CIRCUIT ANALYSIS 2 NOTES

CHAPTER 13 | ECE 231 Study Notes

Magnetic coupling and mutual inductance

Based on Chapter 13, slides 2-19. These notes explain the slide concepts in study form. Added examples

are labeled. Formulas use corrected notation where the slides contain errors.

Self inductance and mutual inductance

An inductor stores energy in a magnetic field. A changing current changes the magnetic flux and produces

an induced voltage. Self inductance describes a coil’s response to its own changing current. Mutual

inductance describes the voltage induced in a second coil by changing current in the first coil, even when

the coils have no electrical connection.

v = L di/dt L = N dΦ/di Unit of L and M: henry (H)

For constant L, flux linkage is NΦ = Li. Coil 1 flux can be separated into leakage/self-linking flux Φ11 and

shared flux Φ12: Φ1 = Φ11 + Φ12. Only shared flux links the second winding. For reciprocal linear coils,

M12 = M21 = M.

Mutual voltage magnitude: |v₂| = M |di₁/dt|

Dot convention and voltage signs

Dots identify corresponding winding polarities. If the reference current enters a dotted terminal on one

winding, the mutual voltage on the other winding is positive at its dot. If the current leaves the dot, the

mutual voltage is negative at the other dot. Actual voltage also follows the sign of the current derivative.

With each voltage referenced positive where its own current enters, use +M when both currents enter dots

or both leave dots. Use -M when one current enters a dot and the other leaves a dot.

v₁ = L₁ di₁/dt ± M di₂/dt

v₂ = L₂ di₂/dt ± M di₁/dt

Coupled coils in series

Series aiding: L_eq = L₁ + L₂ + 2M

Series opposing: L_eq = L₁ + L₂ - 2M

Aiding means the magnetic effects reinforce; opposing means they subtract. Determine the connection

from the dots and current direction, not just the drawing’s shape.

Added example

If L₁ = 4 H, L₂ = 9 H, and M = 3 H, then aiding gives 4 + 9 + 6 = 19 H; opposing gives 4 + 9 - 6 = 7 H.

Circuit Analysis II • Chapters 13–15 1CHAPTER 13 | ECE 231 Study Notes

Energy and linear transformers

Based on Chapter 13, slides 13-23.

Energy and coupling coefficient

w = ½L₁i₁² + ½L₂i₂² ± Mi₁i₂

0 ≤ M ≤ √(L₁L₂) k = M/√(L₁L₂) M = k√(L₁L₂)

Use the positive energy cross term when both reference currents enter dots or both leave dots. A passive

pair cannot have negative stored magnetic energy. This gives the upper bound on M. The coupling

coefficient k ranges from 0 to 1: zero means no magnetic coupling; one means perfect coupling.

Frequency domain equations

For sinusoidal steady state, replace d/dt with jω and use phasors. Angular frequency is ω = 2πf. Winding

resistances contribute ordinary resistive voltage drops.

V₁ = (R₁ + jωL₁)I₁ ± jωMI₂

V₂ = (R₂ + jωL₂)I₂ ± jωMI₁

These equations use passive voltage references at both windings. If the secondary current is instead

defined leaving the dot toward the load, change the associated signs consistently.

Linear transformer and reflected impedance

The primary connects to the source; the secondary connects to the load. A linear transformer has a linear

magnetic core, so inductances are treated as constants. With a load Z_L across the secondary, define Z₂ =

R₂ + jωL₂ + Z_L.

Z_ref = ω²M²/Z₂

Z_in = R₁ + jωL₁ + ω²M²/(R₂ + jωL₂ + Z_L)

The secondary changes the impedance seen at the primary even though the circuits may be electrically

isolated. This linear-transformer result differs from the ideal turns-ratio formula on the next page.

Equivalent T and pi networks

For the positive-mutual two-port convention, the T equivalent has series arms L₁ - M and L₂ - M, with a

shared shunt arm M. These are mathematical equivalent elements; an individual equivalent arm can be

negative.

Let D = L₁L₂ - M². For the corresponding pi network:

input shunt L_A = D/(L₂ - M)

output shunt L_B = D/(L₁ - M)

bridge L_C = D/M

Use limiting cases or the original network if a denominator is zero. Equivalent circuits must reproduce the

same terminal voltages and currents; check the port and dot conventions before using them.

Reliable solution method

1. Assign current arrows and voltage polarities. 2. Mark the dots. 3. Write each self term. 4. Add the mutual

terms with their signs. 5. Apply KVL/KCL, solve, and check units and limiting cases.

Circuit Analysis II • Chapters 13–15 2CHAPTER 13 | ECE 231 Study Notes

Ideal transformers and review

Based on Chapter 13, slides 24-30.

Ideal transformer assumptions

An ideal transformer is lossless, perfectly coupled (k = 1), and has negligible magnetizing current in the

ideal large-inductance limit. Perfect coupling alone does not imply all the other ideal assumptions.

Define n = N₂/N₁. Then |V₂|/|V₁| = n

|I₂|/|I₁| = 1/n Z_in = Z_L/n²

With both voltages positive at the dots: V₂ = nV₁. With both currents entering their dots: I₁ = -nI₂. If

secondary current is defined leaving its dot toward the load, I₁ = nI₂. Always distinguish signed phasor

relations from magnitude ratios.

n > 1 gives a step-up voltage transformer; n < 1 gives a step-down voltage transformer. A voltage increase

comes with a current decrease for the same transmitted power.

Delivered output power = input power, for a lossless transformer

Reflecting quantities across an ideal transformer

Secondary to primary: divide impedances by n²; divide voltage magnitudes by n; multiply current

magnitudes by n. Primary to secondary: multiply impedances by n²; multiply voltages by n; divide currents

by n. Preserve source polarities and dot signs.

Added example

Let N₁ = 200, N₂ = 100, V₁ = 120 V rms, and Z_L = 8 Ω. Then n = 0.5, V₂ = 60 V, I₂ = 60/8 = 7.5 A, Z_in =

8/(0.5)² = 32 Ω, and I₁ = 120/32 = 3.75 A. Both sides transmit 450 W.

Autotransformers and three-phase transformers

An autotransformer uses one tapped winding shared by the input and output. Voltage follows the turns

across the selected terminals. For the slide connection, V₁/V₂ = N₁/(N₁ + N₂). It can step voltage up or down

but does not provide electrical isolation.

Three-phase service can use three single-phase transformers as a bank or one three-phase transformer. A

single three-phase unit is generally smaller and less expensive. Wye and delta connection choices

determine the line-to-phase relationships.

Important correction to the slides

Slide 25 reverses the voltage ratio. With n = N₂/N₁, the correct equation is V₂/V₁ = n for matching dot

polarities. The slide’s step-up/down statements and reflected-impedance rule use this corrected definition.

Check yourself

Can you choose a mutual-voltage sign from the dots, calculate M from k, distinguish series aiding from

opposing, and reflect a load through an ideal transformer?

Circuit Analysis II • Chapters 13–15 3CHAPTER 14 | ECE 231 Study Notes

Frequency response and Bode plots

Based on Chapter 14, slides 2-16.

Transfer functions

Frequency response describes how output magnitude and phase change with frequency. A transfer

function relates output to input under zero initial conditions. Evaluate H(s) at s = jω for sinusoidal

frequency response.

H(jω) = Y(jω)/X(jω)

Voltage gain: V_o/V_i Current gain: I_o/I_i

Transfer impedance: V_o/I_i Transfer admittance: I_o/V_i

Use Z_R = R, Z_L = jωL, and Z_C = 1/(jωC), then apply circuit analysis. For H(s) = N(s)/D(s), zeros are roots

of N and poles are roots of D after cancellation. Poles generally lie in the complex s-plane; they need not

correspond to infinite response at a real driving frequency.

Decibels and logarithmic frequency

Power ratio in dB = 10 log₁₀(P₂/P₁)

Magnitude response in dB = 20 log₁₀|H(jω)|

Use 20 log for a voltage/current amplitude ratio; interpreting it as a power ratio requires equal

corresponding resistances. 0 dB means a ratio of 1; +20 dB means amplitude ×10; -20 dB means

amplitude ×0.1. A decade is a tenfold frequency change.

Bode building blocks

A Bode plot has logarithmic frequency on the horizontal axis. One plot shows magnitude in dB; the other

shows phase. Factor H into standard factors, add their dB magnitudes, and add their phases.

Positive gain K: 20 log₁₀K dB, 0°

Zero at origin jω: +20 dB/decade, +90°

Pole at origin 1/(jω): -20 dB/decade, -90°

A negative real gain adds a 180° phase shift. For a simple zero 1 + jω/ω_z, the asymptotic slope changes

by +20 dB/decade at ω_z. For a simple pole 1/(1 + jω/ω_p), it changes by -20 dB/decade at ω_p.

Simple zero phase: +atan(ω/ω_z)

Simple pole phase: -atan(ω/ω_p)

Approximate each simple phase transition from 0.1 times its corner frequency to 10 times its corner

frequency. At the corner, phase is ±45°. Actual magnitude at a simple zero corner is +3.01 dB relative to

its low-frequency level; at a pole corner it is -3.01 dB.

A quadratic pole pair approaches -40 dB/decade and -180°; a zero pair has the opposite effects. For 1/[1 +

2ζ(jω/ω_n) + (jω/ω_n)²], damping ζ controls peaking. Use a quadrant-aware phase calculation.

Circuit Analysis II • Chapters 13–15 4CHAPTER 14 | ECE 231 Study Notes

Resonance and selectivity

Based on Chapter 14, slides 17-22. Formulas below assume the simple ideal series RLC or parallel RLC

topology described.

Resonance

Inductive and capacitive reactive effects cancel at resonance. Energy moves between the inductor’s

magnetic field and the capacitor’s electric field. Resistance dissipates energy and determines how sharp

the response is.

ω₀ = 1/√(LC) rad/s f₀ = 1/[2π√(LC)] Hz

Series RLC circuit

Z = R + j[ωL - 1/(ωC)]

At ω₀: Z = R, current is maximum, power factor = 1

Below resonance the series circuit is capacitive; above resonance it is inductive. At resonance, V_L and V_C

cancel as phasors, even though each can exceed the source voltage. For the simple series circuit, each

reactive voltage magnitude at resonance equals Q times the source magnitude.

Q = ω₀L/R = 1/(ω₀CR)

B = ω₂ - ω₁ = R/L = ω₀/Q (B in rad/s)

ω₁ = √[(R/2L)² + 1/(LC)] - R/(2L)

ω₂ = √[(R/2L)² + 1/(LC)] + R/(2L)

Parallel RLC circuit

Y = 1/R + j[ωC - 1/(ωL)]

At ω₀: Y = 1/R, impedance is maximum, Z = R

Q = ω₀RC = R/(ω₀L) B = 1/(RC) = ω₀/Q

ω₁,₂ = √[(1/2RC)² + 1/(LC)] ∓ 1/(2RC)

For a fixed source voltage, total input current is minimum at parallel resonance. For a fixed source current,

voltage and resistor power are maximum. Inductor and capacitor branch currents cancel in the input

current.

Quality factor and half-power points

Q = 2π × (maximum stored energy)/(energy lost per cycle)

ω₀ = √(ω₁ω₂) Bandwidth in Hz: BW = B/(2π)

High Q means a narrow bandwidth and greater selectivity. At half-power points, power is half its peak and

the relevant amplitude is 1/√2 ≈ 0.707 of its peak: a -3.01 dB change. Keep Hz and rad/s consistent.

Added example and slide clarification

For series R = 10 Ω, L = 0.1 H, C = 100 µF: ω₀ ≈ 316.23 rad/s, f₀ ≈ 50.33 Hz, Q ≈ 3.16, and B = 100 rad/s ≈

15.92 Hz.

Slide 19 says the transfer-function magnitude is minimum at resonance. That depends on the chosen

output: series impedance is minimum, but current and resistor voltage are maximum. Complex poles alone

also do not guarantee a visible magnitude peak; damping and output choice matter.

Circuit Analysis II • Chapters 13–15 5CHAPTER 14 | ECE 231 Study Notes

Passive and active filters

Based on Chapter 14, slides 23-33.

The four filter types

Lowpass passes DC and lower frequencies while attenuating high frequencies. Highpass attenuates DC and

low frequencies while passing higher frequencies. Bandpass passes an intermediate range. Bandstop, also

called bandreject or notch, attenuates an intermediate range.

First-order passive RC filters

Lowpass, output across C: H(jω) = 1/(1 + jωRC)

Highpass, output across R: H(jω) = jωRC/(1 + jωRC)

ω_c = 1/(RC) f_c = 1/(2πRC)

The lowpass approaches unity at DC and rolls off at -20 dB/decade at high frequency. Its phase goes from

0° toward -90°. The highpass rises at +20 dB/decade below cutoff and approaches unity at high frequency;

its phase goes from +90° toward 0°. At cutoff both have magnitude 0.707 of passband gain.

Series RLC filters

Bandpass, output across R:

H(jω) = R/[R + j(ωL - 1/(ωC))]

Bandstop, output across the series LC pair:

H(jω) = j(ωL - 1/(ωC))/[R + j(ωL - 1/(ωC))]

At resonance, resistor output is maximum and the combined LC output is zero. Bandpass passes between

the half-power frequencies; bandstop suppresses that intermediate band. Real filter edges are gradual,

rather than perfect pass/block boundaries.

Active filters using op amps

Active filters use an amplifier with resistors and capacitors. They can supply gain and buffering and avoid

bulky inductors at low frequencies. Their operation is limited by the op amp’s bandwidth, supply voltage,

and output capability. Passive networks cannot supply net power gain, although reactive voltage

magnification can occur.

Inverting active lowpass:

H(jω) = -(R_f/R_i)/(1 + jωR_fC_f)

ω_c = 1/(R_fC_f)

Inverting active highpass:

H(jω) = -(R_f/R_i)[jωR_iC_i/(1 + jωR_iC_i)]

ω_c = 1/(R_iC_i)

The lowpass uses R_f in parallel with C_f in the feedback path. The highpass uses R_i in series with C_i in

the input path. Slides 30-31 incorrectly call these RL combinations; their diagrams and equations show RC

combinations.

Bandpass can be built by cascading a highpass and lowpass with lower cutoff below upper cutoff.

Bandreject can combine lowpass and highpass paths through a summing amplifier, with a gap between

their passbands. Account for loading unless stages are buffered.

Added example

For R = 10 kΩ and C = 0.01 µF, RC = 10⁻⁴ s, so ω_c = 10,000 rad/s and f_c ≈ 1591.55 Hz. Always convert

µF to farads before multiplying.

Circuit Analysis II • Chapters 13–15 6CHAPTER 15 | ECE 231 Study Notes

Laplace transform essentials

Based on Chapter 15, slides 2-12. Functions in the transform tables are causal unless otherwise stated.

What the transform does

The unilateral Laplace transform converts differential and integral equations into algebraic equations. It

includes initial stored energy and gives both natural and forced response. In contrast, ordinary phasor

analysis focuses on sinusoidal steady state.

F(s) = ∫ from 0⁻ to ∞ of f(t)e^(-st) dt

s = σ + jω

The real part σ accounts for exponential growth/decay; jω represents oscillation. The 0⁻ lower limit

includes switching events at t = 0. Inverse transformation returns the time-domain function.

Core transform pairs

δ(t) ↔ 1 u(t) ↔ 1/s

e^(-at)u(t) ↔ 1/(s + a)

t u(t) ↔ 1/s² tⁿu(t) ↔ n!/s^(n+1)

tⁿe^(-at)u(t) ↔ n!/(s + a)^(n+1)

sin(ωt)u(t) ↔ ω/(s² + ω²)

cos(ωt)u(t) ↔ s/(s² + ω²)

e^(-at)sin(ωt)u(t) ↔ ω/[(s + a)² + ω²]

e^(-at)cos(ωt)u(t) ↔ (s + a)/[(s + a)² + ω²]

Linearity, scaling, and shifts

L{Af(t) + Bg(t)} = AF(s) + BG(s)

L{f(at)} = (1/a)F(s/a), a > 0

L{f(t - a)u(t - a)} = e^(-as)F(s), a ≥ 0

L{e^(-at)f(t)} = F(s + a)

A time delay multiplies the transform by e^(-as). Multiplication by an exponential shifts s. These are

different operations: do not confuse e^(-as), which depends on s, with e^(-at), which depends on time.

Phase-shifted sinusoids

L{sin(ωt + θ)u(t)} = [s sinθ + ω cosθ]/(s² + ω²)

L{cos(ωt + θ)u(t)} = [s cosθ - ω sinθ]/(s² + ω²)

The minus sign in the phase-shifted cosine numerator is important; the slide table prints a plus sign. Derive

these results by expanding the angle sum before transforming.

Added example

For f(t) = 3e^(-2t)u(t) + 4sin(5t)u(t), linearity gives F(s) = 3/(s + 2) + 20/(s² + 25).

Circuit Analysis II • Chapters 13–15 7CHAPTER 15 | ECE 231 Study Notes

Properties and inverse transforms

Based on Chapter 15, slides 6-20.

Differentiation and integration

L{f′(t)} = sF(s) - f(0⁻)

L{f″(t)} = s²F(s) - sf(0⁻) - f′(0⁻)

L{∫₀ᵗ f(τ)dτ} = F(s)/s

L{t f(t)} = -dF(s)/ds

L{f(t)/t} = ∫ from s to ∞ of F(q)dq, when convergent

For an nth derivative, subtract s^(n-1)f(0⁻), s^(n-2)f′(0⁻), and the remaining initial-derivative terms

through f^(n-1)(0⁻). Initial values are part of the transformed equation, not optional extra terms.

Periodic signals and value theorems

If period is T: F(s) = [∫₀ᵀ f(t)e^(-st)dt]/[1 - e^(-sT)]

f(0⁺) = lim as s→∞ of sF(s)

f(∞) = lim as s→0 of sF(s), if the final-value conditions hold

The initial-value theorem assumes an ordinary finite right-hand initial value, without an impulse at the

origin. The final-value theorem requires all poles of sF(s) to be in the open left half-plane. Do not apply it to

growing or sustained oscillatory responses.

Inverse transform procedure

1. If a rational function is improper, perform polynomial division. 2. Factor its denominator. 3. Use partial

fractions. 4. Match each term to a transform pair. 5. Include u(t) and any required delay.

Distinct poles: F(s) = Σ k_i/(s + p_i)

k_i = [(s + p_i)F(s)] evaluated at s = -p_i

f(t) = [Σ k_i e^(-p_i t)]u(t)

For a repeated pole, include every power: A₁/(s + p) + A₂/(s + p)² + ... + A_m/(s + p)^m. The inverse of

A_k/(s + p)^k is A_k t^(k-1)e^(-pt)/(k-1)! times u(t).

Complete the square: s² + as + b = (s + a/2)² + b - a²/4

For complex poles, set α = a/2 and β = √(b - a²/4), when b > a²/4. Rewrite the numerator using s + α to

identify damped cosine and sine terms. This corrects inconsistent coefficients in slide 19.

Added inverse examples

5/[(s + 1)(s + 2)] = 5/(s + 1) - 5/(s + 2), giving 5(e^(-t) - e^(-2t))u(t). Also, (s + 3)/(s² + 4s + 13) = (s +

2)/[(s + 2)² + 9] + 1/[(s + 2)² + 9], giving e^(-2t)[cos(3t) + (1/3)sin(3t)]u(t).

Circuit Analysis II • Chapters 13–15 8CHAPTER 15 | ECE 231 Study Notes

Convolution and circuit solutions

Based on Chapter 15, slides 21-29. The component equations and worked example below are added

applications of the slide properties.

Convolution and impulse response

For a linear time-invariant system with zero initial conditions, the impulse response h(t) determines the

output for any input x(t). Convolution combines the input with shifted copies of that impulse response.

y(t) = x(t) * h(t) = ∫ from -∞ to ∞ of x(λ)h(t - λ)dλ

For causal signals: y(t) = ∫₀ᵗ x(λ)h(t - λ)dλ

Y(s) = X(s)H(s)

The star means convolution, not ordinary pointwise multiplication. To evaluate it graphically: fold h(λ) to

h(-λ), shift to h(t - λ), multiply by x(λ), and integrate over their overlap. Break the answer into time

intervals when overlap changes.

Useful convolution properties

x h = h x

f (x + y) = f x + f * y

f (x y) = (f x) y

f δ = f f δ(t - t₀) = f(t - t₀)

f δ′ = f′ f u = ∫ from -∞ to t of f(λ)dλ

Use s-domain multiplication when transforms and partial fractions are simple. Time-domain convolution is

useful for piecewise waveforms or numerical data. Nonzero initial conditions add a separate zero-input

response.

Circuit equations including initial conditions

Resistor: V(s) = RI(s)

Inductor: V_L(s) = sLI_L(s) - Li_L(0⁻)

Capacitor: I_C(s) = sCV_C(s) - Cv_C(0⁻)

With zero initial energy, the component impedances are R, sL, and 1/(sC). With nonzero initial energy,

retain the initial-condition terms or equivalent sources. The usual continuity rules are i_L(0⁺) = i_L(0⁻) and

v_C(0⁺) = v_C(0⁻), provided no impulse voltage/current forces a jump.

Added example using an initial value

Solve y′ + 2y = 4u(t), with y(0⁻) = 1. Transform: sY - 1 + 2Y = 4/s. Thus Y = (s + 4)/[s(s + 2)] = 2/s - 1/(s +

2). The result for t ≥ 0 is y(t) = 2 - e^(-2t). Its initial value is 1 and its final value is 2, agreeing with the

equation.

Final study checklist

Chapter 13: dots, M and k, energy, turns ratios, and reflected impedance. Chapter 14: transfer functions,

Bode slopes, resonance, Q, bandwidth, and filter output locations. Chapter 15: transform pairs, initial

conditions, shifts, partial fractions, value-theorem conditions, and convolution.

Source set: the supplied Chapter 13 Power Points (30 slides), Chapter 14 Power Points (33 slides), and

Chapter 15 Power Points (29 slides), Fundamentals of Electric Circuits, McGraw-Hill. Notes paraphrase the

sources; added examples and technical clarifications are identified.

Circuit Analysis II • Chapters 13–15 9