Why is it true that the velocity error constant can be read from the Bode diagram at the frequency w=1 rad/s?

I am trying to find a proof as to why we can read the velocity error constant $$K_v$$ of a closed-loop directly from the Bode diagram. That is, why the following is true: $$K_v = \frac{|L(j\omega)|}{\omega}$$ and therefore $$K_v = |L(j\omega)|_{\omega=1}$$. Here $$K_v$$ is the velocity error constant of the closed-loop transfer function of a system with loop transfer function $$L(s)$$.

For the position error constant it is relatively obvious that we can read it from the low-frequency asymptote since by the very definition of $$K_p$$ we take the limit as the frequency approaches zero: $$K_p=\lim_{s\rightarrow 0} L(s)= \lim_{\omega\rightarrow 0} L(j\omega)$$. However, the same reasoning is not readily applicable to $$K_v$$.

The static velocity error constant is obtained from the initial -20db/decade segment of the Bode plot or the extension of that segment.

It can be obtained as either the intersection of the segment or its extension with the $$\omega=1$$ vertical line or the $$0$$ dB horizontal line.

Here are a couple of examples.

For the proof that it can be obtained as the intersection of the segment and the $$\omega=1$$ line, let's start with the definition $$k_v=\lim_{s \to 0}\ s \ L(s)$$ and let $$\omega_\epsilon$$ be a very small value of $$\omega$$ which is much smaller than 1 and from which we can compute $$k_v$$.

That is, $$k_v= |j\ \omega_\epsilon L(j \ \omega_\epsilon)|$$

Consider the -20db/decade segment. It starts say from $$\omega_\epsilon$$ where the magnitude is $$|L(j \ \omega_\epsilon)|$$. When that segment reaches $$\omega=1$$, the frequency has increased by a factor of $$\frac{1}{\omega_\epsilon}$$. So the magnitude should have decreased to

$$20 \log_{10} |L(j \ \omega_\epsilon)| - 20 \log_{10} \frac{1}{\omega_\epsilon}$$

$$=20 \log_{10} |L(j \ \omega_\epsilon)| - 20 \log_{10}| \frac{1}{j \ \omega_\epsilon}|$$

$$=20 \log_{10} |{j \ \omega_\epsilon} L(j \ \omega_\epsilon)|$$

$$= 20 \log_{10}k_v$$

The other proof, that it can be obtained from the intersection of the segment and the $$0$$ dB line is similar. Let's assume that the frequency when it intersects the $$0$$ dB line is $$\bar{\omega }$$.

$$20 \log_{10} |L(j \ \omega_\epsilon)| - 20 \log_{10} \frac{\bar{\omega }}{\omega_\epsilon}=0$$

$$\log_{10} \frac{|L(j \ \omega_\epsilon)|\omega_\epsilon}{\bar{\omega }}=0$$

$$\bar{\omega } = |L(j \ \omega_\epsilon)|\omega_\epsilon=|L(j \ \omega_\epsilon)||j \ \omega_\epsilon|= |j \ \omega_\epsilon L(j \ \omega_\epsilon)|=k_v$$

• Thank you, that makes sense! Oct 30, 2023 at 20:53