Transfer function ($G(s)=Y(s)/U(s)$)

A transfer function packages the response of a linear time-invariant system into a rule for turning a plant input into a plant output. Instead of solving a differential equation separately for every signal, take a Laplace transform, multiply by $G(s)$, then transform back. This also makes it possible to compare slow and fast oscillations through the continuous-time frequency response.

Here consider a continuous-time, causal, finite-dimensional linear time-invariant system with one plant input $u(t)$ and one plant output $y(t)$. With every initial state set to zero, its transfer function is defined byThe ratio $G(s)=Y(s)/U(s)$ is used where $U(s)\ne 0$ in a common Laplace convergence region; $G$ describes the system, independently of that particular plant input. Zero initial state matters: stored motion or another nonzero initial condition can contribute to the plant output even when the plant input vanishes. ^[ref]

The first-order low-pass filter below shows the same transfer function predicting both a step and a sustained sinusoid. Its pole gives a decay rate, its value at zero gives the eventual gain, and its values along the imaginary axis give the continuous-time frequency response. These descriptions make a useful bridge between a physical relaxation model and control theory calculations.

Substitution $s=i\omega$ into an algebraic transfer function can produce finite amplitude and phase even when the causal system has unstable poles. It then describes a sinusoidal particular solution whose growing homogeneous response is not generically absent. A physically settled frequency response requires transient asymptotic stability; for an unstable causal transfer function, its Laplace region of convergence need not include the imaginary axis.

A first-order low-pass filter with positive constant gain $K$ and time constant $\tau >0$ obeys the ordinary differential equationThe derivative term limits how quickly the plant output follows a changing plant input. A larger time constant means slower adjustment. This standard form, with $K=1$, is developed in Derek Rowell, MIT 2.004, lecture 20, section 1.1: first-order systems; the following calculation keeps the gain $K$ explicit.

Write $y_0=y(0)$. The Laplace transform of the derivative is $sY(s)-y_0$, soSetting the initial condition $y_0=0$ gives the transfer function $G(s)=K/(\tau s+1)$. Its single simple pole is $s=-1/\tau$, which has negative real part. The unforced response decays, and the input/output map has BIBO stability. The eventual gain for a constant plant input is $G(0)=K$.

For a unit step, represented by the Heaviside step function $u(t)=H(t)$, the Laplace transform is $U(s)=1/s$. From zero initial condition,Take $K=2$ and $\tau =3\mathrm{s}$, with dimensionless plant input and plant output. Then $y(t)=2(1-e^{-t/(3\mathrm{s})})$ rises toward $2$. After one time constant, at $t=3\mathrm{s}$, it reaches $2(1-e^{-1})\approx 1.2642$, or $63.2\%$ of its final value. At $t=9\mathrm{s}=3\tau$, it reaches $2(1-e^{-3})\approx 1.9004$, or $95.0\%$. These percentages describe the rise from zero, rather than an instantaneous jump to the target.

With a nonzero initial condition, the same step instead givesFor $K=2$ and $\tau =3\mathrm{s}$, compare $y_0=0$ with $y_0=1$. Their plant outputs differ by $e^{-t/(3\mathrm{s})}$ despite having the same transfer function and the same plant input. The extra exponential decay term must be supplied from the initial condition; $G$ and $u$ alone cannot determine it.

The continuous-time frequency response is the frequency response of a continuous-time linear time-invariant system. From zero initial state, a BIBO stable system driven by a sustained sinusoid $u(t)=A\cos (\omega t)$ approaches the steady responseAny decaying initial or switching transient is separate. The modulus $|G(i\omega )|$ is the amplitude gain; the complex argument is the shift in oscillation phase. Thus evaluating a transfer function at $s=i\omega$ measures both how much of an oscillation passes through and how far it lags.

For the first-order low-pass filter, with $K>0$, direct complex number arithmetic givesThe phase uses the inverse tangent branch between $-\pi /2$ and $\pi /2$. Low angular frequencies have gain close to $K$ and little lag. High angular frequencies have gain approximately $K/(\omega \tau )$ and lag approaching $\pi /2$. This is why the model is a first-order low-pass filter: it suppresses fast variation compared with slow variation. Rowell relates frequency response gain and phase to poles and then introduces Bode plots in MIT 2.004, lecture 32, sections 1 and 2.

At the corner angular frequency,For $K=2$ and $\tau =3\mathrm{s}$, ${\omega }_c=1/3\mathrm{rad}{\mathrm{s}}^{-1}$, the gain is $\sqrt{2}\approx 1.4142$, and the lag is $45$ degrees. The angular frequency $\omega$ is measured in radians per second; ordinary frequency $f$ is measured in hertz (cycles per second), with $\omega =2\pi f$. Consequently the same corner is $f_c=1/(2\pi \tau )\approx 0.0531\mathrm{Hz}$, not $1/3\mathrm{Hz}$.

A Bode plot displays the gain and oscillation phase of a frequency response against a logarithmic frequency axis. Gain is often shown as $20{\log }_{10}|G(i\omega )|$ in decibels; phase can use degrees or radians. The first-order low-pass filter loses approximately $20$ decibels per factor of ten in angular frequency well above its corner. ^[ref]

A proper rational transfer function is a rational function whose numerator polynomial has polynomial degree no greater than its denominator polynomial, after common factors are canceled. Every finite-dimensional continuous-time linear time-invariant system of the form $\dot{x}=Ax+Bu$, $y=Cx+Du$ has such a transfer function $G(s)=C(sI-A{)}^{-1}B+D$. ^[ref]

Bounded-input bounded-output stability means that every bounded plant input produces a bounded plant output from zero initial state. For a causal continuous-time linear time-invariant system with a proper rational transfer function, BIBO stability is equivalent to every pole of the reduced transfer function having strictly negative real part. Cancel common factors first: an apparent denominator root may be absent from the reduced transfer function. For $K/(\tau s+1)$ with $\tau >0$, the only pole is $-1/\tau$, so the first-order low-pass filter is BIBO stable. ^[ref]

Internal asymptotic stability also requires every unforced internal state to decay. A canceled factor can hide a growing internal state. For example, consider the linear system of ordinary differential equationsIts zero-state transfer function can be written $(s-1)/((s+1)(s-1))=1/(s+1)$ and has BIBO stability. Yet from zero initial conditions and a unit-step plant input, $x_2(t)=e^t-1$ grows without bound while the plant output remains bounded. The reduced transfer function conceals that second state. Even a transfer function with no visible cancellations can be implemented with additional hidden states; its poles alone do not establish internal asymptotic stability of an arbitrary implementation.

Let $L$ be the open-loop transfer function. The winding of its Nyquist plot around $-1$, together with the number of unstable open-loop poles, determines the number of unstable zeros of $1+L$, which are the closed-loop poles.

The output caused by a persistent step change in the input, with an explicitly specified initial condition.

The time after which a response remains within a specified tolerance of its final value. State both the tolerance and its reference scale.