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)
