Question

A signal expected to be of the form F(t)= 10 sin 15.7t is to be measured with a first-order instrument having a time constant of 50 ms. Write the expected indicated steady response output sig- nal. Is this instrument a good choice for this mea- surement? What is the expected time lag between input and output signal? Plot the output amplitude spectrum. Use y(0) = 0 and K = 1 V/V

Answer 1

`y_(steady)(t) = 0.636times sin(15.7t + 1.507)\\\\`

Time lag = 0.096 sec

Step-by-step explanation:

A first order system with simple sinusoidal signal as input, the output is of the form:

`y = ce^(-t)/( \tau ) + m( \omega)kAsin(15.7t + \phi( \omega))\\\\\omega = 15.7 rad/s\\\\`

For first order instrument :

t = 1 sec

k = 1

`m( \omega )=(1)/(√(1 + (t \omega)^2) ) \\\\=(1)/(√(1 + (1 *15.7)^2) )\\\\=0.0636`

`\phi ( \omega)= -tan^(-1) \tau \omega\\\\=-tan^(-1) (1 *15.7) = -86.36^o = -1.507 rad\\`

`y_(steady) = m( \omega)kAsin(15.7t + \phi( \omega))`

`y_(steady)(t) = 0.0636 * 1 * 10 * sin(15.7t - 1.507)\\\\`

`y_(steady)(t) = 0.636 * sin(15.7t - 1.507)\\\\`

The expected time lag between input and output signal

`\beta_1 = \phi( \omega)/ \omega\\\\= (-1.507)/(15.7) \\\\= 0.096 sec`