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
ReIm22−2−244−4−4θz

The number z = 4·ej0.4 has its angle in radians. Find its imaginary part b.

r = 4θ = 0.4 rad
b =
2.Complex numbersHard
ReIm22−2−244−4−4z₁z₂

One root of z³ + 11z² + 44z + 60 = 0 is z₁ = −4 + j2. The coefficients are real. Find the real root r.

z₁ = −4 + j2
r =
3.AC impedance & phasorsMedium
12 V100 HzR = 50 ΩL = 50 mHI

A resistor and an inductor are connected in parallel across a 12 V (rms), 100 Hz supply. Find the magnitude of the impedance |Z|.

V = 12 V (rms)f = 100 HzR = 50 ΩL = 50 mH
|Z| = Ω
4.AC impedance & phasorsHard
10 V1000 HzR = 100 ΩL = 150 mHC = 4.7 µF+−VC​

A series RLC circuit is connected to a 10 V (rms), 1000 Hz supply. Find the rms voltage VC across the capacitor.

V = 10 V (rms)f = 1000 HzR = 100 ΩL = 150 mHC = 4.7 µF
VC = V
5.AC power & power factorMedium
230 V129 kW, 0.7527 kW, 1I = ?

Two loads are connected in parallel to a 230 V single-phase supply: a 29 kW load at a power factor of 0.75 lagging and a 7 kW heater (unity power factor). Find the total current I drawn from the supply.

V = 230 VP1 = 29 kW, pf1 = 0.75 laggingP2 = 7 kW, pf2 = 1
I = A
6.AC power & power factorHard
P = 10 kWQ1​S1​φ1​QC​

A single-phase load takes 10 kW at a power factor of 0.6 lagging from a 400 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.98. Find the capacitance C needed.

P = 10 kWpf1 = 0.6pf2 = 0.98V = 400 Vf = 60 Hz
C = µF
7.LC resonance, Q factor & bandwidthMedium
V1 kΩ2.2 mH100 nF

A parallel RLC circuit has R = 1 kΩ, L = 2.2 mH and C = 100 nF. Find its quality factor Q.

R = 1 kΩL = 2.2 mHC = 100 nF
Q =
8.LC resonance, Q factor & bandwidthHard
VR = ?100 µHC

A series resonant circuit uses a 100 µH inductor. Which total series resistance R gives a bandwidth of 50 kHz?

L = 100 µHBW = 50 kHz
R = Ω
9.Three-phase powerMedium
ABCVL​ = 480 VNZ = 12 + j16 Ωper phase

A balanced Y-connected load with an impedance of 12 + j16 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 = 12 + j16 Ω per phase
P = kW
10.Three-phase powerHard
P = 300 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 300 kW at a power factor of 0.8 lagging from a 415 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.9 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 300 kWpf1 = 0.8pf2 = 0.9VL = 415 Vf = 60 Hz
Cph = µF
11.RC & RL transientsMedium
+−Vs​12 Vt = 0R = 10 ΩL68 mHi(t)

The inductor carries no current before the switch closes at t = 0. Find the energy W stored in the inductor at t = 15 ms, in mJ.

Vs = 12 VR = 10 ΩL = 68 mHt = 15 ms
W = mJ
12.RC & RL transientsHard
+−Vs​20 Vt = 0R = 3.3 kΩC33 µF+−vC​vC​(0) = 30 V

Just before the switch closes at t = 0, the capacitor holds 30 V with the polarity shown. How long does it take for vC to reach 24 V? Find t.

Vs = 20 VR = 3.3 kΩC = 33 µFvC(0) = 30 V
t = ms
13.Thévenin & NortonMedium
R1​ = 6 ΩR3​ = 20 Ω+−V1​10 VR2​4 ΩAab

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

V1 = 10 VR1 = 6 ΩR2 = 4 ΩR3 = 20 Ω
Vth = V
14.Thévenin & NortonHard
R1​ = 25 ΩR3​ = 4 Ω+−V1​18 VR2​3 ΩAIs​5 ABab

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

V1 = 18 VIs = 5 AR1 = 25 ΩR2 = 3 ΩR3 = 4 Ω
IN = A
15.Bode plots & marginsMedium
vin​L = 20 mHC = 0.1 µFR = 20 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 20 Ω, with L = 20 mH and C = 0.1 µF. Find the centre (resonant) frequency f0 in Hz.

R = 20 ΩL = 20 mHC = 0.1 µF
f0 = Hz
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) = 234/(s(s + 4)(s + 9)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 234/(s(s + 4)(s + 9))
ω180 = rad/s

3Solutions

1.Complex numbers

Answer: b = 1.558

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 = 4·(cos 0.4 + j sin 0.4)

Components (radians)

a = 4·cos 0.4 = 3.684
b = 4·sin 0.4 = 1.558
b = 1.558

Answer

b = 1.558
2.Complex numbers

Answer: r = −3

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 − j2
(z − z₁)(z − z₂) = z² − 2·(−4)·z + ((−4)² + 2²) = z² + 8z + 20

Factor out the quadratic

z³ + 11z² + 44z + 60 = (z² + 8z + 20)(z + 3)

Real root

r = −3

Answer

r = −3

Check with the sum of the roots: z₁ + z₂ + r = −8 + (−3) = −11 = −(11).

3.AC impedance & phasors

Answer: |Z| = 26.6 Ω

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·100·0.05 = 31.42 Ω

Branch currents

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

IR = V/R = 12/50 = 0.24 A
IX = V/XL = 12/31.42 = 0.382 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) = √(0.24² + 0.382²) = 0.4511 A

Impedance

|Z| = V/I = 12/0.4511 = 26.6 Ω

Answer

|Z| = 26.6 Ω
4.AC impedance & phasors

Answer: VC = 0.3704 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 = 6283·0.15 = 942.5 Ω
XC = 1/(ωC) = 1/(6283·4.7·10⁻⁶) = 33.86 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(100² + (908.6)²) = 914.1 Ω

Current

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

I = V/|Z| = 10/914.1 = 0.01094 A

Voltage across the capacitor

VC = I·XC = 0.01094·33.86 = 0.3704 V

Answer

VC = 0.3704 V
5.AC power & power factor

Answer: I = 192 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 of each load

A lagging load takes reactive power (Q > 0); a leading load supplies it (Q < 0).

Q1 = 29·tan(arccos 0.75) = 25.58 kvar
Q2 = 0 = 0 kvar

Totals

P = 29 + 7 = 36 kW
Q = 25.58 + 0 = 25.58 kvar
S = √(36² + 25.58²) = 44.16 kVA

Total current

I = S/V = 44160/230 = 192 A

Answer

I = 192 A
6.AC power & power factor

Answer: C = 187.4 µF

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(53.13°) = 13.33 kvar
Q2 = P·tan φ2 = 10·tan(11.48°) = 2.031 kvar
QC = Q1 − Q2 = 11.3 kvar

Capacitance

A capacitor across the supply delivers Q = V²·ωC.

C = QC/(2πf·V²) = 11303/(2π·60·400²) = 187.4 µF

Answer

C = 187.4 µF
7.LC resonance, Q factor & bandwidth

Answer: Q = 6.742

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).

Resonance frequency

ω₀ = 1/√(LC) = 67420 rad/s

Quality factor

Q = R/(ω₀L) = 1000/(67420·0.0022) = 6.742

Answer

Q = 6.742

The bandwidth is 1/(2πRC) = 1.592 kHz.

8.LC resonance, Q factor & bandwidth

Answer: R = 31.42 Ω

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).

Series bandwidth

BW = R/(2πL)

Resistance

R = 2π·1 × 10⁻⁴·50000 = 31.42 Ω

Answer

R = 31.42 Ω
9.Three-phase power

Answer: P = 6.912 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| = √(12² + 16²) = 20 Ω
Iph = 277.1/20 = 13.86 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·13.86²·12 = 6.912 kW

Answer

P = 6.912 kW

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

10.Three-phase power

Answer: Cph = 409.2 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 = 300·tan(acos 0.8) = 300·0.75 = 225 kvar
Q2 = 300·tan(acos 0.9) = 300·0.4843 = 145.3 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 79.7 kvar

Capacitance per phase

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

V = 415 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 26568/(377·415²) = 409.2 µF

Answer

Cph = 409.2 µF
11.RC & RL transients

Answer: W = 38.77 mJ

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 = 68 mH / 10 Ω = 6.8 ms
i(∞) = Vs/R = 12/10 = 1.2 A

Current at the given time

i(t) = 1.2·(1 − e−15/6.8) = 1.068 A

Stored energy

W = ½·L·i² = ½·68·1.068² = 38.77 mJ

Answer

W = 38.77 mJ
12.RC & RL transients

Answer: t = 99.78 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) = 30 V, vC(∞) = Vs = 20 V
τ = R·C = 3.3 kΩ · 33 µF = 108.9 ms
vC(t) = 20 + (30 − 20)·e−t/τ

Solve for the time

24 = 20 + (10)·e−t/τ → e−t/τ = (24 − 20)/(30 − 20) = 0.4
t = τ·ln((30 − 20)/(24 − 20)) = 108.9·ln(2.5) = 99.78 ms

Answer

t = 99.78 ms
13.Thévenin & Norton

Answer: Vth = 4 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​ = 6 ΩR3​ = 20 ΩSHORTR2​4 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 = 6·4/(6+4) = 2.4 Ω
Rth = (R1 ∥ R2) + R3 = 2.4 + 20 = 22.4 Ω
R1​ = 6 ΩR3​ = 20 Ω+−V1​10 VR2​4 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) = 10·4/(6+4) = 4 V
14.Thévenin & Norton

Answer: IN = 5.289 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​ = 25 ΩR3​ = 4 ΩSHORTR2​3 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 = 25·3/(25+3) = 2.679 Ω
Rth = (R1 ∥ R2) + R3 = 2.679 + 4 = 6.679 Ω
R1​ = 25 ΩR3​ = 4 Ω+−V1​18 VR2​3 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) = 1.929 + 5·2.679 = 15.32 V
Vth = VB = VA + Is·R3 = 15.32 + 5·4 = 35.32 V
IN​5.289 ARth​6.679 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 = 35.32/6.679 = 5.289 A
15.Bode plots & margins

Answer: f0 = 3559 Hz

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.

Resonant frequency

ω₀ = 1/√(LC) = 1/√(0.02·1 × 10⁻⁷) = 22361 rad/s
f₀ = ω₀/(2π) = 3559 Hz

Answer

f0 = 3559 Hz
16.Bode plots & margins

Answer: ω180 = 6 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/4)·(ω/9) = 1 → ω180 = √(4·9) = 6 rad/s

Answer

ω180 = 6 rad/s

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