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 input x(t) = 4·e−6tu(t) is applied to a system with h(t) = e−1tu(t). Find the output y(0.25) at t = 0.25 s.
x(t) = 4·e−6tu(t)h(t) = e−1tu(t)t = 0.25 s
y(0.25) =
2.Convolution & LTI systemsHard
A triangle wave rises and falls linearly between −1 and +1 with period T = 4 ms. Find the amplitude c3 of its Fourier component at 750 Hz.
±1T = 4 msf = 750 Hz
c3 =
3.Laplace transformsMedium
F(s) = (2s + 23)/(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) = (2s + 23)/(s² + 6s + 5)
A =
4.Laplace transformsHard
A system is described by the differential equation y″ + 7y′ + 6y = 3u′ + 20u, with input u(t) and output y(t). Find its transfer function G(s) = Y(s)/U(s) (zero initial conditions) and evaluate the impulse response g(t) at t = 1.2 s. Find g(1.2).
y″ + 7y′ + 6y = 3u′ + 20ut = 1.2 s
g(1.2) =
5.Step responseMedium
A system has the transfer function G(s) = 122/(s² + 3s + 61) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.
G(s) = 122/(s² + 3s + 61)
σ =
6.Step responseHard
The step response of a second-order system without zeros shows 25 % overshoot, and the first peak occurs at t = 0.25 s. Find the natural frequency ωn.
%OS = 25 %tp = 0.25 s
ωn = rad/s
7.Bode plots & marginsMedium
The RL low-pass filter shown has R = 10 Ω, L = 1 mH. A sinusoidal input of 4.8 kHz is applied. Find the gain |H| = |vout/vin| in dB.
R = 10 ΩL = 1 mHf = 4.8 kHz
|H| = dB
8.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 15.5/(s(s + 5)). Find the phase margin PM in degrees.
L(s) = 15.5/(s(s + 5))
PM = °
3Solutions
1.Convolution & LTI systems
Answer: y(0.25) = 0.4445
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).
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)|.
Harmonic number
f₀ = 1/T = 250 Hz → k = 3
Amplitude
c3 = 8·1/(π²·3²) = 0.09006
Answer
c3 = 0.09006
3.Laplace transforms
Answer: A = 5.25
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 = (2s + 23)/(s + 5) |s = −1 = 21/4 = 5.25
Answer
A = 5.25
(B = −3.25, so f(t) = 5.25·e−t − 3.25·e−5t)
4.Laplace transforms
Answer: g(1.2) = 1.024
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).
Transfer function
G(s) = Y(s)/U(s) = (3s + 20)/(s² + 7s + 6)
Partial fractions
s² + 7s + 6 = (s + 1)(s + 6)
A = 3.4, B = −0.4
Impulse response
g(t) = 3.4·e−t − 0.4·e−6t
g(1.2) = 1.024
Answer
g(1.2) = 1.024
5.Step response
Answer: σ = −1.5
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 = 61 → ωₙ = 7.81 rad/s
2ζωₙ = 3 → ζ = 3/(2·7.81) = 0.1921
Result
σ = −ζωₙ = −3/2 = −1.5
Answer
σ = −1.5
6.Step response
Answer: ωn = 13.74 rad/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).
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.
Corner frequency
ωc = R/L = 10/0.001 = 10000 rad/s
fc = 1592 Hz → f/fc = 4800/1592 = 3.016
Gain
|H| = 1/√(1 + 3.016²) = 0.3147
20·log₁₀(0.3147) = −10.04 dB
Answer
|H| = −10.04 dB
8.Bode plots & margins
Answer: PM = 61.43 °
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.