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I am testing a feedback system in which a feedback signal from sensor is applied to a correction network and then compared with threshold values.

I need some clue on how to get around this task, I will be applying DC voltage to the network and will compare its output obtained with the one I calculated theoretically. Here is the transfer function of the network with a natural frequency of 90Hz

$$H(s)=\dfrac{1+0.145s+0.0019s^2}{1+\dfrac{0.8}{150 }s+\dfrac{1}{150²}s²}×\dfrac{1}{1+0.0012s}×\dfrac{1}{1+0.008s}$$

I calculated the output of the network by finding its inverse Laplace with step function as its input. The issue is that the output voltage obtained from solving this is a non-linear function hence changes it at every instant so how am I supposed to test the output for a particular DC input?

Any ideas?

Here is the step response I plotted using Matlab

enter image description here

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  • \$\begingroup\$ Why is it non-linear? \$\endgroup\$ – Andy aka Jun 12 '15 at 12:52
  • \$\begingroup\$ @Andyaka there are exponential terms in v(t) equation \$\endgroup\$ – alexhilton Jun 12 '15 at 13:58
  • \$\begingroup\$ @Chu Exactly voltage will vary with time but my problem is i do not understand at which time stamp to pick output voltage to test the circuit? so that I can compare the calculated result with the observed one. \$\endgroup\$ – alexhilton Jun 12 '15 at 14:00
  • \$\begingroup\$ @alexhilton, taking (4 x time constant) as the transient decay time, the decay times for the various terms are approximately: 4.8ms, 32ms, and 67ms. To be on the safe side you may want to give it 0.1s. (67ms is for the 2nd order TF, using 4/(zeta x wn)) \$\endgroup\$ – Chu Jun 12 '15 at 14:23
  • \$\begingroup\$ @Chu According to system specs damping for a second order damped system is defined as 1.5 are you using damping factor to find settling time? and do you mean that I should consider the output at t tends to infinity? Wont that give step input magnitude at output? Please elaborate \$\endgroup\$ – alexhilton Jun 12 '15 at 14:40
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What you have are: two 1st order lag terms, and a 2nd order underdamped term with the slight complication of a 2nd order numerator.

The general ROT is that the settling time for a 1st order lag is \$4\tau\$ where \$\tau\$ is the time constant (\$\tau\$ is the coefficient of \$s\$ in the general term: \$1/(1+\tau s)\$). The settling time for a 2nd order denominator term is given by: \$4/(\zeta \times \omega_n)\$ where \$\zeta\$ is the damping coefficient and \$\omega_n\$ is the natural frequency.

One standard form for the 2nd order denominator is: \$1 + 2(\zeta/\omega_n)s + s^2/\omega_n^2\$

In your case \$\omega_n=150\$ rad/sec and \$\zeta=0.4\$, hence the settling time for this term is \$4/(0.4 \times 150) = 0.067 = 67\$ms. Note, for this particular application we can ignore the 2nd order numerator as it only has a small effect on the transient generated by the denominator.

The two 1st order terms give settling times: \$4 \times 1.2 = 4.8\$ms, and \$4 \times 8 = 32\$ms. The longest settling time is thus \$67\$ms, after which all the transients may be considered spent.

The steady-state gain (or 'DC gain') of the sensor is unity. This can determined by setting \$s = 0\$ in the TF.

Here's a useful plotting tool for time domain and frequency domain responses from Laplace TFs:

http://sim.okawa-denshi.jp/en/dtool.php

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  • \$\begingroup\$ Please take some time to learn how to format math nicely on this site. It will greatly improve the readability and quality of your answers. There's an extensive tutorial on Math.SE meta and a more specific tutorial for EE.SE on EE.SE meta. \$\endgroup\$ – Null Jun 12 '15 at 15:26
  • \$\begingroup\$ @Null, You're quite right, I've been lazy with this. \$\endgroup\$ – Chu Jun 12 '15 at 15:39
  • \$\begingroup\$ There's a bit of a learning curve, but once you start using MathJax you'll pick it up in pieces and get better and better at it. Also, if you're curious how a particular post formats something, you can click the "edit" button just to look at the post's source. You can pick it up even faster that way. \$\endgroup\$ – Null Jun 12 '15 at 15:51
  • \$\begingroup\$ @Chu Thanks for the detailed explanation. I will now calculate output at 70-100ms. Will you please check the link below I calculated the results here, should I calculate the output using the last equation where t is assumed to be real? wolframalpha.com/input/… \$\endgroup\$ – alexhilton Jun 13 '15 at 5:18
  • \$\begingroup\$ @alexhilton, the Wolfram answer 'assuming t is real' is the one to go for if you need to use the values in another application \$\endgroup\$ – Chu Jun 13 '15 at 8:38

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