# Nichols plot confusion

The transfer function is $$\F(p)=10/(p^2+2p+1).\$$

Using Wolfram Alpha, the Nichols plot shows that the argument $$\\to -\pi\$$ when $$\\omega \to +\infty\$$.

However, calculating the argument on paper gives $$\-\arctan(2w/(1-w^2))\$$ which $$\\to 0\$$ when $$\w \to +\infty\$$.

Where is the error, mathematically?

Probably no error with the math, but realize that the chart is not a grid. It looks like the chart below. So if charting by hand make sure the grid is a Log scale of the magnitude or $$\20\log_{10}(|G(s)|)\$$ on one axis and phase on the other. If the closed loop transfer function is given, then a nichols plot is the open loop version, so it will need to be converted from closed loop to open loop.

• I didn't try plotting a Nichols grid, the problem is that when I plotted a Nichols PLOT of my transfer function, I got a graphical result that didn't match with the calculus on paper (the limit of the argument when w->infinity).... Jan 14, 2021 at 20:58

Look at the real and imaginary parts:

{re, im} = ComplexExpand@ReIm[10/(p^2 + 2 p + 1) /. p -> I w] // Simplify


$$\left\{-\frac{10 \left(w^2-1\right)}{\left(w^2+1\right)^2},-\frac{20 w}{\left(w^2+1\right)^2}\right\}$$

The real part starts off positive, becomes zero when $$\w=1\$$, and then becomes negative. The imaginary part starts off at $$\0\$$ and then becomes negative. Thus the phase starts at the positive real axis, goes through the fourth quadrant, then the first, and ends up on the negative real axis.

Calling ArcTan without cancelling the sign shows this.

Plot[ArcTan[re, im]/Degree, {w, 0, 20}, PlotRange -> All, Frame -> True, FrameLabel -> {"freq.", "phase"}] If you cancel, you lose the sign information and get the result you are getting. This is not correct.

ArcTan[im/re]


$$\tan ^{-1}\left(\frac{2 w}{w^2-1}\right)$$

Plot[%/Degree, {w, 0, 20}, PlotRange -> All, Frame -> True, FrameLabel -> {"freq.", "phase"}] • I really don't understand your answer. The expression of the argument is the expression of the imaginary?? (Of what??) I read the script but I don't understand the result and your analysis.... Jan 14, 2021 at 20:59
• And I'm not understanding your question. "The expression of the argument is the expression of the imaginary"? Please refer to specific things I have in my answer, so I can follow your question. Jan 14, 2021 at 22:25

We have $$F(j\omega)=\frac{10}{-\omega^2+2j\omega+1}=10\frac{(1-\omega^2)-2\omega j} {((1-\omega^2)+2\omega j)((1-\omega^2)-2\omega j)}$$ $$=10\frac{(1-\omega^2)-2\omega j} {(1-\omega^2)^2+4\omega^2} =10\frac{1-\omega^2-2\omega j} {1-2\omega^2+\omega^4+4\omega^2} =10\frac{1-\omega^2-2\omega j} {1+2\omega^2+\omega^4}$$ $$=\frac{10}{(1+\omega^2)^2}(1-\omega^2-2\omega j).$$ Since multiplication by $$\\frac{10}{(1+\omega^2)^2}\$$ does not change the value of an argument, consider $$\\arg(1-\omega^2-2\omega j)=\alpha \$$. The trigonometric form of $$\1-\omega^2-2\omega j\$$ is $$|1-\omega^2-2\omega j| \left( \underbrace{\frac{1-\omega^2}{|1-\omega^2-2\omega j|}}_{\cos\alpha}+ \underbrace{\frac{-2\omega}{|1-\omega^2-2\omega j|}}_{\sin\alpha}j \right).$$ This implies that $$\lim_{\omega\to+\infty}\cos \alpha= \lim_{\omega\to+\infty} \frac{1-\omega^2}{|1-\omega^2-2\omega j|}= -1,$$ $$\lim_{\omega\to+\infty}\sin \alpha= \lim_{\omega\to+\infty} \frac{-2\omega}{|1-\omega^2-2\omega j|}= 0.$$ Hence, calculating the argument on paper does not give $$\0\$$.

• Thanks for the reply, but I don't understand your reasoning. You showed that alpha is |alpha| x (cos/|alpha|+j sin/|alpha|) and proved that cos-> -1 and sin->0 when w->inf. What about |alpha|, the one that's multiplied in the beginning, |alpha| -> inf. When w->inf. But what does all of this tell us about the argument's limit??? Jan 15, 2021 at 13:12
• @A.Bukhari You showed that alpha is |alpha| x (cos/|alpha|+j sin/|alpha|) - no, it is not so. You are confusing the value $1-\omega^2-2\omega j$ and its argument value $\alpha$.
– AVK
Jan 15, 2021 at 13:33
• @A.Bukhari I have updated the answer to make it clearer what $\cos \alpha$ and $\sin\alpha$ are
– AVK
Jan 15, 2021 at 13:36
• @A.Bukhari Multliplication by $|1-\omega^2-2\omega j|$ does not change the argument because it is real and positive.
– AVK
Jan 15, 2021 at 13:38