Linear Systems · Study · FE Electrical and Computer · FE → PE Prep
Linear Systems
5% of exam
Frequency and transient response, resonance, Laplace transforms, transfer functions.
3 concepts
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≥0, to F(s)=∫0∞f(t)e−stdt
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 L{f′}=sF(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→jω
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
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)→∞). 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.
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→∞
Frequency response from H(s)
Set s=jω and H(jω) becomes the steady-state sinusoidal response: ∣H(jω)∣ scales amplitude and ∠H(jω)
Exam strategy
Keep the handbook transform table open and recognize, don't derive. For circuits, redraw with ZL=sL, ZC=1/(sC)
Key equations
Laplace transform definitionF(s)=∫0∞f(t)e−stdt
Worked examples
RL circuit step response in the s-domain
Problem. A series RL circuit (R=5Ω, L=0.5H, zero initial current) has a 10V
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/s2 growth), the FVT gives a wrong finite number — check pole locations first.
•Forgetting the initial condition in the derivative rule:
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.
. 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.
L{f(t)}=F(s)=∫0∞f(t)e−stdt
, which is exactly the assumption baked into a transfer function.
L{f′(t)}=sF(s)−f(0),L{f(t−a)u(t−a)}=e−asF(s)
recovers the steady-state
jωL
and
1/(jωC)
), and it lets you write
H(s)
for any RLC network by inspection.
ZR=R,ZL=sL,ZC=sC1
-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)=X(s)Y(s)=ansn+⋯+a0bmsm+⋯+b0
H(s)=K(s−p1)(s−p2)⋯(s−z1)(s−z2)⋯
limit; the final-value theorem gives
f(∞)
from the
s→0
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+)=s→∞limsF(s),f(∞)=s→0limsF(s)
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(jω)=H(s)s=jω,gain=∣H(jω)∣,phase=∠H(jω)
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→∞)
. 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.
Unilateral transform for t≥0. s=σ+jω is complex frequency.
, the steady-state current via the final-value theorem, and the current at
t=0.2s
.
Solution. s-domain: source 10/s, impedance R+sL=5+0.5s. So I(s)=5+0.5s10/s=s(0.5s+5)10=s(s+10)20.
Final value: i(∞)=lims→0sI(s)=lims→0s+1020=2.00A (pole at s=−10 is stable, so FVT is valid).
Invert: i(t)=2(1−e−t/τ) with τ=L/R=0.1s. At t=0.2s: i=2(1−e−2)=2(0.8647)=1.73A.
Sanity: i(∞)=V/R=10/5=2A matches the FVT; at 2τ the current is 86.5% of 2A → 1.73A, consistent.
i(∞)=s→0lims+1020=2.00A
Poles, zeros, and DC gain of a transfer function
Problem. A system has H(s)=s2+3s+210. 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: s2+3s+2=(s+1)(s+2), so poles at s=−1
y(∞)=H(0)=210=5.00
Initial- and final-value theorems together
Problem. Given F(s)=s2+4s+32s+3, find f(0+) and f(∞) without inverting, then verify by partial fractions.
, 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/(s2+b2) 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.
.
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)⋅s1
, so
y(∞)=lims→0sY(s)=lims→0H(s)=H(0)=5.00
(FVT valid since poles are stable).
Sanity: for a stable system the step settles to the DC gain, so