s) = \frac{K}{(s+1)(s+3)(s+6)}$
Plot the root locus
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1100ECT307122103 Pages 8
Plotted the pole zero locations in s plane 1
Locus on real axis extends to infinity to the left of pole at -6. 1
The other two loci
Starts from $s_1 = -1$ to the left and $s_2 = -3$ to the right. These loci meets at break away
point B and extends to infinity at angle of asymptotes
$\phi_1 = 60^\circ$
$\phi_2 = 300^\circ$ 1
The centroid of the asymptotes is at $\sigma = -3.33$
Break away point : The characteristic equation $1 + GH = 0$
Or $K = p(s) = (s+1)(s+3)(s+6)$
Along the real axis between $s_1 = -1$ and $s_2 = -3$ the break-away point happens at
maximum value of $K = p(s)$
where $\frac{dK}{ds} = 0$.
This is at $s = -1.877$. 1
The phase criterion makes the loci perpendicular to the real axis. 1
Plot of the loci from break away point -1.877. 2
16 a) The effect of adding a pole –
Showing the s-plane pole location for dominant pole and not-dominant pole 2
Dominant pole effects on step response with diagram 3
Non- dominant pole effects on step response 2
b) Transfer function of Proportional controller 1
PI 2
PID controller. 2
Explain their effect in a control system design. 2
Explain their effect in a control system design. 2
Module -4
17 a) Using the Nyquist contour, analyse the following system to obtain the limit of K
for the stability. The system has the open-loop transfer function
G(s)H(s) = K / [s(τ₁s + 1)(τ₂s + 1)]
Also find the expression for gain margin of the system.
Draw the Nyquist contour in s plane: Semicircle of infinite radius enclosing the
right half of s-plane and excluding the s=0 point (since it is a pole). 2
The contour plot in GH plane 3
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Derive the intersection of the plot on the real axis 4
u = -KT₁T₂ / (τ₁ + τ₂)
Found the limit of K for stability in terms of τ₁ and τ₂ equating u=1. 3
Derive gain margin of the system 2
Gain margin of the system = 20 log₁₀(1/|uₚc|) dB
Where d = 1/|uₚc| 2
18 a) State Cauchy's argument principle with the conditions to be applied on the 3.5
contour of mapping.
State the Nyquist criterion of stability on the open loop transfer function of a 3.5
control system.
b) Draw the Bode plots (magnitude and phase) of the system.
G(s)H(s) = K / [s(s + 1)(s + 2)]
Express G(s)H(s) = [K / (2s)] * [1 / (s/1 + 1)] * [1 / (s/2 + 1)]
Draw approximate logarithmic (Bode) plots for each factors
Corner frequencies are 0, 1 and 2 respectively
Add these factors and plot the complete magnitude curve 2
The phase plot: -90 for factor 1/jω, add -45 at corner frequency for other two factors 2
Find the gain margin and phase margin for the Bode plots (magnitude and phase)
Phase margin = 45° 1
Gain margin = 15 dB 1
State the stability using gain and phase margin of the system: System is stable 1
Module -5
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We have a unity feedback control system with closed loop transfer function C of s over R of s equals s plus k over s squared plus a s plus b. First, we find the open loop transfer function using the unity feedback relationship. For unity feedback, the closed loop transfer function equals G of s over one plus G of s. Solving for G of s, we get G of s equals s plus k over s squared plus a minus one s plus b minus k. Next, for steady state error with unit ramp input, we use the final value theorem. The steady state error equals the limit as s approaches zero of s times one over s squared times one over one plus G of s. This simplifies to a minus k over b.
For a second order underdamped system with transfer function omega n squared over s squared plus two zeta omega n s plus omega n squared, we derive the peak time. The unit step response has characteristic equation roots at negative zeta omega n plus or minus j omega d, where omega d is the damped frequency. The time domain response is one minus exponential decay times sine function. The peak occurs when the derivative equals zero, which happens at time t equals pi over omega d. Therefore, the peak time is pi over omega n times square root of one minus zeta squared.
Given the unity feedback system with open loop transfer function G of s equals K over s times s plus one times s plus two. For part one, we determine the system type and order. The system has one pole at the origin, making it a Type 1 system. With three total poles, it is a third order system. For part two, we find the minimum K for steady state error less than zero point two for unit ramp input. For Type 1 systems, the steady state error for ramp input is one over K v, where K v equals the limit as s approaches zero of s times G of s, which equals K over two. Therefore, the steady state error is two over K. For the error to be less than zero point two, we need K greater than ten, so the minimum value is K equals ten.
We analyze the stability of the system with characteristic equation s to the fourth plus two s cubed plus eleven s squared plus eighteen s plus eighteen equals zero using the Routh-Hurwitz criterion. We construct the Routh table by arranging coefficients in rows. The first row contains coefficients of even powers: one, eleven, eighteen. The second row contains odd power coefficients: two, eighteen, zero. For the s squared row, we calculate two times eleven minus one times eighteen divided by two, which equals two. For the s row, we get eighteen. The s zero row is eighteen. Examining the first column, we have one, two, two, eighteen, eighteen. There is one sign change from positive to negative, indicating one pole in the right half plane. Therefore, the system is unstable.
To summarize what we have learned about control systems: Unity feedback systems relate closed loop and open loop transfer functions through simple algebraic relationships. Peak time for second order underdamped systems is pi over omega n times square root of one minus zeta squared. System type is determined by the number of poles at the origin and directly affects steady state error calculations. The Routh-Hurwitz stability criterion uses sign changes in the first column to determine the number of right half plane poles. These fundamental concepts form the foundation for control system analysis and design.