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.

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

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).


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:


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

  • $\begingroup$ 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

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.


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