Question - Answer Set 15-28
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 $e_ss = 1/K_v = \infty$
(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 $e_ss = 1/K_v = \infty$
