Career upgrade: Learn practical AI skills for better jobs and higher pay.
Level up

4.1 Linear Systems & Transfer Functions

Key Takeaways

  • A linear time-invariant (LTI) system obeys superposition and is fully described by its impulse response h(t); the output is y(t) = x(t) * h(t), the convolution of input and impulse response.
  • Convolution in time becomes multiplication in the Laplace/frequency domain: Y(s) = X(s)H(s), so the transfer function is H(s) = Y(s)/X(s) for zero initial conditions.
  • Poles are the roots of the denominator of H(s) and set stability and response speed; zeros are roots of the numerator and shape the frequency response.
  • A continuous-time LTI system is stable when every pole lies in the open left half of the s-plane (negative real part).
  • Frequency response is H(jω) = H(s) evaluated at s = jω; Bode plots show magnitude in decibels (20·log10|H|) and phase versus log frequency.
Last updated: May 2026

Why linear systems matters on the FE

The Linear Systems knowledge area is 5-8 of the 110 questions, and its concepts feed directly into Signal Processing, Control Systems, and Communications. Most items are not derivations from scratch — they ask you to recognize a standard form and pull one fact: a time constant, a pole location, a gain, or a stability verdict. Your first move is always to classify the object in front of you.

A linear time-invariant (LTI) system has two properties. Linearity means it obeys superposition: scaling and adding inputs scales and adds outputs. Time invariance means a delayed input produces an equally delayed output with the same shape. Almost every system on the FE is treated as LTI, which is exactly what lets you use transfer functions and convolution.

Impulse response and convolution

An LTI system is completely characterized by its impulse response h(t) — the output when the input is a unit impulse δ(t). For any input x(t), the output is the convolution:

y(t) = x(t) * h(t) = ∫ x(τ) h(t − τ) dτ

Convolution is the time-domain workhorse, but it is tedious. The exam-smart move is to leave the time domain. In the Laplace domain, convolution becomes ordinary multiplication:

Y(s) = X(s) · H(s)

That single fact is why transfer functions exist and why they dominate FE linear-systems questions.

Transfer functions, poles, and zeros

The transfer function is the Laplace-domain ratio of output to input for zero initial conditions:

H(s) = Y(s)/X(s) = N(s)/D(s)

It is a rational function of s. The zeros are the roots of the numerator N(s) — frequencies where H(s) = 0. The poles are the roots of the denominator D(s) — frequencies where H(s) blows up. Poles govern the natural response and stability; zeros shape gain and notches.

s-plane featureMeaning
Pole in left half-plane (Re < 0)Decaying mode — contributes to stability
Pole in right half-plane (Re > 0)Growing mode — system is unstable
Pole on imaginary axis (Re = 0)Marginally stable / sustained oscillation
Complex-conjugate polesOscillatory (underdamped) response
Pole far from originFaster mode (short time constant)
ZeroFrequency where output is suppressed

Stability rule (continuous-time): an LTI system is stable if and only if every pole of H(s) has a negative real part — all poles strictly in the open left half-plane.

Frequency response and Bode plots

The frequency response is the transfer function evaluated along the imaginary axis:

H(jω) = H(s)|_{s = jω}

It is complex: |H(jω)| is the gain at frequency ω, and ∠H(jω) is the phase shift. A Bode plot displays both versus log frequency — magnitude in decibels, where dB = 20·log10|H(jω)|, and phase in degrees.

Key Bode landmarks worth memorizing:

  • A factor of (1 + jω/ωc) in the denominator is a pole at corner frequency ωc: magnitude rolls off at −20 dB/decade above ωc and the phase drops 90° (centered at ωc, where it is −45°).
  • A numerator factor (1 + jω/ωc) is a zero: +20 dB/decade and +90° phase.
  • A simple low-pass first-order filter H(jω) = 1/(1 + jωRC) has corner ωc = 1/(RC); at ωc the gain is −3 dB (|H| = 1/√2) and phase = −45°.
  • A gain factor K adds a flat 20·log10K dB and no phase.

First- and second-order forms

The two canonical forms cover most FE items. A first-order system H(s) = K/(τs + 1) has a single real pole at s = −1/τ; for a unit step it reaches ~63% of final value after one time constant τ. A second-order system H(s) = ωn²/(s² + 2ζωn s + ωn²) introduces natural frequency ωn and damping ratio ζ. Complex poles (0 < ζ < 1) give overshoot and ringing; ζ ≥ 1 gives no overshoot. Recognizing which form you have is usually the whole problem.

Test Your Knowledge

An LTI system has transfer function H(s) = 10/((s + 2)(s − 3)). Is the system stable, and why?

A
B
C
D
Test Your Knowledge

A first-order low-pass filter has H(jω) = 1/(1 + jωRC) with RC = 1 ms. At the corner (cutoff) frequency, what is the magnitude and phase?

A
B
C
D