Linear Systems

5% of exam

Frequency and transient response, resonance, Laplace transforms, transfer functions.

3 concepts

Exam areas

A. Frequency/transient response

Frequency and Transient Response

Read a circuit's behavior straight from its time constant, natural frequency, damping ratio, and pole locations — the core of every linear-systems question.

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B. Resonance

Resonance in RLC Circuits

Find the resonant frequency, quality factor, bandwidth, and half-power points for series and parallel RLC circuits, and reason about selectivity from Q.

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C. Laplace transforms and transfer functions

Laplace Transforms and Transfer Functions

Move circuits and systems into the s-domain, build the transfer function H(s), read its poles and zeros, and pull endpoints out with the initial- and final-value theorems.

The Laplace transform turns the calculus of linear systems into algebra: differential equations become polynomial equations in

s

, convolutions become products, and initial conditions fold in automatically. Once you are in the

s

-domain, a circuit's entire input-output behavior lives in one rational function — the transfer function

H(s)

— whose poles and zeros tell you everything about stability, transient shape, and frequency response. The FE Reference Handbook (Mathematics — Laplace Transforms) supplies the transform pair table and the initial- and final-value theorems; this concept turns that table into working fluency. Master the half-dozen pairs and three properties below and a wide swath of linear-systems points come almost for free.

The transform and the essential pairs

The (unilateral) Laplace transform maps a time function

f(t)

t\ge0

, to

F(s)=\int_0^\infty f(t)e^{-st}\,dt

. You do not integrate on the exam — you recognize forms in the handbook table. The pairs you must know cold are the impulse, step, ramp, exponential, and damped/undamped sinusoid. Each pair is a building block; partial fractions break a complicated

F(s)

into a sum of these so you can invert term by term.

\mathcal{L}\{f(t)\} = F(s) = \int_0^\infty f(t)\,e^{-st}\,dt

Properties that do the work

Three properties carry most problems. Linearity lets you transform sums termwise. The differentiation property converts a derivative into multiplication by

s

and injects the initial condition — this is how a differential equation becomes algebra. The time-shift property handles delays. Note how

f(0)

enters the derivative rule: zero initial conditions make

\mathcal{L}\{f'\}=sF(s)

, which is exactly the assumption baked into a transfer function.

\mathcal{L}\{f'(t)\} = sF(s) - f(0), \qquad \mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s)

s-domain impedance

Circuit analysis goes algebraic when you replace each element with its

s

-domain impedance: a resistor stays

R

, an inductor becomes

sL

, and a capacitor becomes

1/(sC)

. Then KVL, KCL, series/parallel combination, and voltage division all work exactly as in DC — just with

s

in place of numbers. This is the generalization of phasor impedance (

s\to j\omega

recovers the steady-state

j\omega L

and

1/(j\omega C)

), and it lets you write

H(s)

for any RLC network by inspection.

Z_R = R, \qquad Z_L = sL, \qquad Z_C = \dfrac{1}{sC}

The transfer function

The transfer function is the ratio of output transform to input transform with all initial conditions zero. It is a property of the system, not the input — feed it any

X(s)

and the output is

Y(s)=H(s)X(s)

. For circuits you build

H(s)

from

s

-domain impedances (often a voltage divider); for systems it comes from the governing ODE. Writing it as a ratio of polynomials in

s

exposes its structure at a glance.

H(s) = \dfrac{Y(s)}{X(s)} = \dfrac{b_m s^m + \cdots + b_0}{a_n s^n + \cdots + a_0}

Poles and zeros

Zeros are the roots of the numerator (where

H(s)=0

) and poles are the roots of the denominator (where

H(s)\to\infty

). The poles alone determine the natural response and stability: every left-half-plane pole contributes a decaying mode, so a system is stable iff all poles have negative real part. Zeros shape how strongly each mode is excited and sculpt the frequency response. A factored, pole-zero form is the most readable way to write

H(s)

, and a quick scan of the pole signs answers the stability question instantly.

H(s) = K\,\dfrac{(s-z_1)(s-z_2)\cdots}{(s-p_1)(s-p_2)\cdots}

Initial- and final-value theorems

These two theorems read the endpoints of

f(t)

straight off

F(s)

— no inverse transform needed. The initial-value theorem gives

f(0^+)

from the

s\to\infty

limit; the final-value theorem gives

f(\infty)

from the

s\to0

limit. The FVT has a crucial precondition: it is valid only when

f(t)

actually settles, i.e. all poles of

sF(s)

lie in the open left half-plane (poles on the imaginary axis or right half-plane make

f

oscillate or diverge, and the theorem lies). Used on

sH(s)

for a unit step, the FVT delivers the DC gain

H(0)

f(0^+) = \lim_{s\to\infty} sF(s), \qquad f(\infty) = \lim_{s\to 0} sF(s)

Frequency response from H(s)

Set

s=j\omega

and

H(j\omega)

becomes the steady-state sinusoidal response:

|H(j\omega)|

scales amplitude and

\angle H(j\omega)

shifts phase. This is the bridge from the

s

-domain to Bode plots and filter design — poles pull the magnitude down (rolloff) and add phase lag, zeros push it up and add phase lead. The same

H(s)

that gave you transient shape and stability now hands you the frequency-domain behavior, all from one rational function.

H(j\omega) = H(s)\big|_{s=j\omega}, \qquad \text{gain} = |H(j\omega)|,\;\; \text{phase} = \angle H(j\omega)

Exam strategy

Keep the handbook transform table open and recognize, don't derive. For circuits, redraw with

Z_L=sL

Z_C=1/(sC)

and use a divider to get

H(s)

in two lines. To find a steady-state or final value, reach for the FVT before doing partial fractions — but first confirm the poles are stable, or the answer is garbage. For DC gain, evaluate

H(0)

; for high-frequency behavior,

H(s\to\infty)

. When you must invert, factor the denominator, do partial fractions into table forms, and write the time function term by term. Always check stability by the sign of the real parts of the poles.

Key equations

Laplace transform definition

F(s) = \int_0^\infty f(t)\,e^{-st}\,dt

Unilateral transform for $t\ge0$ . $s=\sigma+j\omega$ is complex frequency.

Core transform pairs

\delta(t)\!\to\!1,\;\; u(t)\!\to\!\tfrac{1}{s},\;\; t\,u(t)\!\to\!\tfrac{1}{s^2},\;\; e^{-at}\!\to\!\tfrac{1}{s+a}

Impulse, unit step, ramp, decaying exponential. The building blocks for partial-fraction inversion.

Sinusoid pairs

\sin(bt)\!\to\!\dfrac{b}{s^2+b^2}, \quad \cos(bt)\!\to\!\dfrac{s}{s^2+b^2}

Damped versions shift $s\to s+a$ : $e^{-at}\sin bt\to b/((s+a)^2+b^2)$ .

Differentiation property

\mathcal{L}\{f'(t)\} = sF(s) - f(0)

Turns an ODE into algebra and carries the initial condition. Second derivative: $s^2F(s)-sf(0)-f'(0)$ .

Integration and time shift

\mathcal{L}\Big\{\!\int_0^t f\,d\tau\Big\}=\dfrac{F(s)}{s}, \;\; \mathcal{L}\{f(t-a)u(t-a)\}=e^{-as}F(s)

Integration divides by $s$ ; a delay $a$ multiplies by $e^{-as}$ (the dead-time factor).

s-domain impedances

Z_R=R, \quad Z_L=sL, \quad Z_C=\dfrac{1}{sC}

Generalized impedances. Apply KVL/KCL/dividers as in DC; $s\to j\omega$ recovers phasor impedance.

Transfer function

H(s)=\dfrac{Y(s)}{X(s)}\Big|_{\text{zero IC}}

Output-to-input ratio with zero initial conditions. Output for any input: $Y(s)=H(s)X(s)$ .

Pole-zero (factored) form

H(s)=K\dfrac{\prod(s-z_i)}{\prod(s-p_j)}

Zeros $z_i$ from numerator roots, poles $p_j$ from denominator roots; poles set stability and natural response.

Initial-value theorem

f(0^+)=\lim_{s\to\infty} sF(s)

Gives the starting value directly from $F(s)$ . Valid when $F(s)$ is strictly proper (no impulse at $t=0$ , so $sF(s)$ has a finite $s\to\infty$ limit).

Final-value theorem

f(\infty)=\lim_{s\to 0} sF(s)

Gives the steady-state value — ONLY if all poles of $sF(s)$ are in the open left half-plane.

DC gain

H(0)=\lim_{s\to0}H(s)

Steady-state gain to a constant input; equals the final value of the step response of a stable system.

Frequency response

H(j\omega)=H(s)\big|_{s=j\omega}

Steady-state sinusoidal transfer: magnitude scales amplitude, angle shifts phase. Basis of Bode plots.

Worked examples

RL circuit step response in the s-domain

Problem. A series $RL$ circuit ( $R=5\,\Omega$ , $L=0.5\,\text{H}$ , zero initial current) has a $10\,\text{V}$ step applied at $t=0$ . Find $I(s)$ , the steady-state current via the final-value theorem, and the current at $t=0.2\,\text{s}$ .

Solution. s-domain: source

10/s

, impedance

R+sL=5+0.5s

. So

I(s)=\dfrac{10/s}{5+0.5s}=\dfrac{10}{s(0.5s+5)}=\dfrac{20}{s(s+10)}

. Final value:

i(\infty)=\lim_{s\to0} sI(s)=\lim_{s\to0}\dfrac{20}{s+10}=2.00\,\text{A}

(pole at

s=-10

is stable, so FVT is valid). Invert:

i(t)=2(1-e^{-t/\tau})

with

\tau=L/R=0.1\,\text{s}

. At

t=0.2\,\text{s}

i=2(1-e^{-2})=2(0.8647)=1.73\,\text{A}

. Sanity:

i(\infty)=V/R=10/5=2\,\text{A}

matches the FVT; at

2\tau

the current is

86.5\%

2\,\text{A}

→

1.73\,\text{A}

, consistent.

i(\infty)=\lim_{s\to0}\dfrac{20}{s+10}=2.00\,\text{A}

Poles, zeros, and DC gain of a transfer function

Problem. A system has $H(s)=\dfrac{10}{s^2+3s+2}$ . Find its poles and zeros, its DC gain, whether it is stable, and the final value of its unit-step response.

Solution. Factor the denominator:

s^2+3s+2=(s+1)(s+2)

, so poles at

s=-1

and

s=-2

; no finite zeros (numerator is constant). DC gain:

H(0)=10/2=5.00

. Stability: both poles have negative real part → stable. Step final value:

Y(s)=H(s)\cdot\tfrac{1}{s}

, so

y(\infty)=\lim_{s\to0}sY(s)=\lim_{s\to0}H(s)=H(0)=5.00

(FVT valid since poles are stable). Sanity: for a stable system the step settles to the DC gain, so

y(\infty)=H(0)=5

— both routes agree.

y(\infty)=H(0)=\dfrac{10}{2}=5.00

Initial- and final-value theorems together

Problem. Given $F(s)=\dfrac{2s+3}{s^2+4s+3}$ , find $f(0^+)$ and $f(\infty)$ without inverting, then verify by partial fractions.

Solution. IVT:

f(0^+)=\lim_{s\to\infty}sF(s)=\lim_{s\to\infty}\dfrac{2s^2+3s}{s^2+4s+3}=2.00

(ratio of leading coefficients). FVT: poles of

s^2+4s+3=(s+1)(s+3)

are

-1,-3

(stable), so

f(\infty)=\lim_{s\to0}sF(s)=\dfrac{0}{3}=0

. Verify:

\dfrac{2s+3}{(s+1)(s+3)}=\dfrac{A}{s+1}+\dfrac{B}{s+3}

with

A=\dfrac{2(-1)+3}{-1+3}=\tfrac12

B=\dfrac{2(-3)+3}{-3+1}=\tfrac32

. So

f(t)=0.5e^{-t}+1.5e^{-3t}

. Check:

f(0)=0.5+1.5=2.00

(matches IVT);

f(\infty)=0

(matches FVT).

f(t)=0.5\,e^{-t}+1.5\,e^{-3t}

Common pitfalls

•Applying the final-value theorem to an unstable or marginally stable $F(s)$ . If any pole of $sF(s)$ is on or right of the imaginary axis (e.g. pure sinusoid, $1/s^2$ growth), the FVT gives a wrong finite number — check pole locations first.
•Forgetting the initial condition in the derivative rule: $\mathcal{L}\{f'\}=sF(s)-f(0)$ , not $sF(s)$ . Transfer functions assume $f(0)=0$ , but transient problems with stored energy do not.
•Using $sL$ for a capacitor or $1/(sC)$ for an inductor. Inductor impedance rises with $s$ ( $sL$ ); capacitor impedance falls with $s$ ( $1/sC$ ).
•Confusing poles and zeros: poles are denominator roots (response blows up, set stability); zeros are numerator roots (response nulls). Only poles determine stability.
•Calling a system with complex poles unstable. Stability is the sign of the real part — complex poles in the left half-plane are stable and simply oscillate as they decay.
•Mismatching transform pairs — using $s/(s^2+b^2)$ for sine instead of cosine. Sine has the $b$ in the numerator; cosine has the $s$ .
•Reading the transfer function as input/output. It is output over input, $Y/X$ , with zero initial conditions.

References

NCEES FE Reference Handbook — Mathematics: Laplace Transforms — Transform pair table, differentiation/integration properties, initial- and final-value theorems.
NCEES FE Reference Handbook — Electrical and Computer Engineering: AC Circuits — Element impedances; $s\to j\omega$ connects s-domain to phasors.
NCEES FE Reference Handbook — Instrumentation, Measurement, and Control: Transfer Functions — First/second-order transfer-function models and block diagrams.
Nilsson & Riedel, Electric Circuits — s-domain circuit analysis and transfer-function derivation.