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
ReIm11−1−122−2−2z₁z₂w

Given z₁ = 3∠55° and z₂ = 1.5∠20°, find the real part of w = z₁/z₂.

z₁ = 3∠55°z₂ = 1.5∠20°
Re(w) =
2.Complex numbersHard
ReIm22−2−244−4−4ABCD

Exactly one of the points A–D is a root of z⁴ = 16∠120°. Click it.

w = 16∠120°n = 4
  1. 1A (2∠−60°)
  2. 2B (3.6∠−15°)
  3. 3C (2∠−15°)
  4. 4D (2∠60°)
3.AC impedance & phasorsMedium
24 V100 HzR = 20 ΩL = 250 mHC = 47 µFI

A series RLC circuit is connected to a 24 V (rms), 100 Hz supply. Find the rms current I.

V = 24 V (rms)f = 100 HzR = 20 ΩL = 250 mHC = 47 µF
I = A
4.AC impedance & phasorsHard
5 Vf0​R = 10 ΩL = 20 mHC = 4.7 µF

A series RLC circuit is driven by a 5 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage VL across the inductor.

V = 5 V (rms)R = 10 ΩL = 20 mHC = 4.7 µF
VL = V
5.AC power & power factorMedium
P = 24 kWQ = ?Sφ

A single-phase motor takes 24 kW at a power factor of 0.85 lagging from a 230 V supply. Find the reactive power Q it draws.

P = 24 kWpf = 0.85V = 230 V
Q = kvar
6.AC power & power factorHard
P = 50 kWQ1​S1​φ1​QC​

A single-phase load takes 50 kW at a power factor of 0.75 lagging from a 230 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.9. Find the new line current Inew.

P = 50 kWpf1 = 0.75pf2 = 0.9V = 230 Vf = 50 Hz
Inew = A
7.LC resonance, Q factor & bandwidthMedium
f|I|0.7071200 kHzf1​ = ?f2​BW

A series RLC circuit resonates at 200 kHz with Q = 1.5. Find the lower half-power frequency f1 exactly, in kHz.

f0 = 200 kHzQ = 1.5
f1 = kHz
8.LC resonance, Q factor & bandwidthHard
C = 470 pFR = 20 ΩL = 220 µH

A charged 470 pF capacitor discharges through a 220 µH coil with a total series resistance of 20 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 = 220 µHC = 470 pFR = 20 Ω
n =
9.Three-phase powerMedium
ABCVL​ = 480 VIL​ = 60 Acos φ = 0.8lagging

A balanced three-phase load draws a line current of 60 A from a supply with a line-to-line voltage of 480 V. The power factor is 0.8 lagging. Find the total real power P drawn by the load in kW.

VL = 480 VIL = 60 Acos φ = 0.8
P = kW
10.Three-phase powerHard
P = 250 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 250 kW at a power factor of 0.8 lagging from a 400 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.92 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.

P = 250 kWpf1 = 0.8pf2 = 0.92VL = 400 Vf = 60 Hz
Cph = µF
11.RC & RL transientsMedium
+−Vs​5 Vt = 0R = 33 ΩL22 mHi(t)

The inductor carries no current before the switch closes at t = 0. Find the current i at t = 1 ms.

Vs = 5 VR = 33 ΩL = 22 mHt = 1 ms
i = A
12.RC & RL transientsHard
+−Vs​24 Vt = 0R = 3.3 kΩC10 µF+−vC​vC​(0) = 20 V

Just before the switch closes at t = 0, the capacitor holds 20 V with the polarity shown. How long does it take for vC to reach 21.2 V? Find t.

Vs = 24 VR = 3.3 kΩC = 10 µFvC(0) = 20 V
t = ms
13.Thévenin & NortonMedium
R1​ = 2 ΩR3​ = 6 Ω+−V1​9 VR2​10 ΩARL​3 ΩIL​B

A load RL = 3 Ω is connected across a–b. Replace the rest of the circuit with its Thévenin equivalent and find the load current IL.

V1 = 9 VR1 = 2 ΩR2 = 10 ΩR3 = 6 ΩRL = 3 Ω
IL = A
14.Thévenin & NortonHard
R1​ = 30 ΩR3​ = 30 Ω+−V1​10 VR2​25 ΩAIs​1.5 ABab

Find the Norton current IN: the current that flows from a to b if the terminals are shorted.

V1 = 10 VIs = 1.5 AR1 = 30 ΩR2 = 25 ΩR3 = 30 Ω
IN = A
15.Bode plots & marginsMedium
|H| (dB)0.10.111101010010010001000−40−20020406080ω (rad/s)φ (°)ω (rad/s)−135−90−450ω = 4

Find the phase ∠G(jω) of G(s) = 32(s + 16)/(s(s + 4)) at ω = 4 rad/s, in degrees.

G(s) = 32(s + 16)/(s(s + 4))ω = 4 rad/s
∠G(jω) = °
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) = K/(s(s + 2)(s + 10)). Find the gain K that gives a gain margin of 8 dB.

L(s) = K/(s(s + 2)(s + 10))GM = 8 dB
K =

3Solutions

1.Complex numbers

Answer: Re(w) = 1.638

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| = 3 / 1.5 = 2
θ = 55° − 20° = 35°

Rectangular form

Re(w) = 2·cos(35°) = 1.638
Im(w) = 2·sin(35°) = 1.147
Re(w) = 1.638

Answer

Re(w) = 1.638
2.Complex numbers

Answer: A (2∠−60°)

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.

Modulus and arguments of the roots

|z| = 2
θ = 30°, 120°, −150°, −60°

The point

A = 2∠−60°

Answer

A (2∠−60°)
3.AC impedance & phasors

Answer: I = 0.1923 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 = 628.3·0.25 = 157.1 Ω
XC = 1/(ωC) = 1/(628.3·47·10⁻⁶) = 33.86 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(20² + (123.2)²) = 124.8 Ω

Current

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

I = V/|Z| = 24/124.8 = 0.1923 A

Answer

I = 0.1923 A
4.AC impedance & phasors

Answer: VL = 32.62 V

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|.

Resonance

At resonance the reactances cancel, so the impedance is just R.

ω₀ = 1/√(LC) = 1/√(0.02·4.7·10⁻⁶) = 3262 rad/s
I = V/R = 5/10 = 0.5 A

Voltage across the inductor

XL = ω₀L = 3262·0.02 = 65.23 Ω
VL = I·XL = 0.5·65.23 = 32.62 V

Answer

VL = 32.62 V

This is Q = 6.523 times the supply voltage.

5.AC power & power factor

Answer: Q = 14.87 kvar

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.

Angle

φ = arccos(0.85) = 31.79°, tan φ = 0.6197

Reactive power

Q = P·tan φ = 24·0.6197 = 14.87 kvar

Answer

Q = 14.87 kvar
6.AC power & power factor

Answer: Inew = 241.5 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 = P·tan φ1 = 50·tan(41.41°) = 44.1 kvar
Q2 = P·tan φ2 = 50·tan(25.84°) = 24.22 kvar
QC = Q1 − Q2 = 19.88 kvar

New line current

Inew = P/(V·cos φ2) = 50000/(230·0.9) = 241.5 A

Answer

Inew = 241.5 A
7.LC resonance, Q factor & bandwidth

Answer: f1 = 144.2 kHz

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).

Square-root term

√(1 + 1/(4·1.5²)) = 1.054

Lower half-power frequency

f1 = 200000·(1.054 − 0.3333) = 144152 Hz = 144.2 kHz

Answer

f1 = 144.2 kHz

The upper one is f₂ = 277.5 kHz; check: f₂ − f₁ = 133.3 kHz = f₀/Q.

8.LC resonance, Q factor & bandwidth

Answer: n = 7.548

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 = 2·0.00022·ln 2/20 = 1.525 × 10⁻⁵ s

Number of cycles

n = t·f0 = 1.525 × 10⁻⁵·494948 = 7.548

Answer

n = 7.548

Shortcut: n = Q·ln 2/π with Q = 34.21.

9.Three-phase power

Answer: P = 39.91 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 = √3·480·60 = 49.88 kVA

Real power

P = S·cos φ = 49.88·0.8 = 39.91 kW

Answer

P = 39.91 kW
10.Three-phase power

Answer: Cph = 1343 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 = 250·tan(acos 0.8) = 250·0.75 = 187.5 kvar
Q2 = 250·tan(acos 0.92) = 250·0.426 = 106.5 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 81 kvar

Capacitance per phase

In a Y bank each capacitor is connected between a line and the star point.

V = 400/√3 = 230.9 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 27000/(377·230.9²) = 1343 µF

Answer

Cph = 1343 µF
11.RC & RL transients

Answer: i = 0.1177 A

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 = 22 mH / 33 Ω = 0.6667 ms
i(∞) = Vs/R = 5/33 = 0.1515 A

Current at the given time

i(t) = i(∞)·(1 − e−t/τ) = 0.1515·(1 − e−1/0.6667) = 0.1177 A

Answer

i = 0.1177 A
12.RC & RL transients

Answer: t = 11.77 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

The capacitor voltage cannot jump, so it starts at its stored value and heads for the source voltage.

vC(0) = 20 V, vC(∞) = Vs = 24 V
τ = R·C = 3.3 kΩ · 10 µF = 33 ms
vC(t) = 24 + (20 − 24)·e−t/τ

Solve for the time

21.2 = 24 + (−4)·e−t/τ → e−t/τ = (21.2 − 24)/(20 − 24) = 0.7
t = τ·ln((20 − 24)/(21.2 − 24)) = 33·ln(1.429) = 11.77 ms

Answer

t = 11.77 ms
13.Thévenin & Norton

Answer: IL = 0.7031 A

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​ = 2 ΩR3​ = 6 ΩSHORTR2​10 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 = 2·10/(2+10) = 1.667 Ω
Rth = (R1 ∥ R2) + R3 = 1.667 + 6 = 7.667 Ω
R1​ = 2 ΩR3​ = 6 Ω+−V1​9 VR2​10 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) = 9·10/(2+10) = 7.5 V
Rth​ = 7.667 Ω+−Vth​7.5 VRL​3 ΩIL​

Connect the load to the Thévenin equivalent

The Thévenin source and R_th in series behave exactly like the original circuit at a–b, so the load sees a simple series loop.

IL = Vth/(Rth + RL) = 7.5/(7.667 + 3) = 0.7031 A
14.Thévenin & Norton

Answer: IN = 1.604 A

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​ = 30 ΩR3​ = 30 ΩSHORTR2​25 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 = 30·25/(30+25) = 13.64 Ω
Rth = (R1 ∥ R2) + R3 = 13.64 + 30 = 43.64 Ω
R1​ = 30 ΩR3​ = 30 Ω+−V1​10 VR2​25 ΩAIs​1.5 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) = 4.545 + 1.5·13.64 = 25 V
Vth = VB = VA + Is·R3 = 25 + 1.5·30 = 70 V
IN​1.604 ARth​43.64 Ωab

Norton equivalent

A Norton source is the same circuit seen as a current source in parallel with R_th. Its current is the short-circuit current at a–b.

IN = Vth/Rth = 70/43.64 = 1.604 A
15.Bode plots & margins

Answer: ∠G(jω) = −121 °

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(4/16) − 90° − atan(4/4) = 14.04° − 90° − 45°

Result

∠G(j4) = −121°

Answer

∠G(jω) = −121°
16.Bode plots & margins

Answer: K = 95.55

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/10) = 1 → ω180 = √(2·10) = 4.472 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 4)·√(ω² + 100)) = K/(2·10·(2 + 10)) = K/240

Required gain

240/K = 108/20 = 2.512
K = 240/2.512 = 95.55

Answer

K = 95.55

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