1
\$\begingroup\$

I have been given the following Nyquist polar plot of an unknown transfer function, along with the fact that this system is openloop stable:

enter image description here

Now I can directly see this is a type 1 system. - So there is a non-zero, limited error when given a constant input. A simple system that shows a similar shape is for example (constants are different of course). $$H_1 = \frac{1}{s+1}\frac{1}{s^2 + s + 1}$$

Now it is asked from me to determine -from the graph- what the steady state error will be, if this system is inputted into a unity feedback loop: $$H_2 = \frac{H_1}{1+H_1}$$

|improve this question
\$\endgroup\$
  • \$\begingroup\$ Do you know how the steady-state error is defined - and how it can be simply found? \$\endgroup\$ – LvW Aug 8 '16 at 12:31
1
\$\begingroup\$

Hints: -

  • What is the open loop gain at DC (frequency = 0 Hz)?
  • What is the closed loop DC gain after applying unity gain negative feedback?
  • Can you tell the system will be stable when the loop is closed? If you can, what will the phase margin be approximately?
  • What is the gain margin?
  • Do you understand how these margins work in a bode plot?

After further questions about bode versus Nyquist I'm showing a simple equivalence: -

enter image description here

Can you see how the phase margin is seen on the Nyquist plot compared to how it is seen (easier of course) on the Bode plot?

Can you see that the unity gain circle in the Nyquist plot can predict the phase margin and gain margin?

|improve this answer
\$\endgroup\$
  • \$\begingroup\$ DC? Sorry this is a system describing a mass-spring system. Do you mean by DC a unit step/zero frequency input? I understand how everything works in a bode plot, I just can't seem to figure out how to to read these polar plots. - I know the system is closed loop stable since it doesn't go "over" the point "-1+0j" but that's about it? \$\endgroup\$ – paul23 Aug 8 '16 at 12:53
  • \$\begingroup\$ @paul23 DC in your system means constant steady state, I mean apply any input which after sometimes becomes a constant -yes a Heaviside would do- . Then wait till transient is over, you know it will since your system is stable, and finally have a look to output. Displace your input say 1cm, wait for transient to be over, measure output displacement. \$\endgroup\$ – carloc Aug 8 '16 at 14:22
  • \$\begingroup\$ @paul23 - hehe - this is an EE site and DC is jargon for 0 Hz - I've added something that hopefully will help but it seems the other answer has spoon fed you the goods! \$\endgroup\$ – Andy aka Aug 8 '16 at 15:06
  • \$\begingroup\$ @Andyaka Wait, that clears everything up for me, thanks. However I now wonder, does that mean I can never have phase instability while gain is still stable? \$\endgroup\$ – paul23 Aug 9 '16 at 0:43
  • \$\begingroup\$ @paul23 in more complex transfer functions you can have either or both. \$\endgroup\$ – Andy aka Aug 9 '16 at 7:01
0
\$\begingroup\$

This is a type 0 system. (If it is type 1 or higher the plot will go to infinity). Thus, there is a finite steady-state error for a unit step input and infinite error for ramp and higher order inputs.

From the plot it can be seen that the DC gain is \$K=3\$. The steady state error to a unit-step is \$\frac{1}{K+1}=\frac{1}{4}\$. (for the derivation of this formula see Ogata).

|improve this answer
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.