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
ReIm22−2−244−4−4z₁z₂w

Given z₁ = 4∠100° and z₂ = 2∠45°, find the imaginary part of w = z₁/z₂.

z₁ = 4∠100°z₂ = 2∠45°
Im(w) =
2.Complex numbersHard
ReIm11−1−122−2−2θz

Find the imaginary part of (−2 − j2)³.

z = −2 − j2n = 3
Im(z³) =
3.AC impedance & phasorsMedium
100 V60 HzR = 10 ΩL = 20 mHI

A resistor and an inductor are connected in parallel across a 100 V (rms), 60 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).

V = 100 V (rms)f = 60 HzR = 10 ΩL = 20 mH
φ = °
4.AC impedance & phasorsHard
10 V1000 HzR = 80 ΩL = 20 mHC = 2.2 µFI

A series RLC circuit is connected to a 10 V (rms), 1000 Hz supply. Find the rms current I.

V = 10 V (rms)f = 1000 HzR = 80 ΩL = 20 mHC = 2.2 µF
I = A
5.AC power & power factorMedium
240 V110 kW, 0.626 kW, 0.95I

Two loads are connected in parallel to a 240 V single-phase supply: a 10 kW load at a power factor of 0.6 lagging and a 6 kW load at a power factor of 0.95 lagging. Find the overall power factor cos φ seen by the supply.

V = 240 VP1 = 10 kW, pf1 = 0.6 laggingP2 = 6 kW, pf2 = 0.95 lagging
cos φ =
6.AC power & power factorHard
P = 5 kWQ1​S1​φ1​QC​

A single-phase load takes 5 kW at a power factor of 0.7 lagging from a 400 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.92. Find the new line current Inew.

P = 5 kWpf1 = 0.7pf2 = 0.92V = 400 Vf = 50 Hz
Inew = A
7.LC resonance, Q factor & bandwidthMedium
V22 kΩ47 mH1 nF

A parallel RLC circuit has R = 22 kΩ, L = 47 mH and C = 1 nF. Find its quality factor Q.

R = 22 kΩL = 47 mHC = 1 nF
Q =
8.LC resonance, Q factor & bandwidthHard
C = 1 µFR = 1 ΩL = 4.7 mH

A charged 1 µF capacitor discharges through a 4.7 mH coil with a total series resistance of 1 Ω, so the oscillation slowly dies out. Its amplitude decays as e−Rt/(2L). How many oscillation cycles n pass before the amplitude has halved? (The losses are small, so the frequency is f₀.)

L = 4.7 mHC = 1 µFR = 1 Ω
n =
9.Three-phase powerMedium
ABCVL​ = 690 VIL​ = 12 Acos φ = 0.88lagging

A balanced three-phase load draws a line current of 12 A from a supply with a line-to-line voltage of 690 V. The power factor is 0.88 lagging. Find the total real power P drawn by the load in kW.

VL = 690 VIL = 12 Acos φ = 0.88
P = kW
10.Three-phase powerHard
P = 200 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 200 kW at a power factor of 0.7 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 0.95 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 200 kWpf1 = 0.7pf2 = 0.95VL = 415 Vf = 60 Hz
QC = kvar
11.RC & RL transientsMedium
+−Vs​10 Vt = 0R = 68 ΩL22 mHi(t)

The inductor carries no current before the switch closes at t = 0. Find the voltage vL across the inductor at t = 1 ms.

Vs = 10 VR = 68 ΩL = 22 mHt = 1 ms
vL = V
12.RC & RL transientsHard
+−Vs​5 Vt = 0R = 47 kΩC?+−vC​

A timer starts charging an uncharged capacitor through a resistor from a 5 V supply at t = 0. It must trigger when the capacitor voltage reaches 4 V, exactly 500 ms after the start. With R = 47 kΩ, find the capacitance C needed.

Vs = 5 VVth = 4 VT = 500 msR = 47 kΩ
C = µF
13.Thévenin & NortonMedium
R1​ = 12 ΩR3​ = 8 Ω+−V1​6 VR2​6 ΩAab

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

V1 = 6 VR1 = 12 ΩR2 = 6 ΩR3 = 8 Ω
Vth = V
14.Thévenin & NortonHard
R1​ = 25 ΩR3​ = 30 Ω+−V1​6 VR2​12 ΩAIs​2.5 ABab

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

V1 = 6 VIs = 2.5 AR1 = 25 ΩR2 = 12 ΩR3 = 30 Ω
Vth = V
15.Bode plots & marginsMedium
vin​R = 22 kΩC = 47 nFvout​

The RC low-pass filter shown has R = 22 kΩ, C = 47 nF. A sinusoidal input of 770 Hz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 22 kΩC = 47 nFf = 770 Hz
φ = °
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 + 5)). Find the gain K that gives a gain margin of 2 dB.

L(s) = K/(s(s + 4)(s + 5))GM = 2 dB
K =

3Solutions

1.Complex numbers

Answer: Im(w) = 1.638

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.

Modulus and argument of w

|w| = 4 / 2 = 2
θ = 100° − 45° = 55°

Rectangular form

Re(w) = 2·cos(55°) = 1.147
Im(w) = 2·sin(55°) = 1.638
Im(w) = 1.638

Answer

Im(w) = 1.638
2.Complex numbers

Answer: Im(z³) = −16

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 = (−2 − j2)(−2 − j2) = j8
z³ = (j8)(−2 − j2) = 16 − j16

Check with De Moivre

z = 2.828∠−135° → z³ = 22.63∠−45°
Im(z³) = −16

Answer

Im(z³) = −16
3.AC impedance & phasors

Answer: φ = 52.98 °

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·60·0.02 = 7.54 Ω

Branch currents

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

IR = V/R = 100/10 = 10 A
IX = V/XL = 100/7.54 = 13.26 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) = √(10² + 13.26²) = 16.61 A

Phase angle

The inductor current lags V, so the total current lags and the angle is positive.

φ = arctan(IX/IR) = arctan(13.26/10) = 52.98°

Answer

φ = 52.98°
4.AC impedance & phasors

Answer: I = 0.104 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·1000 = 6283 rad/s
XL = ωL = 6283·0.02 = 125.7 Ω
XC = 1/(ωC) = 1/(6283·2.2·10⁻⁶) = 72.34 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(80² + (53.32)²) = 96.14 Ω

Current

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

I = V/|Z| = 10/96.14 = 0.104 A

Answer

I = 0.104 A
5.AC power & power factor

Answer: cos φ = 0.7226

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 = 10·tan(arccos 0.6) = 13.33 kvar
Q2 = 6·tan(arccos 0.95) = 1.972 kvar

Totals

P = 10 + 6 = 16 kW
Q = 13.33 + 1.972 = 15.31 kvar
S = √(16² + 15.31²) = 22.14 kVA

Overall power factor

cos φ = P/S = 16/22.14 = 0.7226 lagging

Answer

cos φ = 0.7226
6.AC power & power factor

Answer: Inew = 13.59 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 = P·tan φ1 = 5·tan(45.57°) = 5.101 kvar
Q2 = P·tan φ2 = 5·tan(23.07°) = 2.13 kvar
QC = Q1 − Q2 = 2.971 kvar

New line current

Inew = P/(V·cos φ2) = 5000/(400·0.92) = 13.59 A

Answer

Inew = 13.59 A
7.LC resonance, Q factor & bandwidth

Answer: Q = 3.209

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) = 145865 rad/s

Quality factor

Q = R/(ω₀L) = 22000/(145865·0.047) = 3.209

Answer

Q = 3.209

The bandwidth is 1/(2πRC) = 7.234 kHz.

8.LC resonance, Q factor & bandwidth

Answer: n = 15.13

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).

Half-life of the amplitude

t = 2·0.0047·ln 2/1 = 0.006516 s

Number of cycles

n = t·f0 = 0.006516·2322 = 15.13

Answer

n = 15.13

Shortcut: n = Q·ln 2/π with Q = 68.56.

9.Three-phase power

Answer: P = 12.62 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.

Apparent power

S = √3·VL·IL = √3·690·12 = 14.34 kVA

Real power

P = S·cos φ = 14.34·0.88 = 12.62 kW

Answer

P = 12.62 kW
10.Three-phase power

Answer: QC = 138.3 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 = 200·tan(acos 0.7) = 200·1.02 = 204 kvar
Q2 = 200·tan(acos 0.95) = 200·0.3287 = 65.74 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 138.3 kvar

Answer

QC = 138.3 kvar
11.RC & RL transients

Answer: vL = 0.4546 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 = 22 mH / 68 Ω = 0.3235 ms
i(∞) = Vs/R = 10/68 = 0.1471 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 = 10·e−1/0.3235 = 0.4546 V

Answer

vL = 0.4546 V
12.RC & RL transients

Answer: C = 6.61 µF

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.

Required time constant

4 = 5·(1 − e−500/τ) → τ = 500/ln(5/(5 − 4))
τ = 500/1.609 = 310.7 ms

Capacitance

kΩ times µF gives milliseconds.

C = τ/R = 310.7/47 = 6.61 µF

Answer

C = 6.61 µF
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​ = 12 ΩR3​ = 8 ΩSHORTR2​6 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 = 12·6/(12+6) = 4 Ω
Rth = (R1 ∥ R2) + R3 = 4 + 8 = 12 Ω
R1​ = 12 ΩR3​ = 8 Ω+−V1​6 VR2​6 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) = 6·6/(12+6) = 2 V
14.Thévenin & Norton

Answer: Vth = 97.22 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​ = 25 ΩR3​ = 30 ΩSHORTR2​12 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 = 25·12/(25+12) = 8.108 Ω
Rth = (R1 ∥ R2) + R3 = 8.108 + 30 = 38.11 Ω
R1​ = 25 ΩR3​ = 30 Ω+−V1​6 VR2​12 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) = 1.946 + 2.5·8.108 = 22.22 V
Vth = VB = VA + Is·R3 = 22.22 + 2.5·30 = 97.22 V
15.Bode plots & margins

Answer: φ = −78.7 °

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.

Corner frequency

ωc = 1/(RC) = 1/(22000·4.7 × 10⁻⁸) = 967.1 rad/s
fc = 153.9 Hz → f/fc = 770/153.9 = 5.003

Phase

φ = − atan(5.003) = −78.7°

Answer

φ = −78.7°
16.Bode plots & margins

Answer: K = 143

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/5) = 1 → ω180 = √(4·5) = 4.472 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 16)·√(ω² + 25)) = K/(4·5·(4 + 5)) = K/180

Required gain

180/K = 102/20 = 1.259
K = 180/1.259 = 143

Answer

K = 143

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