Transfer Functions and System Response¶

Exam-ready answer

For a lumped, linear time-invariant system, the transfer function is the ratio of the Laplace transform of output to input with every independent initial condition set to zero:

Input and output reference directions must be declared; reversing either direction changes the corresponding sign. The units of \(G\) are output units divided by input units, for example \((\text{rad/s})/\text{V}\) for a motor speed plant.

If an LTI differential equation is

its zero-state Laplace transform gives

The roots \(z_i\) are zeros and \(p_k\) are poles. Important properties are: \(G\) describes system dynamics independently of a particular input; cascade blocks multiply, parallel blocks add, and a negative-feedback loop reduces to \(G/(1+GH)\); the impulse response is \(g(t)=\mathcal L^{-1}\{G(s)\}\); poles determine natural modes and stability, while zeros shape forced response. Frequency response follows by evaluating \(G(j\omega)\) when the imaginary axis is in the region of convergence.

Worked derivation: for \(\tau\dot c+c=Kr\) with \(\tau>0\) s, zero initial conditions give \((\tau s+1)C=KR\), hence

A unit step produces \(c(t)=K(1-e^{-t/\tau})u(t)\); at \(t=\tau\), it reaches \(0.632K\), checking the pole \(-1/\tau\).

Advantages: differential equations become algebraic; interconnections, steady state, transients, poles/zeros and frequency response are handled systematically. Limitations: a transfer function omits nonzero initial-state response, is not directly valid for nonlinear or time-varying systems, and a scalar form does not capture MIMO interactions or internal state structure. Pole-zero cancellation can hide an unstable internal mode, and parameter uncertainty, saturation and delay must be modeled explicitly. State-space or nonlinear models are used when these limitations matter.

Practice target: 9 minutes; derive the polynomial ratio from a differential equation and separate properties from modeling limitations.

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Assume an LTI SISO system with zero initial conditions. At the negative-feedback summing junction, define the reference as positive and the feedback signal as negative:

Substitution and collection of the output terms give

The product \(L(s)=G(s)H(s)\) is loop transfer. The characteristic equation

sets the closed-loop poles and therefore stability and natural response. If the summing sign were positive, the denominator would be \(1-GH\); the sign must come from the diagram, not memory.

The same result follows from standard block reduction: a single negative-feedback loop around forward block \(G\) reduces to \(G/(1+GH)\).

It also follows from Mason's gain formula. In the signal-flow graph, the forward path is \(P_1=G\) and the loop gain is \(L_1=-GH\); hence \(\Delta=1-L_1=1+GH\), \(\Delta_1=1\), and \(T=P_1\Delta_1/\Delta=G/(1+GH)\). For larger graphs,

where \(\Delta\) contains loop and non-touching-loop products.

Check: with \(G=10/(s+2)\) and unity feedback, \(T=10/(s+12)\). The pole moves from \(-2\) to \(-12\), DC gain changes from \(5\) to \(10/12\), and the first-order response becomes faster. Reduction and Mason analysis assume linear signal relations; nonlinear saturation or a moved summing/take-off point without the required compensating block invalidates the algebra.

Practice target: 9 minutes; derive from signal equations first, then identify the characteristic equation and one independent reduction check.

Exam-ready answer

Write a proper rational transfer function as

A zero \(z_i\) makes the numerator zero and blocks or reshapes response at that complex frequency. A pole \(p_k\) makes the denominator zero and corresponds to a natural mode \(e^{p_kt}\). The horizontal coordinate \(\sigma=\operatorname{Re}s\) has units s\(^{-1}\) and sets exponential growth/decay; the vertical coordinate \(\omega=\operatorname{Im}s\) is oscillation frequency in rad/s.

For a standard second-order denominator,

Poles nearest the imaginary axis generally dominate the long transient. For \(0<\zeta<1\), \(t_s\approx4/(\zeta\omega_n)\) s for the 2% criterion and \(M_p=e^{-\pi\zeta/\sqrt{1-\zeta^2}}\).

Zeros do not set asymptotic stability but alter rise, overshoot and phase. A nearby LHP zero can speed response and increase overshoot; a RHP zero causes nonminimum-phase behavior, often an initial motion opposite the final value and a bandwidth limitation. Exact cancellation is unsafe: uncertainty leaves a residual mode, and cancelling an unstable pole can hide internal instability.

Check: \(s^2+4s+25\) has \(\omega_n=5\) rad/s, \(\zeta=0.4\), and poles \(-2\pm j4.58\), so it is stable and underdamped. The approximations give \(t_s\approx2\) s and \(M_p\approx25.4\%\). A zero at \(+1\) would not move those poles but would introduce inverse response. These ideas support controller design, model reduction and transient prediction, subject to the accuracy of the pole-zero model.

Practice target: 9 minutes; mark pole regions on an s-plane and connect each location to its time-domain mode before discussing zeros.

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A physical transfer function is obtained by choosing positive input/output directions, writing conservation laws with passive signs, applying the Laplace transform with zero initial conditions, and forming output over input.

Series RLC system. Let current \(i(t)\) flow from the positive terminal of input voltage \(v_i(t)\) through \(R\), \(L\) and \(C\), and let capacitor voltage \(v_c(t)\) be positive at the current-entering plate. KVL gives

Substituting current,

With zero capacitor voltage and inductor current initially,

and therefore

Here \(R\) is in ohms, \(L\) in henries, \(C\) in farads and the voltage ratio is dimensionless. Comparing with \(\omega_n^2/(s^2+2\zeta\omega_ns+\omega_n^2)\) gives

Translational mechanical system. Let positive displacement \(x(t)\) and applied force \(F(t)\) point to the right. The spring, viscous damper and inertia oppose positive motion with forces \(-Kx\), \(-B\dot x\) and \(-M\ddot x\). Newton's law gives

Under zero initial displacement and velocity,

Here \(M\) is kg, \(B\) is N·s/m, \(K\) is N/m, and \(X/F\) is m/N. The natural parameters are

Using charge \(q\) for the series circuit, \(L\ddot q+R\dot q+q/C=v_i\), which has the force-voltage analogy \(L\leftrightarrow M\), \(R\leftrightarrow B\), \(1/C\leftrightarrow K\), \(q\leftrightarrow x\), and \(v_i\leftrightarrow F\).

Worked check: for \(R=2\,\Omega\), \(L=0.5\) H and \(C=0.1\) F,

Thus \(\omega_n=\sqrt{20}=4.47\) rad/s and \(\zeta=4/(2\sqrt{20})=0.447\); the DC gain is one, consistent with the capacitor becoming open circuit at steady DC. Both models are second order because two independent energy-storage mechanisms exist: electric/magnetic energy in \(C,L\), and kinetic/elastic energy in \(M,K\).

The RLC model applies to filters and resonant networks; the mechanical model applies to suspensions, instruments and servo loads. Ideal lumped, linear elements omit parasitics, Coulomb friction, backlash, saturation and distributed modes. Nonzero initial energy adds separate response terms and is not part of the transfer function.

Practice target: 18–20 minutes; state signs before KVL/Newton equations, derive both ratios, and compare their natural frequency and damping.

Exam-ready answer

Consider a separately excited armature-controlled DC motor with constant field flux. Choose positive armature voltage \(v_a\), current \(i_a\), electromagnetic torque and angular speed \(\omega\) in the same motoring sense. Back emf opposes the applied voltage, and positive load torque \(T_L\) opposes rotation. The linear relations are

Here \(R_a\) is in \(\Omega\), \(L_a\) in H, \(K_b\) in V·s/rad, \(K_t\) in N·m/A, \(J\) in kg·m\(^2\), \(B\) in N·m·s/rad, \(\omega\) in rad/s and torque in N·m. Neglecting initial current and speed, the Laplace equations are

For the voltage-to-speed transfer set \(T_L=0\). From the mechanical equation,

Substitute into the electrical equation:

Therefore

Since \(\Omega=s\Theta\) under zero initial position,

The term \(K_tK_b\) is electromechanical negative feedback: increased speed raises opposing back emf, reducing current and torque. Retaining load torque and setting \(V_a=0\) gives the signed disturbance transfer

where the minus sign follows from the declared opposing \(T_L\) direction.

Worked check: take \(R_a=2\,\Omega\), \(L_a=0.5\) H, \(J=0.02\) kg·m\(^2\), \(B=0.10\) N·m·s/rad, and \(K_t=K_b=0.10\) in SI units. Then

The DC speed gain is \(0.10/0.21=0.476\) (rad/s)/V; substitution into the steady electrical and mechanical equations gives the same value.

The model supports speed/position servo design, drive simulation and disturbance analysis. It assumes constant field, linear magnetic circuit, viscous friction and rigid inertia; armature reaction, saturation, Coulomb friction, backlash, converter delay and load elasticity require extensions. If \(L_a\) is negligible relative to the mechanical time scale, the speed model may be approximated as first order, but that approximation must be justified.

Practice target: 18–20 minutes; declare torque signs, eliminate armature current step by step, and include both speed and position forms plus the load-torque check.