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) = 5 for 0 ≤ t < 3 s is convolved with the pulse h(t) = 1 for 0 ≤ t < 4 s. Find the area Ay under y(t) = x(t) ∗ h(t).
x(t) = 5, 0 ≤ t < 3 sh(t) = 1, 0 ≤ t < 4 s
Ay =
2.Convolution & LTI systemsHard
The input x(t) = 2·cos(20t) is applied to a system with h(t) = 2·e−10tu(t). Find the amplitude Ym of the steady-state output.
h(t) = 2e−10tu(t)x(t) = 2cos(20t)
Ym =
3.Laplace transformsMedium
F(s) = (6s + 19)/(s² + 6s + 5). Its inverse transform has the form f(t) = A·e−t + B·e−5t. Find the coefficient A by partial fractions.
F(s) = (6s + 19)/(s² + 6s + 5)
A =
4.Laplace transformsHard
Solve y″ + 3y′ + 2y = 0 with y(0) = 5 and y′(0) = 12 using the Laplace transform, and find y(0.5).
y″ + 3y′ + 2y = 0y(0) = 5y′(0) = 12t = 0.5 s
y(0.5) =
5.Step responseMedium
A system has the transfer function G(s) = 67/(s² + 6s + 67) and receives a unit step input. Find the percent overshoot %OS of the step response.
G(s) = 67/(s² + 6s + 67)
%OS = %
6.Step responseHard
A mass m = 0.5 kg hangs on a spring k = 400 N/m with a viscous damper c = 4 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 20 N is suddenly applied. Find the damping ratio ζ.
m = 0.5 kgk = 400 N/mc = 4 N·s/mF = 20 N
ζ =
7.Bode plots & marginsMedium
Find the phase ∠G(jω) of G(s) = 31(s + 19)/(s(s + 1)) at ω = 3 rad/s, in degrees.
G(s) = 31(s + 19)/(s(s + 1))ω = 3 rad/s
∠G(jω) = °
8.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 3.9/(s(s + 1)). Find the gain crossover frequency ωc, where |L(jω)| = 1.
L(s) = 3.9/(s(s + 1))
ωc = rad/s
3Solutions
1.Convolution & LTI systems
Answer: Ay = 60
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).
Areas of the inputs
∫x dt = 5·3 = 15
∫h dt = 1·4 = 4
Area of the output
Ay = 15·4 = 60
Answer
Ay = 60
2.Convolution & LTI systems
Answer: Ym = 0.1789
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)|.
Frequency response
H(jω) = 2/(10 + jω)
Gain at the input frequency
|H(j20)| = 2/√(10² + 20²) = 0.08944
Ym = 2·0.08944 = 0.1789
Answer
Ym = 0.1789
3.Laplace transforms
Answer: A = 3.25
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).
Factor the denominator
s² + 6s + 5 = (s + 1)(s + 5)
F(s) = A/(s + 1) + B/(s + 5)
Cover-up rule
Multiply by (s + 1) and set s = −1: the B-term vanishes.
A = (6s + 19)/(s + 5) |s = −1 = 13/4 = 3.25
Answer
A = 3.25
(B = 2.75, so f(t) = 3.25·e−t + 2.75·e−5t)
4.Laplace transforms
Answer: y(0.5) = 7.09
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
s²Y − 5s − 12 + 3(sY − 5) + 2Y = 0
Y(s) = (5s + 27)/((s + 1)(s + 2))
Partial fractions
A = 22/1 = 22
B = 17/(−1) = −17
Result
y(t) = 22·e−t − 17·e−2t
y(0.5) = 7.09
Answer
y(0.5) = 7.09
Check: y(0) = A + B = 5.
5.Step response
Answer: %OS = 29.01 %
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).
Compare with the standard form
The denominator must match s² + 2ζωₙs + ωₙ².
ωₙ² = 67 → ωₙ = 8.185 rad/s
2ζωₙ = 6 → ζ = 6/(2·8.185) = 0.3665
Result
%OS = 100·e−ζπ/√(1−ζ²) = 100·e−1.238 = 29.01 %
Answer
%OS = 29.01 %
6.Step response
Answer: ζ = 0.1414
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).
Natural frequency and damping ratio
ωₙ = √(k/m) = √(400/0.5) = 28.28 rad/s
ζ = c/(2√(km)) = 4/(2·√(400·0.5)) = 0.1414
Result
ζ = 0.1414
Answer
ζ = 0.1414
7.Bode plots & margins
Answer: ∠G(jω) = −152.6 °
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.
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.