By using the open loop transfer function for a control system, how would you then sketch the Nyquist plot by hand?

I'm aware you'd substitute any \$s\$ term for \$j\omega\$ and then rationalise the denominator by moving the \$j\$ terms to the numerator. But from this point forward I am unable to get to the point where I can plot the rough shape of the Nyquist plot. Any break-down of the steps would be greatly appreciated.

  • \$\begingroup\$ Can you draw a Bode plot by hand? \$\endgroup\$
    – John D
    Jan 19, 2017 at 23:15
  • \$\begingroup\$ Yes i can for certain systems, control is quite a new topic for me which i'm studying in my final year of engineering. I need to be able to plot Nyquist plots for an exam next week and i'm struggling to get my head round them \$\endgroup\$ Jan 19, 2017 at 23:17
  • 1
    \$\begingroup\$ So assuming you draw the Bode plot you have gain and phase information- How would you translate that information to a Nyqvist plot? \$\endgroup\$
    – John D
    Jan 19, 2017 at 23:33
  • \$\begingroup\$ Using the bode you could get phase values for corresponding gain values and plot them on the complex plane taking the phase angle from the real axis. But we won't have access to that information to sketch the plot we need to be able to determine quickly its general shape and where it crosses the axes and weather or not it approaches from a negative or positive quadrant \$\endgroup\$ Jan 20, 2017 at 0:01
  • \$\begingroup\$ This sounds like a different question to what you actually asked. What it sounds like you need is to be able to quickly recognise how a given transfer function would be illustrated as using a Nyquist plot, rather than calculating each point. Perhaps you should update your question. \$\endgroup\$
    – loudnoises
    Jan 30, 2018 at 8:07

2 Answers 2


A system with smooth rolloff and numerous poles will become a spiral, if plotted on a polar plot.

If you have only 90 degree phase shift, then the low-freq gain, assumed to be huge, starts far out from the center point (the gain-is-zero point), rotates over to the 90 degree region and plunges down to ZERO. There is no stability issue here.

The classic purpose of Nyquist is to examine how this polar plot behaves near the [Mag=1/Phase=-180 degree] point. For stability, perhaps marginal and ringy but stable, the plot needs to be inside the magnitude=1 point at 180 degrees.


As you say, get the transfer function in the form a + jb. This may involve multiplying the numerator and denominator by the complex conjugate of the denominator.

Once you have the transfer function in the form a + jb then draw up a table of values for the magnitude and phase of the transfer function as w increases from 0 to infinity.





Once you have your table of magnitude and phase values you can plot them on a complex plane (GH plane).

0 degrees starts at the positive real axis and increasing phase lag goes clockwise. Magnitude is the distance from the origin.

Once you have plotted your table of values, join them all up and you have your Nyquist plot.

But if you just need to ascertain if the closed loop system is stable then derive the closed loop transfer function from the open loop transfer function (this is straight forward for a unity feedback system) and examine the denominator to see if there are any closed loop poles on the rhs of the s-plane.

By drawing the Nyquist plot you will not only be able to see if the system is stable but also, if it is stable, how far from instability it is. That is to say you will be able to determine gain and phase margins from the plot.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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