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
ReIm55−5−51010−10−10θz

Write z = 12∠125° in rectangular form a + jb and give the real part a.

r = 12θ = 125°
a =
2.Complex numbersHard
ReIm22−2−244−4−4z₁z₂

One root of z³ − 10z² + 49z − 100 = 0 is z₁ = 3 + j4. The coefficients are real. Find the real root r.

z₁ = 3 + j4
r =
3.AC impedance & phasorsMedium
100 V100 HzR = 50 ΩL = 250 mHC = 100 µFI

A series RLC circuit is connected to a 100 V (rms), 100 Hz supply. Find the power factor cos φ of the circuit.

V = 100 V (rms)f = 100 HzR = 50 ΩL = 250 mHC = 100 µF
cos φ =
4.AC impedance & phasorsHard
10 V200 HzR = 30 ΩL = 20 mHC = 10 µF

A series RLC circuit is connected to a 10 V (rms), 200 Hz supply. Find the resonant frequency f0 of the circuit.

V = 10 V (rms)f = 200 HzR = 30 ΩL = 20 mHC = 10 µF
f0 = Hz
5.AC power & power factorMedium
240 VR + jXP = 1632 W8 A

An inductive load takes 8 A and 1632 W from a 240 V (rms) single-phase supply. Model it as a resistance R in series with a reactance X. Find X.

V = 240 VI = 8 AP = 1632 W
X = Ω
6.AC power & power factorHard
240 V60 Hzload13 kW, 0.65C100 µFI

A single-phase load takes 13 kW at a power factor of 0.65 lagging from a 240 V, 60 Hz supply. A 100 µF capacitor is connected in parallel with the load. Find the new power factor cos φ seen by the supply.

P = 13 kWpf1 = 0.65V = 240 Vf = 60 HzC = 100 µF
cos φ =
7.LC resonance, Q factor & bandwidthMedium
f|I|0.707150 kHzf1​ = ?f2​BW

A series RLC circuit resonates at 50 kHz with Q = 3. Find the lower half-power frequency f1 exactly, in kHz.

f0 = 50 kHzQ = 3
f1 = kHz
8.LC resonance, Q factor & bandwidthHard
V100 ΩLC = ?

Design a series RLC filter with a total resistance of 100 Ω, a resonance frequency of 200 kHz and a bandwidth of 8 kHz. Find the capacitance C in nF.

R = 100 Ωf0 = 200 kHzBW = 8 kHz
C = nF
9.Three-phase powerMedium
ABCVL​ = 480 VIL​ = ?M11 kW, η = 95%cos φ = 0.8

A three-phase induction motor delivers 11 kW of mechanical output power at full load. At this load its efficiency is 95% and its power factor is 0.8 lagging. The supply has a line-to-line voltage of 480 V. Find the line current IL.

Pout = 11 kWη = 95%cos φ = 0.8VL = 480 V
IL = A
10.Three-phase powerHard
P = 50 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 50 kW at a power factor of 0.78 lagging from a 690 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.92 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 50 kWpf1 = 0.78pf2 = 0.92VL = 690 Vf = 50 Hz
QC = kvar
11.RC & RL transientsMedium
R = 15 Ωt = 0L68 mHi(t)

The inductor carries a steady current of 3 A. At t = 0 the switch closes the loop through R and the source is removed, so the inductor current decays through R. How long does it take for the current to fall to 1.5 A? Find t.

i(0) = 3 AR = 15 ΩL = 68 mH
t = ms
12.RC & RL transientsHard
+−Vs​9 VR1​ = 15 ΩR2​ = 22 Ωopens at t = 0L68 mHi(t)

The switch across R2 has been closed for a long time, so the inductor current is steady. At t = 0 the switch opens. Find the inductor current i at t = 5 ms.

Vs = 9 VR1 = 15 ΩR2 = 22 ΩL = 68 mHt = 5 ms
i = A
13.Thévenin & NortonMedium
R1​ = 3 ΩR3​ = 6 Ω+−V1​15 VR2​5 ΩAab

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

V1 = 15 VR1 = 3 ΩR2 = 5 ΩR3 = 6 Ω
IN = A
14.Thévenin & NortonHard
R1​ = 6 ΩR3​ = 2 Ω+−V1​24 VR2​2 ΩAIs​3 ABab

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

V1 = 24 VIs = 3 AR1 = 6 ΩR2 = 2 ΩR3 = 2 Ω
IN = A
15.Bode plots & marginsMedium
vin​R = 1 kΩC = 10 nFvout​

The RC low-pass filter shown has R = 1 kΩ, C = 10 nF. A sinusoidal input of 3.2 kHz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 1 kΩC = 10 nFf = 3.2 kHz
φ = °
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 + 4)(s + 8)). Find the gain K that gives a gain margin of 5 dB.

L(s) = K/(s(s + 4)(s + 8))GM = 5 dB
K =

3Solutions

1.Complex numbers

Answer: a = −6.883

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.

Components

a = 12·cos(125°) = −6.883
b = 12·sin(125°) = 9.83

Rectangular form

z = −6.883 + j9.83
a = −6.883

Answer

a = −6.883
2.Complex numbers

Answer: r = 4

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.

Conjugate pair

z₂ = 3 − j4
(z − z₁)(z − z₂) = z² − 2·3·z + (3² + 4²) = z² − 6z + 25

Factor out the quadratic

z³ − 10z² + 49z − 100 = (z² − 6z + 25)(z − 4)

Real root

r = 4

Answer

r = 4

Check with the sum of the roots: z₁ + z₂ + r = 6 + 4 = 10 = −(−10).

3.AC impedance & phasors

Answer: cos φ = 0.3339

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

XL = 157.1 Ω, XC = 15.92 Ω

Impedance

|Z| = √(50² + (141.2)²) = 149.8 Ω

Power factor

cos φ = R/|Z| = 50/149.8 = 0.3339
The circuit is inductive, so the power factor is lagging.

Answer

cos φ = 0.3339
4.AC impedance & phasors

Answer: f0 = 355.9 Hz

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 inductive and capacitive reactances cancel: ωL = 1/(ωC). The supply frequency does not matter.

f0 = 1/(2π√(LC)) = 1/(2π·√(0.02·10·10⁻⁶)) = 355.9 Hz

Answer

f0 = 355.9 Hz
5.AC power & power factor

Answer: X = 15.8 Ω

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.

Resistance from the real power

R = P/I² = 1632/8² = 25.5 Ω

Reactance from the impedance

|Z| = V/I = 240/8 = 30 Ω
X = √(|Z|² − R²) = √(30² − 25.5²) = 15.8 Ω

Answer

X = 15.8 Ω
6.AC power & power factor

Answer: cos φ = 0.7064

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 = 13·tan(arccos 0.65) = 15.2 kvar
QC = V²·2πf·C = 240²·2π·60·100·10⁻⁶ = 2171 var = 2.171 kvar
Q2 = 15.2 − 2.171 = 13.03 kvar

New apparent power

S2 = √(13² + 13.03²) = 18.4 kVA

New power factor

cos φ = P/S2 = 13/18.4 = 0.7064

Answer

cos φ = 0.7064
7.LC resonance, Q factor & bandwidth

Answer: f1 = 42.36 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/(4·3²)) = 1.014

Lower half-power frequency

f1 = 50000·(1.014 − 0.1667) = 42356 Hz = 42.36 kHz

Answer

f1 = 42.36 kHz

The upper one is f₂ = 59.02 kHz; check: f₂ − f₁ = 16.67 kHz = f₀/Q.

8.LC resonance, Q factor & bandwidth

Answer: C = 0.3183 nF

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).

Inductance from the bandwidth

L = R/(2π·BW) = 100/(2π·8000) = 0.001989 H

Capacitance from the resonance frequency

C = 1/((2π·200000)²·0.001989) = 3.183 × 10⁻¹⁰ F = 0.3183 nF

Answer

C = 0.3183 nF

Q = f₀/BW = 25.

9.Three-phase power

Answer: IL = 17.41 A

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.

Electrical input power

Pin = Pout/η = 11/0.95 = 11.58 kW

Line current

IL = Pin/(√3·VL·cos φ) = 11579/(√3·480·0.8) = 17.41 A

Answer

IL = 17.41 A
10.Three-phase power

Answer: QC = 18.81 kvar

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 = 50·tan(acos 0.78) = 50·0.8023 = 40.11 kvar
Q2 = 50·tan(acos 0.92) = 50·0.426 = 21.3 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 18.81 kvar

Answer

QC = 18.81 kvar
11.RC & RL transients

Answer: t = 3.142 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

τ = L/R = 68 mH / 15 Ω = 4.533 ms

Solve for the time

1.5 = 3·e−t/τ → t = τ·ln(3/1.5)
t = 4.533·ln(2) = 3.142 ms

Answer

t = 3.142 ms
12.RC & RL transients

Answer: i = 0.2667 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.

Initial value, final value, time constant

Before t = 0 the closed switch bypasses R₂; afterwards R₁ and R₂ are both in the loop.

i(0) = 9/15 = 0.6 A
i(∞) = 9/(15 + 22) = 0.2432 A
τ = L/R = 68/37 = 1.838 ms
i(t) = 0.2432 + (0.6 − 0.2432)·e−t/τ

Current at the given time

i(5) = 0.2432 + (0.3568)·e−5/1.838 = 0.2667 A

Answer

i = 0.2667 A
13.Thévenin & Norton

Answer: IN = 1.19 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​ = 3 ΩR3​ = 6 ΩSHORTR2​5 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 = 3·5/(3+5) = 1.875 Ω
Rth = (R1 ∥ R2) + R3 = 1.875 + 6 = 7.875 Ω
R1​ = 3 ΩR3​ = 6 Ω+−V1​15 VR2​5 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) = 15·5/(3+5) = 9.375 V
IN​1.19 ARth​7.875 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 = 9.375/7.875 = 1.19 A
14.Thévenin & Norton

Answer: IN = 4.714 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​ = 6 ΩR3​ = 2 ΩSHORTR2​2 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 = 6·2/(6+2) = 1.5 Ω
Rth = (R1 ∥ R2) + R3 = 1.5 + 2 = 3.5 Ω
R1​ = 6 ΩR3​ = 2 Ω+−V1​24 VR2​2 ΩAIs​3 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) = 6 + 3·1.5 = 10.5 V
Vth = VB = VA + Is·R3 = 10.5 + 3·2 = 16.5 V
IN​4.714 ARth​3.5 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 = 16.5/3.5 = 4.714 A
15.Bode plots & margins

Answer: φ = −11.37 °

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 = 1/(RC) = 1/(1000·1 × 10⁻⁸) = 100000 rad/s
fc = 15915 Hz → f/fc = 3200/15915 = 0.2011

Phase

φ = − atan(0.2011) = −11.37°

Answer

φ = −11.37°
16.Bode plots & margins

Answer: K = 215.9

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/8) = 1 → ω180 = √(4·8) = 5.657 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 16)·√(ω² + 64)) = K/(4·8·(4 + 8)) = K/384

Required gain

384/K = 105/20 = 1.778
K = 384/1.778 = 215.9

Answer

K = 215.9

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