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
ReIm22−2−244−4−466−6−6θz

The number z = 6·e−j0.5 has its angle in radians. Find its imaginary part b.

r = 6θ = −0.5 rad
b =
2.Complex numbersHard
ReIm11−1−122−2−2z₀z₁z₂z₃z₄z₅

The roots of zⁿ = w with n = 6 and w = 64∠180° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, …, 5. Find the argument of z₂ in degrees, between −180° and 180°.

w = 64∠180°n = 6k = 2
θ₂ = °
3.AC impedance & phasorsMedium
10 V1000 HzR = 80 ΩL = 20 mHC = 100 µF

A series RLC circuit is connected to a 10 V (rms), 1000 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).

V = 10 V (rms)f = 1000 HzR = 80 ΩL = 20 mHC = 100 µF
φ = °
4.AC impedance & phasorsHard
50 V100 HzR = 15 ΩL = 250 mHC = 4.7 µFI

A series RLC circuit is connected to a 50 V (rms), 100 Hz supply. Find the rms current I.

V = 50 V (rms)f = 100 HzR = 15 ΩL = 250 mHC = 4.7 µF
I = A
5.AC power & power factorMedium
400 V14 kW, 0.8525 kW, 0.7I

Two loads are connected in parallel to a 400 V single-phase supply: a 4 kW load at a power factor of 0.85 lagging and a 5 kW load at a power factor of 0.7 leading. Find the overall power factor cos φ seen by the supply.

V = 400 VP1 = 4 kW, pf1 = 0.85 laggingP2 = 5 kW, pf2 = 0.7 leading
cos φ =
6.AC power & power factorHard
240 V50 Hzload9 kW, 0.6C220 µFI

A single-phase load takes 9 kW at a power factor of 0.6 lagging from a 240 V, 50 Hz supply. A 220 µF capacitor is connected in parallel with the load. Find the new power factor cos φ seen by the supply.

P = 9 kWpf1 = 0.6V = 240 Vf = 50 HzC = 220 µF
cos φ =
7.LC resonance, Q factor & bandwidthMedium
V22 kΩ1 mH1 nF

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

R = 22 kΩL = 1 mHC = 1 nF
Q =
8.LC resonance, Q factor & bandwidthHard
f|I|0.7071f0​ = ?2 kHz3 kHzBW

The half-power frequencies of a resonance curve are 2 kHz and 3 kHz. Find the resonance frequency as their geometric mean, and then the quality factor Q.

f1 = 2 kHzf2 = 3 kHz
Q =
9.Three-phase powerMedium
ABCVL​ = 415 VIL​ = 150 Acos φ = 0.8lagging

A balanced three-phase load draws a line current of 150 A from a supply with a line-to-line voltage of 415 V. The power factor is 0.8 lagging. Find the total real power P drawn by the load in kW.

VL = 415 VIL = 150 Acos φ = 0.8
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 415 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.95 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 80 kWpf1 = 0.7pf2 = 0.95VL = 415 Vf = 50 Hz
QC = kvar
11.RC & RL transientsMedium
+−Vs​10 Vt = 0R1​ = 4.7 kΩR2​ = 1 kΩC100 µF+−vC​

The source Vₛ charges the capacitor through R₁, and R₂ is connected in parallel with C. The capacitor is uncharged when the switch closes at t = 0. Find the capacitor voltage vC at t = 30 ms.

Vs = 10 VR1 = 4.7 kΩR2 = 1 kΩC = 100 µFt = 30 ms
vC = V
12.RC & RL transientsHard
+−Vs​5 Vt = 0R = ?C2.2 µF+−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 3.16 V, exactly 25 ms after the start. With C = 2.2 µF, find the resistance R needed.

Vs = 5 VVth = 3.16 VT = 25 msC = 2.2 µF
R = kΩ
13.Thévenin & NortonMedium
R1​ = 6 ΩR3​ = 8 Ω+−V1​18 VR2​10 ΩAab

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

V1 = 18 VR1 = 6 ΩR2 = 10 ΩR3 = 8 Ω
IN = A
14.Thévenin & NortonHard
R1​ = 4 ΩR3​ = 6 Ω+−V1​18 VR2​8 ΩAIs​2 ABab

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

V1 = 18 VIs = 2 AR1 = 4 ΩR2 = 8 ΩR3 = 6 Ω
Vth = V
15.Bode plots & marginsMedium
vin​L = 50 mHC = 0.1 µFR = 50 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 50 Ω, with L = 50 mH and C = 0.1 µF. Find the centre (resonant) frequency f0 in Hz.

R = 50 ΩL = 50 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) = K/(s(s + 2)(s + 7)). Find the gain K that gives a gain margin of 8 dB.

L(s) = K/(s(s + 2)(s + 7))GM = 8 dB
K =

3Solutions

1.Complex numbers

Answer: b = −2.877

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 = 6·(cos −0.5 + j sin −0.5)

Components (radians)

a = 6·cos −0.5 = 5.265
b = 6·sin −0.5 = −2.877
b = −2.877

Answer

b = −2.877
2.Complex numbers

Answer: θ₂ = 150 °

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

641/6 = 2

Argument of the root

θ₂ = (180° + 360°·2)/6 = 150°

Argument

θ₂ = 150°

Answer

θ₂ = 150°

The 6 roots lie on the circle of radius 2, 60° apart.

3.AC impedance & phasors

Answer: φ = 57.19 °

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/(6283·100·10⁻⁶) = 1.592 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(80² + (124.1)²) = 147.6 Ω

Phase angle

φ = arctan((XL − XC)/R) = arctan(124.1/80) = 57.19°

Answer

φ = 57.19°
4.AC impedance & phasors

Answer: I = 0.2745 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·100 = 628.3 rad/s
XL = ωL = 628.3·0.25 = 157.1 Ω
XC = 1/(ωC) = 1/(628.3·4.7·10⁻⁶) = 338.6 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(15² + (−181.5)²) = 182.2 Ω

Current

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

I = V/|Z| = 50/182.2 = 0.2745 A

Answer

I = 0.2745 A
5.AC power & power factor

Answer: cos φ = 0.9601

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 = 4·tan(arccos 0.85) = 2.479 kvar
Q2 = −5·tan(arccos 0.7) = −5.101 kvar

Totals

P = 4 + 5 = 9 kW
Q = 2.479 + (−5.101) = −2.622 kvar
S = √(9² + (−2.622)²) = 9.374 kVA

Overall power factor

cos φ = P/S = 9/9.374 = 0.9601 leading

Answer

cos φ = 0.9601
6.AC power & power factor

Answer: cos φ = 0.7466

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 = 9·tan(arccos 0.6) = 12 kvar
QC = V²·2πf·C = 240²·2π·50·220·10⁻⁶ = 3981 var = 3.981 kvar
Q2 = 12 − 3.981 = 8.019 kvar

New apparent power

S2 = √(9² + 8.019²) = 12.05 kVA

New power factor

cos φ = P/S2 = 9/12.05 = 0.7466

Answer

cos φ = 0.7466
7.LC resonance, Q factor & bandwidth

Answer: Q = 22

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) = 1000000 rad/s

Quality factor

Q = R/(ω₀L) = 22000/(1000000·0.001) = 22

Answer

Q = 22

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

8.LC resonance, Q factor & bandwidth

Answer: Q = 2.449

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

f0 = √(2000·3000) = 2449 Hz

Bandwidth

BW = 3000 − 2000 = 1000 Hz

Quality factor

Q = 2449/1000 = 2.449

Answer

Q = 2.449
9.Three-phase power

Answer: P = 86.26 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 = √3·415·150 = 107.8 kVA

Real power

P = S·cos φ = 107.8·0.8 = 86.26 kW

Answer

P = 86.26 kW
10.Three-phase power

Answer: QC = 55.32 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 = 80·tan(acos 0.7) = 80·1.02 = 81.62 kvar
Q2 = 80·tan(acos 0.95) = 80·0.3287 = 26.29 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 55.32 kvar

Answer

QC = 55.32 kvar
11.RC & RL transients

Answer: vC = 0.5351 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.

Thévenin equivalent seen by the capacitor

Short the source to find Rth; open the capacitor to find the voltage it will settle at.

Rth = R1 ∥ R2 = 4.7·1/(4.7 + 1) = 0.8246 kΩ
vC(∞) = 10·1/(4.7 + 1) = 1.754 V

Time constant

τ = Rth·C = 0.8246·100 = 82.46 ms

Charging curve

vC(t) = 1.754·(1 − e−30/82.46) = 0.5351 V

Answer

vC = 0.5351 V
12.RC & RL transients

Answer: R = 11.37 kΩ

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

3.16 = 5·(1 − e−25/τ) → τ = 25/ln(5/(5 − 3.16))
τ = 25/0.9997 = 25.01 ms

Resistance

kΩ times µF gives milliseconds.

R = τ/C = 25.01/2.2 = 11.37 kΩ

Answer

R = 11.37 kΩ
13.Thévenin & Norton

Answer: IN = 0.9574 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​ = 8 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 + 8 = 11.75 Ω
R1​ = 6 ΩR3​ = 8 Ω+−V1​18 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) = 18·10/(6+10) = 11.25 V
IN​0.9574 ARth​11.75 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 = 11.25/11.75 = 0.9574 A
14.Thévenin & Norton

Answer: Vth = 29.33 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​ = 4 ΩR3​ = 6 ΩSHORTR2​8 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 = 4·8/(4+8) = 2.667 Ω
Rth = (R1 ∥ R2) + R3 = 2.667 + 6 = 8.667 Ω
R1​ = 4 ΩR3​ = 6 Ω+−V1​18 VR2​8 ΩAIs​2 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) = 12 + 2·2.667 = 17.33 V
Vth = VB = VA + Is·R3 = 17.33 + 2·6 = 29.33 V
15.Bode plots & margins

Answer: f0 = 2251 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/√(0.05·1 × 10⁻⁷) = 14142 rad/s
f₀ = ω₀/(2π) = 2251 Hz

Answer

f0 = 2251 Hz
16.Bode plots & margins

Answer: K = 50.16

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/7) = 1 → ω180 = √(2·7) = 3.742 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 4)·√(ω² + 49)) = K/(2·7·(2 + 7)) = K/126

Required gain

126/K = 108/20 = 2.512
K = 126/2.512 = 50.16

Answer

K = 50.16

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