# Relation between open-loop and closed-loop plots

Discussion in 'Homework Help' started by Triton, Sep 30, 2011.

1. ### Triton Thread Starter New Member

Sep 30, 2011
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Hello. I am not sure if this forum would be the correct place to ask this question in (if it is not then I apologize for it) but here goes:

I have an exercise where I'm supposed to match different plots with each other (five open-loop bode plots, five closed-loop bode plots and five step response plots). Basically I am supposed to match an open-loop plot with the correct closed-loop plot and it's corresponding step response plot. I'm not really looking for the answers as I already have the full solution with a description (that gets me confused). Instead I'm having a hard time seeing the various relations between these plots which prevents me from actually solving the exercise.

I've had a similar exercise before where I was supposed to match bode plots (wasn't specified whether it was of an open-loop or closed-loop system...I assume the latter) and I didn't have problems at that one as I could relate several factors with each other (bandwidth with rise time, resonance peak with overshoots, static gain (DC gain?) with steady state value etc.)

I find it hard when I have three different plots to match together. More specifically I would say that I find it difficult to relate the open-loop plots with the closed-loop plots. Also I am still somewhat confused to what the exact relation between the steady state value of a time step response and the static gain in a Bode plot is. More than often in this type of exercise one is told to observe the static gain of the bode plots as they can easily identify the possible combinations but I don't really see how.

I've attached the picture of the plots. Sorry for the really bad quality but it's the best I can get at this moment.

Any help would be greatly appreciated.

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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It's a challenging problem. It's a matter of doing the slow slog.

I'd start with case D on the left. I'm not sure whether you are familiar with the 's' [s=jω] domain forms but ....

This looks like a single pole system with a transfer function

G=a/(s+a)

The closed loop form will be (assuming unity feedback)

Gc=a/(s+2a)

The most likely closed loop candidate would be item E

For a unit step input the 's' domain output response would be

Vout=a/(s(s+2a))

which translates to a time domain form

Vout(t)=0.5(1-e^(-2at))

For which the likely step response candidate would be C

That's about the only process I can think of.

Perhaps now consider some second order options.

I'd look at [unit?] step response E which would be typical for an underdamped second order system of type

$G_c(s)=\frac{{\omega_0}^2}{s^2+2as+{\omega_0}^2}$

One could then back convert this to the open loop form - presumably with unity feedback(??). However I'm finding it difficult to identify an open loop form that would match with the aforementioned conversion.

Last edited: Oct 1, 2011
3. ### Triton Thread Starter New Member

Sep 30, 2011
5
0
I checked the exercise again and it does not mention whether it is a unity feedback or not nor if it is a unit step response or not. I am familiar with the s=jw domain thing and it's relation to the frequency response however.

I checked around the net a bit regarding the DC gain of a system. The DC gain is merely the value of the transfer function as s approach zero which would be equal as to when t approach infinity according to the final value theorem so at least that has been cleared for me.

I managed (with some very minor problems) to relate most of the closed-loop bode plots to the step responses correctly based on several factors. However then it is a matter of relating the remaining open-loop plots with them. Based on the DC gain only I have the following relation:

G_o(iw) = open-loop system at frequency w
G_c(iw) = closed-loop system at frequency w

If I want the gain of the closed-loop system at a certain frequency w I would have the following relation:

G_c(iw) = |G_o(iw)|/|1+G_o(iw)|

If G_o have large values at low frequencies this would give me a value close to 1 for G_c. The only open-loop plot with high values as the frequency gets lower is B, the corresponding closed-loop plot would be C as it pretty much start at 1 (compared to the other plots) for low frequencies and based on the relation I mentioned above regarding the DC gain we would have a time response that settles at 1 so B-C-E. That is the only open-loop bode plot I've managed to relate to so far.

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
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OK - some further comments ....

You didn't pick me up on my error regarding case D open loop. Since this has a total 180° phase shift it will be a second order (not first order) case of the general form

G(s)=a*b/[(s+a)*(s+b)]

and hence

Gc(s)=ab/(s^2+(a+b)s+2ab) - looking like closed loop case E

which would yield a step response of the type C.

So the result would be the same as I had proposed. i.e. D-E-C.

I'd agree with your B-C-E reasoning. The confusion for me was the phase plot. But I guess I was stuck on a second order system - which may not be the case.

I would probably also link the others as

C-A-B.

A-D-D

E-B-A

Last edited: Oct 1, 2011
5. ### Triton Thread Starter New Member

Sep 30, 2011
5
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I might as well post the solution here so it can be compared.

A-E-C
B-C-E
C-A-B
D-D-D
E-B-A

So three right out of five. I am not sure though how you exactly came to the conclusion of C-A-B or E-B-A. Here's my assumption though:

C-A-B. The step response B goes toward zero as t goes toward infinity, the corresponding closed-loop bode plot would be a gain that goes toward zero as s approach zero and the only one doing so is closed-loop plot A. For the gain of it to be zero the gain of the open-loop plot would obviously have to be zero when w->0

I'm not sure though if this is how you got the answer for it.

As for E-B-A...I'm clueless on that one. To begin with I can't use the method of looking at the gains as I did previously. The step response of A settles down at a value closer to zero while the closed-loop plot B is closer to 1 for lower frequencies so:

DC gain = lim_t->inf (y(t)) = lim_s->0 (s*Y(s))

doesn't seem to really add up here unless I am being stupid and missing something obvious.

Also why is it again that a phase shift of 180° corresponded to a second order case?

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
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At least I correctly matched the closed loop Bode plot and the step responses.

Yes I agree with your C-A-B reasoning.

Regarding E-B-A I looked at the frequency at which the open loop Bode magnitude plots started to drop away. The highest damped natural frequency component in the step response occurs in step response A. The highest roll off frequency in the open loop Bode plot occurs in E. Ditto for closed loop B Bode plot.

I can probably see why D-D-D is correct rather than A-D-D.

Given an open DC gain of unity is required by the D open loop response, one could reason that a possible (under-damped) function is

$G(s)=\frac{a^2+b^2}{(s+a+jb)(s+a-jb)}$

which with unity feedback translates to a closed loop form

$G_c(s)=\frac{a^2+b^2}{(s+a+jb)(s+a-jb)+(a^2+b^2)}$

which will have DC gain of

$\frac{a^2+b^2}{2(a^2+b^2)}=0.5$

therefore pointing to unit step response D.

The problem for me is reconciling the Closed Loop response choice of D - which doesn't show a likely DC gain of 0.5. I guess one has to think about other possible transfer functions and feedback factors - given these are unspecified.

It's more likely the transfer function for open loop A is of the (damped) form

$G(s)=\frac{K}{(s+a)(s+b)}$

with K≠ ab

Leading to a closed loop form E and step response C

As I indicated earlier this isn't a trivial exercise and requires some careful thinking. I haven't spent much time thinking about such problems in many years. Perhaps others might be able to offer a clearer perspective.

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I've attached a 'concrete' example for a B-C-E type connection of Bode plots & step response.

You'll note my curiosity about the open loop phase plot with respect to plot B on the Figure 5.5a you attached. Your post figure shows a phase shift starting at ~ -90° at low frequency and then trending to less than -180° (perhaps -270°) at high frequency, whereas in my plot the angle shift is within -90° to -180°, a result that makes more sense to me.

The open loop transfer function was set as

$G(s)=\frac{25}{s*(s+2)}$

• ###### B-C-E type.png
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Last edited: Oct 2, 2011
8. ### Triton Thread Starter New Member

Sep 30, 2011
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I can see how step response A is the highest damped one and by excluding several of the closed-loop plots I can relate it to closed-loop plot B. Also I do see that open-loop plot E does have the highest roll of frequency...but I don't see how you relate it to the corresponding step response and closed-loop plot, I think this is where my particular weakness lies in this exercise. Basically what factors could I compare between the open-loop bode plot and the closed-loop plot (together with step response). For example does the roll of frequency in the open-loop plot relate to the roll of frequency in the closed-loop plot? Or when I look at the open-loop plot, should I then relate it to a step response and then relate the step response to the closed-loop bode plot? Or perhaps I should try and see what kind of system the open-loop bode plot describes (the way you did, damped/under-damped) and then try to go from there? I guess there isn't exactly one method of solving this which makes this question hard to answer.

Thanks by the way for your help so far (and thanks for the example you gave of B-C-E, it helps!)

If you are interested then I could write down the description given in the solution.

Also there are so many relations between various factors that things start to get really confusing at this stage (not in this particular exercise but in general). It is challenging

----------
Of course one way of relating the open-loop plot to the closed-loop plot was through the dc gain that I mentioned but only this relation is not sufficient, obviously there are more factors than just the DC gain and as seen this method does not always give an answer.

Last edited: Oct 2, 2011
9. ### Triton Thread Starter New Member

Sep 30, 2011
5
0
Well I think I may have found the answer to one of the questions I asked in some old notes. It seems that the gain cross over frequency in the open-loop bode plot should be close to the value of the bandwidth in the closed-loop bode plot.

Furthermore an additional relation would be the decrease in phase margin in the open-loop bode plot. From what I remember a smaller phase margin meant that one got closer to an unstable closed-loop system if one were to look at a Nyquist plot of an open-loop system. So the smaller the phase margin is the larger the overshoot in the step response plot would be and thus a larger resonance peak in the closed-loop bode plot. Also this should mean that the phase margin should affect the settling time in the step response as well.

There's also the steady state error which would depend on the DC gain of both the open-loop and closed-loop but since the exercise doesn't really mention what kind of step response it is I don't think it helps much in this case but I could be wrong.

This is my assumptions on the relations between them so far, which of course could be entirely wrong. What do you think of it?

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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I thought an informative exercise might be to 'create' an example for the case C-A-B.

As a start I assumed a time domain step response type for type B as

$h(t)=exp^{-at}sin(\omega t+\phi)$

After some 'fiddling' with values I took a simple function

$h(t)=exp^{-3t}sin(t+\frac{\pi}{20})$

which gave a similar response form to that shown.

If this is a unit step response it's possible to specify the corresponding 's' domain form as

$H(s)=\frac{cos(\frac{\pi}{20})+(s+3)sin(\frac{\pi}{20})}{(s+3)^2+1}$

for which the corresponding closed loop transfer function would be

$G_c(s)=sH(s)$

$Gc(s) =\frac{ 1.4569917s + 0.1564345s^2}{10 + 6s + s^2}$

It's possible to show this has a maximum value of 0.2579552 at a frequency of 0.564Hz.

The Closed Loop bode magnitude plot A shows a maximum of 1.0. So the Function Gc(s) needs to be adjusted (multiplied) by a factor of 1/0.2579552 = 3.8766428 to 'normalize' the maximum magnitude.

So the adjusted closed loop function then becomes

$Gc'(s) =\frac{ 3.8766428(1.4569917s + 0.1564345s^2)}{10 + 6s + s^2}$

or

$Gc'(s)=\frac{ 5.6482365s + 0.6064405s^2 }{10 + 6s + s^2}$

This then leads to an open loop transfer function which hopefully looks something like option C.

This is found from G=Gc'/(1-Gc') and becomes

$G(s)=\frac{14.351673s + 1.5409121s^2}{ 25.409121 + 0.8938000s + s^2 }$

Using the above G(s) as the starting point, the corresponding open loop bode plot, closed loop bode plot and unit step response are shown in the png attachment. The forms look reasonably similar to sequence C-A-B in figure 5.5a

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11. ### exzacklyright New Member

Nov 19, 2011
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I've tried reading through all this and am still way lost... I got a question like this on my test coming up. Is there any SIMPLE way to go about comparing the 3 graphs?

Got a higher quality image triton?