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You Are On Multi Choice Question Bank SET 1373

68651. For a first order system having transfer function , the unit impulse response is




68652. A stepper motor is





68653. In a two phase ac servomotor rotor resistance is R and rotor reactance is X. The speed curve will be linear if


68654. In the given figure, the input frequency is such that R = XC. Then





68655. From the noise point of view, bandwidth should





68656. First column elements of Routh's tabulation are 3, 5, , 2. It means that there





68657. When a unit step voltage drives a lag network the output





68658. Use the Collective noun:Fish





68659. A system has its two poles on the negative real axis and one pair of poles lies on jω axis. The system is





68660. In the given figure the input frequency is such that R << Xc, then





68661. In the given figure, of potentiometer V0 = Vi (R0/Ri) only when





68662. Bellows converts





68663. 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





68664. Consider the systems with following open loop transfer functions If unity feedback is used, the increasing order of time taken for unit step response to settle is





68665. The phase margin and damping ratio have no relation.



68666. For the transport lag G(jω) = e-jωT, the magnitude is always equal to





68667. The log magnitude curve for a constant gain K is a





68668. The compensator of the given figure is a





68669. If error voltage is e(t), integral square error =



68670. A lag compensator is essentially a





68671. Stepper motors find applications in





68672. For type 2 system, the magnitude and phase angle of the term (jω)2 in the denominator, at ω = 0, are respectively





68673. In an integral controller





68674. In Bode diagram (log magnitude plot) the factor in the transfer function gives a line having slope





68675. In the given figure the input is x(t) = A sin ωt. The steady state output y(t) =





68676. In the given figure x6 =





68677. A negative feedback system has . The closed loop system is stable for





68678. For the control system in the given figure, the value of K for critical damping is





68679. Bode magnitude plot is drawn between





68680. For the system in the given figure,


68681. In the given figure, if R = XC, voltage gain is





68682. The system in the given figure, has





68683. The given figure shows a pole zero diagram. The transfer function G(j1) is





68684. For very low frequencies, v0/vi in the given figure equals





68685. In a minimum phase system





68686. The first column of a Routh array isHow many roots of the corresponding characteristic equation are in left half s-plane?





68687. For the given figure, time constant RC = t . Then



68688. The entries in the first column of Routh array of a fourth order are 5, 2, - 0.1, 2, 1. The number of poles in the right half plane are





68689. In Bode diagram (log magnitude plot) the factor (jω)n in the transfer function gives a line having slope




68690. The magnitude of transport lag factor is always zero.



68691. The primary function of lag compensator is to provide sufficient





68692. For G(jω) =





68693. The transient response of a second order system is given by for 5% criterion the settling time is





68694. The polar plot of a transfer function passes through (-1, 0) point. The gain margin is





68695. For the system in the given figure, the transfer function C(s)/R(s) is





68696. For the system in the given figure, the characteristic equation is





68697. In control systems the magnitude of error voltage





68698. The slope of log-magnitude asymptote changes by - 40 dB/ decade at a frequency ω1. This means that





68699. The transfer function can be for





68700. The system of the given figure





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