Transfer function and ODE

The Core Rule

For a linear, time-invariant (LTI) system with zero initial conditions, the Laplace transform replaces derivatives and integrals with powers of $s$:

$$\frac{d^n y}{dt^n} ;\longleftrightarrow; s^n Y(s), \qquad \underbrace{\int\cdots\int}_{n}, y,dt^n ;\longleftrightarrow; \frac{Y(s)}{s^n}$$

The order of a system is the highest derivative acting on the output — equivalently, the degree of the denominator polynomial in $H(s)$, i.e. the number of poles. Integral terms work the same way but in reverse: they add negative powers of $s$, which get cleared into the denominator once you multiply through.

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Table 1 — Differentiation-Based Systems (by Order)

Order	Time-domain ODE	Transfer Function $H(s) = \dfrac{Y(s)}{U(s)}$
Zero-order (static gain)	$y(t) = K,u(t)$	$H(s) = K$
First-order	$\tau\dot{y} + y = K,u$	$H(s) = \dfrac{K}{\tau s + 1}$
First-order, general coefficients	$a_1\dot{y} + a_0 y = b_0 u$	$H(s) = \dfrac{b_0}{a_1 s + a_0}$
Second-order, standard form	$\ddot{y} + 2\zeta\omega_n\dot{y} + \omega_n^2 y = K\omega_n^2 u$	$H(s) = \dfrac{K\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}$
Second-order, general coefficients	$a_2\ddot{y}+a_1\dot{y}+a_0y = b_0 u$	$H(s) = \dfrac{b_0}{a_2 s^2 + a_1 s + a_0}$
Second-order with a zero	$a_2\ddot{y}+a_1\dot{y}+a_0y = b_1\dot{u}+b_0u$	$H(s) = \dfrac{b_1 s + b_0}{a_2 s^2 + a_1 s + a_0}$
Pure differentiator	$y = K\dot{u}$	$H(s) = Ks$
General $n$-th order LTI	$a_n y^{(n)} + \dots + a_0 y = b_m u^{(m)} + \dots + b_0 u$	$H(s) = \dfrac{b_m s^m + \dots + b_0}{a_n s^n + \dots + a_0}$

$\zeta$ and $\omega_n$ are just a repackaging of the general second-order coefficients — chosen because they map to physically meaningful quantities (damping ratio, natural frequency) instead of raw numbers:

$$\omega_n = \sqrt{\frac{a_0}{a_2}}, \qquad \zeta = \frac{a_1}{2\sqrt{a_0 a_2}}$$

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Table 2 — Adding Integration

Type	ODE / Integro-differential equation	Transfer Function $H(s)$
Pure integrator	$\dot{y} = K u$	$H(s) = \dfrac{K}{s}$
Double integrator	$\ddot{y} = K u$	$H(s) = \dfrac{K}{s^2}$
First-order + integral action	$\dot{y} + a_0 y + a_{-1}\displaystyle\int y,dt = b_0 u$	$H(s) = \dfrac{b_0 s}{s^2 + a_0 s + a_{-1}}$
Second-order + integral (series RLC, current $i$)	$L\dot{i} + Ri + \dfrac{1}{C}\displaystyle\int i,dt = v(t)$	$H(s) = \dfrac{s}{Ls^2 + Rs + \frac{1}{C}}$
PI controller	$u(t) = K_p e(t) + K_i\displaystyle\int e,dt$	$C(s) = K_p + \dfrac{K_i}{s} = \dfrac{K_p s + K_i}{s}$
PID controller	$u(t) = K_p e + K_i\displaystyle\int e,dt + K_d\dot{e}$	$C(s) = \dfrac{K_d s^2 + K_p s + K_i}{s}$
General rule — clearing $1/s$ terms	any equation with a $\int^{(k)} y,dt$ term	multiply numerator and denominator by $s^k$ to get a proper polynomial ratio

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Why $1/s$ Specifically?

It falls straight out of the Laplace integration property — just as $s^n$ represents $n$ derivatives, $s^{-n}$ represents $n$ integrations. A system’s order isn’t really “how many derivatives” — it’s the net power of $s$ once fractions are cleared.

One practical consequence: a pure integrator puts a pole at $s = 0$, which is why $1/s$ and $1/s^2$ systems are marginally stable (or worse) on their own — a constant input never settles, it ramps or accelerates forever. That’s the mathematical reason integral-only control is never used alone in practice; it’s always paired with proportional (and often derivative) action, as in the PI/PID rows above.

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The Fully General Form

Putting differentiation and integration together, any LTI system can be written as:

$$\sum_{k=-p}^{n} a_k \frac{d^k y}{dt^k} = \sum_{j=-q}^{m} b_j \frac{d^j u}{dt^j}$$

where negative $k, j$ denote integrations. Laplace-transforming and clearing the lowest negative power ($s^p$ in the denominator, $s^q$ in the numerator) gives the general transfer function:

$$H(s) = \frac{b_m s^{m+q} + \dots + b_{-q}}{a_n s^{n+p} + \dots + a_{-p}}$$

Every row in the two tables above is a special case of this — set $p = q = 0$ for the pure differentiation table, or let $p, q > 0$ to bring in integral terms.