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

Answer:

$y_{steady}(t) = 0.636times sin(15.7t + 1.507)$

Time lag =  0.096 sec

Explanation:

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

$y = ce^\frac{-t}{ \tau } + m( \omega)kAsin(15.7t + \phi( \omega))\\\\\omega = 15.7 rad/s$

For first order instrument :

t = 1 sec

k = 1

$m( \omega )=\frac{1}{\sqrt{1 + (t \omega)^2} } \\\\=\frac{1}{\sqrt{1 + (1 *15.7)^2} }\\\\=0.0636$

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

$y_{steady} = m( \omega)kAsin(15.7t + \phi( \omega))$

$y_{steady}(t) = 0.0636 \times 1 \times 10 \times sin(15.7t - 1.507)$

$y_{steady}(t) = 0.636 \times sin(15.7t - 1.507)$

The expected time lag between input and output signal

$\beta_1 = \phi( \omega)/ \omega\\\\= \frac{-1.507}{15.7} \\\\= 0.096 sec$