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
ReIm55−5−5θz

Find the argument θ of z = −4 − j8 in degrees, as the principal value between −180° and 180°.

z = −4 − j8
θ = °
2.Complex numbersHard
ReIm22−2−244−4−4z₁z₂

One root of z³ + 12z² + 57z + 100 = 0 is z₁ = −4 + j3. The coefficients are real. Find the real root r.

z₁ = −4 + j3
r =
3.AC impedance & phasorsMedium
Vf0​ = 2000 HzRL = 5 mHC = ?

A series RLC circuit with L = 5 mH must resonate at 2000 Hz. Find the capacitance C needed.

f0 = 2000 HzL = 5 mH
C = µF
4.AC impedance & phasorsHard
10 Vf0​R = 20 ΩL = 100 mHC = 10 µF

A series RLC circuit is driven by a 10 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage VL across the inductor.

V = 10 V (rms)R = 20 ΩL = 100 mHC = 10 µF
VL = V
5.AC power & power factorMedium
240 VZ3 + j4 ΩI

A load with impedance Z = 3 + j4 Ω is connected to a 240 V (rms) single-phase supply. Find the real power P it absorbs.

V = 240 VZ = 3 + j4 Ω
P = W
6.AC power & power factorHard
P = 10 kWQ1​S1​φ1​QC​

A single-phase load takes 10 kW at a power factor of 0.7 lagging from a 240 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.95. Find the reactive power QC the capacitor must supply.

P = 10 kWpf1 = 0.7pf2 = 0.95V = 240 Vf = 50 Hz
QC = kvar
7.LC resonance, Q factor & bandwidthMedium
C = 100 nF+−4 VL = 2.2 mHiL​

A 100 nF capacitor charged to 5 V oscillates with an ideal 2.2 mH inductor. Find the energy WL stored in the inductor at the moment the capacitor voltage is 4 V, in nJ.

C = 100 nFV0 = 5 VvC = 4 V
WL = nJ
8.LC resonance, Q factor & bandwidthHard
V5 Ω4.7 mH4.7 nF500 mA

A series RLC circuit (R = 5 Ω, L = 4.7 mH, C = 4.7 nF) is driven at resonance with a current amplitude of 500 mA. Find the energy Ecycle dissipated in R during one cycle, in µJ.

R = 5 ΩL = 4.7 mHC = 4.7 nFIm = 500 mA
Ecycle = µJ
9.Three-phase powerMedium
ABCVL​ = 415 VNZ = 18 + j24 Ωper phase

A balanced Y-connected load with an impedance of 18 + j24 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 415 V. Find the total real power P absorbed by the load in kW.

VL = 415 VZ = 18 + j24 Ω per phase
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 0.72 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 0.98 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.

P = 60 kWpf1 = 0.72pf2 = 0.98VL = 690 Vf = 50 Hz
Cph = µF
11.RC & RL transientsMedium
+−Vs​9 Vt = 0R1​ = 2.2 kΩR2​ = 4.7 kΩC68 µF+−vC​

The source Vₛ charges the capacitor through R₁, and R₂ is connected in parallel with C. The capacitor is uncharged when the switch closes at t = 0. Find the capacitor voltage vC at t = 200 ms.

Vs = 9 VR1 = 2.2 kΩR2 = 4.7 kΩC = 68 µFt = 200 ms
vC = V
12.RC & RL transientsHard
+−Vs​5 Vt = 0R = 6.8 kΩC100 µF+−vC​vC​(0) = 3 V

Just before the switch closes at t = 0, the capacitor holds 3 V with the polarity shown. How long does it take for vC to reach 4.2 V? Find t.

Vs = 5 VR = 6.8 kΩC = 100 µFvC(0) = 3 V
t = ms
13.Thévenin & NortonMedium
R1​ = 15 ΩR3​ = 12 Ω+−V1​30 VR2​15 ΩAab

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

V1 = 30 VR1 = 15 ΩR2 = 15 ΩR3 = 12 Ω
Vth = V
14.Thévenin & NortonHard
R1​ = 8 ΩR3​ = 20 Ω+−V1​5 VR2​4 ΩAIs​1 ABab

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

V1 = 5 VIs = 1 AR1 = 8 ΩR2 = 4 ΩR3 = 20 Ω
Vth = V
15.Bode plots & marginsMedium
vin​C = 10 nFR = 1 kΩvout​

The RC high-pass filter shown has R = 1 kΩ, C = 10 nF. A sinusoidal input of 32 kHz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 1 kΩC = 10 nFf = 32 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 + 2)(s + 7)). Find the gain K that gives a gain margin of 5 dB.

L(s) = K/(s(s + 2)(s + 7))GM = 5 dB
K =

3Solutions

1.Complex numbers

Answer: θ = −116.6 °

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.

Quadrant

a = −4, b = −8 → III

Reference angle

α = atan(8/4) = 63.43°

Argument

θ = −(180° − α) = −116.6°

Answer

θ = −116.6°

In polar form z = 8.944∠−116.6°.

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 = −4 − j3
(z − z₁)(z − z₂) = z² − 2·(−4)·z + ((−4)² + 3²) = z² + 8z + 25

Factor out the quadratic

z³ + 12z² + 57z + 100 = (z² + 8z + 25)(z + 4)

Real root

r = −4

Answer

r = −4

Check with the sum of the roots: z₁ + z₂ + r = −8 + (−4) = −12 = −(12).

3.AC impedance & phasors

Answer: C = 1.267 µF

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 condition

At resonance the inductive and capacitive reactances cancel: ωL = 1/(ωC).

ω₀ = 2π·2000 = 12566 rad/s
C = 1/(ω₀²·L) = 1/(12566²·0.005) = 1.267 µF

Answer

C = 1.267 µF
4.AC impedance & phasors

Answer: VL = 50 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/√(0.1·10·10⁻⁶) = 1000 rad/s
I = V/R = 10/20 = 0.5 A

Voltage across the inductor

XL = ω₀L = 1000·0.1 = 100 Ω
VL = I·XL = 0.5·100 = 50 V

Answer

VL = 50 V

This is Q = 5 times the supply voltage.

5.AC power & power factor

Answer: P = 6912 W

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.

Current

|Z| = √(3² + 4²) = 5 Ω
I = V/|Z| = 240/5 = 48 A

Real power

P = I²·R = 48²·3 = 6912 W

Answer

P = 6912 W
6.AC power & power factor

Answer: QC = 6.915 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.

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 = 10·tan(45.57°) = 10.2 kvar
Q2 = P·tan φ2 = 10·tan(18.19°) = 3.287 kvar
QC = Q1 − Q2 = 6.915 kvar

Answer

QC = 6.915 kvar
7.LC resonance, Q factor & bandwidth

Answer: WL = 450 nJ

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.

Total energy

½·1 × 10⁻⁷·5² = 1.25 × 10⁻⁶ J

Energy left in the capacitor

½·1 × 10⁻⁷·4² = 8 × 10⁻⁷ J

Energy in the inductor

WL = 4.5 × 10⁻⁷ J = 450 nJ

Answer

WL = 450 nJ
8.LC resonance, Q factor & bandwidth

Answer: Ecycle = 18.46 µJ

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).

Stored energy

W = ½·0.0047·0.5² = 0.0005875 J

Quality factor

Q = (1/5)·√(0.0047/4.7 × 10⁻⁹) = 200

Energy lost per cycle

Ecycle = 2π·W/Q = 2π·0.0005875/200 = 1.846 × 10⁻⁵ J = 18.46 µJ
check: ½·0.5²·5/33863 = 1.846 × 10⁻⁵ J

Answer

Ecycle = 18.46 µJ
9.Three-phase power

Answer: P = 3.445 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.

Phase voltage

Vph = 415/√3 = 239.6 V

Phase current

|Z| = √(18² + 24²) = 30 Ω
Iph = 239.6/30 = 7.987 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·7.987²·18 = 3.445 kW

Answer

P = 3.445 kW

Check: cos φ = R/|Z| = 0.6, and √3·VL·IL·cos φ gives the same power.

10.Three-phase power

Answer: Cph = 305.2 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 0.72) = 60·0.9639 = 57.83 kvar
Q2 = 60·tan(acos 0.98) = 60·0.2031 = 12.18 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 45.65 kvar

Capacitance per phase

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

V = 690/√3 = 398.4 V, ω = 2π·50 = 314.2 rad/s
Cph = (QC/3)/(ω·V²) = 15216/(314.2·398.4²) = 305.2 µF

Answer

Cph = 305.2 µF
11.RC & RL transients

Answer: vC = 5.269 V

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.

Thévenin equivalent seen by the capacitor

Short the source to find Rth; open the capacitor to find the voltage it will settle at.

Rth = R1 ∥ R2 = 2.2·4.7/(2.2 + 4.7) = 1.499 kΩ
vC(∞) = 9·4.7/(2.2 + 4.7) = 6.13 V

Time constant

τ = Rth·C = 1.499·68 = 101.9 ms

Charging curve

vC(t) = 6.13·(1 − e−200/101.9) = 5.269 V

Answer

vC = 5.269 V
12.RC & RL transients

Answer: t = 623.1 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) = 3 V, vC(∞) = Vs = 5 V
τ = R·C = 6.8 kΩ · 100 µF = 680 ms
vC(t) = 5 + (3 − 5)·e−t/τ

Solve for the time

4.2 = 5 + (−2)·e−t/τ → e−t/τ = (4.2 − 5)/(3 − 5) = 0.4
t = τ·ln((3 − 5)/(4.2 − 5)) = 680·ln(2.5) = 623.1 ms

Answer

t = 623.1 ms
13.Thévenin & Norton

Answer: Vth = 15 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​ = 15 ΩR3​ = 12 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 = 15·15/(15+15) = 7.5 Ω
Rth = (R1 ∥ R2) + R3 = 7.5 + 12 = 19.5 Ω
R1​ = 15 ΩR3​ = 12 Ω+−V1​30 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) = 30·15/(15+15) = 15 V
14.Thévenin & Norton

Answer: Vth = 24.33 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​ = 20 ΩSHORTR2​4 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 = 8·4/(8+4) = 2.667 Ω
Rth = (R1 ∥ R2) + R3 = 2.667 + 20 = 22.67 Ω
R1​ = 8 ΩR3​ = 20 Ω+−V1​5 VR2​4 ΩAIs​1 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) = 1.667 + 1·2.667 = 4.333 V
Vth = VB = VA + Is·R3 = 4.333 + 1·20 = 24.33 V
15.Bode plots & margins

Answer: φ = 26.44 °

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 = 32000/15915 = 2.011

Phase

φ = 90° − atan(2.011) = 26.44°

Answer

φ = 26.44°
16.Bode plots & margins

Answer: K = 70.86

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/7) = 1 → ω180 = √(2·7) = 3.742 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 4)·√(ω² + 49)) = K/(2·7·(2 + 7)) = K/126

Required gain

126/K = 105/20 = 1.778
K = 126/1.778 = 70.86

Answer

K = 70.86

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