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)t22h(t)t44

The pulse x(t) = 2 for 0 ≤ t < 2 s is convolved with h(t), a triangle that falls linearly from 4 at t = 0 to 0 at t = 4 s. Find the area Ay under y(t) = x(t) ∗ h(t).

x(t) = 2, 0 ≤ t < 2 sh(0) = 4, h(4) = 0
Ay =
2.Convolution & LTI systemsHard
δ(t)y′ + 1y = 2xh(t)

A system obeys y′(t) + 1·y(t) = 2·x(t). Find its impulse response at t = 1 s, h(1).

y′ + 1y = 2xt = 1 s
h(1) =
3.Laplace transformsMedium
ReIm−1−2−3−4−5−6−7−81

F(s) = (6s + 29)/(s² + 11s + 28). Its inverse transform has the form f(t) = A·e−4t + B·e−7t. Find the coefficient A by partial fractions.

F(s) = (6s + 29)/(s² + 11s + 28)
A =
4.Laplace transformsHard
ReIm−1−2−3−4−51

A system is described by the differential equation y″ + 6y′ + 10y = u′ + 5u, 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 = 0.4 s. Find g(0.4).

y″ + 6y′ + 10y = u′ + 5ut = 0.4 s
g(0.4) =
5.Step responseMedium
t (s)y(t)1ts​ = ?

A system has the transfer function G(s) = 30/(s² + 2s + 30) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 30/(s² + 2s + 30)
ts = s
6.Step responseHard
R(s)+Ks(s + 8)G(s)C(s)−

A unity negative feedback loop has G(s) = K/(s(s + 8)). Find the gain K that gives a step response with 5 % overshoot.

G(s) = K/(s(s + 8))%OS = 5 %
K =
7.Bode plots & marginsMedium
vin​L = 10 mHC = 0.22 µFR = 5 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 5 Ω, with L = 10 mH and C = 0.22 µF. Find the −3 dB bandwidth BW in Hz.

R = 5 ΩL = 10 mHC = 0.22 µF
BW = Hz
8.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−60−40−200204060ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​ = ?

A unity-feedback loop has the open-loop transfer function L(s) = 88/(s(s + 8)). Find the gain crossover frequency ωc, where |L(jω)| = 1.

L(s) = 88/(s(s + 8))
ωc = rad/s

3Solutions

1.Convolution & LTI systems

Answer: Ay = 32

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 = 2·2 = 4
∫h dt = ½·4·4 = 8

Area of the output

Ay = 4·8 = 32

Answer

Ay = 32
2.Convolution & LTI systems

Answer: h(1) = 0.7358

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)|.

Transfer function

H(s) = Y(s)/X(s) = 2/(s + 1)

Impulse response

h(t) = 2·e−1tu(t)
h(1) = 2·e−1 = 0.7358

Answer

h(1) = 0.7358
3.Laplace transforms

Answer: A = 1.667

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 + 11s + 28 = (s + 4)(s + 7)
F(s) = A/(s + 4) + B/(s + 7)

Cover-up rule

Multiply by (s + 4) and set s = −4: the B-term vanishes.

A = (6s + 29)/(s + 7) |s = −4 = 5/3 = 1.667

Answer

A = 1.667
(B = 4.333, so f(t) = 1.667·e−4t + 4.333·e−7t)
4.Laplace transforms

Answer: g(0.4) = 0.512

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) = (s + 5)/(s² + 6s + 10)

Partial fractions

s² + 6s + 10 = (s + 3)² + 1²
G(s) = 1·(s + 3)/((s + 3)² + 1²) + (2/1)·1/((s + 3)² + 1²)

Impulse response

g(t) = e−3t·(cos 1t + 2·sin 1t)
g(0.4) = 0.512

Answer

g(0.4) = 0.512
5.Step response

Answer: ts = 4 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 = 30 → ωₙ = 5.477 rad/s
2ζωₙ = 2 → ζ = 2/(2·5.477) = 0.1826

Result

ζωₙ = 1 → ts ≈ 4/1 = 4 s

Answer

ts = 4 s
6.Step response

Answer: K = 33.6

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) = K/(s² + 8s + K)

Damping ratio from the overshoot

ζ = −ln(0.05)/√(π² + ln²(0.05)) = 0.6901

Gain

ωₙ = 8/(2ζ) = 5.796 rad/s
K = ωₙ² = 33.6

Answer

K = 33.6
7.Bode plots & margins

Answer: BW = 79.58 Hz

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/√(0.01·2.2 × 10⁻⁷) = 21320 rad/s
f₀ = ω₀/(2π) = 3393 Hz

Bandwidth

BW = R/L = 5/0.01 = 500 rad/s = 79.58 Hz

Answer

BW = 79.58 Hz
8.Bode plots & margins

Answer: ωc = 7.851 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.

Gain crossover

ω⁴ + 64ω² − 7744 = 0
ω² = (−64 + √(4096 + 4·7744))/2 = 61.64
ωc = 7.851 rad/s

Answer

ωc = 7.851 rad/s

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