Signals and Systems: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

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1Formulas and methods

Convolution & LTI systems

Discrete convolution
y[n] = Σ x[k]·h[n − k]
Continuous convolution
y(t) = ∫ x(τ)·h(t − τ) dτ
Length and sum
Ly = Lx + Lh − 1, Σy = Σx·Σh
Start, end and area
t1 + t3 … t2 + t4, ∫y = ∫x·∫h
Step into e^(−at)u(t)
y(t) = (A/a)(1 − e−at)
First-order system
y′ + a·y = b·x → h(t) = b·e−atu(t)
First-order recursion
y[n] = a·y[n − 1] + b·x[n] → h[n] = b·aⁿ
Sinusoid through h(t)
|H(jω)|, ∠H(jω); H(jω) = b/(a + jω)
Fourier series amplitudes
square ±A: 4A/(kπ), odd k; triangle ±A: 8A/(π²k²), odd k; pulse train: A·D and (2A/(kπ))|sin kπD|

Method

  1. Sketch both signals and note where each one is nonzero.
  2. Flip h and slide it: for each output time, find the range where both signals overlap.
  3. Multiply and add (discrete) or multiply and integrate (continuous) over that overlap only.
  4. 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
x(t)t−20h(t)t34

The signal x(t) is nonzero only between t = −2 s and t = 0 s, and h(t) only between t = 3 s and t = 4 s. Find the time tend at which y(t) = x(t) ∗ h(t) ends.

x: −2 … 0 sh: 3 … 4 s
tend = s
2.Convolution & LTI systemsHard
x(t)t (ms)12410

A square wave switches between 10 and 0 with period T = 2 ms (half a period at each level). Find the amplitude c7 of its Fourier component at 3500 Hz.

0 … 10T = 2 msf = 3500 Hz
c7 =
3.Laplace transformsMedium
t (s)f(t)t = 0.6f(0.6) = ?

Find f(t) for F(s) = 45/(s(s + 2)(s + 3)) by partial fractions, and evaluate f(0.6).

F(s) = 45/(s(s + 2)(s + 3))t = 0.6 s
f(0.6) =
4.Laplace transformsHard
t (s)y(t)t = 0.2y(0.2) = ?

A system obeys y″ + 7y′ + 12y = 60·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 1 for t ≥ 0. Use the Laplace transform to find the output y(0.2).

y″ + 7y′ + 12y = 60·u(t)u(t) = 1y(0) = 0, y′(0) = 0t = 0.2 s
y(0.2) =
5.Step responseMedium
t (s)y(t)3tp​ = ?

A system has the transfer function G(s) = 216/(s² + 8s + 72) and receives a unit step input. Find the peak time tp of the step response.

G(s) = 216/(s² + 8s + 72)
tp = s
6.Step responseHard
k = 400 N/mc = 8 N·s/mm = 1 kgF = 5 Nx(t)

A mass m = 1 kg hangs on a spring k = 400 N/m with a viscous damper c = 8 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 5 N is suddenly applied. Find the damping ratio ζ.

m = 1 kgk = 400 N/mc = 8 N·s/mF = 5 N
ζ =
7.Bode plots & marginsMedium
vin​L = 22 mHR = 47 Ωvout​

The RL low-pass filter shown has R = 47 Ω, L = 22 mH. A sinusoidal input of 170 Hz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 47 ΩL = 22 mHf = 170 Hz
φ = °
8.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−80−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−270−180−900ω180​ = ?

A unity-feedback loop has the open-loop transfer function L(s) = 140/(s(s + 4)(s + 10)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 140/(s(s + 4)(s + 10))
ω180 = rad/s

3Solutions

1.Convolution & LTI systems

Answer: tend = 4 s

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).

Start and end

tstart = −2 + 3 = 1 s
tend = 0 + 4 = 4 s

Answer

tend = 4 s

Answer

tend = 4 s
2.Convolution & LTI systems

Answer: c7 = 0.9095

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 = 500 Hz → k = 3500/500 = 7

Amplitude

c7 = 4·5/(7π) = 0.9095

Answer

c7 = 0.9095
3.Laplace transforms

Answer: f(0.6) = 3.203

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).

Partial fractions

F(s) = A/s + B/(s + 2) + C/(s + 3)
A = 45/(2·3) = 7.5
B = 45/((−2)·1) = −22.5
C = 45/((−3)·(−1)) = 15

Inverse transform

f(t) = 7.5 − 22.5·e−2t + 15·e−3t

Evaluate

f(0.6) = 7.5 − 22.5·e−2·0.6 + 15·e−3·0.6 = 3.203

Answer

f(0.6) = 3.203
4.Laplace transforms

Answer: y(0.2) = 0.7637

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

Y(s) = 60/(s(s + 3)(s + 4))

Partial fractions

A = 60/(3·4) = 5
B = 60/((−3)·1) = −20
C = 60/((−4)·(−1)) = 15

Output

y(t) = 5 − 20·e−3t + 15·e−4t
y(0.2) = 0.7637

Answer

y(0.2) = 0.7637
5.Step response

Answer: tp = 0.4198 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).

Compare with the standard form

The denominator must match s² + 2ζωₙs + ωₙ².

ωₙ² = 72 → ωₙ = 8.485 rad/s
2ζωₙ = 8 → ζ = 8/(2·8.485) = 0.4714

Result

ωd = 8.485·√(1 − 0.4714²) = 7.483 rad/s → tp = π/ωd = 0.4198 s

Answer

tp = 0.4198 s
6.Step response

Answer: ζ = 0.2

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) = √(400/1) = 20 rad/s
ζ = c/(2√(km)) = 8/(2·√(400·1)) = 0.2

Result

ζ = 0.2

Answer

ζ = 0.2
7.Bode plots & margins

Answer: φ = −26.56 °

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 = 47/0.022 = 2136 rad/s
fc = 340 Hz → f/fc = 170/340 = 0.5

Phase

φ = − atan(0.5) = −26.56°

Answer

φ = −26.56°
8.Bode plots & margins

Answer: ω180 = 6.325 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.

Phase crossover

(ω/4)·(ω/10) = 1 → ω180 = √(4·10) = 6.325 rad/s

Answer

ω180 = 6.325 rad/s

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