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Having the Transfer Function of a discrete system as such: $$H(z) = \frac{0.8}{z(z-0.8)}$$ I am asked to find the Steady State Gain of the system. I have the solution and it simply states:
Steady State Gain:
$$H(1) = \frac{0.8}{1-0.8} = 4$$ I have no idea why this is the case, and the book, the slides, the solutions, my (limited) notes and the internet are seemingly empty on the subject. Is the Steady State Gain of a system always the outcome of the Transfer Function applied to 1? That just sounds ridiculous, especially since I'm not finding any references to it online.
I was chased out of mathoverflow with this question, those guys really hate homework... Then again, who doesn't.

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The "mechanical" result of just plugging in $z = 1$ into the transfer response is essentially a product of two facts. The steady-state gain is (usually, I believe) defined as the (magnitude of the) limiting response as $t \to \infty$ of the system to a unit-step input.

The so-called final-value theorem states that, if the limit $\lim_{n \to \infty} y(n)$ exists, then $\lim_{n \to \infty} y(n) = \lim_{z \to 1} (1-z^{-1}) Y(z)$ where $y(n)$ is the time-domain output and $Y(z)$ is its corresponding $z$-transform.

Now, for the steady-state gain, the input is a unit-step function, so $x(n) = 1$ for each $n \geq 0$. Hence, $$ X(z) = \sum_{n=0}^\infty x(n) z^{-n} = \sum_{n=0}^\infty z^{-n} = \frac{1}{1-z^{-1}} . $$

Using the transfer equation, we get that the $z$-transform of the output is $$ Y(z) = X(z) H(z) = \frac{H(z)}{1-z^{-1}} . $$

(Assuming that the limit $\lim_{n\to\infty} y(n)$ exists) we have that $$ \lim_{n \to \infty} y(n) = \lim_{z \to 1} (1-z^{-1}) Y(z) = \lim_{z \to 1}\, H(z) . $$

The left-hand side is the steady-state value of a step-response (i.e., it is the value of the response as time goes to $\infty$ of a one-unit constant input), and so the steady-state gain is $|\lim_{n \to \infty} y(n)| = \lim_{n \to \infty} |y(n)|$.

Technically, you need to check that the limit exists (which I've tried to emphasize). It seems to me that a sufficient condition would be that all the poles of the transfer response be strictly inside the unit circle. (Caveat lector: I haven't checked that closely at all.)

If this does not sufficiently clarify things, you might try doing Google searches on terms like "dc gain" and "final-value theorem", which are closely related to what you want.

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I asked this question at another stack exchange site. It's fairly simple:

  1. Steady state gain is the gain the systems has when DC is applied to it, which has a frequency of f=0 or omega = 0
  2. The variable z in the z-transform is defined as z = r * exp(j*omega). Set omega to 0 and you have z = r
  3. r = 1 because to get the frequency response, we have to evaluate the z-transform on the unit circle

This is a very brief summary for the answer that has been given here:

https://dsp.stackexchange.com/questions/46337/what-does-g1-1-say-about-a-system#46339

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The term Steady state gain comes up when your input function is the unit step function $u(t)=1$. If your input is the unit step function, then the gain is the system's value at steady state, $t= \infty$. The steady state value is also called the final value.

The Final Value Theorem lets you calculate this steady state value quite easily: $\lim_{t \to \infty} y(t) = \lim_{z \to 0} z*Y(z)$, where $y(t)$ is in the time domain and $Y(z)$ is in the frequency domain.

So if your transfer function is $H(z) = \frac{Y(z)}{X(z)} = \frac{.8}{z(z-.8)}$, you can find $y(t\rightarrow \infty)$ with $\lim_{z \to 0} z*Y(z)$.

First, solve for $Y(z)$ which is your output of the system:

$Y(z) = H(z)*X(z) = \frac{.8}{z(z-.8)}*X(z)=\frac{.8}{z(z-.8)}*\frac{1}{z}$

(recall that $X(z)=\frac{1}{s}$ because $x{t} = u(t)$, the unit step function.)

Now multiplying $Y(z)$ by $z$ to get $\lim_{z \to 0} \frac{z*.8}{z^2(z-.8)}$

One of the $z$-terms cancel and you get $\lim_{z \to 0} \frac{.8}{z(z-.8)}$

You can't plug in $z=0$ to the denominator or you'll get an undefined value. The reason this doesn't work is because the Final Value theorem doesn't work in this case. The roots of the transfer function need to be negative, and here they're positive (with a root of $z=.8$). IDK how to solve it any other way but that's how you would have done it...

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  • $\begingroup$ I'm not an expert but this looks like the final value theorem for continuous-time (s-Laplace domain) not discrete time (z-Transform). $\endgroup$
    – Bill
    Aug 31, 2020 at 2:16
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find steady state gain of transfer function H(s), we let s=0.

since z=exp(sT), to find steady state gain of H(z), let z=exp(0)=1.

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  • $\begingroup$ Why $z=exp(sT)$? $\endgroup$
    – mpiktas
    Aug 17, 2012 at 8:29

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