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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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
ReIm11−1−122−2−233−3−3θz

Write z = 4∠35° in rectangular form a + jb and give the real part a.

r = 4θ = 35°
a =
2.Complex numbersHard
ReIm11−1−122−2−233−3−3z₀z₁z₂z₃z₄

The roots of zⁿ = w with n = 5 and w = 243∠−120° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, …, 4. Find the real part of z₃.

w = 243∠−120°n = 5k = 3
Re(z₃) =
3.AC impedance & phasorsMedium
120 V500 HzR = 80 ΩL = 200 mHC = 4.7 µFI

A series RLC circuit is connected to a 120 V (rms), 500 Hz supply. Find the rms current I.

V = 120 V (rms)f = 500 HzR = 80 ΩL = 200 mHC = 4.7 µF
I = A
4.AC impedance & phasorsHard
100 V50 HzR1​ = 30 ΩR2​ = 20 ΩC = 47 µFI

A resistor R1 is connected in series with a parallel combination of R2 and a capacitor C, across a 100 V (rms), 50 Hz supply. Find the rms current I.

V = 100 V (rms)f = 50 HzR1 = 30 ΩR2 = 20 ΩC = 47 µF
I = A
5.AC power & power factorMedium
240 VZ9 + j12 ΩI

A load with impedance Z = 9 + j12 Ω is connected to a 240 V (rms) single-phase supply. Find the real power P it absorbs.

V = 240 VZ = 9 + j12 Ω
P = W
6.AC power & power factorHard
P = 15 kWQ1​S1​φ1​QC​

A single-phase load takes 15 kW at a power factor of 0.6 lagging from a 230 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.98. Find the reactive power QC the capacitor must supply.

P = 15 kWpf1 = 0.6pf2 = 0.98V = 230 Vf = 60 Hz
QC = kvar
7.LC resonance, Q factor & bandwidthMedium
V5 Ω100 µH2.2 nF

A series RLC circuit has R = 5 Ω, L = 100 µH and C = 2.2 nF. Find its quality factor Q.

R = 5 ΩL = 100 µHC = 2.2 nF
Q =
8.LC resonance, Q factor & bandwidthHard
V50 ΩLC = ?

Design a series RLC filter with a total resistance of 50 Ω, a resonance frequency of 200 kHz and a bandwidth of 5 kHz. Find the capacitance C in nF.

R = 50 Ωf0 = 200 kHzBW = 5 kHz
C = nF
9.Three-phase powerMedium
ABCVL​ = 208 VZ = 7 + j24 Ωper phase

A balanced Δ-connected load with an impedance of 7 + j24 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 208 V. Find the total real power P absorbed by the load in kW.

VL = 208 VZ = 7 + j24 Ω per phase
P = kW
10.Three-phase powerHard
P = 80 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 80 kW at a power factor of 0.7 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 0.9 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 80 kWpf1 = 0.7pf2 = 0.9VL = 400 Vf = 50 Hz
Cph = µF
11.RC & RL transientsMedium
+−Vs​24 Vt = 0R = 10 ΩL33 mHi(t)

The inductor carries no current before the switch closes at t = 0. Find the voltage vL across the inductor at t = 5 ms.

Vs = 24 VR = 10 ΩL = 33 mHt = 5 ms
vL = V
12.RC & RL transientsHard
+−Vs​24 Vt = 0R = 1 kΩC10 µF+−vC​vC​(0) = 12 V

Just before the switch closes at t = 0, the capacitor holds 12 V with the polarity shown. Find the capacitor voltage vC at t = 5 ms.

Vs = 24 VR = 1 kΩC = 10 µFvC(0) = 12 Vt = 5 ms
vC = V
13.Thévenin & NortonMedium
R1​ = 15 ΩR3​ = 8 Ω+−V1​9 VR2​20 ΩAab

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

V1 = 9 VR1 = 15 ΩR2 = 20 ΩR3 = 8 Ω
IN = A
14.Thévenin & NortonHard
R1​ = 15 ΩR3​ = 3 Ω+−V1​6 VR2​30 ΩAIs​4 ABab

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

V1 = 6 VIs = 4 AR1 = 15 ΩR2 = 30 ΩR3 = 3 Ω
Vth = V
15.Bode plots & marginsMedium
|H| (dB)0.10.111101010010010001000−40−200204060ω (rad/s)φ (°)ω (rad/s)−135−90−450ω = 5

Find the phase ∠G(jω) of G(s) = 28(s + 18)/(s(s + 12)) at ω = 5 rad/s, in degrees.

G(s) = 28(s + 18)/(s(s + 12))ω = 5 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) = 18/(s(s + 2)(s + 6)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 18/(s(s + 2)(s + 6))
ω180 = rad/s

3Solutions

1.Complex numbers

Answer: a = 3.277

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 = 4·cos(35°) = 3.277
b = 4·sin(35°) = 2.294

Rectangular form

z = 3.277 + j2.294
a = 3.277

Answer

a = 3.277
2.Complex numbers

Answer: Re(z₃) = −2.934

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 of the roots

2431/5 = 3

Argument of the root

θ₃ = (−120° + 360°·3)/5 = 192° → −168°

Rectangular form

z₃ = 3·(cos −168° + j sin −168°) = −2.934 − j0.6237
Re(z₃) = −2.934

Answer

Re(z₃) = −2.934

The 5 roots lie on the circle of radius 3, 72° apart.

3.AC impedance & phasors

Answer: I = 0.2119 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·500 = 3142 rad/s
XL = ωL = 3142·0.2 = 628.3 Ω
XC = 1/(ωC) = 1/(3142·4.7·10⁻⁶) = 67.73 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(80² + (560.6)²) = 566.3 Ω

Current

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

I = V/|Z| = 120/566.3 = 0.2119 A

Answer

I = 0.2119 A
4.AC impedance & phasors

Answer: I = 2.053 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|.

Reactance

XC = 1/(2π·50·47·10⁻⁶) = 67.73 Ω

Parallel combination

Multiply top and bottom by the complex conjugate to split Zp into real and imaginary parts.

Zp = R₂·(−jXC)/(R₂ − jXC) = (R₂·XC² − jR₂²·XC)/(R₂² + XC²)
Zp = 18.4 − j5.432 Ω

Total impedance

Z = 30 + 18.4 − j5.432 = 48.4 − j5.432 Ω
|Z| = √(48.4² + 5.432²) = 48.7 Ω

Current

I = V/|Z| = 100/48.7 = 2.053 A

Answer

I = 2.053 A
5.AC power & power factor

Answer: P = 2304 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| = √(9² + 12²) = 15 Ω
I = V/|Z| = 240/15 = 16 A

Real power

P = I²·R = 16²·9 = 2304 W

Answer

P = 2304 W
6.AC power & power factor

Answer: QC = 16.95 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 = 15·tan(53.13°) = 20 kvar
Q2 = P·tan φ2 = 15·tan(11.48°) = 3.046 kvar
QC = Q1 − Q2 = 16.95 kvar

Answer

QC = 16.95 kvar
7.LC resonance, Q factor & bandwidth

Answer: Q = 42.64

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).

Characteristic impedance

√(L/C) = √(1 × 10⁻⁴/2.2 × 10⁻⁹) = 213.2 Ω

Quality factor

Q = 213.2/5 = 42.64

Answer

Q = 42.64

The resonance frequency is 339.3 kHz, so BW = f₀/Q = 7.958 kHz.

8.LC resonance, Q factor & bandwidth

Answer: C = 0.3979 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) = 50/(2π·5000) = 0.001592 H

Capacitance from the resonance frequency

C = 1/((2π·200000)²·0.001592) = 3.979 × 10⁻¹⁰ F = 0.3979 nF

Answer

C = 0.3979 nF

Q = f₀/BW = 40.

9.Three-phase power

Answer: P = 1.454 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 = 208 V

Phase current

|Z| = √(7² + 24²) = 25 Ω
Iph = 208/25 = 8.32 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·8.32²·7 = 1.454 kW

Answer

P = 1.454 kW

Check: cos φ = R/|Z| = 0.28, and √3·VL·IL·cos φ gives the same power.

10.Three-phase power

Answer: Cph = 284.3 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 = 80·tan(acos 0.7) = 80·1.02 = 81.62 kvar
Q2 = 80·tan(acos 0.9) = 80·0.4843 = 38.75 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 42.87 kvar

Capacitance per phase

In a Δ bank each capacitor is connected between two lines.

V = 400 V, ω = 2π·50 = 314.2 rad/s
Cph = (QC/3)/(ω·V²) = 14290/(314.2·400²) = 284.3 µF

Answer

Cph = 284.3 µF
11.RC & RL transients

Answer: vL = 5.275 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.

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 / 10 Ω = 3.3 ms
i(∞) = Vs/R = 24/10 = 2.4 A

Inductor voltage

At t = 0 the whole supply voltage appears across the inductor; it then decays to zero.

vL(t) = Vs·e−t/τ = 24·e−5/3.3 = 5.275 V

Answer

vL = 5.275 V
12.RC & RL transients

Answer: vC = 16.72 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 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) = 12 V, vC(∞) = Vs = 24 V
τ = R·C = 1 kΩ · 10 µF = 10 ms
vC(t) = 24 + (12 − 24)·e−t/τ

Voltage at the given time

vC(5) = 24 + (−12)·e−5/10 = 16.72 V

Answer

vC = 16.72 V
13.Thévenin & Norton

Answer: IN = 0.3103 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​ = 8 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 = 15·20/(15+20) = 8.571 Ω
Rth = (R1 ∥ R2) + R3 = 8.571 + 8 = 16.57 Ω
R1​ = 15 ΩR3​ = 8 Ω+−V1​9 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) = 9·20/(15+20) = 5.143 V
IN​0.3103 ARth​16.57 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 = 5.143/16.57 = 0.3103 A
14.Thévenin & Norton

Answer: Vth = 56 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​ = 3 ΩSHORTR2​30 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 = 15·30/(15+30) = 10 Ω
Rth = (R1 ∥ R2) + R3 = 10 + 3 = 13 Ω
R1​ = 15 ΩR3​ = 3 Ω+−V1​6 VR2​30 ΩAIs​4 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 + 4·10 = 44 V
Vth = VB = VA + Is·R3 = 44 + 4·3 = 56 V
15.Bode plots & margins

Answer: ∠G(jω) = −97.1 °

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(5/18) − 90° − atan(5/12) = 15.52° − 90° − 22.62°

Result

∠G(j5) = −97.1°

Answer

∠G(jω) = −97.1°
16.Bode plots & margins

Answer: ω180 = 3.464 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.

Phase crossover

(ω/2)·(ω/6) = 1 → ω180 = √(2·6) = 3.464 rad/s

Answer

ω180 = 3.464 rad/s

ExamProblems · examproblems.com/courses/circuit-analysis-2