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 $K_v$.


1 Answer 1


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.

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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 $\bar{\omega }$.

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

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

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

  • $\begingroup$ Thank you, that makes sense! $\endgroup$
    – ddmcp
    Oct 30, 2023 at 20:53

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