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 − j7 in degrees, as the principal value between −180° and 180°.

z = −6 − j7
θ = °
2.Complex numbersHard
ReIm0.50.5−0.5−0.511−1−1z₁z₂

Solve z² + 2z + 2 = 0. The roots are a ± jb; find b > 0.

p = 2q = 2
b =
3.AC impedance & phasorsMedium
120 V500 HzR = 40 ΩL = 20 mHC = 100 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 = 40 ΩL = 20 mHC = 100 µF
I = A
4.AC impedance & phasorsHard
VfR = 100 ΩL = 50 mHC = 0.022 µF

A series RLC circuit has R = 100 Ω, L = 50 mH and C = 0.022 µF. Find its bandwidth B (the width of the band between the half-power frequencies), in Hz.

R = 100 ΩL = 50 mHC = 0.022 µF
B = Hz
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 apparent power S.

V = 120 VZ = 12 + j9 Ω
S = VA
6.AC power & power factorHard
230 V1000 HzRs​ = 5 ΩXs​ = j10 ΩZL​ = ?

A 230 V (rms), 1000 Hz source has an internal impedance Zs = 5 + j10 Ω. The load impedance ZL can be chosen freely. Find the maximum real power Pmax that can be delivered to the load.

Vs = 230 V (rms)f = 1000 HzZs = 5 + j10 Ω
Pmax = W
7.LC resonance, Q factor & bandwidthMedium
f|I|0.70715 kHzf1​ = ?f2​BW

A series RLC circuit resonates at 5 kHz with Q = 2.5. Find the lower half-power frequency f1 exactly, in kHz.

f0 = 5 kHzQ = 2.5
f1 = kHz
8.LC resonance, Q factor & bandwidthHard
V50 ΩLC = ?

Design a series RLC filter with a total resistance of 50 Ω, a resonance frequency of 500 kHz and a bandwidth of 25 kHz. Find the capacitance C in nF.

R = 50 Ωf0 = 500 kHzBW = 25 kHz
C = nF
9.Three-phase powerMedium
ABCVL​ = 415 VIL​ = ?M18.5 kW, η = 90%cos φ = 0.9

A three-phase induction motor delivers 18.5 kW of mechanical output power at full load. At this load its efficiency is 90% and its power factor is 0.9 lagging. The supply has a line-to-line voltage of 415 V. Find the line current IL.

Pout = 18.5 kWη = 90%cos φ = 0.9VL = 415 V
IL = A
10.Three-phase powerHard
P = 300 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 300 kW at a power factor of 0.65 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 0.98 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.

P = 300 kWpf1 = 0.65pf2 = 0.98VL = 400 Vf = 50 Hz
Cph = µF
11.RC & RL transientsMedium
+−Vs​20 Vt = 0R = 68 ΩL47 mHi(t)

The inductor carries no current before the switch closes at t = 0. How long after the switch closes does the current reach 50% of its final value? Find t.

Vs = 20 VR = 68 ΩL = 47 mH
t = ms
12.RC & RL transientsHard
+−Vs​24 VR1​ = 33 ΩR2​ = 5 Ωopens at t = 0L100 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 = 24 VR1 = 33 ΩR2 = 5 ΩL = 100 mHt = 5 ms
i = A
13.Thévenin & NortonMedium
R1​ = 6 ΩR3​ = 12 Ω+−V1​6 VR2​10 ΩAab

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

V1 = 6 VR1 = 6 ΩR2 = 10 ΩR3 = 12 Ω
IN = A
14.Thévenin & NortonHard
R1​ = 30 ΩR3​ = 30 Ω+−V1​30 VR2​10 ΩAIs​5 ABRL​6 ΩIL​

A load RL = 6 Ω is connected across a–b. Use a Thévenin equivalent to find the power PL delivered to the load.

V1 = 30 VIs = 5 AR1 = 30 ΩR2 = 10 ΩR3 = 30 ΩRL = 6 Ω
PL = W
15.Bode plots & marginsMedium
|H| (dB)0.010.010.10.1111010100100−20020406080ω (rad/s)φ (°)ω (rad/s)−135−90−450ω = 0.5

Find the phase ∠G(jω) of G(s) = 17(s + 9)/(s(s + 7)) at ω = 0.5 rad/s, in degrees.

G(s) = 17(s + 9)/(s(s + 7))ω = 0.5 rad/s
∠G(jω) = °
16.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​

A unity-feedback loop has the open-loop transfer function L(s) = 20/(s(s + 5)). Find the phase margin PM in degrees.

L(s) = 20/(s(s + 5))
PM = °

3Solutions

1.Complex numbers

Answer: θ = −130.6 °

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 = −6, b = −7 → III

Reference angle

α = atan(7/6) = 49.4°

Argument

θ = −(180° − α) = −130.6°

Answer

θ = −130.6°

In polar form z = 9.22∠−130.6°.

2.Complex numbers

Answer: b = 1

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.

Discriminant

p² − 4q = 4 − 8 = −4 < 0

Quadratic formula

z = (−2 ± j√4)/2 = −1 + j, −1 − j

Answer

b = 1

Answer

b = 1
3.AC impedance & phasors

Answer: I = 1.671 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 = 3142·0.02 = 62.83 Ω
XC = 1/(ωC) = 1/(3142·100·10⁻⁶) = 3.183 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(40² + (59.65)²) = 71.82 Ω

Current

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

I = V/|Z| = 120/71.82 = 1.671 A

Answer

I = 1.671 A
4.AC impedance & phasors

Answer: B = 318.3 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|.

Resonant frequency

f0 = 1/(2π·√(0.05·0.022·10⁻⁶)) = 4799 Hz

Quality factor

Q compares the reactance at resonance with the resistance: Q = ω₀L/R.

Q = (1/R)·√(L/C) = (1/100)·√(0.05/(0.022·10⁻⁶)) = 15.08

Bandwidth

B = f0/Q = 4799/15.08 = 318.3 Hz

Answer

B = 318.3 Hz
5.AC power & power factor

Answer: S = 960 VA

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

Apparent power

S = V·I = 120·8 = 960 VA

Answer

S = 960 VA
6.AC power & power factor

Answer: Pmax = 2645 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.

Conjugate match

The load reactance cancels the source reactance, and the load resistance equals the source resistance.

ZL = Zs* = 5 − j10 Ω

Maximum power

I = Vs/(2Rs) = 230/10 = 23 A
Pmax = I²·RL = Vs²/(4Rs) = 230²/(4·5) = 2645 W

Answer

Pmax = 2645 W
7.LC resonance, Q factor & bandwidth

Answer: f1 = 4.099 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/(4·2.5²)) = 1.02

Lower half-power frequency

f1 = 5000·(1.02 − 0.2) = 4099 Hz = 4.099 kHz

Answer

f1 = 4.099 kHz

The upper one is f₂ = 6.099 kHz; check: f₂ − f₁ = 2 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) = 50/(2π·25000) = 0.0003183 H

Capacitance from the resonance frequency

C = 1/((2π·500000)²·0.0003183) = 3.183 × 10⁻¹⁰ F = 0.3183 nF

Answer

C = 0.3183 nF

Q = f₀/BW = 20.

9.Three-phase power

Answer: IL = 31.77 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 = 18.5/0.9 = 20.56 kW

Line current

IL = Pin/(√3·VL·cos φ) = 20556/(√3·415·0.9) = 31.77 A

Answer

IL = 31.77 A
10.Three-phase power

Answer: Cph = 5766 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 = 300·tan(acos 0.65) = 300·1.169 = 350.7 kvar
Q2 = 300·tan(acos 0.98) = 300·0.2031 = 60.92 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 289.8 kvar

Capacitance per phase

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

V = 400/√3 = 230.9 V, ω = 2π·50 = 314.2 rad/s
Cph = (QC/3)/(ω·V²) = 96607/(314.2·230.9²) = 5766 µF

Answer

Cph = 5766 µF
11.RC & RL transients

Answer: t = 0.4791 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 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 = 47 mH / 68 Ω = 0.6912 ms
i(∞) = Vs/R = 20/68 = 0.2941 A

Solve for the time

0.5 = 1 − e−t/τ → t = −τ·ln(1 − 0.5)
t = −0.6912·ln(0.5) = 0.4791 ms

Answer

t = 0.4791 ms
12.RC & RL transients

Answer: i = 0.6459 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) = 24/33 = 0.7273 A
i(∞) = 24/(33 + 5) = 0.6316 A
τ = L/R = 100/38 = 2.632 ms
i(t) = 0.6316 + (0.7273 − 0.6316)·e−t/τ

Current at the given time

i(5) = 0.6316 + (0.09569)·e−5/2.632 = 0.6459 A

Answer

i = 0.6459 A
13.Thévenin & Norton

Answer: IN = 0.2381 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​ = 12 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 = 6·10/(6+10) = 3.75 Ω
Rth = (R1 ∥ R2) + R3 = 3.75 + 12 = 15.75 Ω
R1​ = 6 ΩR3​ = 12 Ω+−V1​6 VR2​10 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) = 6·10/(6+10) = 3.75 V
IN​0.2381 ARth​15.75 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 = 3.75/15.75 = 0.2381 A
14.Thévenin & Norton

Answer: PL = 120.6 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​ = 30 ΩR3​ = 30 ΩSHORTR2​10 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 = 30·10/(30+10) = 7.5 Ω
Rth = (R1 ∥ R2) + R3 = 7.5 + 30 = 37.5 Ω
R1​ = 30 ΩR3​ = 30 Ω+−V1​30 VR2​10 ΩAIs​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) = 7.5 + 5·7.5 = 45 V
Vth = VB = VA + Is·R3 = 45 + 5·30 = 195 V
Rth​ = 37.5 Ω+−Vth​195 VRL​6 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) = 195/(37.5 + 6) = 4.483 A
PL = IL²·RL = 4.483²·6 = 120.6 W
15.Bode plots & margins

Answer: ∠G(jω) = −90.91 °

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(0.5/9) − 90° − atan(0.5/7) = 3.18° − 90° − 4.086°

Result

∠G(j0.5) = −90.91°

Answer

∠G(jω) = −90.91°
16.Bode plots & margins

Answer: PM = 56.34 °

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 − 400 = 0
ω² = (−25 + √(625 + 4·400))/2 = 11.08
ωc = 3.329 rad/s

Phase margin

∠L(jωc) = −90° − atan(3.329/5) = −123.7°
PM = 180° + ∠L = 56.34°

Answer

PM = 56.34°

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