2.2 Complex Numbers & Laplace Transforms

Complex numbers and phasors

Alternating-current (AC) analysis is complex-number arithmetic. A sinusoid v(t) = V_m cos(ωt + φ) is represented by the phasor V = V_m∠φ, a single complex number that drops the time dependence at a fixed frequency. NCEES phasors typically use the amplitude (peak) convention in the Handbook, though some problems use RMS magnitude; read the stem carefully.

A complex number has two interchangeable forms:

• Rectangular: z = a + jb (electrical engineering uses j, not i, to avoid clashing with current).
• Polar: z = r∠θ, where r = sqrt(a² + b²) and θ = atan2(b, a).

Euler's identity ties them together: e^(jθ) = cosθ + j sinθ, so z = r e^(jθ) = r(cosθ + j sinθ).

Worked example (rectangular to polar): Convert z = 3 + j4. The magnitude is r = sqrt(3² + 4²) = sqrt(25) = 5; the angle is θ = atan2(4, 3) = 53.13°. So z = 5∠53.13°. To divide (8∠60°)/(2∠20°), divide magnitudes and subtract angles: 4∠40°. To multiply, do the reverse: (3∠15°)(2∠25°) = 6∠40°. The biggest time sink is converting back and forth by hand, so drill the rectangular↔polar keys on your NCEES-approved calculator until they are automatic.

A few traps recur. The atan key on a calculator returns angles in only two quadrants, so a number like z = -1 - j1 with both parts negative lies in the third quadrant (angle -135° or 225°), not the +45° that a naive arctangent gives — always check the quadrant against the signs of a and b. The complex conjugate z* = a - jb flips the sign of the imaginary part (and the angle), and z·z* = a² + b² = r² is a fast way to get magnitude squared without a square root.

Complex impedance

In the phasor domain, Ohm's law becomes V = I Z, where impedance Z is complex and combines in series and parallel exactly like resistance:

• Resistor: Z_R = R (angle 0 degrees).
• Inductor: Z_L = jωL (impedance angle +90 degrees, so current lags voltage by 90 degrees).
• Capacitor: Z_C = 1/(jωC) = -j/(ωC) (angle -90 degrees, so current leads voltage by 90 degrees).

The mnemonic ELI the ICE man captures it: in an inductor (L) voltage E leads current I; in a capacitor (C) current I leads voltage E. At resonance in a series RLC circuit, Z_L and Z_C cancel and the impedance is purely resistive at ω₀ = 1/sqrt(LC).

Worked example (impedance): A 50 µF capacitor at ω = 377 rad/s (60 Hz) has Z_C = -j/(ωC) = -j/((377)(50×10⁻⁶)) = -j/0.01885 = -j53.1 Ω, i.e. 53.1∠-90° Ω. A 0.1 H inductor at the same frequency has Z_L = jωL = j(377)(0.1) = j37.7 Ω. Note ω = 2πf, so a stem that gives 60 Hz means ω = 2π(60) = 377 rad/s; forgetting the 2π factor is one of the most common AC errors. Series impedances add directly (Z = Z_R + Z_L + Z_C, complex addition in rectangular form), while parallel impedances combine by the reciprocal rule 1/Z = 1/Z₁ + 1/Z₂, easiest done by converting each term, summing, and inverting.

Laplace transforms

The Laplace transform F(s) = ∫ f(t) e^(-st) dt converts a time-domain differential equation into an algebraic equation in s, solves it, and inverts back. This is the backbone of linear systems and control systems questions. Two properties do most of the work:

• Derivative: L{f'(t)} = s F(s) - f(0).
• Integral: L{∫ f dt} = F(s)/s.

For zero initial conditions, differentiation becomes multiplication by s and integration becomes division by s, which is why a transfer function H(s) = Output(s)/Input(s) captures system behavior. Substituting s = jω gives the steady-state frequency response, connecting Laplace directly to Bode plots and phasors.

These pairs are in the FE Reference Handbook. The exam skill is recognizing which row matches the stem, not memorizing the integral definition. A common trap is forgetting the f(0) term in the derivative property when initial conditions are nonzero.

Two theorems speed up FE problems. The final value theorem gives the steady-state value without inverting: lim(t→∞) f(t) = lim(s→0) sF(s), valid only when the system is stable. The initial value theorem gives the t = 0⁺ value: lim(t→0) f(t) = lim(s→∞) sF(s). Worked example: For F(s) = 5/[s(s+2)], the final value is lim(s→0) s·5/[s(s+2)] = 5/2 = 2.5. To invert a transform with a denominator that factors, use partial fractions to split it into the simple table rows, then read each term back to the time domain — e.g. 5/[s(s+2)] = 2.5/s - 2.5/(s+2), giving f(t) = 2.5(1 - e^(-2t))u(t), the classic charging curve.