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)t12h(t)t01

The signal x(t) is nonzero only between t = 1 s and t = 2 s, and h(t) only between t = 0 s and t = 1 s. Find the time tstart at which y(t) = x(t) ∗ h(t) starts.

x: 1 … 2 sh: 0 … 1 s
tstart = s
2.Convolution & LTI systemsHard
δ[n]3+h[n]0.4·z⁻¹+

A system obeys y[n] = 0.4·y[n − 1] + 3·x[n] and starts at rest. Find its impulse response sample h[4].

y[n] = 0.4y[n − 1] + 3x[n]x[n] = δ[n]
h[4] =
3.Laplace transformsMedium
ReIm−224

Find f(t) for F(s) = 16/(s² + 6s + 25) by completing the square, and evaluate f(0.4).

F(s) = 16/(s² + 6s + 25)t = 0.4 s
f(0.4) =
4.Laplace transformsHard
t (s)y(t)t = 0.25y(0.25) = ?

A system obeys y″ + 9y′ + 20y = 80·u(t), with y(0) = 4 and y′(0) = 0. At t = 0 a step input u(t) = 5 is applied. Use the Laplace transform to find y(0.25).

y″ + 9y′ + 20y = 80·u(t)y(0) = 4y′(0) = 0u(t) = 5t = 0.25 s
y(0.25) =
5.Step responseMedium
t (s)y(t)1

A system has the transfer function G(s) = 87/(s² + 6s + 87) and receives a unit step input. Find the damping ratio ζ.

G(s) = 87/(s² + 6s + 87)
ζ =
6.Step responseHard
R(s)+50s(s + 8)G(s)C(s)−

A unity negative feedback loop has G(s) = 50/(s(s + 8)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).

G(s) = 50/(s(s + 8))
ts = s
7.Bode plots & marginsMedium
vin​L = 50 mHC = 4.7 µFR = 50 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 50 Ω, with L = 50 mH and C = 4.7 µF. Find the quality factor Q.

R = 50 ΩL = 50 mHC = 4.7 µF
Q =
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) = 78/(s(s + 4)(s + 7)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 78/(s(s + 4)(s + 7))
ω180 = rad/s

3Solutions

1.Convolution & LTI systems

Answer: tstart = 1 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 = 1 + 0 = 1 s
tend = 2 + 1 = 3 s

Answer

tstart = 1 s

Answer

tstart = 1 s
2.Convolution & LTI systems

Answer: h[4] = 0.0768

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] = 3
h[1] = 1.2
h[2] = 0.48
h[3] = 0.192
h[4] = 0.0768

Result

h[4] = 3·(0.4)4 = 0.0768

Answer

h[4] = 0.0768
3.Laplace transforms

Answer: f(0.4) = 1.204

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).

Complete the square

s² + 6s + 25 = (s + 3)² + 16 = (s + 3)² + 4²

Match the numerator

F(s) = (16/4)·4/((s + 3)² + 4²)

Inverse transform

f(t) = e−3t·(4·sin 4t)
f(0.4) = e−3·0.4·(4·sin(4·0.4)) = 1.204

Answer

f(0.4) = 1.204
4.Laplace transforms

Answer: y(0.25) = 8.906

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 + 9s + 20)·Y(s) = 4s + 36 + 400/s
Y(s) = (4s² + 36s + 400)/(s(s + 4)(s + 5))

Partial fractions

A = 400/(4·5) = 20
B = 320/((−4)·1) = −80
C = 320/((−5)·(−1)) = 64

Result

y(t) = 20 − 80·e−4t + 64·e−5t
y(0.25) = 8.906

Answer

y(0.25) = 8.906

Check: y(0) = A + B + C = 4.

5.Step response

Answer: ζ = 0.3216

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 = 87 → ωₙ = 9.327 rad/s
2ζωₙ = 6 → ζ = 6/(2·9.327) = 0.3216

Result

ζ = 0.3216

Answer

ζ = 0.3216
6.Step response

Answer: ts = 1 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).

Closed-loop transfer function

T(s) = 50/(s² + 8s + 50): ωₙ = 7.071 rad/s, ζ = 0.5657

Settling time

ζωₙ = 8/2 = 4
ts ≈ 4/(ζωₙ) = 1 s

Answer

ts = 1 s

For this loop ζωₙ = a/2 whatever K is: raising K adds overshoot but does not shorten the settling time.

7.Bode plots & margins

Answer: Q = 2.063

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/√(0.05·4.7 × 10⁻⁶) = 2063 rad/s
f₀ = ω₀/(2π) = 328.3 Hz

Quality factor

BW = R/L = 50/0.05 = 1000 rad/s = 159.2 Hz
Q = ω₀/BW = 2063/1000 = 2.063

Answer

Q = 2.063
8.Bode plots & margins

Answer: ω180 = 5.292 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/7) = 1 → ω180 = √(4·7) = 5.292 rad/s

Answer

ω180 = 5.292 rad/s

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