CONTROL SYSTEM

CS.1) The transfer function of a linear system is
(a) ratio of the output Vo(t) to input Vi(t)
(b) ratio of derivative of output to input
(c) ratio of Laplace transform of output to input with all initial conditions set to zero
(d) none of above

CS.2)The open loop DC gain of unity negative feedback system with closed loop transfer function
$\frac{s+4}{s^{2} + 7s + 13}$ is
(a)$ {4\over13}$
(b)$ {4\over9}$
(c)4
(d)13
Answer is (b)
Solution :The closed loop transfer function is given by :${G(s)\over( 1 + G(s)H(s))}$
$= {G(s)\over 1 + G(s)H(s)}$ ... Eq(1)
$= {s+4 \over s^{2}+7s+13}$
$H(s) = 1$ for unity feedback
$\therefore $, Eq (1) reduces to ${G(s)\over(1+G(s))} = {s+4\over(s^{2}+7s+13)}$ ... Eq(2)
To simplify this, take the reciprocal of Eq (2)
${(1 + G(s))\over G(s)} = {s^{2} + 7s + 13\over s+4 }$
${1\over G(s)} = {(s^2 + 7s+13)\over(s+4)} -1 $
${1 \over G(s)} ={(s^2 +7s + 13 ) - (s+4)\over(s+4)}$
${1\over G(s)} = {s^2+6s+9\over(s+4)}$
$G(s) = {(s+4)\over(s^2+6s+9)}$
For DC gain , s = 0 $\therefore, G(s) = 4/9$

CS.3) The gain margin of system under loop unity negative feedback G(s)H(s) = ${100 \over s(s+10)^2}$
(a)0 dB
(b)20 dB
(c)26 dB
(d)46 dB

Answer is C) 26 dB

Solution : Gain Margin under closed loop unity negative feedback is =
$G(s)H(s) = {100 \over s(s+10)^2} ... Eq(1)$
Phase $\phi = -90° - 2arctan(ω/10)$
Phase cross - over frequency = -180°
$-180° = -90° - 2arctan(ω/10)$
$90° = 2arctan(ω/10)$
$tan(45°) = ω/10 \therefore ω = 10$
In Eq(1) , put s = jω
$\therefore G(jω)H(jω) = 100/jω(jω+2)^{2}$
$|G(jω)H(jω)| = {100\over ω(ω2 + 100)}$...Eq(2)
Substituting for ω = 10,Eq (2) becomes, ${100\over 10(100+100)} = 1/20 $
Gain Margin = $1 \over |G(jω)H(jω)|$ = 1/(1/20) = 20
In dB, G.M (in dB) = 20 log1020 = 26dB

CS.4) Output of a linear time invariant control system is c(t) for a certain input r(t). If r(t) is modified by passing it through a block whose transfer function is $e^{-s}$ and then applied to the system, the modified output of the system would be
(a) $c(t-1)u(t-1)$
(b) $c(t)u(t-1)$
(c) ${c(t)\over(1+e^t)}$
(d) $c(t)\over(1+e^-t)$
Answer :(b)


CS.5)In the signal flow graph y/x equals
(a)3
(b)5/2
(c)2
(d)0

CS.6) Whether a linear system is stable or unstable (a) is a property of the system only
(b) depends on the input function only
(c) either (a) or (b)
(d) both (a) and (b)

Answer is (a)

CS.7) A system is described by the following differential equation: $y''(t)+ 3y'(t)+2y = x(t)$ is initially at rest.For input x(t) = 2u(t) the output y(t) is
(a)$(1-2e^{-t} +e^-{2t})u(t)$
(b)$(1+2e^t+e^{2t})u(t-1)$
(c)$(.5 -e^t + 0.5e^{2t})u(t)$
(d)$none of above$

CS.8) For a first order system having transfer function ${1\over(1+sT)}$, the unit impulse response is
(a) $e^{-t/T}$
(b) $Te^{-t/T}$
(c) ${1\over T} e^{-t/T}$
(d) $T^2 e^{-t/T}$
Answer is (c)
Solution: $TF = {1\over 1+sT}$ $ = {1/T\over 1/T+s}$
$ = \frac{1}{T}\left(\frac{1}{s+1}\right) $
$ = {1\over T} e^{-t/T}$

CS.9) From noise point of view, bandwidth should be
(a) large
(b) infinite
(c) not too large
(d) as large as possible
Answer is (c)

CS.10) First column elements of Routh's tabulation are 3,-4,-1/4,6,2. It means that there is/are
(a) two roots on the RHS of s-plane
(b) two roots on the LHS of s-plane
(c) three roots on the RHS of s-plane
(d) one root on the RHS of s-plane

Answer is (a)
Explanation:Consider one pair of elements each time i.e.,Sign change (from positive to negative) 3 and -4
-4 & -1/4 : No sign change
-1/4 & 6 : Sign change (from negative to positive)
6 & 2 : No sign change:
This indicates that there are two sign changes in the first column of Routh's tabulation. Thus we have two roots on the RHS of s-plane.

CS.11) For the given figure, C(s)/R(s) is
(a)${4(s+2) \over ((s+2)(s+1))} $
(b)${(s-3)\over ((s+2)(s+1))} $
(c)${(9s+13)\over ((s+2)(s+1))}$
(d)${1 \over ((s+2)(s+1))}$

Answer is (b)
CS.12)For a system transfer function represented by 1/(s+1)  input is x(t) and output y(t).
For x(t) = sin(t)u(t)
In steady state response the output y(t) is given by:
(a)$ {1 \over √{2}} sin(t-\pi/4)$
(b)$ {1 \over √{2}} sin(t+\pi/4)$
(c)$ sin t - cos t$
(d)$1-e^-t cos t$

Answer is (a)

CS.13)The unit impulse response of a system $h(t) = e^{-t} $, t > 0 & t=0 .For this system, the steady state value of output for unit step input is:
(a)1
(b)-1
(c)0
(d)infinite

Answer is (a)

CS.14)The unit step response of a system starting from rest is given by: $c(t) = 1-e^{-2t}$, t > 0 & t=0.The transfer function of system
(a)1/s
(b)1/(2+s)
(c)2/(2+s)
(d)0

Answer is (c)

CS.15)The transfer function of a tachometer is of form
(a)Ks
(b)${K \over s}$
(c)${K \over s+1}$
(d)${Ks \over 2}$

Answer is (b)

CS.16) Steady state error of a feedback control system with an acceleration input becomes finite in
(a) type 0 system
(b) type 1 system
(c) type 2 system
(d) type 3 system

Answer is (c)

CS.17) A system has its two poles on negative real axis and one pair of poles lie on jw axis. The system is
(a) stable
(b) unstable
(c) limitedly stable
(d) either (a) or (c)

Answer is (c)

CS.18) Assertion (A): The steady state response, of a stable, linear, time invariant system, to sinusoidal input depends on initial conditions. Reason (R): Frequency response, in steady state, is obtained by replacing s in the transfer function by jω
(a) Both A and R are correct and R is correct explanation of A
(b) Both A and R are correct but R is not correct explanation of A
(c) A is correct but R is wrong
(d) R is correct but A is wrong

Answer is (d)

CS.19)The open loop transfer function with unity feed back are given below for different systems. The unstable system is
(a)${2 \over (s+2)}$
(b)${2 \over s^{2}(s+2)}$
(c)${2 \over s(s+2)}$
(d)${2(s+1)\over s(s+2)}$

Answer is (b)
Explanation:Stability can be seen from Characteristic Equation:If in the polynomial equation there is a discontinuity of powers(exponent) of the variable, then the System is Considered UNSTABLE
In Option (b) this is evident $s^3+2s^2+2=0$

CS.20) For the transport lag G(jw) = e^(-jwt), the magnitude is always equal to
(a) 0
(b) 1
(c) 10
(d) 0.5

Answer is (b)

CS.21) For the characteristic equation:$s^4+Ks^3+s^2+s+1=0$ the range of K for stability is
(a)K>0
(b)K>1
(c)K<1
(d)none of above

Answer is (d)
Explanation : Construct Roth's array:
$$ \begin{array}{c|lcr} n & \text{Left} & \text{Center} & \text{Right} \\ \hline s^{4} & 1 & 1 & 1 \\ s^{3} & K & 1 & \\ s^{2} & {K-1 \over K} & 1 & \\ s^{1} & {K-1 - K^{2} \over K} & & \\ s^{0} & 1 \end{array} $$ For stability of system $ \left. \begin{array}{l} \ K >0 & \\ \frac{K}{K-1} > 0 & \\ \end{array} \right\} $ = K>1 also , by substituting K > 1 in equation $ frac{K-1-K^{2}}{K-1} > 0 $, the equation gives negative value.Since all three conditions cannot be satisfied simultaneously.Hence (d).


CS.22) A second order system exibhits 100% overshoot.It's damping coefficient is
(a)0
(b)1
(c)< 1
(d)> 1

Answer is (a)
Explanation:For a second order system , Peak OverShoot is $M_p = 100 $ %
$M_p = e^{\frac{-\zeta}{\sqrt{1-\zeta^2}}}$
$e^{\frac{-\zeta}{\sqrt{1-\zeta^2}}} = 100% = 1 $
Taking Logarithm on both sides,$\frac{-\zeta}{\sqrt{1-\zeta^2}} = ln(1) = 0$
This implies, $\zeta = 0$
This indicates no damping

CS.23) The log magnitude curve for a constant gain K is a
(a) horizontal straight line
(b) horizontal straight line of magnitude 20logK decibels
(c) an inclined line having slope K
(d) an inclined line having slope -K

Answer is ( )

CS.24) The closed loop transfer function of a control system is given as $T(s)= \frac{s-5}{(s+2)(s+3)}$.This is
(a) an unstable system
(b) an uncontrollable system
(c) a minimum phase system
(d) a non-minimum phase system

Answer is (d)
All poles are on right side.But, the zero is on the right half zero.Hence,its a non-minimum phase system

CS.25) The transfer function of a system is $T(s)=\frac{5}{(s+3)(s+6)}$.The damping ratio and natural frequency are respectively
(a) 4.24,2.12 rad/s
(b) 2.12,4.24 rad/s
(c) 4.24,1.06 rad/s
(d) 1.06,4.24 rad/s

Answer is (d)
The transfer function of a system is $T(s) = \frac{5}{s^2 + 9s +18}$
Characteristic Eqn. $s^2 + 9s +18 = 0$ STANDARD Characteristic Eqn. $s^2 + 2\zeta\omega_n s + \omega_n^{2} = 0$
Comparing both Std & Characteristic Characteristic Eqn. gives $\omega_n^{2} = 18$
$2\zeta\omega_n = 9$ .On Solving , $\omega_n = 4.24$ rad/sec & $\zeta = 1.06$

CS.26) A system with zero initial conditions has the closed loop transfer function $ T(s) =s^2 + \frac{4}{(s+1)(s+4)}$.The system output is zero at the frequency______rad/s
(a) 2
(b) 1
(c) 0
(d) none of above

Answer is (a)
The transfer function of system is $T(s) = \frac{s^2+4}{(s+1)(s+4)} $
$T(j\omega) = \frac{(j\omega)^{2}+4}{(j\omega+1)(j\omega+4)} $
If system output is zero, $|T(j\omega)| = \frac{|4-\omega^2|}{|(j\omega+4)(j\omega+1|} = 0 $
$ - \omega^2 + 4 = 0, \omega = 2 rad/sec $

CS.27)In Bode diagram (log magnitude plot) the factor (1/j$\omega$) in the transfer function gives a line having slope
(a) -20 dB per octave
(b) -10 dB per octave
(c) -6 dB per octave
(d) -2 dB per octave

Answer is ( )

CS.28)A system has position error constant $K_p = 3$. The steady error for input of 8tu(t) is
(a) 2.67
(b) 2
(c) $\infty$
(d) 0

Answer is (c)
Positional Error Constant $K_p = 3$ , Input Signal = $r(t) = 8tu(t)$
$R(s) = \frac{8}{s^2}$ , Input signal is Ramp type, velocity error constant $K_v = 0$
Steady State Error $e_ss = 1/K_v = \infty$