How to Find the DC Gain of a Transfer Function (Examples Included)

Contents

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Key learnings:

DC Gain Definition: DC gain is the ratio of the steady-state output to the steady-state input of a control system when given a step input.
Transfer Function: A transfer function represents the relationship between a control system’s input and output using the Laplace transform.
Final Value Theorem: The final value theorem helps find the DC gain by evaluating the transfer function at zero for continuous systems.
Continuous vs. Discrete Systems: DC gain calculations differ between continuous (using G(s)) and discrete systems (using G(z)), but the principles remain similar.
Practical Examples: Examples of first-order systems illustrate how to apply these concepts to find the DC gain in real scenarios.

What is a Transfer Function?

DC gain of a transfer function is the steady-state output divided by a constant input, for a stable linear control system. For a continuous system you evaluate G(s) at s = 0. For a discrete system you evaluate G(z) at z = 1.

A transfer function represents a linear time-invariant (LTI) system in the Laplace or z variable. It is not “Linear Transform Inverse.”

The input (reference or plant input) is transformed to the output through G(s) or G(z). That map holds only for an LTI model with the stated initial-condition assumption.

With zero initial conditions, the Laplace Transform of the input is $R(s)$ and the output is $C(s)$ . The ratio C(s)/R(s) is G(s).

The stored line below writes that ratio as G(s) = C(s)/R(s).

$\begin{align*} G(s)=\frac{C(s)}{R(s)}\end{align*}$

What is DC Gain?

Steady-state gain (DC gain) is that output/input ratio after transients die. For a stable system it is a finite real number. It can be negative. If the system has a free integrator, the step response does not settle and the ratio is not a finite DC gain.

Give a stable system a step and the output settles to a constant. Unstable or marginally stable models do not give a usable finite DC gain this way.

DC gain is (settled output) / (step size). For a unit step that is just the settled output value.

The final value theorem gives that settled value. For a unit step, lim s→0 s · G(s) · (1/s) = G(0), so DC gain is G(0) when every pole of sG(s)/s that the theorem needs is in the open left half-plane.

Time Response of First Order Systems

System order is the order of the governing differential equation (or the number of poles of a strictly proper G(s), after cancelling common factors). First-order models are the usual teaching case.

A common first-order form is G(s) = b0 / (s + a0). The stored line below writes G(s) = G(s)/R(s). Read that as G(s) = C(s)/R(s) = b0 / (s + a0).

$\begin{align*}G(s)=\frac{G(s)}{R(s)} = \frac{b_{0}}{s+ a_{0}}\end{align*}$

$G(s)$ is often rewritten in time-constant form:

$\begin{align*}\frac{K}{\tau s+1} = \frac{b_{0}}{s+a_{0}}\end{align*}$

Matching coefficients, the usual identities are a0 = 1/τ and b0 = K/τ:

$\begin{align*} a {0}=\frac{1}{\tau} \; \; \; \; b {0}=\frac{K}{\tau} \end{align*}$

$\tau$ is the time constant. K is the DC gain: G(0) = K for this first-order form.

How to Find the DC Gain of a Transfer Function

Again: DC gain = settled output / constant input. For a unit step it equals the settled output.

Do this for both continuous and discrete linear time-invariant (LTI) systems. LTI does not mean “Linear Transform Inverse.”

Continuous LTI:

(1) $\begin{equation*} G(s)=\frac{Y(s)}{U(s)}\end{equation*}$

Discrete LTI:

(2) $\begin{equation*} G(z)=\frac{Y(z)}{U(z)}\end{equation*}$

Use the final value theorem on the unit-step response Y(s) = G(s)/s.

(3) $\begin{equation*} L\left ( y_{step(t)} \right )=G(s)\frac{1}{s}\end{equation*}$

(4) $\begin{equation*}DC\; \; Gain = \lim_{t\rightarrow \infty }y_{step(t)}\end{equation*}$

(5) $\begin{equation*} DC\; \; Gain = \lim_{s\rightarrow 0 }s\left [ G(s)\frac{1}{s} \right ]\end{equation*}$

The theorem needs $G(s)$ to be such that the step response settles: all poles of G(s) in the open left half-plane (no jω poles, no right-half poles).

The next stored line still has an extra s and reads lim s→0 s[G(s)]. That extra s is a slip. After the 1/s of the step cancels, DC gain is G(0), as in the G(s)|s=0 line that follows.

(6) $\begin{equation*}DC\; \; Gain = \lim_{s\rightarrow 0 }s\left [ G(s)\right ]\end{equation*}$

Continuous result after the theorem (not the theorem statement itself):

(7) $\begin{equation*}\frac{y(\infty)}{u(\infty)} = G(s)_{s=0}=G(0)\end{equation*}$

Discrete result: evaluate G(z) at z = 1, which is the DC point on the unit circle.

(8) $\begin{equation*}\frac{y(\infty)}{u(\infty)} = G(z)_{z=1}=G(1)\end{equation*}$

A free integrator (pole at s = 0, or at z = 1) makes G(0) or G(1) unbounded, so the step output ramps and the finite DC-gain ratio does not exist. The stored $\infty$ is that case.

DC gain is y_ss / u_ss, not input divided by the derivative of the output. The next headings only show how to write a time derivative in s or z. They are not a second formula for DC gain.

Differentiation in the Continuous Domain

In s, a time derivative multiplies by s (zero initial conditions).

(9) $\begin{equation*}\frac{\dot{Y(s)}}{U(s)}= sG(s)\end{equation*}$

where $\dot{Y(s)}$ is the Laplace transform of $\dot{y(t)}$ .

Differentiation in the Discrete Domain

In discrete time a first difference approximates the derivative.

(10) $\begin{equation*}\dot{y(k)}=\frac{y_{k}-y_{k-1}}{T}\end{equation*}$

(11) $\begin{equation*}\dot{Y(z)}=\frac{Y(z)-z^{-1}Y(z)}{T}\end{equation*}$

(12) $\begin{equation*}\dot{Y(z)}=Y(z)\left [\frac{ ^{1-z^{-1}}}{T} \right ]\end{equation*}$

(13) $\begin{equation*}\dot{Y(z)}=Y(z)\left [\frac{z-1}{T_{z}} \right ]\end{equation*}$

So a discrete derivative multiplies by $\frac{z-1}{T_{z}}$ (the stored T_z is T·z). That still is not DC gain.

Numerical Examples To Find DC Gain

Example 1

Example 1. Take

$\begin{align*} H(s) =\frac{Y(s)}{U(s)} = \frac{12}{(s+2)(s+10)}\end{align*}$

Correct method: H(0) = 12 / (2 × 10) = 0.6. Or FVT on the unit step: lim s→0 s · H(s) · (1/s) = H(0) = 0.6. The two stored lines below drop that 1/s, keep an extra s and replace (s+10) with 3. Do not use the printed 2.

$\begin{align*}\lim_{t\rightarrow \infty}y(t)= \lim_{s\rightarrow 0}s\times \frac{12}{(s+2)(s+10)}\end{align*}$

$\begin{align*}\lim_{t\rightarrow \infty}y(t)= \lim_{s\rightarrow 0}s\times \frac{12}{2\times 3}=2\end{align*}$

For a unit step, DC gain equals the settled output. That value is 0.6, not the printed 2 in the next line.

DC Gain = $\frac{2}{1}=2$

Use this finite ratio only when the step response actually settles.

Example 2

Example 2. First-order lag:

$\begin{align*}G(s)=\frac{K}{\tau s+1}\end{align*}$

Unit-step response (inverse Laplace of G(s)/s):

$\begin{align*}y_{step}(t)=L^{-1}\left [\frac{K}{(\tau s+1)s} \right ]\end{align*}$

$\begin{align*}y_{step}(t)=L^{-1}\left [ K\left ( \frac{1}{s}-\frac{\tau }{\tau s+1} \right ) \right ]\end{align*}$

FVT then gives y_ss = K, which matches G(0) = K. This example is consistent.

$\begin{align*}y_{ss}=\lim_{t\rightarrow \infty }y_{step}(t)= \lim_{s\rightarrow 0}\frac{K}{(\tau s+1)s}s = K\end{align*}$