I am given a half Nyquist plot as below

enter image description here

The dashed line stands for the asymptote of the locus and the poles are 0 or negative at real parts, besides there is no double root in the transfer function.

I learned the steps of plotting the Nyquist plot from an open-loop transfer function. Yet, I have no idea how to reverse this process by the steps I know.

I noticed that the interception with the real axis is -0.5, and the center of this circle-like plot is -0.25, plus this plot is usually a third-order transfer function and I believe there is a pole at 0.

As I know, the plot will intersect with the real axis only if the imaginary part is equal to 0 in the equation, how should I derive the function?

  • \$\begingroup\$ A Nyquist plot doesn't (intentionally) contain frequency information so, it's impossible to do. Presumably the arrow indicates frequency increasing. \$\endgroup\$
    – Andy aka
    Commented Jul 13, 2023 at 11:09
  • 1
    \$\begingroup\$ Hmm...that is thought-provoking. Because I am facing a situation where I have to figure out the transfer function based on the diagram and check its stability based on the coefficients of the characteristic equation. \$\endgroup\$
    – Sonamu
    Commented Jul 13, 2023 at 12:39

2 Answers 2


I think, it should be possible to find the corresponding transfer function H(s) - however, first separately for the real as well as imaginary part.

We know that

  • H(s) will be of 3rd order (with a pol at s=0).

  • Hence: H(s)=K/[s(1+As+Bs²)]

  • R[H(w=0)]=-0.25

  • Im [H(w=0)]=-infinite

  • Im[H(w=wi)]=0 and and R[H(w=wi)]=-0.5

  • R[H(w=infinite)]=Im[H(w=infinite)]=0

So we have three unknowns and three equations to find K, A and B. However, at first we have to split H(jw) into the real as well as imaginary part.

  • \$\begingroup\$ Yes, I think it should be solvable. How can I be sure that the $$H(s) = K/[s(Bs^2+As+1)]$$ instead of $$H(s) = K/[s(Bs^2+As+C)]$$ \$\endgroup\$
    – Sonamu
    Commented Jul 13, 2023 at 12:56
  • \$\begingroup\$ If I replace H(s) with H(jω), it will look like this $$ \begin{align*} H(jω) & = \dfrac{-K}{(Bω^2)+j(Aω^3-ω)} \\ & = \dfrac{-K}{(Bω^2)+j(Aω^3-ω)} \cdot \dfrac{(Bω^2)-j(Aω^3-ω)}{(Bω^2)-j(Aω^3-ω)} \\ & = \dfrac{-K[(Bω^2)-j(Aω^3-ω)]}{(Bω^2)^2+(Aω^3-ω)^2} \end{align*} $$ But, how can I list the equation based on the conditions listed? It looks like they are undefined or converges to 0. \$\endgroup\$
    – Sonamu
    Commented Jul 13, 2023 at 12:59
  • \$\begingroup\$ First question: Dividing the whole expression by "C" we get the "1" in the denomionator. Of course, all constants (A,B,K) change their values. Your 2nd comment: I think that A and B are exchanged in your expression. \$\endgroup\$
    – LvW
    Commented Jul 13, 2023 at 14:33
  • \$\begingroup\$ Yes, I unintentionally changed them. I will try to follow your ideas to find the exact open-loop transfer function. \$\endgroup\$
    – Sonamu
    Commented Jul 16, 2023 at 8:29

The transfer function starts with an infinite value with phase \$-90 {}^{\circ}\$. Therefore it has a pole at the origin.

When the frequency is infinity it's phase is \$-270 {}^{\circ}\$. This implies that it has two other poles.

Thus the general form of the transfer function is $$tf=\frac{k}{s (s+p1) (s+p2)}$$

Substitute \$s=i\ w\$ and split it into its real and imaginary parts.

{re, im} = ComplexExpand[ReIm[tf /. s -> I w]]

$$\left\{-\frac{k \text{p1}}{\left(\text{p1}^2+w^2\right) \left(\text{p2}^2+w^2\right)}-\frac{k \text{p2}}{\left(\text{p1}^2+w^2\right) \left(\text{p2}^2+w^2\right)},\\\frac{k w}{\left(\text{p1}^2+w^2\right) \left(\text{p2}^2+w^2\right)}-\frac{k \text{p1} \text{p2}}{w \left(\text{p1}^2+w^2\right) \left(\text{p2}^2+w^2\right)}\right\}$$

Now we have three conditions. The real part at zero frequency is -0.25, and the real and imaginary parts at some frequency \$w\$ are -0.5 and 0.

eqs = {(re /. w -> 0) == -0.25, re == -0.5, im == 0}

$$\left\{-\frac{k}{\text{p1}^2 \text{p2}}-\frac{k}{\text{p1} \text{p2}^2}=-0.25,\\-\frac{k \text{p1}}{\left(\text{p1}^2+w^2\right) \left(\text{p2}^2+w^2\right)}-\frac{k \text{p2}}{\left(\text{p1}^2+w^2\right) \left(\text{p2}^2+w^2\right)}=-0.5,\\\frac{k w}{\left(\text{p1}^2+w^2\right) \left(\text{p2}^2+w^2\right)}-\frac{k \text{p1} \text{p2}}{w \left(\text{p1}^2+w^2\right) \left(\text{p2}^2+w^2\right)}=0\right\}$$

These are 3 equations in 4 unknowns \$\{k, p1, p2, w\}\$. We cannot determine an unique transfer function unless we know the frequency at which the plot crosses the \$\{-0.5, 0\}\$ point.

If it crosses at \$w=1\$, the transfer function will be

Chop@ExpandDenominator[tf /. NSolve[Join[eqs /. w -> 1, {k > 0}], {k, p1, p2}][[1]]]

$$\frac{0.353553}{s^3+0.707107 s^2+1. s}$$

If it's \$w=2\$, the solution will be

Chop@ExpandDenominator[tf /. NSolve[Join[eqs /. w -> 2, {k > 0}], {k, p1, p2}][[1]]]

$$\frac{2.82843}{s^3+1.41421 s^2+4. s}$$

So there are infinite possibilities.

  • 2
    \$\begingroup\$ Why did you not use the THREE contants k, A and B (instead of the poles p1 and p2)? From the logic point of view there cannot exist "infinite possibilities" which lead to the shown curve. \$\endgroup\$
    – LvW
    Commented Jul 13, 2023 at 14:40
  • \$\begingroup\$ The procedure looks reasonable, but I agree with @LvW that there should not exist infinite possibilities. \$\endgroup\$
    – Sonamu
    Commented Jul 16, 2023 at 8:32
  • \$\begingroup\$ Plot the Nyquist curve of both the transfer functions I have shown and see if you get the same result. \$\endgroup\$ Commented Jul 16, 2023 at 13:07

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