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A signal suspected to be of the nominal form y(t) = 5 sin 1000 mV is measured by a first-order instrument having a time constant of 100 ms and K 1 V/V. It is then to be passed through a second-order amplifier having a K= 100 V/V, a natural frequency of 15,000 Hz, and a damping ratio of 0.8. What is the expected form of the output signal, y(t)? Estimate the dynamic error and phase lag in the output. Is this system a good choice here? If not, do you have any suggestions?

Respuesta :

Sorry no sudgestions.

Explanation:

Answer:

dynamic error , C(s) = K ( 1- e-T/t )

output y(t) = e-qw + [ Ae(1-q2)1/2 wt + A2e(1-q2)1/2 wt ]

Explanation:

Given that,

Y(t) = 5 Sin 1000t mV

t = 100 ms

K = 1v/v

Second amplifier , K = 100 V/V

Natural frequency , f = 15000 Hz

damping ratio = 0.8

Y (t) = ?

the given equation is in the form of x = X sin ( wt + \phi)

f = 1/T

  dynamic error , C(s) = K ( 1- e-T/t )

                                = 100 ( 1- e-100 / t )

    y = 5 sin 1000t mV

  Y = amplitude = 5

     f = 1/t

  t = 1/f = 1/ 5000 = 500 ms

  C(s) = 100 ( 1- e-100/500 )

         = 18.12

damping ratio ,q= 0.8

output y(t) = e-qw + [ Ae(1-q2)1/2 wt + A2e(1-q2)1/2 wt ]

                = 54

Yes it is a good choice because of having the under damped system.

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