# Find the transfer function given responses

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

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.

• I would plot the responses on a bode plot to make visualizing things easier. – Andy aka Jul 21 '17 at 11:52
• 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. – John Katsantas Jul 21 '17 at 11:56
• 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!! – Andy aka Jul 21 '17 at 12:18

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):

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 $s=jw$.

• 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. – SuperGeo Aug 1 '17 at 20:08