Transfer Functions and Time Response

This section develops input-output models from physical laws, combines those models with block diagrams and signal-flow graphs, and then uses them to predict transient and steady-state response. Pole and zero locations are treated separately in the section.

Transfer Function

For a linear time-invariant (LTI) system with all initial conditions set to zero, the transfer function is

$$\boxed{G(s)=\frac{C(s)}{R(s)}}.$$

Here $R(s)=\mathcal{L}\{r(t)\}$ is the input transform, $C(s)=\mathcal{L}\{c(t)\}$ is the output transform, and $s=\sigma+j\omega$ is the complex frequency.

The usual transfer-function model assumes a linear, time-invariant, lumped-parameter system and zero initial conditions. For a SISO differential equation

$$\begin{aligned} a_n\frac{\mathrm{d}^n c}{\mathrm{d}t^n} +a_{n-1}\frac{\mathrm{d}^{n-1}c}{\mathrm{d}t^{n-1}}+\cdots+a_1\frac{\mathrm{d}c}{\mathrm{d}t}+a_0c ={}&b_m\frac{\mathrm{d}^m r}{\mathrm{d}t^m} +b_{m-1}\frac{\mathrm{d}^{m-1}r}{\mathrm{d}t^{m-1}}+\cdots\\ &+b_1\frac{\mathrm{d}r}{\mathrm{d}t}+b_0r, \end{aligned}$$

the zero-state Laplace transform gives

Caution

Key Formula — General transfer-function form

$$\frac{C(s)}{R(s)}= \frac{b_ms^m+b_{m-1}s^{m-1}+\cdots+b_1s+b_0} {a_ns^n+a_{n-1}s^{n-1}+\cdots+a_1s+a_0}.$$

Advantages	Limitations
Converts a differential equation into an algebraic relation in $s$.	Does not directly represent nonlinear or time-varying behavior.
Gives the external input-output relation directly.	Zero-state definition omits the response due to nonzero initial conditions.
Poles and zeros expose stability and transient characteristics.	Does not reveal internal state variables or physical realization.
Supports block-diagram and frequency-response analysis.	Different physical systems can have the same transfer function.
Is independent of the particular applied input signal.	The scalar ratio is primarily a SISO description; MIMO systems require a transfer matrix.

Advantages and limitations of a transfer-function model

Note

Example — Transfer function from a differential equation

Find the transfer function when

$$\frac{\mathrm{d}^2c}{\mathrm{d}t^2}+5\frac{\mathrm{d}c}{\mathrm{d}t}+6c =2\frac{\mathrm{d}r}{\mathrm{d}t}+4r.$$

With zero initial conditions,

$$\begin{aligned} (s^2+5s+6)C(s)&=(2s+4)R(s),\\ \therefore\quad \frac{C(s)}{R(s)}&= \boxed{\frac{2s+4}{s^2+5s+6}}. \end{aligned}$$

Mathematical Modeling of Physical Systems

A model follows from an element law plus a compatibility or balance equation. Taking the Laplace transform only after writing the physical equations makes signs, units, and initial-condition assumptions visible.

Electrical elements

Element	Time-domain relation	Impedance $Z(s)=V/I$
Resistor $R$	$v(t)=Ri(t)$	$R$
Inductor $L$	$v(t)=L\,\mathrm{d}i(t)/\mathrm{d}t$	$Ls$
Capacitor $C$	$i(t)=C\,\mathrm{d}v(t)/\mathrm{d}t$	$1/(Cs)$

Electrical element relations under zero initial conditions

For the series RLC network of the figure, KVL and the capacitor voltage give

$$V_i(s)=\left(R+Ls+\frac{1}{Cs}\right)I(s), \qquad V_o(s)=\frac{I(s)}{Cs}.$$

Hence

$$\frac{V_o(s)}{V_i(s)}=\frac{1}{LCs^2+RCs+1}.$$

Translational mechanical elements

Element	Force relation	Coefficient for $F(s)/X(s)$
Mass $M$	$F=M\,\mathrm{d}^2x/\mathrm{d}t^2$	$Ms^2$
Viscous damper $B$	$F=B\,\mathrm{d}x/\mathrm{d}t$	$Bs$
Linear spring $K$	$F=Kx$	$K$

Translational mechanical element relations

Newton’s law for the mass-spring-damper system is

$$M\ddot{x}(t)+B\dot{x}(t)+Kx(t)=F(t).$$

Therefore

$$(Ms^2+Bs+K)X(s)=F(s), \qquad \boxed{\frac{X(s)}{F(s)}=\frac{1}{Ms^2+Bs+K}}.$$

Electrical and mechanical models obtained from element laws and force balance.

Rotational mechanical elements

The rotational counterparts of mass, force, and displacement are moment of inertia $J$, torque $T$, and angular displacement $\theta$.

Element	Torque relation
Moment of inertia $J$	$T=J\,\mathrm{d}^2\theta/\mathrm{d}t^2=J\,\mathrm{d}\omega/\mathrm{d}t$
Viscous friction $B$	$T=B\,\mathrm{d}\theta/\mathrm{d}t=B\omega$
Torsional spring $K$	$T=K\theta$

Rotational mechanical element relations

For a single inertia with viscous friction and a torsional spring,

$$J\ddot{\theta}+B\dot{\theta}+K\theta=T(t),$$

so

$$\boxed{\frac{\Theta(s)}{T(s)}=\frac{1}{Js^2+Bs+K}}, \qquad \frac{\Omega(s)}{T(s)}=\frac{s}{Js^2+Bs+K}.$$

Electrical-mechanical analogies

Force-voltage (impedance) analogy	Force-current (mobility) analogy
$F\leftrightarrow V$	$F\leftrightarrow I$
Velocity $v\leftrightarrow i$	Velocity $v\leftrightarrow V$
$M\leftrightarrow L$	$M\leftrightarrow C$
$B\leftrightarrow R$	$B\leftrightarrow 1/R$ (conductance)
$K\leftrightarrow 1/C$	$K\leftrightarrow 1/L$

Force-voltage and force-current analogies

Armature-Controlled DC Motor

Assume constant field flux, linear operation, viscous shaft friction, and zero initial conditions. Let $R_a$ and $L_a$ be armature resistance and inductance, $K_b$ the back-emf constant, $K_t$ the torque constant, $J$ the total reflected inertia, $B$ the viscous-friction coefficient, and $T_L$ a load torque defined positive when it opposes positive rotation.

The electrical and electromechanical relations are

$$\begin{aligned} V_a(t)&=R_ai_a(t)+L_a\frac{\mathrm{d}i_a}{\mathrm{d}t}+e_b(t),\\ e_b(t)&=K_b\omega(t)=K_b\frac{\mathrm{d}\theta}{\mathrm{d}t},\\ T_m(t)&=K_ti_a(t). \end{aligned}$$

The signed mechanical balance is

$$J\frac{\mathrm{d}\omega}{\mathrm{d}t}+B\omega=T_m-T_L.$$

In the $s$-domain,

$$\begin{aligned} V_a(s)&=(L_as+R_a)I_a(s)+K_b\Omega(s),\\ (Js+B)\Omega(s)&=K_tI_a(s)-T_L(s). \end{aligned}$$

Eliminating $I_a(s)$ from the equation and the equation gives

$$\begin{aligned} I_a(s)&=\frac{V_a(s)-K_b\Omega(s)}{L_as+R_a},\\ \bigl[(L_as+R_a)(Js+B)+K_bK_t\bigr]\Omega(s) &=K_tV_a(s)-(L_as+R_a)T_L(s). \end{aligned}$$

Define the motor denominator

$$D_m(s)=(L_as+R_a)(Js+B)+K_bK_t.$$

Superposition then gives the complete speed relation

Caution

Key Formula — Motor speed with command and disturbance inputs

$$\Omega(s)=\frac{K_t}{D_m(s)}V_a(s) -\frac{L_as+R_a}{D_m(s)}T_L(s).$$

Consequently,

$$\begin{aligned} \left.\frac{\Omega(s)}{V_a(s)}\right|_{T_L=0} &=\boxed{\frac{K_t}{(L_as+R_a)(Js+B)+K_bK_t}},\\ \left.\frac{\Theta(s)}{V_a(s)}\right|_{T_L=0} &=\boxed{\frac{K_t}{s\bigl[(L_as+R_a)(Js+B)+K_bK_t\bigr]}},\\ \left.\frac{\Omega(s)}{T_L(s)}\right|_{V_a=0} &=\boxed{-\frac{L_as+R_a}{(L_as+R_a)(Js+B)+K_bK_t}}. \end{aligned}$$

The minus sign in the last expression is a consequence of the chosen opposing-load convention. If $L_a$ is negligible,

$$\frac{\Omega(s)}{V_a(s)}\approx \boxed{\frac{K_t}{R_aJs+R_aB+K_bK_t}}.$$

Tip

Exam Focus — Motor signs and SI units

$K_t$ has units $\mathrm{N\,m/A}$ and $K_b$ has units $\mathrm{V\,s/rad}$. They are numerically equal in a self-consistent SI model, but they play different physical roles. Write $T_m-T_L$ before transforming; changing the reference direction changes the disturbance sign, not the motor physics.

Armature-controlled DC motor model with back-emf feedback, load-torque disturbance, speed output, and position integrator.

Block-Diagram Reduction

A block diagram represents functional input-output relations and the direction of signal flow. Reduction preserves the relation between the selected external input and output.

Connection	Equivalent transfer	Condition
Cascade	$G_1G_2$	The same signal passes through both blocks
Parallel paths	$G_1+G_2$	Branch outputs enter an additive summing point
Negative feedback	$G/(1+GH)$	Feedback enters the comparator with a minus sign
Positive feedback	$G/(1-GH)$	Feedback enters the comparator with a plus sign

Basic block-diagram equivalents

Basic cascade, parallel, and feedback reduction rules.

Moving a summing or take-off point across a block requires compensation:

• moving a sum from before $G$ to after $G$ multiplies its side path by $G$;

• moving a sum from after $G$ to before $G$ multiplies its side path by $1/G$;

• moving a take-off from before $G$ to after $G$ inserts $1/G$ in the branch;

• moving a take-off from after $G$ to before $G$ inserts $G$ in the branch.

In every case, verify the signal carried by the moved branch rather than memorizing the picture alone.

Compensating gains when a summing or take-off point is moved across a block.

Reduction strategy

1. Combine obvious cascade blocks.

2. Combine parallel blocks that share the same input and summing point.

3. Reduce the innermost feedback loop first.

4. Move summing or take-off points only when a loop cannot otherwise be isolated.

5. Repeat until one equivalent block remains, then check limiting cases and signs.

Worked nested reduction

In the figure, $G_2$ has the inner negative feedback $H_1$, while the complete forward path has outer negative feedback $H_2$. First reduce the inner loop:

$$G_{\mathrm{in}}=\frac{G_2}{1+G_2H_1}.$$

The resulting cascade is

$$G_{\mathrm{eq}}=G_1G_{\mathrm{in}}G_3 =\frac{G_1G_2G_3}{1+G_2H_1}.$$

Finally reduce the outer loop:

$$\begin{aligned} \frac{C(s)}{R(s)} &=\frac{G_{\mathrm{eq}}}{1+G_{\mathrm{eq}}H_2}\\ &=\boxed{\frac{G_1G_2G_3} {1+G_2H_1+G_1G_2G_3H_2}}. \end{aligned}$$

Worked reduction of nested negative-feedback loops from the inner loop outward.

Signal-Flow Graphs and Mason’s Gain Formula

A signal-flow graph (SFG) represents a set of linear algebraic relations by signal nodes and directed branches whose transmittances multiply the signal at the branch tail.

Term	Meaning
Node	A system variable or signal
Branch	Directed connection from one node to another
Branch gain	Transmittance multiplying the signal along a branch
Source (input) node	Node having outgoing branches but no incoming branch
Sink (output) node	Node having incoming branches but no outgoing branch
Forward path	Source-to-sink path that does not pass through any node more than once
Loop	Closed path that begins and ends at one node without repeating any other node
Non-touching loops	Loops that share no node

Signal-flow graph terminology

Note

Law — Mason’s gain formula

If $P_k$ is the gain of the $k$th forward path, then

$$\boxed{T=\frac{C(s)}{R(s)}= \frac{\displaystyle\sum_{k=1}^{N}P_k\Delta_k}{\Delta}}.$$

The graph determinant is

$$\Delta=1-\sum_i L_i +\sum_{i<j}L_iL_j -\sum_{i<j<\ell}L_iL_jL_\ell+\cdots,$$

where a product is included only when all loops in that product are mutually non-touching. The cofactor $\Delta_k$ is formed by deleting every loop that touches forward path $P_k$ and evaluating the determinant of what remains.

Mason procedure

1. List every forward path and calculate each path gain $P_k$.

2. List every individual loop and its signed loop gain $L_i$.

3. Identify all pairs, triples, and higher groups of non-touching loops.

4. Form the graph determinant $\Delta$ with alternating signs.

5. Form each path cofactor $\Delta_k$ after removing loops that touch $P_k$.

6. Substitute in Mason’s formula and simplify.

For the figure, the two forward paths are

$$P_1=G_1G_2G_3,\qquad P_2=G_1G_4G_5.$$

The loop gains are $L_1=-G_2H_1$ on nodes $\{x_2,x_3\}$ and $L_2=-H_2$ at $x_4$. Because these loops do not touch,

$$\begin{aligned} \Delta&=1-(L_1+L_2)+L_1L_2\\ &=1+G_2H_1+H_2+G_2H_1H_2. \end{aligned}$$

Path $P_1$ touches $L_1$ but not $L_2$, whereas $P_2$ touches both loops. Thus

$$\Delta_1=1-L_2=1+H_2,\qquad \Delta_2=1,$$

and

$$\boxed{\frac{x_5}{x_1}= \frac{G_1G_2G_3(1+H_2)+G_1G_4G_5} {1+G_2H_1+H_2+G_2H_1H_2}}.$$

Signal-flow graph with two forward paths and two non-touching loops, evaluated using Mason’s formula.

Standard Test Signals

Standard inputs make the performance of different systems directly comparable.

Input	Time function	Laplace transform	Main use
Impulse	$\delta(t)$	$1$	Reveals the impulse response directly
Step	$u(t)$	$1/s$	Tests set-point regulation
Ramp	$t\,u(t)$	$1/s^2$	Tests constant-velocity tracking
Parabolic	$t^2u(t)/2$	$1/s^3$	Tests constant-acceleration tracking
Sinusoid	$\sin(\omega t)u(t)$	$\omega/(s^2+\omega^2)$	Tests frequency response

Unit test signals and their one-sided Laplace transforms

Time Response

The output is commonly separated as

$$c(t)=c_{\mathrm{tr}}(t)+c_{\mathrm{ss}}(t),$$

where the transient component contains decaying natural modes and the steady-state component remains after those modes vanish.

First-order unit-step response

The normalized stable first-order prototype is

Caution

Key Formula — First-order prototype

$$G(s)=\frac{1}{Ts+1},\qquad T>0,$$

where $T$ is the time constant.

For a unit step $R(s)=1/s$,

$$\begin{aligned} C(s)&=\frac{1}{s(Ts+1)} =\frac{1}{s}-\frac{1}{s+1/T},\\ c(t)&=\boxed{1-e^{-t/T}},\qquad t\geq0. \end{aligned}$$

At $t=T$ the remaining error is $e^{-1}$, so the response has completed $63.2\%$ of its total change.

$t$	$T$	$2T$	$3T$	$4T$	$5T$
$c(t)$	$0.632$	$0.865$	$0.950$	$0.982$	$0.993$

First-order step-response landmarks

Solving $c(t)=q$ gives $t_q=-T\ln(1-q)$. Therefore the common timing specifications are

$$\begin{aligned} t_d&=T\ln2=0.693T &&\text{(time to $50\%$)},\\ t_r&=t_{90}-t_{10}=T\ln9=2.197T\approx2.2T,\\ t_s(2\%)&=-T\ln(0.02)=3.912T\approx4T,\\ t_s(5\%)&=-T\ln(0.05)=2.996T\approx3T. \end{aligned}$$

The first-order prototype is monotonic, so it has no finite peak time or overshoot.

Second-order prototype and poles

The unity-DC-gain second-order closed-loop prototype is

Caution

Key Formula — Second-order prototype

$$\frac{C(s)}{R(s)}= \frac{\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2},$$

where $\omega_n>0$ is the undamped natural frequency and $\zeta$ is the damping ratio.

The characteristic roots are

$$s_{1,2}=-\zeta\omega_n \pm\omega_n\sqrt{\zeta^2-1}.$$

For $0<\zeta<1$ this becomes

$$s_{1,2}=-\zeta\omega_n\pm j\omega_d, \qquad \boxed{\omega_d=\omega_n\sqrt{1-\zeta^2}}.$$

Their geometry is shown in the figure.

Underdamped unit-step development

With $R(s)=1/s$,

$$\begin{aligned} C(s)&=\frac{\omega_n^2} {s(s^2+2\zeta\omega_ns+\omega_n^2)}\\ &=\frac{1}{s}- \frac{s+2\zeta\omega_n}{s^2+2\zeta\omega_ns+\omega_n^2}. \end{aligned}$$

Completing the square in the quadratic denominator and inverting term by term gives

$$\begin{aligned} c(t)&=1-e^{-\zeta\omega_nt} \left[\cos(\omega_dt)+ \frac{\zeta}{\sqrt{1-\zeta^2}}\sin(\omega_dt)\right]\\ &=\boxed{1- \frac{e^{-\zeta\omega_nt}}{\sqrt{1-\zeta^2}} \sin(\omega_dt+\phi)}, \end{aligned}$$

where

$$\phi=\cos^{-1}\zeta =\tan^{-1}\!\left(\frac{\sqrt{1-\zeta^2}}{\zeta}\right), \qquad 0<\zeta<1.$$

Damping cases

Damping ratio	Poles	Unit-step behavior
$\zeta<0$	At least one pole in RHP	Growing, unstable response
$\zeta=0$	$\pm j\omega_n$	Undamped oscillation, $c(t)=1-\cos\omega_nt$; not BIBO stable
$0<\zeta<1$	Complex-conjugate LHP pair	Decaying oscillation with overshoot
$\zeta=1$	Repeated pole at $-\omega_n$	Critical response $1-(1+\omega_nt)e^{-\omega_nt}$
$\zeta>1$	Two real LHP poles	Slow, nonoscillatory overdamped response

Second-order damping cases

For $\zeta>1$, let

$$p_{1,2}=-\omega_n\left(\zeta\mp\sqrt{\zeta^2-1}\right).$$

Then the overdamped unit-step response can be written

$$c(t)=1+\frac{p_2e^{p_1t}-p_1e^{p_2t}}{p_1-p_2}.$$

Unit-step responses of the normalized second-order prototype for representative damping ratios.

Second-order timing specifications

Let $c(\infty)=1$ and $0<\zeta<1$. The standard measures are listed below.

Specification	Symbol	Definition or result
Delay time	$t_d$	First time the response reaches $50\%$ of final value; normally found numerically
Rise time	$t_r$	First $0$–$100\%$ crossing: $(\pi-\phi)/\omega_d$
Peak time	$t_p$	First maximum: $\pi/\omega_d$
Maximum overshoot	$M_p$	$(c_{\max}-c_\infty)/c_\infty=e^{-\pi\zeta/\sqrt{1-\zeta^2}}$
Percent overshoot	$\%OS$	$100e^{-\pi\zeta/\sqrt{1-\zeta^2}}$
Settling time	$t_s$	Time after which the response remains inside a specified tolerance band

Underdamped second-order time-domain specifications

The oscillatory error is bounded by

$$|1-c(t)|\leq \frac{e^{-\zeta\omega_nt}}{\sqrt{1-\zeta^2}}.$$

Thus an envelope-based tolerance estimate is

$$t_s(\varepsilon)= \frac{-\ln\!\left(\varepsilon\sqrt{1-\zeta^2}\right)} {\zeta\omega_n}.$$

The frequently used textbook approximations are

Caution

Key Formula — Common settling-time approximations

$$t_s(2\%)\approx\frac{4}{\zeta\omega_n}, \qquad t_s(5\%)\approx\frac{3}{\zeta\omega_n}.$$

They neglect the factor $1/\sqrt{1-\zeta^2}$ and round the logarithm, so they are design estimates rather than exact crossing times; they are most useful for ordinary underdamped designs away from $\zeta=0$ and $\zeta=1$.

Increasing $\zeta$ generally reduces overshoot, but beyond critical damping it slows the dominant real mode. At fixed $\zeta$, increasing $\omega_n$ scales all times downward. The product $\zeta\omega_n$ sets the exponential decay rate, so moving dominant poles farther left shortens settling time.

First-order time constant and second-order delay, rise, peak, overshoot, and settling constructions.

Note

Example — Second-order response specifications

A standard second-order system has $\zeta=0.5$ and $\omega_n=10\,\mathrm{rad/s}$. Find $\omega_d$, $t_p$, percent overshoot, and the approximate $2\%$ settling time.

$$\begin{aligned} \omega_d&=10\sqrt{1-0.5^2}=8.66\,\mathrm{rad/s},\\ t_p&=\frac{\pi}{8.66}=0.363\,\mathrm{s},\\ \%OS&=100e^{-\pi(0.5)/\sqrt{1-0.5^2}}=16.3\%,\\ t_s(2\%)&\approx\frac{4}{0.5(10)}=0.80\,\mathrm{s}. \end{aligned}$$

Using the full envelope expression gives about $0.811\,\mathrm{s}$; the standard rounded result is $0.80\,\mathrm{s}$.

Steady-State Error and Error Constants

If the limit exists, the steady-state error is

$$\boxed{e_{ss}=\lim_{t\to\infty}e(t)}.$$

For a negative-feedback system with loop transfer $L(s)=G(s)H(s)$,

$$E(s)=\frac{R(s)}{1+L(s)}.$$

For unity feedback, $H(s)=1$ and hence $L(s)=G(s)$.

The final-value theorem gives

Caution

Key Formula — Final-value evaluation

$$e_{ss}=\lim_{s\to0}sE(s) =\lim_{s\to0}\frac{sR(s)}{1+L(s)}.$$

[!NOTE]

Law — Final-value theorem condition

The limit above is valid only when every pole of $sE(s)$ lies strictly in the open LHP. In particular, do not use a static error-constant formula to claim a finite error for an unstable closed loop or a loop with persistent oscillation. Establish closed-loop stability first.

For the standard unit polynomial inputs, define the static error constants

$$\boxed{K_p=\lim_{s\to0}L(s)},\qquad \boxed{K_v=\lim_{s\to0}sL(s)},\qquad \boxed{K_a=\lim_{s\to0}s^2L(s)}.$$

Input	$R(s)$	Relevant constant	$e_{ss}$
Unit step	$1/s$	$K_p=\lim_{s\to0}L(s)$	$1/(1+K_p)$
Unit ramp	$1/s^2$	$K_v=\lim_{s\to0}sL(s)$	$1/K_v$
Unit parabolic	$1/s^3$	$K_a=\lim_{s\to0}s^2L(s)$	$1/K_a$

Static error constants for a stable negative-feedback loop

Here $1/0$ denotes an unbounded tracking error and $1/\infty=0$, provided the closed loop satisfies the final-value condition.

System type

The system type is the number of uncancelled poles at the origin in the loop transfer $L(s)=G(s)H(s)$. For finite, nonzero low-frequency gain apart from those integrators:

Type	Origin poles	Step	Ramp	Parabolic
	0	$1/(1+K_p)$, finite	$\infty$	$\infty$
1	1	$0$	$1/K_v$, finite	$\infty$
2	2	$0$	$0$	$1/K_a$, finite

System type and unit-input steady-state error

Increasing type improves low-frequency tracking accuracy, but each added integrator contributes phase lag and can reduce relative stability. It is therefore not valid to improve error performance without rechecking the closed-loop characteristic equation.

Note

Example — Steady-state error of a Type 1 system

For negative unity feedback with

$$G(s)=\frac{10}{s(s+2)},$$

find the error constants and the unit-step and unit-ramp errors. The closed-loop characteristic polynomial is $s(s+2)+10=s^2+2s+10$, whose roots have real part $-1$; the final-value theorem is therefore applicable.

$$\begin{aligned} K_p&=\lim_{s\to0}\frac{10}{s(s+2)}=\infty, &e_{ss,\mathrm{step}}&=\frac{1}{1+K_p}=0,\\ K_v&=\lim_{s\to0}\frac{10}{s+2}=5, &e_{ss,\mathrm{ramp}}&=\frac{1}{K_v}=0.2,\\ K_a&=\lim_{s\to0}\frac{10s}{s+2}=0, &e_{ss,\mathrm{parabolic}}&=\infty. \end{aligned}$$

The last result is included to complete the error-constant classification, although only step and ramp errors were requested.

Quick Revision

Topic	Key result
Transfer function	$G(s)=C(s)/R(s)$ for an LTI zero-state model
Pole and zero	Denominator and numerator roots; see the section
Stable pole region	All uncancelled poles strictly in the LHP for BIBO stability
Mass-spring-damper	$X/F=1/(Ms^2+Bs+K)$
DC motor speed	$\Omega/V_a=K_t/[(L_as+R_a)(Js+B)+K_bK_t]$
DC motor disturbance	$\Omega/T_L=-(L_as+R_a)/[(L_as+R_a)(Js+B)+K_bK_t]$
Cascade and parallel	$G_1G_2$ and $G_1+G_2$
Negative feedback	$G/(1+GH)$
Mason formula	$T=\sum_kP_k\Delta_k/\Delta$
First-order step	$c(t)=1-e^{-t/T}$
First-order timing	$t_r(10$–$90\%)\approx2.2T$, $t_s(2\%)\approx4T$
Second-order prototype	$\omega_n^2/(s^2+2\zeta\omega_ns+\omega_n^2)$
Damped frequency	$\omega_d=\omega_n\sqrt{1-\zeta^2}$
Peak time	$t_p=\pi/\omega_d$
Percent overshoot	$100e^{-\pi\zeta/\sqrt{1-\zeta^2}}$
Approximate $2\%$ settling	$t_s\approx4/(\zeta\omega_n)$
Error constants	$K_p=\lim L$, $K_v=\lim sL$, $K_a=\lim s^2L$ as $s\to0$
System type	Number of uncancelled pure integrators in $L(s)=G(s)H(s)$

Transfer-function and response revision table

Note

Key Point — Quick review

• A transfer function is a zero-state input-output description, not a complete internal-state model.

• Poles generate natural modes and decide BIBO stability; zeros reshape the forced response and may create inverse response.

• Dominant poles are the slow, appreciably excited poles nearest the imaginary axis, not merely the nearest symbols on a plot.

• Standard second-order formulas apply to the normalized prototype and use clearly stated rise- and settling-band conventions.

• System type predicts polynomial tracking error only after closed-loop stability and the final-value theorem conditions have been verified.

The four worked examples are distributed by topic: transfer-function formation, pole-zero interpretation in the section, second-order timing, and steady-state error.