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
ReIm11−1−122−2−233−3−344−4−455−5−566−6−677−7−788−8−8zABCD

The diagram shows z = −2 + j3. With w = 0.5∠90°, click the point that represents z/w.

z = −2 + j3w = 0.5∠90°
  1. 1A (3 + j2)
  2. 2B (−6 − j4)
  3. 3C (6 + j4)
  4. 4D (−4 + j6)
2.Complex numbersHard
ReIm11−1−122−2−233−3−3z₁z₂

One root of z³ − z² − 17z + 65 = 0 is z₁ = 3 + j2. The coefficients are real. Find the real root r.

z₁ = 3 + j2
r =
3.AC impedance & phasorsMedium
50 V100 HzR = 20 ΩL = 150 mHC = 47 µFI

A series RLC circuit is connected to a 50 V (rms), 100 Hz supply. Find the power factor cos φ of the circuit.

V = 50 V (rms)f = 100 HzR = 20 ΩL = 150 mHC = 47 µF
cos φ =
4.AC impedance & phasorsHard
10 Vf0​R = 15 ΩL = 200 mHC = 10 µF

A series RLC circuit is driven by a 10 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage VL across the inductor.

V = 10 V (rms)R = 15 ΩL = 200 mHC = 10 µF
VL = V
5.AC power & power factorMedium
P = 23 kWQSφ

A single-phase motor takes 23 kW at a power factor of 0.9 lagging from a 230 V supply. Find the line current I.

P = 23 kWpf = 0.9V = 230 V
I = A
6.AC power & power factorHard
P1​ = 42 kWQ1​S1​φ

A 100 kVA single-phase transformer supplies a load of 42 kW at a power factor of 0.7 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 = 42 kWpf1 = 0.7
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
VR = ?2.2 mHC

A series resonant circuit uses a 2.2 mH inductor. Which total series resistance R gives a bandwidth of 20 kHz?

L = 2.2 mHBW = 20 kHz
R = Ω
9.Three-phase powerMedium
ABCVL​ = 415 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 415 V. Find the total real power P absorbed by the load in kW.

VL = 415 VZ = 12 + j16 Ω per phase
P = kW
10.Three-phase powerHard
P = 150 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 150 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 = 150 kWpf1 = 0.65pf2 = 0.98VL = 400 Vf = 50 Hz
Cph = µF
11.RC & RL transientsMedium
R = 22 Ωt = 0L100 mHi(t)

The inductor carries a steady current of 0.5 A. At t = 0 the switch closes the loop through R and the source is removed, so the inductor current decays through R. How long does it take for the current to fall to 0.25 A? Find t.

i(0) = 0.5 AR = 22 ΩL = 100 mH
t = ms
12.RC & RL transientsHard
+−Vs​20 Vopens at t = 0R1​ = 2.2 kΩR2​ = 1 kΩi2​C33 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 current i2 through R2 at t = 30 ms.

Vs = 20 VR1 = 2.2 kΩR2 = 1 kΩC = 33 µFt = 30 ms
i2 = mA
13.Thévenin & NortonMedium
R1​ = 20 ΩR3​ = 20 Ω+−V1​10 VR2​5 ΩAab

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

V1 = 10 VR1 = 20 ΩR2 = 5 ΩR3 = 20 Ω
Vth = V
14.Thévenin & NortonHard
R1​ = 30 ΩR3​ = 8 Ω+−V1​5 VR2​25 ΩAIs​1 ABab

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

V1 = 5 VIs = 1 AR1 = 30 ΩR2 = 25 ΩR3 = 8 Ω
IN = A
15.Bode plots & marginsMedium
vin​L = 20 mHC = 10 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 = 10 µF. Find the centre (resonant) frequency f0 in Hz.

R = 20 ΩL = 20 mHC = 10 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) = K/(s(s + 4)(s + 10)). Find the gain K that gives a gain margin of 10 dB.

L(s) = K/(s(s + 4)(s + 10))GM = 10 dB
K =

3Solutions

1.Complex numbers

Answer: C (6 + j4)

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.

Effect of w

|z/w| = |z|/0.5, arg(z/w) = arg z − 90°

The point

z/w = 6 + j4 → C

Answer

C (6 + j4)
2.Complex numbers

Answer: r = −5

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 = 3 − j2
(z − z₁)(z − z₂) = z² − 2·3·z + (3² + 2²) = z² − 6z + 13

Factor out the quadratic

z³ − z² − 17z + 65 = (z² − 6z + 13)(z + 5)

Real root

r = −5

Answer

r = −5

Check with the sum of the roots: z₁ + z₂ + r = 6 + (−5) = 1 = −(−1).

3.AC impedance & phasors

Answer: cos φ = 0.3144

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

XL = 94.25 Ω, XC = 33.86 Ω

Impedance

|Z| = √(20² + (60.39)²) = 63.61 Ω

Power factor

cos φ = R/|Z| = 20/63.61 = 0.3144
The circuit is inductive, so the power factor is lagging.

Answer

cos φ = 0.3144
4.AC impedance & phasors

Answer: VL = 94.28 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/√(0.2·10·10⁻⁶) = 707.1 rad/s
I = V/R = 10/15 = 0.6667 A

Voltage across the inductor

XL = ω₀L = 707.1·0.2 = 141.4 Ω
VL = I·XL = 0.6667·141.4 = 94.28 V

Answer

VL = 94.28 V

This is Q = 9.428 times the supply voltage.

5.AC power & power factor

Answer: I = 111.1 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.

Line current

Only the real part of the power does work, so divide by the power factor.

I = P/(V·cos φ) = 23000/(230·0.9) = 111.1 A

Answer

I = 111.1 A
6.AC power & power factor

Answer: ΔP = 48.35 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 = 42·tan(arccos 0.7) = 42.85 kvar

Largest total real power

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

P = √(Smax² − Q1²) = √(100² − 42.85²) = 90.35 kW

Extra real power

ΔP = 90.35 − 42 = 48.35 kW

Answer

ΔP = 48.35 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: R = 276.5 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·0.0022·20000 = 276.5 Ω

Answer

R = 276.5 Ω
9.Three-phase power

Answer: P = 5.167 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 = 415/√3 = 239.6 V

Phase current

|Z| = √(12² + 16²) = 20 Ω
Iph = 239.6/20 = 11.98 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·11.98²·12 = 5.167 kW

Answer

P = 5.167 kW

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

10.Three-phase power

Answer: Cph = 2883 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 = 150·tan(acos 0.65) = 150·1.169 = 175.4 kvar
Q2 = 150·tan(acos 0.98) = 150·0.2031 = 30.46 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 144.9 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) = 48304/(314.2·230.9²) = 2883 µF

Answer

Cph = 2883 µF
11.RC & RL transients

Answer: t = 3.151 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

τ = L/R = 100 mH / 22 Ω = 4.545 ms

Solve for the time

0.25 = 0.5·e−t/τ → t = τ·ln(0.5/0.25)
t = 4.545·ln(2) = 3.151 ms

Answer

t = 3.151 ms
12.RC & RL transients

Answer: i2 = 2.518 mA

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) = 20·1/(2.2 + 1) = 6.25 V

Time constant after the switch opens

τ = R2·C = 1·33 = 33 ms

Discharge

vC(t) = 6.25·e−30/33 = 2.518 V
i2 = vC/R2 = 2.518/1 = 2.518 mA

Answer

i2 = 2.518 mA
13.Thévenin & Norton

Answer: Vth = 2 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​ = 20 ΩR3​ = 20 ΩSHORTR2​5 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 = 20·5/(20+5) = 4 Ω
Rth = (R1 ∥ R2) + R3 = 4 + 20 = 24 Ω
R1​ = 20 ΩR3​ = 20 Ω+−V1​10 VR2​5 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) = 10·5/(20+5) = 2 V
14.Thévenin & Norton

Answer: IN = −0.895 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​ = 30 ΩR3​ = 8 ΩSHORTR2​25 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 = 30·25/(30+25) = 13.64 Ω
Rth = (R1 ∥ R2) + R3 = 13.64 + 8 = 21.64 Ω
R1​ = 30 ΩR3​ = 8 Ω+−V1​5 VR2​25 ΩAIs​1 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) = 2.273 + (−1)·13.64 = −11.36 V
Vth = VB = VA + Is·R3 = −11.36 + (−1)·8 = −19.36 V
IN​0.895 ARth​21.64 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 = −19.36/21.64 = −0.895 A
15.Bode plots & margins

Answer: f0 = 355.9 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 × 10⁻⁵) = 2236 rad/s
f₀ = ω₀/(2π) = 355.9 Hz

Answer

f0 = 355.9 Hz
16.Bode plots & margins

Answer: K = 177.1

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.

Phase crossover

(ω/4)·(ω/10) = 1 → ω180 = √(4·10) = 6.325 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 16)·√(ω² + 100)) = K/(4·10·(4 + 10)) = K/560

Required gain

560/K = 1010/20 = 3.162
K = 560/3.162 = 177.1

Answer

K = 177.1

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