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We have an open loop system with an input u , a transfer function G(s) and an output y. We apply the following inputs $$u_i(t)=\sin(ω_it), i=1,2,...6 $$ and get the following responses enter image description here

These are of the form $$y(t)=Y_1\sin(ω_it+ψ_i)$$ Seeing some of the graphs I though G(s) is a differentiator but this doesn't hold true for every graph. The solution manual gives $$G(s)=\frac{s}{(s+10)^2}$$ Any ideas ?

We have $$G(s)U(s)=Y(s)$$ Taking the Laplace transforms : $$G(s)\frac{ω_i}{s^2+ω_i^2}=Y_1\frac{s\sinψ_i+ω_i\cosψ_i}{s^2+ω_i^2} \\ G(s)ω_i=Y_1(s\sinψ_i+ω_i\cosψ_i)$$ Doesn't seem like the (s+10)^2 denominator will show up.

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    \$\begingroup\$ I would plot the responses on a bode plot to make visualizing things easier. \$\endgroup\$ – Andy aka Jul 21 '17 at 11:52
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    \$\begingroup\$ This is straight from an exam sheet and gives way too little points if correct so the answer must be straightforward and simple. I don't see that though. \$\endgroup\$ – John Katsantas Jul 21 '17 at 11:56
  • \$\begingroup\$ All the same I think it is worth showing it to justify your claim that visualizing it doesn't help. After all, it's unlikely I'm going to draw it!! \$\endgroup\$ – Andy aka Jul 21 '17 at 12:18
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Assuming that the order of six the plots is

(i=1)   (i=2)
(i=3)   (i=4)
(i=5)   (i=6)

And that:

  • From the first plot, the initial phase is 90º leading (a regular sine would start at 0º)
  • In the fourth plot, phase lags 90º, so phase become 0º
  • In the firth plot, it lags an additional 90º, so phase become 180º

Then, also considering the amplitudes, we could sketch a bode plot (like Andy aka suggested - took 5 minutes):

bodesketch

From this, it seems that structure is an derivator (DC gain = 0, positive magnitude slope and phase 90º leading) with two poles (same slope but descending, adds 180º of lag):

$$ Y(s) = \frac{as}{(s+b)(s+c)} U(s) $$

(It could actually be a double/triple/"n" integrator with four/six/"2n" poles, you can discover that by checking if the ascending or descending slope of the gain plot is 20 or 40 or 20*n db/decade).

You can then discover the coefficients by substituting \$Y(s)\$ and \$U(s)\$ in the preceding equation by the inputs and outputs (outputs will be the magnitude gain plus the phase lag), also substituting \$s=jw\$.

Hopefully then you will arrive at your answer.

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  • \$\begingroup\$ That's a very detailed answer . Thank you . I had trouble drawing the Bode plot actually. I'm used to sketching the Bode plot for a transfer function but we have 6 responses here and I didn't think I could combine them in a single plot. I would Laplace transform y(t)/u(t) for one of the situations and setting s=jω and converting to dBs I would draw the Bode. Maybe I have some blanks here. Would you mind explaining how you drew the amplitude Bode plot and why we combine the 6 situations ? It's not like the first one is from zero to ω=1 the second from ω=1 to ω=10 etc in order to add them. \$\endgroup\$ – John Katsantas Aug 1 '17 at 19:46
  • \$\begingroup\$ Hi John, normally when given a transfer function, we plot the magnitude and phase for many possible frequencies (\$w\$ from 0 to infinity if we wish) because we know the"general rule"(the TF). Here you are given the opposite problem: You have 6 plots, from which you derive (mag/phase) of ONLY 6 points (frequencies, or as you say "situations") in the bode plot - the six circles in my magnitude plot (there should be also six circles in the phase plot but I drew it continuously), from which you should derive the transfer function that has a bode plot that actually passes through those points. \$\endgroup\$ – SuperGeo Aug 1 '17 at 20:08
  • \$\begingroup\$ So NO, the first of the plots wont give us the behavior of w=1 until w=10, only the behavior AT the frequency. But, considering that the order of the system is not something absurd, we know that there is not much to happen between one point and another and we can "connect the points" to obtain the shape of the bode plot. \$\endgroup\$ – SuperGeo Aug 1 '17 at 20:13
  • \$\begingroup\$ So the six points in your plot are at each ω_i. Now everything is clear. Thanks Geo , it's nice finding people helping on control systems. I always have trouble finding what I need on this subject online. \$\endgroup\$ – John Katsantas Aug 1 '17 at 20:51

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