# Deriving a transfer function from Nyquist plot

I am given a half Nyquist plot and exercise as below

I know how to plot a Nyquist curve for a given transfer function but I have trouble doing it the other way.

I stumbled upon this post Deriving transfer function from nyquist plot and it seems intuitive to split the function in real and imaginary parts, write down special points of interest and create a equation system of some sort to calculate n, m and bring a, b and k into relation with eachother and name a combination of the three that makes $$\ L(s) \$$ match the given plot.

I don’t know if $$\ L(s) = G_1(s) \cdot G_2(s) \$$ is supposed to be stable in this exercise, but it is given that $$\ G_2(s) \$$ is BIBO stable. Assumed that $$\ L(s) \$$ is meant to be stable, we can observe that theres 3π circulations around the critical point $$\ (-1/0) \$$ and according to the Nyquist criterium $$\ L(s) \$$ must have one pole on the imaginary axis and one pole in the left open plane. But somehow, according top the post linked above, $$\ L(s) \$$ needs to be of order 3, so we have two poles(?) / a double pole(?) instead of a single one in the left open plane.

Points of interest:

$$\ L(s) \$$ will be of 3rd order (with a pole at $$\ s=0 \$$).

$$\ L(s) = \frac{k}{s(s²+as+b)} \$$

$$\ Im[L(w=0)] = -infinite \$$

$$\ Im[L(w=wi)] = 0 \$$ and $$\ R[L(w=wi)] = -10 \$$

$$\ R[L(w=infinite)] = Im[L(w=infinite)] = 0 \$$

$$\ L(s) \$$ with $$\ s = jw \$$ split into real and imaginary part should look like this

By playing around in Wolfram Alpha I found $$\ n=1, m=1, a=1, b=1, k=10 \$$ looks similar to the given plot, but how is this exercise supposed to be solved?

• $G_2(s)$ being BIBO stable means that $a,b>0$. From that x-axis crossing (and the particularity of the two transfer functions) we have that either $3(n+m)=2 \, mod \, 4$ or $3(n+m)=0 \, mod \, 4$ depending on the sign of $k$. We can also determine that $10 = \frac{|k|}{a^m b^{(n+m)/2}}$
– jDAQ
Commented Aug 20, 2023 at 21:29
• @jDAQ Thanks for your comment! How did you come up with the last equation? Same way as i did, but keeping it more general than i did with having $n=m=1$ at this point? And just out of curiosity: How did you come up with the mod formulas? I sadly dont understand how you got from the x-axis crossing point and the particularity of the two transfer functions to these two $mod$ conditions. Wow! Commented Aug 20, 2023 at 21:54

I claim that the following two transfer functions have identical Nyquist plots.

$$F_1(s) = \frac{1}{s+1}$$ $$F_2(s) = \frac{-1}{s^3-1}$$

The Nyquist plot for both is a unit diameter semi-circle going clockwise from (1,0) to (0,0)

The Bode plots for $$\F_1(s)\$$ and $$\F_2(s)\$$, on the other hand, are different.

From any (non-constant) transfer function, a different transfer function can be generated which has the same Nyquist plot. Thus, without further assumptions, or further information, it is impossible to determine a unique transfer function from a Nyquist plot.

In practice, however, one may know, or might assume, that a transfer function takes a certain form.

With the current problem, we are given some restraints on the transfer function.

1. We are told that the transfer function takes the form:

$$F(s) = \frac{k}{s^n(s^2+as+b)^m}$$

1. We are told that

$$F_2(s) = \frac{k}{(s^2+as+b)^m}$$

is BIBO stable

1. We are told that the Nyquist plot intersects the x axis at (-10,0)

The requirement that $$\F_2\$$ be BIBO stable entails that the real part of roots of $$\s^2 + as + b\$$ must all be negative.

We will rewrite $$\s^2 + as + b\$$ in terms of its roots:

$$s^2 + as + b = (s-r_1)(s-r_2) = s^2 -(r_1+r_2)s + r_1r_2$$

If the roots are real and both negative, then

$$a^2 \ge 4b$$ $$b=r_1r_2>0$$ $$a=-(r_1+r_2)>0$$

If the roots are complex, and the real parts are negative, then

$$a^2 \le 4b$$ $$a>0$$ $$b\ge\frac{\sqrt{a}}{2}>0$$

So, in either case, for $$\F_2(s)\$$ to be BIBO stable,

$$a>0$$ $$b>0$$

• Thanks for your answer. Maybe i did not state it clear enough but the exercise is only to give exact values for m and n, while k, a, b are only ment to be set in relation, so that you are able to name one of the many correct combinations of values for k, a, b. Have you seen my approach above? What do you think of it and do you maybe see explanations / proof for steps and assumptions, i could not find any for? Commented Aug 20, 2023 at 14:59
• I claim that if (m,n) = (1,1) has solutions, then there are other pairs (m',n') that also have solutions. It may take me some time to find them, but I will try upon request. I think it is easiest to just make the assumption that (m,n)=(1,1) because they are low value exponents, and it is a reasonable assumption both for practical circuits, and also for academic problems. Commented Aug 20, 2023 at 15:15
• Oh, now i get what you tried to tell me with your claim. Well, that answers my question how to find the solution for n and m. I also checked the exercise once more and the exercise is to just find a valid combination of n, m, a, b and k, so the nyquist plot intersects the real axis in (-10/0) and matches the given plot. So you can just go with n=1 and m=1, but how would you know that this yields a valid solution immediately after reading the exercise? It would be very great if you could check my approach / solution and solve my unclarities or if you come up with your own solution. Thanks!!!! Commented Aug 20, 2023 at 15:24
• To be clear: You can just show me how to get a valid combination of a,b,k or give me feedback on my approach. Thanks a lot for your dedication in helping me out! Commented Aug 20, 2023 at 15:33

As Math Keeps Me Busy stated, n=m=1 seems to be the easiest, most trivial solution, although there sure are other valid combinations. However, since the goal is just to find a valid combination for this system that has a matching nyquist plot.

I think $$\ L(s)\$$ is meant to be stable as a whole (?), since we are given the information that $$\ G_2(s) \$$ is BIBO stable, which only becomes useful, if $$\ G_1(s) \$$ is stable as well.

Because then we can make use of the nyquist criterium to see how many poles (there are no zeros in $$\ G_1(s) \$$ or $$\ G_2(s)\$$) our transfer function should have.

Please feel free to share other perspectives on this. Maybe there lies information in "$$\ G_2(s) \$$ is BIBO stable", which i am not aware of.

As if this wasnt enough unclarity i am unsure if real poles count once or by their multiplicity for the nyquist criterium. Maybe this whole stabilty approach is wrong.

Assuming $$\ L(s)\$$ is stable, we can obtain information with help of the nyquist criterium. Since we have $$\ 2π(p_a + \frac{p_r}{2}) = 3π \$$, $$\ L(s)\$$ must have one pole in the origin ($$\ p_r=1\$$) and another one in the open left half plane ($$\ p_a=1\$$), right?

(I know $$\ p_r=3\$$ and $$\ p_a=0\$$ fulfill the nyquist criterium as well and i have no clue why this solution can be ignored)

Thus $$\ L(s) = G_1(s) \cdot G_2(s) = \frac{k}{s(s^2+as+b)} \$$

Let $$\ s=jw \$$ and split $$\ L(jw) \$$ into real and imaginary parts

• $$\ Re[L(jw)]= \frac{-kaw^2}{a^2w^4+(wb-w^3)^2} \$$

• $$\ Im[L(jw)]= \frac{-k(wb-w^3)}{a^2w^4+(wb-w^3)^2} \$$

The only useful information in the plot is

$$\ L(j\tilde{w})= \frac{-ka\tilde{w}^2-k(\tilde{w}b-\tilde{w}^3)}{a^2\tilde{w}^4+(\tilde{w}b-\tilde{w}^3)^2} = -10 \$$ for a fixed frequency $$\ \tilde{w} \$$ This implies $$\ k\neq0 \$$, since else $$\ Re[L(jw)]= 0 \neq -10 \$$ Thus $$\ b\overset ! = \tilde{w}^2\$$ in order to have $$\ Im[L(jw)]= 0 \$$

We are left with the real part. which has to equal -10

$$\ Re[L(j\tilde{w})]= \frac{-kaw^2}{a^2w^4+(wb-w^3)^2} \overset ! = -10\$$ or respectively with $$\ b\overset ! = \tilde{w}^2\$$

$$\ Re[L(j\tilde{w})]= \frac{-kab}{a^2b^2} = \frac{-k}{ab} = -10\$$

A solution is given by $$\ k=10, a=b=1 \$$

$$\ L(s)= \frac{10}{s(s^2+s+1)} \$$

I would be happy if you guys could help me improve this solution and get rid of unclarities and assumptions.