Signals and Systems: cram pack

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

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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)t11h(t)t25

The pulse x(t) = 1 for 0 ≤ t < 1 s (zero elsewhere) is applied to a system with the impulse response h(t) = 5 for 0 ≤ t < 2 s (zero elsewhere). Find the output y(1) at t = 1 s.

x(t) = 1, 0 ≤ t < 1 sh(t) = 5, 0 ≤ t < 2 st = 1 s
y(1) =
2.Convolution & LTI systemsHard
x(t)t (ms)0.5122−2

A triangle wave rises and falls linearly between −2 and +2 with period T = 1 ms. Find the amplitude c3 of its Fourier component at 3000 Hz.

±2T = 1 msf = 3000 Hz
c3 =
3.Laplace transformsMedium
t (s)f(t)t = 0.4f(0.4) = ?

Find f(t) for F(s) = 10/(s(s + 3)(s + 5)) by partial fractions, and evaluate f(0.4).

F(s) = 10/(s(s + 3)(s + 5))t = 0.4 s
f(0.4) =
4.Laplace transformsHard
t (s)y(t)t = 0.75y(0.75) = ?

A system obeys y″ + 6y′ + 5y = 5·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 10 for t ≥ 0. Use the Laplace transform to find the output y(0.75).

y″ + 6y′ + 5y = 5·u(t)u(t) = 10y(0) = 0, y′(0) = 0t = 0.75 s
y(0.75) =
5.Step responseMedium
t (s)y(t)2

A system has the transfer function G(s) = 52/(s² + 3s + 26) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.

G(s) = 52/(s² + 3s + 26)
σ =
6.Step responseHard
R(s)+Ks(s + 10)G(s)C(s)−

A unity negative feedback loop has G(s) = K/(s(s + 10)). Find the gain K that gives a step response with 15 % overshoot.

G(s) = K/(s(s + 10))%OS = 15 %
K =
7.Bode plots & marginsMedium
|H| (dB)0.10.111101010010010001000−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−135−90−450ω = 5

Find the phase ∠G(jω) of G(s) = (s + 14)/(s(s + 9)) at ω = 5 rad/s, in degrees.

G(s) = (s + 14)/(s(s + 9))ω = 5 rad/s
∠G(jω) = °
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) = 62/(s(s + 2)(s + 9)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 62/(s(s + 2)(s + 9))
ω180 = rad/s

3Solutions

1.Convolution & LTI systems

Answer: y(1) = 5

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).

Overlap of the flipped pulse

h(t − τ) is the pulse of h flipped and shifted so that it ends at τ = t.

τ ∈ [0, 1] → length 1 s

Integral

y(1) = 1·5·1 = 5

Answer

y(1) = 5

y(t) is a trapezoid: it rises from 0, stays at 5 and falls back to 0 at t = 3 s.

2.Convolution & LTI systems

Answer: c3 = 0.1801

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 = 1000 Hz → k = 3

Amplitude

c3 = 8·2/(π²·3²) = 0.1801

Answer

c3 = 0.1801
3.Laplace transforms

Answer: f(0.4) = 0.3

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 + 3) + C/(s + 5)
A = 10/(3·5) = 0.6667
B = 10/((−3)·2) = −1.667
C = 10/((−5)·(−2)) = 1

Inverse transform

f(t) = 0.6667 − 1.667·e−3t + e−5t

Evaluate

f(0.4) = 0.6667 − 1.667·e−3·0.4 + e−5·0.4 = 0.3

Answer

f(0.4) = 0.3
4.Laplace transforms

Answer: y(0.75) = 4.154

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) = 50/(s(s + 1)(s + 5))

Partial fractions

A = 50/(1·5) = 10
B = 50/((−1)·4) = −12.5
C = 50/((−5)·(−4)) = 2.5

Output

y(t) = 10 − 12.5·e−t + 2.5·e−5t
y(0.75) = 4.154

Answer

y(0.75) = 4.154
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 = 26 → ωₙ = 5.099 rad/s
2ζωₙ = 3 → ζ = 3/(2·5.099) = 0.2942

Result

σ = −ζωₙ = −3/2 = −1.5

Answer

σ = −1.5
6.Step response

Answer: K = 93.56

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 + 10s + K)

Damping ratio from the overshoot

ζ = −ln(0.15)/√(π² + ln²(0.15)) = 0.5169

Gain

ωₙ = 10/(2ζ) = 9.672 rad/s
K = ωₙ² = 93.56

Answer

K = 93.56
7.Bode plots & margins

Answer: ∠G(jω) = −99.4 °

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.

Add the angles

∠G = atan(5/14) − 90° − atan(5/9) = 19.65° − 90° − 29.05°

Result

∠G(j5) = −99.4°

Answer

∠G(jω) = −99.4°
8.Bode plots & margins

Answer: ω180 = 4.243 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/2)·(ω/9) = 1 → ω180 = √(2·9) = 4.243 rad/s

Answer

ω180 = 4.243 rad/s

ExamProblems · examproblems.com/courses/signals-and-systems