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 = −6 − j8 in degrees, as the principal value between −180° and 180°.

z = −6 − j8
θ = °
2.Complex numbersHard
ReIm11−1−122−2−2θz

Find the real part of (−1 − j2)³.

z = −1 − j2n = 3
Re(z³) =
3.AC impedance & phasorsMedium
Vf0​ = 60 HzRL = 10 mHC = ?

A series RLC circuit with L = 10 mH must resonate at 60 Hz. Find the capacitance C needed.

f0 = 60 HzL = 10 mH
C = µF
4.AC impedance & phasorsHard
120 V1000 HzR = 30 ΩL = 50 mHC = 10 µF+−VC​

A series RLC circuit is connected to a 120 V (rms), 1000 Hz supply. Find the rms voltage VC across the capacitor.

V = 120 V (rms)f = 1000 HzR = 30 ΩL = 50 mHC = 10 µF
VC = V
5.AC power & power factorMedium
120 VZ12 + j9 ΩI

A load with impedance Z = 12 + j9 Ω is connected to a 120 V (rms) single-phase supply. Find the real power P it absorbs.

V = 120 VZ = 12 + j9 Ω
P = W
6.AC power & power factorHard
400 V60 Hzload25 kW, 0.6C330 µFI = ?

A single-phase load takes 25 kW at a power factor of 0.6 lagging from a 400 V, 60 Hz supply. A 330 µF capacitor is connected in parallel with the load. Find the new line current I.

P = 25 kWpf1 = 0.6V = 400 Vf = 60 HzC = 330 µF
I = A
7.LC resonance, Q factor & bandwidthMedium
C = 220 pF+−5 VL = 470 µHiL​

A 220 pF capacitor charged to 10 V oscillates with an ideal 470 µH inductor. Find the energy WL stored in the inductor at the moment the capacitor voltage is 5 V, in nJ.

C = 220 pFV0 = 10 VvC = 5 V
WL = nJ
8.LC resonance, Q factor & bandwidthHard
C = 2.2 nFr = 2 ΩL = 22 mH

A parallel tank circuit is made of a 2.2 nF capacitor and a 22 mH coil with a winding resistance of 2 Ω. At resonance it behaves like a resistor, its dynamic resistance RD = L/(C·r). Find RD in kΩ.

L = 22 mHC = 2.2 nFr = 2 Ω
RD = kΩ
9.Three-phase powerMedium
ABCVL​ = 480 VZ = 4 + j3 Ωper phase

A balanced Δ-connected load with an impedance of 4 + j3 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 = 4 + j3 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 0.7 lagging from a 480 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.98 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 50 kWpf1 = 0.7pf2 = 0.98VL = 480 Vf = 60 Hz
QC = kvar
11.RC & RL transientsMedium
+−Vs​24 Vt = 0R1​ = 10 kΩR2​ = 6.8 kΩC47 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 = 50 ms.

Vs = 24 VR1 = 10 kΩR2 = 6.8 kΩC = 47 µFt = 50 ms
vC = V
12.RC & RL transientsHard
+−Vs​24 Vt = 0R = 6.8 kΩC220 µF+−vC​vC​(0) = 4 V

Just before the switch closes at t = 0, the capacitor holds 4 V with the polarity shown. How long does it take for vC to reach 18 V? Find t.

Vs = 24 VR = 6.8 kΩC = 220 µFvC(0) = 4 V
t = ms
13.Thévenin & NortonMedium
R1​ = 3 ΩR3​ = 2 Ω+−V1​10 VR2​20 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 = 10 VR1 = 3 ΩR2 = 20 ΩR3 = 2 ΩRL = 3 Ω
IL = A
14.Thévenin & NortonHard
R1​ = 20 ΩR3​ = 3 Ω+−V1​30 VR2​8 ΩAIs​2.5 ABRL​3 ΩIL​

A load RL = 3 Ω is connected across a–b. Use a Thévenin equivalent to find the power PL delivered to the load.

V1 = 30 VIs = 2.5 AR1 = 20 ΩR2 = 8 ΩR3 = 3 ΩRL = 3 Ω
PL = W
15.Bode plots & marginsMedium
|H| (dB)0.10.111101010010010001000−40−200204060ω (rad/s)φ (°)ω (rad/s)−90−450ω = 3

Find the phase ∠G(jω) of G(s) = 25(s + 6)/(s(s + 14)) at ω = 3 rad/s, in degrees.

G(s) = 25(s + 6)/(s(s + 14))ω = 3 rad/s
∠G(jω) = °
16.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−60−40−200204060ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​ = ?

A unity-feedback loop has the open-loop transfer function L(s) = 100/(s(s + 5)). Find the gain crossover frequency ωc, where |L(jω)| = 1.

L(s) = 100/(s(s + 5))
ωc = rad/s

3Solutions

1.Complex numbers

Answer: θ = −126.9 °

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 = −6, b = −8 → III

Reference angle

α = atan(8/6) = 53.13°

Argument

θ = −(180° − α) = −126.9°

Answer

θ = −126.9°

In polar form z = 10∠−126.9°.

2.Complex numbers

Answer: Re(z³) = 11

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.

Powers step by step

z² = (−1 − j2)(−1 − j2) = −3 + j4
z³ = (−3 + j4)(−1 − j2) = 11 + j2

Check with De Moivre

z = 2.236∠−116.6° → z³ = 11.18∠10.3°
Re(z³) = 11

Answer

Re(z³) = 11
3.AC impedance & phasors

Answer: C = 703.6 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·60 = 377 rad/s
C = 1/(ω₀²·L) = 1/(377²·0.01) = 703.6 µF

Answer

C = 703.6 µF
4.AC impedance & phasors

Answer: VC = 6.372 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|.

Reactances

ω = 2πf = 2π·1000 = 6283 rad/s
XL = ωL = 6283·0.05 = 314.2 Ω
XC = 1/(ωC) = 1/(6283·10·10⁻⁶) = 15.92 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(30² + (298.2)²) = 299.7 Ω

Current

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

I = V/|Z| = 120/299.7 = 0.4003 A

Voltage across the capacitor

VC = I·XC = 0.4003·15.92 = 6.372 V

Answer

VC = 6.372 V
5.AC power & power factor

Answer: P = 768 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| = √(12² + 9²) = 15 Ω
I = V/|Z| = 120/15 = 8 A

Real power

P = I²·R = 8²·12 = 768 W

Answer

P = 768 W
6.AC power & power factor

Answer: I = 70.95 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 = 25·tan(arccos 0.6) = 33.33 kvar
QC = V²·2πf·C = 400²·2π·60·330·10⁻⁶ = 19905 var = 19.91 kvar
Q2 = 33.33 − 19.91 = 13.43 kvar

New apparent power

S2 = √(25² + 13.43²) = 28.38 kVA

New line current

I = S2/V = 28378/400 = 70.95 A

Answer

I = 70.95 A
7.LC resonance, Q factor & bandwidth

Answer: WL = 8.25 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

½·2.2 × 10⁻¹⁰·10² = 1.1 × 10⁻⁸ J

Energy left in the capacitor

½·2.2 × 10⁻¹⁰·5² = 2.75 × 10⁻⁹ J

Energy in the inductor

WL = 8.25 × 10⁻⁹ J = 8.25 nJ

Answer

WL = 8.25 nJ
8.LC resonance, Q factor & bandwidth

Answer: RD = 5000 kΩ

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).

Dynamic resistance

RD = 0.022/(2.2 × 10⁻⁹·2) = 5 × 10⁶ Ω = 5000 kΩ

Check with Q

Q = (1/r)·√(L/C) = 1581, Q²·r = 5 × 10⁶ Ω

Answer

RD = 5000 kΩ
9.Three-phase power

Answer: P = 110.6 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 = VL = 480 V

Phase current

|Z| = √(4² + 3²) = 5 Ω
Iph = 480/5 = 96 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·96²·4 = 110.6 kW

Answer

P = 110.6 kW

Check: cos φ = R/|Z| = 0.8, and √3·VL·IL·cos φ gives the same power.

10.Three-phase power

Answer: QC = 40.86 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 0.7) = 50·1.02 = 51.01 kvar
Q2 = 50·tan(acos 0.98) = 50·0.2031 = 10.15 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 40.86 kvar

Answer

QC = 40.86 kvar
11.RC & RL transients

Answer: vC = 2.245 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 = 10·6.8/(10 + 6.8) = 4.048 kΩ
vC(∞) = 24·6.8/(10 + 6.8) = 9.714 V

Time constant

τ = Rth·C = 4.048·47 = 190.2 ms

Charging curve

vC(t) = 9.714·(1 − e−50/190.2) = 2.245 V

Answer

vC = 2.245 V
12.RC & RL transients

Answer: t = 1801 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) = 4 V, vC(∞) = Vs = 24 V
τ = R·C = 6.8 kΩ · 220 µF = 1496 ms
vC(t) = 24 + (4 − 24)·e−t/τ

Solve for the time

18 = 24 + (−20)·e−t/τ → e−t/τ = (18 − 24)/(4 − 24) = 0.3
t = τ·ln((4 − 24)/(18 − 24)) = 1496·ln(3.333) = 1801 ms

Answer

t = 1801 ms
13.Thévenin & Norton

Answer: IL = 1.143 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​ = 2 ΩSHORTR2​20 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 = 3·20/(3+20) = 2.609 Ω
Rth = (R1 ∥ R2) + R3 = 2.609 + 2 = 4.609 Ω
R1​ = 3 ΩR3​ = 2 Ω+−V1​10 VR2​20 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) = 10·20/(3+20) = 8.696 V
Rth​ = 4.609 Ω+−Vth​8.696 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) = 8.696/(4.609 + 3) = 1.143 A
14.Thévenin & Norton

Answer: PL = 20.15 W

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​ = 20 ΩR3​ = 3 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 = 20·8/(20+8) = 5.714 Ω
Rth = (R1 ∥ R2) + R3 = 5.714 + 3 = 8.714 Ω
R1​ = 20 ΩR3​ = 3 Ω+−V1​30 VR2​8 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) = 8.571 + 2.5·5.714 = 22.86 V
Vth = VB = VA + Is·R3 = 22.86 + 2.5·3 = 30.36 V
Rth​ = 8.714 Ω+−Vth​30.36 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) = 30.36/(8.714 + 3) = 2.591 A
PL = IL²·RL = 2.591²·3 = 20.15 W
15.Bode plots & margins

Answer: ∠G(jω) = −75.53 °

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(3/6) − 90° − atan(3/14) = 26.57° − 90° − 12.09°

Result

∠G(j3) = −75.53°

Answer

∠G(jω) = −75.53°
16.Bode plots & margins

Answer: ωc = 9.396 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 − 10000 = 0
ω² = (−25 + √(625 + 4·10000))/2 = 88.28
ωc = 9.396 rad/s

Answer

ωc = 9.396 rad/s

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