What is the physical meaning of using negative frequencies while making Nyquist plots?

I know that we do a mapping from $s-$ plane to $f(s)$ plane, and we move along a contour. What I don't get is that Is the Nyquist plot purely a mathematical construct? Is there no relation between the frequency on plot and actual frequency? If there is, then what is the point of plotting negative frequencies,Is it just so that we could count the number of encirclements?

I am not able to grasp the idea behind it.

  • 1
    $\begingroup$ Negative frequencies and complex numbers are only to keep undergrads busy so that professors and grad students can get on with their research. $\endgroup$ Commented Apr 10, 2016 at 14:21

2 Answers 2


A complete Nyquist diagram can tell if the closed loop system is stable. Namely when going from negative infinite imaginary frequency to positive infinite imaginary frequency and back again to negative infinite imaginary frequency over the D contour you encircle the entire right half plane on the input.

When drawing the Nyquist diagram of a system, then each unstable zero and pole will cause a clockwise- and anticlockwise-encirclement of the origin respectively.

If you write the open loop as a quotient of two polynomials

$$ L(s) = C(s) H(s) = \frac{N(s)}{D(s)}, $$

then the closed loop becomes

$$ \frac{L(s)}{1 + L(s)} = \frac{N(s)}{D(s) + N(s)}. $$

If you draw the Nyquist diagram of $L(s)$ and look at the encirclements of the minus one point you are basically looking at the unstable zeros and poles of

$$ 1 + L(s) = \frac{D(s) + N(s)}{D(s)}. $$

The number of clockwise encirclements of the minus one point will be equal to the number of unstable roots of $D(s)+N(s)$ minus the number of unstable roots of $D(s)$. The first are also the number of unstable poles of the closed loop system and the second are the number of unstable poles of the open loop system.

For every physical system the part of Nyquist diagram of the negative frequencies is a mirror image of the positive frequencies. So it is not required to draw them both, but it can help identify the total number of encirclements.

PS: Also be aware of encirclements you get infinity. For example $1/s^3$ does have one clockwise encirclement of the minus one point.

  • $\begingroup$ Yes I understand the encirclements help decide the stability. So the negative frequencies don't actually correspond to anything, right? $\endgroup$
    – Manish
    Commented Apr 12, 2016 at 9:49
  • $\begingroup$ @Manish The negative frequencies correspond to "half" (if you forget about the D contour) of the encirclement of the right half plane, so also cause half of the encirclements (if they are present) of the minus one point. If you consider symmetry (so double the number of encirclements you see) then you are correct that you do not have to draw it to know if the closed loop is stable. But the negative frequencies still correspond to half of the encirclements, so it would be wrong to say that they do not correspond to anything. $\endgroup$
    – fibonatic
    Commented Apr 12, 2016 at 10:51

The negative frequencies are not physical, neither do they convey any new information. The value at the negative frequency is the complex conjugate, and hence the plot is symmetric about the real axis. If you know one you can get the other as a conjugate.

The value in plotting the negative values (or going around the entire contour) is in doing the stability analysis.


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