Sketch both signals and note where each one is nonzero.
Flip h and slide it: for each output time, find the range where both signals overlap.
Multiply and add (discrete) or multiply and integrate (continuous) over that overlap only.
Check the result: the output starts at the sum of the start times, ends at the sum of the end times, and its area or sum is the product of the input areas or sums.
2Practice problems
1.Convolution & LTI systemsMedium
The pulse x(t) = 3 for 0 ≤ t < 1 s (zero elsewhere) is applied to a system with the impulse response h(t) = 5 for 0 ≤ t < 3 s (zero elsewhere). Find the output y(3.5) at t = 3.5 s.
x(t) = 3, 0 ≤ t < 1 sh(t) = 5, 0 ≤ t < 3 st = 3.5 s
y(3.5) =
2.Convolution & LTI systemsHard
A system obeys y[n] = 0.9·y[n − 1] + 2·x[n] and starts at rest. Find its impulse response sample h[3].
y[n] = 0.9y[n − 1] + 2x[n]x[n] = δ[n]
h[3] =
3.Laplace transformsMedium
Find f(t) for F(s) = (3s + 8)/(s² + 4s + 40) by completing the square, and evaluate f(1).
F(s) = (3s + 8)/(s² + 4s + 40)t = 1 s
f(1) =
4.Laplace transformsHard
A system obeys y″ + 10y′ + 24y = 48·u(t), with y(0) = 1 and y′(0) = 0. At t = 0 a step input u(t) = 5 is applied. Use the Laplace transform to find y(0.25).
A system has the transfer function G(s) = 76/(s² + 5s + 38) and receives a unit step input. Find the peak time tp of the step response.
G(s) = 76/(s² + 5s + 38)
tp = s
6.Step responseHard
The step response of a second-order system without zeros shows 50 % overshoot, and the first peak occurs at t = 0.1 s. Find the natural frequency ωn.
%OS = 50 %tp = 0.1 s
ωn = rad/s
7.Bode plots & marginsMedium
In the series RLC band-pass filter shown the output is taken across R = 20 Ω, with L = 100 mH and C = 0.1 µF. Find the centre (resonant) frequency f0 in Hz.
R = 20 ΩL = 100 mHC = 0.1 µF
f0 = Hz
8.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 14/(s(s + 2)). Find the gain crossover frequency ωc, where |L(jω)| = 1.
L(s) = 14/(s(s + 2))
ωc = rad/s
3Solutions
1.Convolution & LTI systems
Answer: y(3.5) = 7.5
Continuous convolution: y(t) = ∫ x(τ)·h(t − τ) dτ. Flip h, slide it to position t and integrate the product of the overlapping parts. For causal signals the integral runs from 0 to t. Useful results: the area of y is the area of x times the area of h; if x lasts from t₁ to t₂ and h from t₃ to t₄, y lasts from t₁ + t₃ to t₂ + t₄. For h(t) = e−atu(t) and a step of height A, y(t) = (A/a)(1 − e−at).
Overlap of the flipped pulse
h(t − τ) is the pulse of h flipped and shifted so that it ends at τ = t.
τ ∈ [0.5, 1] → length 0.5 s
Integral
y(3.5) = 3·5·0.5 = 7.5
Answer
y(3.5) = 7.5
y(t) is a trapezoid: it rises from 0, stays at 15 and falls back to 0 at t = 4 s.
2.Convolution & LTI systems
Answer: h[3] = 1.458
Impulse response and Fourier series: the impulse response h(t) describes an LTI system completely: for y′ + a·y = b·x it is h(t) = b·e−atu(t), and for y[n] = a·y[n − 1] + b·x[n] it is h[n] = b·aⁿu[n]. A sinusoid cos ωt passes through with gain |H(jω)| and phase ∠H(jω), where H is the Fourier transform of h. A periodic signal is a sum of harmonics at k/T: a square wave of ±A has only odd harmonics with amplitudes 4A/(kπ), a triangle wave of peak A has 8A/(π²k²), and a pulse train of height A and duty cycle D has the mean A·D and the harmonics (2A/(kπ))·|sin(kπD)|.
Iterate
h[0] = 2
h[1] = 1.8
h[2] = 1.62
h[3] = 1.458
Result
h[3] = 2·(0.9)3 = 1.458
Answer
h[3] = 1.458
3.Laplace transforms
Answer: f(1) = 0.3772
Laplace transform: F(s) = ∫₀∞ f(t)·e−st dt turns derivatives into multiplications by s: ℒ{f′} = sF(s) − f(0), ℒ{f″} = s²F(s) − s·f(0) − f′(0). Standard pairs: 1 ↔ 1/s, e−at ↔ 1/(s + a), t·e−at ↔ 1/(s + a)², sin ωt ↔ ω/(s² + ω²), cos ωt ↔ s/(s² + ω²), and a shift s → s + σ multiplies the time function by e−σt. To invert a rational F(s), split it into partial fractions: the residue of a simple pole p is A = [(s − p)·F(s)] at s = p. Final value theorem: f(∞) = lims→0 s·F(s), valid only when all poles of s·F(s) lie in the left half-plane. Initial value theorem: f(0⁺) = lims→∞ s·F(s).
Laplace transform: F(s) = ∫₀∞ f(t)·e−st dt turns derivatives into multiplications by s: ℒ{f′} = sF(s) − f(0), ℒ{f″} = s²F(s) − s·f(0) − f′(0). Standard pairs: 1 ↔ 1/s, e−at ↔ 1/(s + a), t·e−at ↔ 1/(s + a)², sin ωt ↔ ω/(s² + ω²), cos ωt ↔ s/(s² + ω²), and a shift s → s + σ multiplies the time function by e−σt. To invert a rational F(s), split it into partial fractions: the residue of a simple pole p is A = [(s − p)·F(s)] at s = p. Final value theorem: f(∞) = lims→0 s·F(s), valid only when all poles of s·F(s) lie in the left half-plane. Initial value theorem: f(0⁺) = lims→∞ s·F(s).
Transform the equation
(s² + 10s + 24)·Y(s) = 1s + 10 + 240/s
Y(s) = (s² + 10s + 240)/(s(s + 4)(s + 6))
Partial fractions
A = 240/(4·6) = 10
B = 216/((−4)·2) = −27
C = 216/((−6)·(−2)) = 18
Result
y(t) = 10 − 27·e−4t + 18·e−6t
y(0.25) = 4.084
Answer
y(0.25) = 4.084
Check: y(0) = A + B + C = 1.
5.Step response
Answer: tp = 0.5575 s
Step response: a first-order system G(s) = K/(τs + 1) answers a step of size U with y(t) = K·U·(1 − e−t/τ): it reaches 63.2 % of the final value K·U after one time constant τ and 98 % after 4τ. A second-order system G(s) = K·ωₙ²/(s² + 2ζωₙs + ωₙ²) has the poles s = −ζωₙ ± jωd with ωd = ωₙ√(1 − ζ²). For 0 < ζ < 1 the step response overshoots: percent overshoot %OS = 100·e−ζπ/√(1−ζ²), peak time tp = π/ωd, and the 2 % settling time ts ≈ 4/(ζωₙ). Compare the denominator with s² + 2ζωₙs + ωₙ² to read off ωₙ and ζ. For a mass–spring–damper ωₙ = √(k/m) and ζ = c/(2√(km)); for a series RLC circuit ωₙ = 1/√(LC) and ζ = (R/2)·√(C/L).
Step response: a first-order system G(s) = K/(τs + 1) answers a step of size U with y(t) = K·U·(1 − e−t/τ): it reaches 63.2 % of the final value K·U after one time constant τ and 98 % after 4τ. A second-order system G(s) = K·ωₙ²/(s² + 2ζωₙs + ωₙ²) has the poles s = −ζωₙ ± jωd with ωd = ωₙ√(1 − ζ²). For 0 < ζ < 1 the step response overshoots: percent overshoot %OS = 100·e−ζπ/√(1−ζ²), peak time tp = π/ωd, and the 2 % settling time ts ≈ 4/(ζωₙ). Compare the denominator with s² + 2ζωₙs + ωₙ² to read off ωₙ and ζ. For a mass–spring–damper ωₙ = √(k/m) and ζ = c/(2√(km)); for a series RLC circuit ωₙ = 1/√(LC) and ζ = (R/2)·√(C/L).
Frequency response: substitute s = jω in the transfer function. The magnitude in decibels is 20·log₁₀|H(jω)| and the phase is the sum of the angles of the numerator factors minus those of the denominator factors. A first-order RC or RL filter has its corner (−3 dB) frequency at ωc = 1/(RC) or R/L, and fc = ωc/(2π); a low-pass filter gives |H| = 1/√(1 + (ω/ωc)²) and φ = −atan(ω/ωc). In the asymptotic (straight-line) Bode plot each pole bends the magnitude down by 20 dB/decade at its corner frequency, each zero bends it up by 20 dB/decade, and an integrator 1/s gives −20 dB/decade at all frequencies. Gain crossover ωc: |L(jω)| = 1 (0 dB); phase margin PM = 180° + ∠L(jωc). Phase crossover ω180: ∠L(jω) = −180°; gain margin GM = −20·log₁₀|L(jω180)| dB. The closed loop is stable when both margins are positive.
Resonant frequency
ω₀ = 1/√(LC) = 1/√(0.1·1 × 10⁻⁷) = 10000 rad/s
f₀ = ω₀/(2π) = 1592 Hz
Answer
f0 = 1592 Hz
8.Bode plots & margins
Answer: ωc = 3.485 rad/s
Frequency response: substitute s = jω in the transfer function. The magnitude in decibels is 20·log₁₀|H(jω)| and the phase is the sum of the angles of the numerator factors minus those of the denominator factors. A first-order RC or RL filter has its corner (−3 dB) frequency at ωc = 1/(RC) or R/L, and fc = ωc/(2π); a low-pass filter gives |H| = 1/√(1 + (ω/ωc)²) and φ = −atan(ω/ωc). In the asymptotic (straight-line) Bode plot each pole bends the magnitude down by 20 dB/decade at its corner frequency, each zero bends it up by 20 dB/decade, and an integrator 1/s gives −20 dB/decade at all frequencies. Gain crossover ωc: |L(jω)| = 1 (0 dB); phase margin PM = 180° + ∠L(jωc). Phase crossover ω180: ∠L(jω) = −180°; gain margin GM = −20·log₁₀|L(jω180)| dB. The closed loop is stable when both margins are positive.