Consider a system where: $ \frac{Y(s)}{U(s)}=\frac{k}{s} $. Is the gain k, or infinity? Given that to a step response, steady state is never reached, but it increases linearly with no end. Thanks!
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$\begingroup$ If by gain, you mean the gain at 0 frequency, then yes the gain is $\infty$. But books often refer to gain of pure integrators at value of $s=1$. In that context, gain is $k$. $\endgroup$– AJNCommented Oct 24, 2022 at 5:10
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1 Answer
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There are multiple definitions for the gain of a transfer function $G(s)$. The two most common are the zero-pole-gain and the (static) DC-gain.
The DC-gain is obtained by $$K_{DC} = G(s) \mid_0 \quad \text{or} \quad K_{DC} = \lim_{s\to 0} G(s),$$ depending on the definition you follow.
The zero-pole-gain follows from writing the transfer function in the general zero-pole-gain model $$G(s) = K_{zp} \frac{(s-z_1)(s-z_2)\ldots(s-z_m)}{(s-p_1)(s-p_2)\ldots(s-p_m)},$$ where $z_i$ are the zero's, $p_j$ the poles and $K_{zp}$ zero-pole-gain, with $i = 1 \ldots, m$ and $j=1\ldots n$.
For example $$ G(s) = \frac{2s^2+10s+8}{s^2+5s+6} = 2\frac{(s+1)(s+4)}{(s+2)(s+3)},$$ has a $K_{DC} = \frac{8}{6} = \frac{4}{3} $ and $K_{zp} = 2$.
For your system, the integrator $G(s)=\frac{k}{s}$, the zero-pole-gain is $K_{zp}= k$. Depending on the definition, the DC-gain does not exists as you cannot divide by zero, or if you use the limit results in $K_{DC} = \infty$.
