Circuit Analysis 2: AC circuits: cram pack

Formula sheets, the key method steps for every topic and 16 practice problems, with full solutions at the back.

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1Formulas and methods

Complex numbers

Rectangular form
z = a + jb, j² = −1, z̄ = a − jb
Modulus and argument
|z| = √(a² + b²), tan θ = b/a (check the quadrant)
Polar and exponential form
z = r∠θ = r(cos θ + j sin θ) = r·ejθ
Division
z₁/z₂ = z₁z̄₂/|z₂|²
Product and quotient
r₁∠θ₁ · r₂∠θ₂ = r₁r₂∠(θ₁ + θ₂), r₁∠θ₁ / r₂∠θ₂ = (r₁/r₂)∠(θ₁ − θ₂)
De Moivre
(r∠θ)ⁿ = rⁿ∠nθ
n-th roots
zk = R1/n∠((φ + 360°·k)/n), k = 0 … n − 1
Quadratic with p² < 4q
z = −p/2 ± j√(q − p²/4)

Method

  1. Decide which form suits the operation: rectangular a + jb for adding and subtracting, polar r∠θ for multiplying, dividing, powers and roots.
  2. Multiply out with j² = −1; divide by multiplying the top and bottom by the conjugate of the denominator.
  3. Convert with r = √(a² + b²) and θ from a sketch of the point (check the quadrant), and back with a = r cos θ, b = r sin θ.
  4. For powers use De Moivre, (r∠θ)ⁿ = rⁿ∠nθ; for zⁿ = R∠φ list all n roots R1/n∠((φ + 360°k)/n).

AC impedance & phasors

Impedances
ZR = R, ZL = jωL, ZC = 1/(jωC)
Angular frequency
ω = 2πf
Magnitude and angle
|Z| = √(R² + X²), φ = atan(X/R)
Phasor Ohm’s law
V = I·Z

Method

  1. Convert the frequency: ω = 2πf.
  2. Find each reactance: XL = ωL and XC = 1/(ωC).
  3. Add impedances like resistors, with Z = R + jX.
  4. Find the magnitude |Z| = √(R² + X²) and the angle φ = atan(X/R), then use Ohm’s law with phasors.

AC power & power factor

Real power
P = V·I·cos φ
Reactive power
Q = V·I·sin φ
Apparent power
S = V·I = √(P² + Q²)
Power factor
pf = cos φ = P/S
Correction capacitor
QC = P(tan φ1 − tan φ2), C = QC/(ωV²)

LC resonance, Q factor & bandwidth

Natural frequency
ω0 = 1/√(LC), f0 = 1/(2π√(LC))
Energy
W = ½CV0² = ½Cv² + ½Li²
Peak current
Imax = V0√(C/L)
Quality factor
Q = 2π·(energy stored)/(energy lost per cycle)
Series RLC
Q = (1/R)√(L/C), BW = R/(2πL)
Parallel RLC
Q = R√(C/L), BW = 1/(2πRC)
Bandwidth
BW = f2 − f1 = f0/Q, f0 = √(f1f2)
Half-power frequencies
f1,2 = f0(√(1 + 1/(4Q²)) ∓ 1/(2Q))
Tank with coil resistance r
RD = L/(C·r) = Q²·r
Decay of the amplitude
e−Rt/(2L)

Method

  1. Find the resonance frequency first: ω₀ = 1/√(LC) and f₀ = ω₀/(2π).
  2. Decide whether the circuit is series or parallel: series Q = (1/R)√(L/C), parallel Q = R√(C/L).
  3. Use BW = f₀/Q for the bandwidth, or the shortcuts R/(2πL) (series) and 1/(2πRC) (parallel).
  4. For energy questions, use that the total energy ½CV₀² stays constant in an ideal LC circuit and swings between C and L every quarter period.

Three-phase power

Y connection
VL = √3·Vph, IL = Iph
Δ connection
VL = Vph, IL = √3·Iph
Total real power
P = √3·VL·IL·cos φ = 3·Iph²·R
Apparent and reactive power
S = √3·VL·IL, Q = P·tan φ
Correction capacitors
QC = P(tan φ1 − tan φ2), C = QC/(3ωV²)
Voltage across each capacitor
Δ: VL Y: VL/√3
Motor input
Pin = Pout/η

RC & RL transients

Time constant
τ = RC | τ = L/R
Response
x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ
Capacitor at DC
open circuit; vC cannot jump
Inductor at DC
short circuit; iL cannot jump
Settled
t ≈ 5τ

Method

  1. Find the initial value just after switching: a capacitor voltage and an inductor current cannot jump.
  2. Find the final value long after switching: a capacitor acts as an open circuit, an inductor as a short circuit.
  3. Find the time constant: τ = RC or τ = L/R, using the resistance seen by the capacitor or inductor.
  4. Use x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ.

Thévenin & Norton

Thévenin voltage
Vth = Vopen circuit
Thévenin resistance
Rth: all sources off, seen from a–b
Norton current
IN = Vth/Rth
Load current
IL = Vth/(Rth + RL)
Maximum power transfer
RL = Rth, Pmax = Vth²/(4Rth)

Method

  1. Remove the load and mark the two terminals a and b.
  2. Find the open-circuit voltage between a and b: this is Vth.
  3. Switch off all independent sources and find the resistance seen from a–b: this is Rth.
  4. Put the load back on the equivalent circuit: Vth in series with Rth.
  5. For the Norton equivalent, convert: IN = Vth/Rth, with RN = Rth in parallel.

2Practice problems

1.Complex numbersMedium
ReIm22−2−244−4−4z₁z₂w

Given z₁ = 2.5∠100° and z₂ = 0.5∠20°, find the imaginary part of w = z₁/z₂.

z₁ = 2.5∠100°z₂ = 0.5∠20°
Im(w) =
2.Complex numbersHard
ReIm11−1−122−2−2θz

Use De Moivre's theorem to find the imaginary part of z⁷ for z = 1 − j√3.

z = 1 − j√3n = 7
Im(z⁷) =
3.AC impedance & phasorsMedium
24 V100 HzR = 60 ΩL = 20 mHC = 100 µF

A series RLC circuit is connected to a 24 V (rms), 100 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).

V = 24 V (rms)f = 100 HzR = 60 ΩL = 20 mHC = 100 µF
φ = °
4.AC impedance & phasorsHard
100 V100 HzR = 30 ΩL = 150 mHC = 10 µFI

A series RLC circuit is connected to a 100 V (rms), 100 Hz supply. Find the rms current I.

V = 100 V (rms)f = 100 HzR = 30 ΩL = 150 mHC = 10 µF
I = A
5.AC power & power factorMedium
230 V12 kW, 0.75219 kW, 0.95I = ?

Two loads are connected in parallel to a 230 V single-phase supply: a 2 kW load at a power factor of 0.75 lagging and a 19 kW load at a power factor of 0.95 leading. Find the total current I drawn from the supply.

V = 230 VP1 = 2 kW, pf1 = 0.75 laggingP2 = 19 kW, pf2 = 0.95 leading
I = A
6.AC power & power factorHard
400 V60 Hzload24 kW, 0.65C100 µFI = ?

A single-phase load takes 24 kW at a power factor of 0.65 lagging from a 400 V, 60 Hz supply. A 100 µF capacitor is connected in parallel with the load. Find the new line current I.

P = 24 kWpf1 = 0.65V = 400 Vf = 60 HzC = 100 µF
I = A
7.LC resonance, Q factor & bandwidthMedium
tvC​iL​T/4 = ?T/2TL = 2.2 µHC = 470 nF

In an ideal LC circuit with L = 2.2 µH and C = 470 nF, the capacitor is fully charged at t = 0. How long does it take until all the energy is in the inductor? Give t in µs.

L = 2.2 µHC = 470 nF
t = µs
8.LC resonance, Q factor & bandwidthHard
C = 220 nFR = 5 ΩL = 22 mH

A charged 220 nF capacitor discharges through a 22 mH coil with a total series resistance of 5 Ω, so the oscillation slowly dies out. Its amplitude decays as e−Rt/(2L). How many oscillation cycles n pass before the amplitude has halved? (The losses are small, so the frequency is f₀.)

L = 22 mHC = 220 nFR = 5 Ω
n =
9.Three-phase powerMedium
ABCVL​ = 208 VIL​ = 30 Acos φ = 0.92lagging

A balanced three-phase load draws a line current of 30 A from a supply with a line-to-line voltage of 208 V. The power factor is 0.92 lagging. Find the total real power P drawn by the load in kW.

VL = 208 VIL = 30 Acos φ = 0.92
P = kW
10.Three-phase powerHard
P = 60 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 60 kW at a power factor of 0.7 lagging from a 415 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.98 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 60 kWpf1 = 0.7pf2 = 0.98VL = 415 Vf = 50 Hz
Cph = µF
11.RC & RL transientsMedium
+−Vs​6 Vt = 0R = 15 ΩL68 mHi(t)

The inductor carries no current before the switch closes at t = 0. How long after the switch closes does the current reach 90% of its final value? Find t.

Vs = 6 VR = 15 ΩL = 68 mH
t = ms
12.RC & RL transientsHard
+−Vs​24 VR1​ = 47 ΩR2​ = 22 Ωcloses at t = 0L150 mHi(t)

The switch across R2 has been open for a long time, so the inductor current is steady. At t = 0 the switch closes and shorts out R2. How long after t = 0 does the inductor current reach 0.413 A? Find t.

Vs = 24 VR1 = 47 ΩR2 = 22 ΩL = 150 mH
t = ms
13.Thévenin & NortonMedium
R1​ = 6 ΩR3​ = 20 Ω+−V1​9 VR2​15 ΩAab

Find the Thévenin voltage Vth seen from terminals a–b.

V1 = 9 VR1 = 6 ΩR2 = 15 ΩR3 = 20 Ω
Vth = V
14.Thévenin & NortonHard
R1​ = 8 ΩR3​ = 2 Ω+−V1​6 VR2​8 ΩAIs​4 ABab

Find the Thévenin voltage Vth seen from terminals a–b.

V1 = 6 VIs = 4 AR1 = 8 ΩR2 = 8 ΩR3 = 2 Ω
Vth = V
15.Bode plots & marginsMedium
vin​L = 4.7 mHR = 10 Ωvout​

The RL low-pass filter shown has R = 10 Ω, L = 4.7 mH. A sinusoidal input of 1.4 kHz is applied. Find the gain |H| = |vout/vin| in dB.

R = 10 ΩL = 4.7 mHf = 1.4 kHz
|H| = dB
16.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) = 334/(s(s + 5)(s + 7)). Find the gain margin GM in dB.

L(s) = 334/(s(s + 5)(s + 7))
GM = dB

3Solutions

1.Complex numbers

Answer: Im(w) = 4.924

Polar form: z = r∠θ = r(cos θ + j sin θ) = r·ejθ, with r = |z| and θ = arg z. Back to rectangular form: a = r cos θ, b = r sin θ. The argument needs the quadrant: θ = atan(b/a) only when a > 0; for a < 0 add or subtract 180°. In polar form multiplication is easy: multiply the moduli and add the arguments; for division divide the moduli and subtract the arguments.

Modulus and argument of w

|w| = 2.5 / 0.5 = 5
θ = 100° − 20° = 80°

Rectangular form

Re(w) = 5·cos(80°) = 0.8682
Im(w) = 5·sin(80°) = 4.924
Im(w) = 4.924

Answer

Im(w) = 4.924
2.Complex numbers

Answer: Im(z⁷) = −110.9

De Moivre and roots: (r∠θ)ⁿ = rⁿ∠nθ. The equation zⁿ = R∠φ has exactly n roots, all with modulus R1/n, at the arguments (φ + 360°·k)/n for k = 0, 1, …, n − 1: they lie evenly spaced on a circle, 360°/n apart. A quadratic z² + pz + q = 0 with p² < 4q has the complex roots z = −p/2 ± j√(q − p²/4); a polynomial with real coefficients always has its complex roots in conjugate pairs a ± jb.

Polar form

z = 1 − j√3 = 2∠−60°

De Moivre

z⁷ = (2)⁷∠−420° = 128∠−60°

Rectangular form

z⁷ = 128·(cos −60° + j sin −60°) = 64 − j110.9
Im(z⁷) = −110.9

Answer

Im(z⁷) = −110.9
3.AC impedance & phasors

Answer: φ = −3.195 °

AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.

Reactances

ω = 2πf = 2π·100 = 628.3 rad/s
XL = ωL = 628.3·0.02 = 12.57 Ω
XC = 1/(ωC) = 1/(628.3·100·10⁻⁶) = 15.92 Ω

Impedance

In series the impedances add as complex numbers: Z = R + j(XL − XC).

|Z| = √(R² + (XL − XC)²) = √(60² + (−3.349)²) = 60.09 Ω

Phase angle

φ = arctan((XL − XC)/R) = arctan(−3.349/60) = −3.195°

Answer

φ = −3.195°
4.AC impedance & phasors

Answer: I = 1.399 A

AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.

Reactances

ω = 2πf = 2π·100 = 628.3 rad/s
XL = ωL = 628.3·0.15 = 94.25 Ω
XC = 1/(ωC) = 1/(628.3·10·10⁻⁶) = 159.2 Ω

Impedance

In series the impedances add as complex numbers: Z = R + j(XL − XC).

|Z| = √(R² + (XL − XC)²) = √(30² + (−64.91)²) = 71.5 Ω

Current

The same current flows through every element of a series circuit.

I = V/|Z| = 100/71.5 = 1.399 A

Answer

I = 1.399 A
5.AC power & power factor

Answer: I = 93.36 A

AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.

Reactive power of each load

A lagging load takes reactive power (Q > 0); a leading load supplies it (Q < 0).

Q1 = 2·tan(arccos 0.75) = 1.764 kvar
Q2 = −19·tan(arccos 0.95) = −6.245 kvar

Totals

P = 2 + 19 = 21 kW
Q = 1.764 + (−6.245) = −4.481 kvar
S = √(21² + (−4.481)²) = 21.47 kVA

Total current

I = S/V = 21473/230 = 93.36 A

Answer

I = 93.36 A
6.AC power & power factor

Answer: I = 81.44 A

AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.

Reactive power before and after

The real power P does not change; only the reactive power drawn from the supply does.

Q1 = 24·tan(arccos 0.65) = 28.06 kvar
QC = V²·2πf·C = 400²·2π·60·100·10⁻⁶ = 6032 var = 6.032 kvar
Q2 = 28.06 − 6.032 = 22.03 kvar

New apparent power

S2 = √(24² + 22.03²) = 32.58 kVA

New line current

I = S2/V = 32576/400 = 81.44 A

Answer

I = 81.44 A
7.LC resonance, Q factor & bandwidth

Answer: t = 1.597 µs

LC resonance: a charged capacitor connected to an inductor drives a current that charges it the other way: the energy swings between the electric field of C (½Cv²) and the magnetic field of L (½Li²) at the natural frequency f0 = 1/(2π√(LC)). Without resistance the total energy ½CV0² stays constant, so the peak current is Imax = V0√(C/L). One period is T = 1/f0; the energy moves from C to L in a quarter of it.

Period

T = 2π√(2.2 × 10⁻⁶·4.7 × 10⁻⁷) = 6.389 µs

Quarter period

t = T/4 = 1.597 µs

Answer

t = 1.597 µs
8.LC resonance, Q factor & bandwidth

Answer: n = 13.95

Q factor and bandwidth: in a resonant circuit Q = 2π·(energy stored)/(energy lost per cycle). For series RLC Q = ω0L/R = (1/R)√(L/C); for parallel RLC Q = R/(ω0L) = R√(C/L). The bandwidth between the half-power frequencies f1 and f2 (where the power is half and the current or voltage 1/√2 of its peak) is BW = f2 − f1 = f0/Q, which is R/(2πL) in series and 1/(2πRC) in parallel. The resonance frequency is the geometric mean f0 = √(f1f2), and f1,2 = f0(√(1 + 1/(4Q²)) ∓ 1/(2Q)). Losses also make the oscillation decay: in series RLC the amplitude falls as e−Rt/(2L).

Half-life of the amplitude

t = 2·0.022·ln 2/5 = 0.0061 s

Number of cycles

n = t·f0 = 0.0061·2288 = 13.95

Answer

n = 13.95

Shortcut: n = Q·ln 2/π with Q = 63.25.

9.Three-phase power

Answer: P = 9.943 kW

Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ.

Apparent power

S = √3·VL·IL = √3·208·30 = 10.81 kVA

Real power

P = S·cos φ = 10.81·0.92 = 9.943 kW

Answer

P = 9.943 kW
10.Three-phase power

Answer: Cph = 302.1 µF

Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ. Power factor correction: capacitors supply reactive power locally, QC = P·(tan φ1 − tan φ2). Each of the three capacitors supplies QC/3 = V²·ω·C, where V is the voltage across that capacitor: VL in a Δ bank, VL/√3 in a Y bank.

Reactive power before and after

The real power is unchanged, so each power factor fixes the reactive power: Q = P·tan φ.

Q1 = 60·tan(acos 0.7) = 60·1.02 = 61.21 kvar
Q2 = 60·tan(acos 0.98) = 60·0.2031 = 12.18 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 49.03 kvar

Capacitance per phase

In a Δ bank each capacitor is connected between two lines.

V = 415 V, ω = 2π·50 = 314.2 rad/s
Cph = (QC/3)/(ω·V²) = 16343/(314.2·415²) = 302.1 µF

Answer

Cph = 302.1 µF
11.RC & RL transients

Answer: t = 10.44 ms

First-order transients: in a circuit with one capacitor or one inductor, every voltage and current moves exponentially from its initial value to its final value: x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ, with τ = RC or τ = L/R. After about 5τ the circuit has settled.

Time constant and final current

Long after the switch closes the inductor behaves like a short circuit, so the final current is set by R alone.

τ = L/R = 68 mH / 15 Ω = 4.533 ms
i(∞) = Vs/R = 6/15 = 0.4 A

Solve for the time

0.9 = 1 − e−t/τ → t = −τ·ln(1 − 0.9)
t = −4.533·ln(0.1) = 10.44 ms

Answer

t = 10.44 ms
12.RC & RL transients

Answer: t = 1.632 ms

First-order transients: in a circuit with one capacitor or one inductor, every voltage and current moves exponentially from its initial value to its final value: x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ, with τ = RC or τ = L/R. After about 5τ the circuit has settled.

Initial value, final value, time constant

Before t = 0 both resistors are in the loop; afterwards the closed switch bypasses R₂.

i(0) = 24/(47 + 22) = 0.3478 A
i(∞) = 24/47 = 0.5106 A
τ = L/R = 150/47 = 3.191 ms
i(t) = 0.5106 + (0.3478 − 0.5106)·e−t/τ

Solve for the time

e−t/τ = (0.413 − 0.5106)/(0.3478 − 0.5106) = 0.5997
t = 3.191·ln(1.668) = 1.632 ms

Answer

t = 1.632 ms
13.Thévenin & Norton

Answer: Vth = 6.429 V

Thévenin’s theorem: any linear two-terminal network behaves like a single voltage source Vth in series with a resistance Rth. Norton’s theorem gives the same network as a current source IN = Vth/Rth in parallel with Rth.

R1​ = 6 ΩR3​ = 20 ΩSHORTR2​15 ΩAabRth​

Find Rth with every source switched off

Replace voltage sources with shorts and current sources with opens, then look into the circuit from a–b.

R1 ∥ R2 = 6·15/(6+15) = 4.286 Ω
Rth = (R1 ∥ R2) + R3 = 4.286 + 20 = 24.29 Ω
R1​ = 6 ΩR3​ = 20 Ω+−V1​9 VR2​15 ΩAab+−Vth​

Find Vth, the open-circuit voltage

With a–b open, no current flows through R₃, so there is no drop across it and V_b = V_A. R₁ and R₂ form a voltage divider.

Vth = V1·R2/(R1+R2) = 9·15/(6+15) = 6.429 V
14.Thévenin & Norton

Answer: Vth = 27 V

Thévenin’s theorem: any linear two-terminal network behaves like a single voltage source Vth in series with a resistance Rth. Norton’s theorem gives the same network as a current source IN = Vth/Rth in parallel with Rth.

R1​ = 8 ΩR3​ = 2 ΩSHORTR2​8 ΩAOPENBabRth​

Find Rth with every source switched off

Replace voltage sources with shorts and current sources with opens, then look into the circuit from a–b.

R1 ∥ R2 = 8·8/(8+8) = 4 Ω
Rth = (R1 ∥ R2) + R3 = 4 + 2 = 6 Ω
R1​ = 8 ΩR3​ = 2 Ω+−V1​6 VR2​8 ΩAIs​4 ABab+−Vth​

Find Vth, the open-circuit voltage

With a–b open, the only path for Iₛ is through R₃ into node A. Find V_A by superposition of V₁ and Iₛ, then add the drop across R₃.

VA = V1·R2/(R1+R2) + Is·(R1∥R2) = 3 + 4·4 = 19 V
Vth = VB = VA + Is·R3 = 19 + 4·2 = 27 V
15.Bode plots & margins

Answer: |H| = −12.58 dB

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 = 10/0.0047 = 2128 rad/s
fc = 338.6 Hz → f/fc = 1400/338.6 = 4.134

Gain

|H| = 1/√(1 + 4.134²) = 0.2351
20·log₁₀(0.2351) = −12.58 dB

Answer

|H| = −12.58 dB
16.Bode plots & margins

Answer: GM = 1.99 dB

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/5)·(ω/7) = 1 → ω180 = √(5·7) = 5.916 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 25)·√(ω² + 49)) = K/(5·7·(5 + 7)) = K/420

Gain margin

GM = 20·log₁₀(420/334) = 1.99 dB

Answer

GM = 1.99 dB

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