0
$\begingroup$

This is a line from a paper I've been reading:

The static gain of the closed loop system must be $1$ ($G(1) = 1$) [...]

First of all: I know what gain is, but isn't gain dependent upon frequency? And why does $G(1)$ do the trick of determining the static gain? And what is "static" gain, anyway?

We are talking about a discrete-time closed loop system and $G(z)$ is its transfer function.

$\endgroup$
  • $\begingroup$ The static gain of a closed loop should be 1 usually but does not need to. $\endgroup$ – Ben Jan 12 '18 at 15:16
2
$\begingroup$

Static gain refers to the DC gain. Namely, it would be the ratio of the output and the input under steady state condition.

Due to DC corresponding to $\omega=0$, in the $z$-domain DC would correspond to $z=1$ because $z=re^{j\omega}$, with $r=1$ (i.e. you have to evaluate your transfer function $G(z)$ in the unit circle to get the frequency response).

EDIT:

The frequency response of a system is represented in the Fourier domain, $G(\omega)$. The DTFT of a given discrete sequence can be calculated as:

$$G_F(\omega)=\sum_{n=-\infty}^{\infty}g[n]e^{-j\omega n}$$

On the other hand, the $z$-transform is defined as:

$$G_Z(z)=\sum_{n=-\infty}^{\infty}g[n]z^{-n}$$

It's easy to see that if we evaluate $z=e^{j\omega}$ (the unit circle), then the $z$-transform returns the frequency response.

$\endgroup$
  • $\begingroup$ Thanks for sharing Tendero! Can you provide some detail on why r = 1, or, G(z) has to be evaluated on the unit circle to get the frequency response? $\endgroup$ – Alon Jan 12 '18 at 15:04
  • $\begingroup$ @JoschKraus Check my edit and see if that clears that out for you. $\endgroup$ – Tendero Jan 12 '18 at 15:10
1
$\begingroup$

In $G(z)$, $z$ denotes a complex variable, interpreted here on the unit-circle as $z=e^{i \omega)}$. When $z=1$, this corresponds to the $0$-frequency, or constant signals. But sometimes, people write this as $G(i\omega)$, as in the definition for the static gain in Transfer Functions:

The number $G(0)$ is called the static gain of the system because it tells the ratio of the output and the input under steady state condition.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.