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
ReIm0.50.5−0.5−0.511−1−11.51.5−1.5−1.5θz

The number z = 1.5·e−j0.5 has its angle in radians. Find its real part a.

r = 1.5θ = −0.5 rad
a =
2.Complex numbersHard
ReIm11−1−122−2−233−3−3ABCD

Exactly one of the points A–D is a root of z⁶ = 64∠120°. Click it.

w = 64∠120°n = 6
  1. 1A (2∠−70°)
  2. 2B (2∠−100°)
  3. 3C (2∠100°)
  4. 4D (3.6∠−70°)
3.AC impedance & phasorsMedium
100 V60 HzR = 20 ΩL = 50 mHI

A resistor and an inductor are connected in parallel across a 100 V (rms), 60 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).

V = 100 V (rms)f = 60 HzR = 20 ΩL = 50 mH
φ = °
4.AC impedance & phasorsHard
1 Vf0​R = 50 ΩL = 20 mHC = 2.2 µF

A series RLC circuit is driven by a 1 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage VL across the inductor.

V = 1 V (rms)R = 50 ΩL = 20 mHC = 2.2 µF
VL = V
5.AC power & power factorMedium
P = 40 kWQ = ?Sφ

A single-phase motor takes 40 kW at a power factor of 0.7 lagging from a 230 V supply. Find the reactive power Q it draws.

P = 40 kWpf = 0.7V = 230 V
Q = kvar
6.AC power & power factorHard
P1​ = 40 kWQ1​S1​φ

A 100 kVA single-phase transformer supplies a load of 40 kW at a power factor of 0.85 lagging. How much extra real power ΔP can be added as a resistive (unity power factor) load before the transformer reaches its kVA rating?

Smax = 100 kVAP1 = 40 kWpf1 = 0.85
ΔP = kW
7.LC resonance, Q factor & bandwidthMedium
V1 Ω4.7 mH2.2 µF

A series RLC circuit has R = 1 Ω, L = 4.7 mH and C = 2.2 µF. Find its bandwidth BW in Hz.

R = 1 ΩL = 4.7 mHC = 2.2 µF
BW = Hz
8.LC resonance, Q factor & bandwidthHard
V5 Ω47 µH100 nF500 mA

A series RLC circuit (R = 5 Ω, L = 47 µH, C = 100 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 = 47 µHC = 100 nFIm = 500 mA
Ecycle = µJ
9.Three-phase powerMedium
ABCVL​ = 480 VNZ = 3 + j4 Ωper phase

A balanced Y-connected load with an impedance of 3 + j4 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 480 V. Find the total real power P absorbed by the load in kW.

VL = 480 VZ = 3 + j4 Ω per phase
P = kW
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 400 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.95 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 = 0.95VL = 400 Vf = 50 Hz
QC = kvar
11.RC & RL transientsMedium
+−Vs​5 Vt = 0R = 50 ΩL33 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 = 50 ΩL = 33 mHt = 1 ms
i = A
12.RC & RL transientsHard
+−Vs​30 Vopens at t = 0R1​ = 2.2 kΩR2​ = 2.2 kΩC22 µF+−vC​

The switch has been closed for a long time, so the circuit is in steady state. At t = 0 the switch opens. Find the capacitor voltage vC at t = 15 ms.

Vs = 30 VR1 = 2.2 kΩR2 = 2.2 kΩC = 22 µFt = 15 ms
vC = V
13.Thévenin & NortonMedium
R1​ = 12 ΩR3​ = 4 Ω+−V1​9 VR2​10 ΩAab

Find the Thévenin resistance Rth seen from terminals a–b.

V1 = 9 VR1 = 12 ΩR2 = 10 ΩR3 = 4 Ω
Rth = Ω
14.Thévenin & NortonHard
R1​ = 15 ΩR3​ = 6 Ω+−V1​15 VR2​5 ΩAIs​2.5 ABRL​20 ΩIL​

A load RL = 20 Ω is connected across a–b. Replace the rest of the circuit with its Thévenin equivalent and find the load current IL.

V1 = 15 VIs = 2.5 AR1 = 15 ΩR2 = 5 ΩR3 = 6 ΩRL = 20 Ω
IL = A
15.Bode plots & marginsMedium
|H| (dB)0.010.010.10.111101010010010001000−20020406080100ω (rad/s)φ (°)ω (rad/s)−135−90−450ω = 0.5

Find the phase ∠G(jω) of G(s) = 44(s + 13)/(s(s + 4)) at ω = 0.5 rad/s, in degrees.

G(s) = 44(s + 13)/(s(s + 4))ω = 0.5 rad/s
∠G(jω) = °
16.Bode plots & marginsHard
|H| (dB)0.10.111101010010010001000−60−40−200204060ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​ = ?

A unity-feedback loop has the open-loop transfer function L(s) = 210/(s(s + 5)). Find the gain crossover frequency ωc, where |L(jω)| = 1.

L(s) = 210/(s(s + 5))
ωc = rad/s

3Solutions

1.Complex numbers

Answer: a = 1.316

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.

Euler's formula

z = 1.5·(cos −0.5 + j sin −0.5)

Components (radians)

a = 1.5·cos −0.5 = 1.316
b = 1.5·sin −0.5 = −0.7191
a = 1.316

Answer

a = 1.316
2.Complex numbers

Answer: B (2∠−100°)

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 = 20°, 80°, 140°, −160°, −100°, −40°

The point

B = 2∠−100°

Answer

B (2∠−100°)
3.AC impedance & phasors

Answer: φ = 46.7 °

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|.

Reactance

XL = ωL = 2π·60·0.05 = 18.85 Ω

Branch currents

Both branches are connected straight across the supply, so each sees the full voltage V.

IR = V/R = 100/20 = 5 A
IX = V/XL = 100/18.85 = 5.305 A

Total current

The resistor current is in phase with V and the reactive current is 90° out of phase, so they add like the sides of a right triangle.

I = √(IR² + IX²) = √(5² + 5.305²) = 7.29 A

Phase angle

The inductor current lags V, so the total current lags and the angle is positive.

φ = arctan(IX/IR) = arctan(5.305/5) = 46.7°

Answer

φ = 46.7°
4.AC impedance & phasors

Answer: VL = 1.907 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/√(0.02·2.2·10⁻⁶) = 4767 rad/s
I = V/R = 1/50 = 0.02 A

Voltage across the inductor

XL = ω₀L = 4767·0.02 = 95.35 Ω
VL = I·XL = 0.02·95.35 = 1.907 V

Answer

VL = 1.907 V

This is Q = 1.907 times the supply voltage.

5.AC power & power factor

Answer: Q = 40.81 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(0.7) = 45.57°, tan φ = 1.02

Reactive power

Q = P·tan φ = 40·1.02 = 40.81 kvar

Answer

Q = 40.81 kvar
6.AC power & power factor

Answer: ΔP = 56.88 kW

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 the existing load

Q1 = 40·tan(arccos 0.85) = 24.79 kvar

Largest total real power

At full rating S = Smax, and Q is unchanged by the resistive load.

P = √(Smax² − Q1²) = √(100² − 24.79²) = 96.88 kW

Extra real power

ΔP = 96.88 − 40 = 56.88 kW

Answer

ΔP = 56.88 kW
7.LC resonance, Q factor & bandwidth

Answer: BW = 33.86 Hz

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).

Bandwidth in rad/s

R/L = 1/0.0047 = 212.8 rad/s

Bandwidth in Hz

BW = 212.8/(2π) = 33.86 Hz = 33.86 Hz

Answer

BW = 33.86 Hz

With f₀ = 1.565 kHz this gives Q = 46.22.

8.LC resonance, Q factor & bandwidth

Answer: Ecycle = 8.514 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 = ½·4.7 × 10⁻⁵·0.5² = 5.875 × 10⁻⁶ J

Quality factor

Q = (1/5)·√(4.7 × 10⁻⁵/1 × 10⁻⁷) = 4.336

Energy lost per cycle

Ecycle = 2π·W/Q = 2π·5.875 × 10⁻⁶/4.336 = 8.514 × 10⁻⁶ J = 8.514 µJ
check: ½·0.5²·5/73413 = 8.514 × 10⁻⁶ J

Answer

Ecycle = 8.514 µJ
9.Three-phase power

Answer: P = 27.65 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 = 480/√3 = 277.1 V

Phase current

|Z| = √(3² + 4²) = 5 Ω
Iph = 277.1/5 = 55.43 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·55.43²·3 = 27.65 kW

Answer

P = 27.65 kW

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

10.Three-phase power

Answer: QC = 23.68 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 0.95) = 50·0.3287 = 16.43 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 23.68 kvar

Answer

QC = 23.68 kvar
11.RC & RL transients

Answer: i = 0.07802 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 = 33 mH / 50 Ω = 0.66 ms
i(∞) = Vs/R = 5/50 = 0.1 A

Current at the given time

i(t) = i(∞)·(1 − e−t/τ) = 0.1·(1 − e−1/0.66) = 0.07802 A

Answer

i = 0.07802 A
12.RC & RL transients

Answer: vC = 11 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.

Initial capacitor voltage

In steady state the capacitor is an open circuit, so R₁ and R₂ form a voltage divider. The capacitor voltage cannot jump when the switch opens.

vC(0) = 30·2.2/(2.2 + 2.2) = 15 V

Time constant after the switch opens

τ = R2·C = 2.2·22 = 48.4 ms

Discharge

vC(t) = 15·e−15/48.4 = 11 V

Answer

vC = 11 V
13.Thévenin & Norton

Answer: Rth = 9.455 Ω

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​ = 12 ΩR3​ = 4 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 = 12·10/(12+10) = 5.455 Ω
Rth = (R1 ∥ R2) + R3 = 5.455 + 4 = 9.455 Ω
14.Thévenin & Norton

Answer: IL = 0.9454 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​ = 15 ΩR3​ = 6 ΩSHORTR2​5 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 = 15·5/(15+5) = 3.75 Ω
Rth = (R1 ∥ R2) + R3 = 3.75 + 6 = 9.75 Ω
R1​ = 15 ΩR3​ = 6 Ω+−V1​15 VR2​5 ΩAIs​2.5 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) = 3.75 + 2.5·3.75 = 13.13 V
Vth = VB = VA + Is·R3 = 13.13 + 2.5·6 = 28.13 V
Rth​ = 9.75 Ω+−Vth​28.13 VRL​20 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) = 28.13/(9.75 + 20) = 0.9454 A
15.Bode plots & margins

Answer: ∠G(jω) = −94.92 °

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(0.5/13) − 90° − atan(0.5/4) = 2.203° − 90° − 7.125°

Result

∠G(j0.5) = −94.92°

Answer

∠G(jω) = −94.92°
16.Bode plots & margins

Answer: ωc = 14.07 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.

Gain crossover

ω⁴ + 25ω² − 44100 = 0
ω² = (−25 + √(625 + 4·44100))/2 = 197.9
ωc = 14.07 rad/s

Answer

ωc = 14.07 rad/s

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