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) = 1 for 0 ≤ t < 2 s (zero elsewhere) is applied to a system with h(t) = e−4tu(t). Find the output y(2.5) at t = 2.5 s, after the pulse has ended.
x(t) = 1, 0 ≤ t < 2 sh(t) = e−4tu(t)t = 2.5 s
y(2.5) =
2.Convolution & LTI systemsHard
A system obeys y[n] = −0.8·y[n − 1] + 3·x[n] and starts at rest. The input is the unit step x[n] = u[n]. Find y[2].
y[n] = −0.8y[n − 1] + 3x[n]x[n] = u[n]
y[2] =
3.Laplace transformsMedium
Find f(t) for F(s) = (4s + 29)/(s² + 8s + 41) by completing the square, and evaluate f(0.25).
F(s) = (4s + 29)/(s² + 8s + 41)t = 0.25 s
f(0.25) =
4.Laplace transformsHard
Solve y″ + 5y′ + 6y = 0 with y(0) = 3 and y′(0) = 1 using the Laplace transform, and find y(0.3).
y″ + 5y′ + 6y = 0y(0) = 3y′(0) = 1t = 0.3 s
y(0.3) =
5.Step responseMedium
A system has the transfer function G(s) = 102/(s² + 8s + 51) and receives a unit step input. Find the peak time tp of the step response.
G(s) = 102/(s² + 8s + 51)
tp = s
6.Step responseHard
A series RLC circuit with R = 124 Ω, L = 500 mH and C = 47 µF is connected to a 10 V step at t = 0. The capacitor is uncharged and no current flows before the step. Find the peak capacitor voltage vC,max.
R = 124 ΩL = 500 mHC = 47 µFV = 10 V
vC,max = V
7.Bode plots & marginsMedium
Find the exact magnitude |G(jω)| of G(s) = 32/((s + 2)(s + 4)) at ω = 0.5 rad/s, in dB.
G(s) = 32/((s + 2)(s + 4))ω = 0.5 rad/s
|G(jω)| = dB
8.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 32/(s(s + 4)(s + 5)). Find the gain margin GM in dB.
L(s) = 32/(s(s + 4)(s + 5))
GM = dB
3Solutions
1.Convolution & LTI systems
Answer: y(2.5) = 0.03382
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).
Output at the end of the pulse
y(2) = (1/4)(1 − e−8) = 0.2499
Decay after the pulse
y(2.5) = 0.2499·e−4·0.5 = 0.03382
Answer
y(2.5) = 0.03382
2.Convolution & LTI systems
Answer: y[2] = 2.52
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
y[0] = 3
y[1] = 0.6
y[2] = 2.52
Result
y[2] = 3·(1 − (−0.8)3)/(1 − (−0.8)) = 2.52
Answer
y[2] = 2.52
3.Laplace transforms
Answer: f(0.25) = 1.372
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
s²Y − 3s − 1 + 5(sY − 3) + 6Y = 0
Y(s) = (3s + 16)/((s + 2)(s + 3))
Partial fractions
A = 10/1 = 10
B = 7/(−1) = −7
Result
y(t) = 10·e−2t − 7·e−3t
y(0.3) = 2.642
Answer
y(0.3) = 2.642
Check: y(0) = A + B = 3.
5.Step response
Answer: tp = 0.531 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.
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.