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2.1 Complex Impedance, Phasors, and AC Circuit Analysis

Key Takeaways

  • Impedance (ZZ) combines real resistance (RR) and frequency-dependent reactance (XX), expressed in rectangular form as Z=R+jXZ = R + jX or polar form as ∣Z∣∠θ|Z| \angle \theta.

  • Inductive reactance is represented as positive imaginary (+jXL+jX_L) where voltage leads current by 90∘90^\circ, while capacitive reactance is negative imaginary (−jXC-jX_C) where voltage lags current by 90∘90^\circ.

  • Phasor analysis models sinusoidal voltages and currents as rotating vectors in the complex plane, using vector addition to determine total circuit voltages, branch currents, and phase angles.

  • Parallel AC circuits are solved efficiently using admittance (Y=1/Z=G+jBY = 1/Z = G + jB), where conductance G=R/(R2+X2)G = R/(R^2 + X^2) and susceptance B=−X/(R2+X2)B = -X/(R^2 + X^2), inverting the sign of reactive components.

  • An antenna feedpoint impedance of 50+j25 Ω50 + j25\ \Omega has 50 Ω50\ \Omega total input resistance, including radiation and loss resistance, plus inductive reactance. Adding −j25 Ω-j25\ \Omega series capacitive reactance cancels that reactance and gives a 50 Ω50\ \Omega resistive load.

Last updated: November 2026

2.1 Complex Impedance, Phasors, and AC Circuit Analysis

In direct current (DC) circuits, opposition to current flow is governed purely by resistance (RR), where Ohm's law defines the relationship between potential difference and charge drift. In alternating current (AC) and radio frequency (RF) networks, current opposition becomes significantly more complex. Reactive components—inductors and capacitors—store and return energy to the circuit on alternating half-cycles, introducing a time-domain phase shift between current and voltage. To analyze such circuits rigorously, radio engineers employ complex impedance (ZZ), unifying dissipative resistance and energy-storing reactance into a single mathematical framework.


The Nature of Impedance and Reactance

Impedance (ZZ), measured in ohms (Ω\Omega), is the total opposition offered by an electrical circuit to the flow of alternating current at a specific frequency. Unlike simple resistance, which dissipates electrical energy irreversibly as heat, impedance encompasses two distinct components:

  1. Resistance (RR): The in-phase, dissipative component of impedance, independent of frequency in idealized conductors.
  2. Reactance (XX): The quadrature (90∘90^\circ out-of-phase), non-dissipative component of impedance, arising from energy stored alternately in magnetic fields (inductive reactance, XLX_L) or electric fields (capacitive reactance, XCX_C).

Reactance is frequency-dependent and defined by the fundamental AC relationships:

XL=2πfL=ωLX_L = 2\pi f L = \omega L XC=12πfC=1ωCX_C = \frac{1}{2\pi f C} = \frac{1}{\omega C}

where ff is frequency in hertz (Hz\text{Hz}), ω=2πf\omega = 2\pi f is angular frequency in radians per second (rad/s\text{rad/s}), LL is inductance in henrys (H\text{H}), and CC is capacitance in farads (F\text{F}).


Complex Mathematical Representation: Rectangular and Polar Forms

Because resistive and reactive voltages are 90∘90^\circ out of phase with one another, they cannot be summed algebraically. Instead, impedance is expressed as a complex quantity using the imaginary operator jj (where j=−1j = \sqrt{-1}, adopted in electrical engineering to avoid confusion with instantaneous current ii).

Rectangular Form

In rectangular notation, impedance is partitioned into real and imaginary terms:

Z=R+jXZ = R + jX
  • A pure inductor has ZL=+jXLZ_L=+jX_L; current lags voltage by 90∘90^\circ.
  • A pure capacitor has ZC=−jXCZ_C=-jX_C; current leads voltage by 90∘90^\circ.
  • A series RLC branch has Z=R+j(XL−XC)Z=R+j(X_L-X_C). With resistance present, its phase depends on the reactance-to-resistance ratio, rather than being fixed at 90∘90^\circ.

Polar Form

Impedance can also be expressed in polar coordinates, defining its total scalar magnitude (∣Z∣|Z|) and phase angle (θ\theta):

Z=∣Z∣∠θZ = |Z| \angle \theta

The magnitude represents the absolute ratio of peak (or RMS) voltage to peak (or RMS) current:

∣Z∣=R2+X2|Z| = \sqrt{R^2 + X^2}

The phase angle θ\theta represents the temporal phase displacement of the voltage waveform relative to the current waveform:

θ=arctan⁡(XR)\theta = \arctan\left(\frac{X}{R}\right)

To convert from polar form back to rectangular components:

R=∣Z∣cos⁡(θ)R = |Z| \cos(\theta) X=∣Z∣sin⁡(θ)X = |Z| \sin(\theta)

When θ>0∘\theta > 0^\circ, the circuit is net inductive and voltage leads current. When θ<0∘\theta < 0^\circ, the circuit is net capacitive and voltage lags current. When θ=0∘\theta = 0^\circ, the circuit is purely resistive (X=0X = 0).


Phasor Diagrams and Vector Analysis

A phasor is a complex vector representing a sinusoidal physical quantity whose length corresponds to the waveform's peak or RMS magnitude, and whose angular orientation relative to the horizontal positive real axis corresponds to its phase angle.

       +j (Imaginary Axis - Inductive Reactance / Voltage)
               ^
               |
         +jX_L |         * Z = |Z| ∠ θ = R + jX_L
         (+jV_L)        /|
               |       / |
               |      /  |
               |     /   | +jX (Net Inductive)
               |    /    |
               |   / θ   |
   -Real ------+----------> +Real (Resistance R / In-phase Voltage V_R)
               |   \ -θ  |
               |    \    |
               |     \   | -jX (Net Capacitive)
               |      \  |
         -jX_C |       \ |
         (-jV_C)        \|
               |         * Z = |Z| ∠ -θ = R - jX_C
               v
       -j (Imaginary Axis - Capacitive Reactance / Voltage)

In AC circuit analysis:

  • The reference vector is placed along the positive real axis (0∘0^\circ). In series circuits, current II is identical throughout every element and serves as the reference phasor (I=∣I∣∠0∘I = |I| \angle 0^\circ).
  • The voltage drop across the resistor (VR=IRV_R = I R) lies in phase with current along the real axis.
  • The voltage drop across the inductor (VL=IXLV_L = I X_L) leads current by +90∘+90^\circ and is plotted along the positive imaginary (+j+j) axis.
  • The voltage drop across the capacitor (VC=IXCV_C = I X_C) lags current by −90∘-90^\circ and is plotted along the negative imaginary (−j-j) axis.
  • Total applied generator voltage is the vector resultant:
Vtotal=VR+j(VL−VC)=VR2+(VL−VC)2∠arctan⁡(VL−VCVR)V_{\text{total}} = V_R + j(V_L - V_C) = \sqrt{V_R^2 + (V_L - V_C)^2} \angle \arctan\left(\frac{V_L - V_C}{V_R}\right)

Series AC Circuit Analysis: R-L, R-C, and R-L-C Networks

In a series circuit, impedances add directly in rectangular form:

Ztotal=Z1+Z2+Z3+⋯+ZnZ_{\text{total}} = Z_1 + Z_2 + Z_3 + \dots + Z_n

For a series R-L-C network:

Ztotal=R+jXL−jXC=R+j(XL−XC)Z_{\text{total}} = R + jX_L - jX_C = R + j(X_L - X_C)

Three operational regimes exist:

  1. XL>XCX_L > X_C (Inductive regime): Net reactance is positive (+jX+jX). Circuit voltage leads circuit current by phase angle θ=arctan⁡[(XL−XC)/R]\theta = \arctan[(X_L - X_C)/R].
  2. XC>XLX_C > X_L (Capacitive regime): Net reactance is negative (−jX-jX). Circuit voltage lags circuit current by phase angle θ=−arctan⁡[(XC−XL)/R]\theta = -\arctan[(X_C - X_L)/R].
  3. XL=XCX_L = X_C (Resonant condition): Net reactance cancels completely (X=0X = 0). Impedance is purely resistive (Z=R∠0∘Z = R \angle 0^\circ), current reaches its maximum possible value, and voltage is in phase with current.

Step-by-Step Worked Example: Series R-L-C Network at HF

An amateur radio transmitter output filter section contains a 40 Ω40\ \Omega non-inductive resistor, a 1.50 μH1.50\ \mu\text{H} inductor, and a 600 pF600\text{ pF} mica capacitor connected in series across an RF source operating at f=7.10 MHzf = 7.10\text{ MHz} with an RMS terminal voltage of 50.0 V50.0\text{ V}.

Step 1: Calculate Inductive Reactance (XLX_L):

XL=2πfL=2π×(7.10×106 Hz)×(1.50×10−6 H)≈66.92 ΩX_L = 2\pi f L = 2\pi \times (7.10 \times 10^6\text{ Hz}) \times (1.50 \times 10^{-6}\text{ H}) \approx 66.92\ \Omega

Step 2: Calculate Capacitive Reactance (XCX_C):

XC=12πfC=12π×(7.10×106 Hz)×(600×10−12 F)≈37.36 ΩX_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times (7.10 \times 10^6\text{ Hz}) \times (600 \times 10^{-12}\text{ F})} \approx 37.36\ \Omega

Step 3: Determine Net Reactance (XX):

X=XL−XC=66.92 Ω−37.36 Ω=+29.56 Ω(net inductive)X = X_L - X_C = 66.92\ \Omega - 37.36\ \Omega = +29.56\ \Omega \quad (\text{net inductive})

Step 4: Formulate Rectangular and Polar Impedance:

Z=40+j29.56 ΩZ = 40 + j29.56\ \Omega ∣Z∣=R2+X2=402+29.562=1600+873.79=2473.79≈49.74 Ω|Z| = \sqrt{R^2 + X^2} = \sqrt{40^2 + 29.56^2} = \sqrt{1600 + 873.79} = \sqrt{2473.79} \approx 49.74\ \Omega θ=arctan⁡(29.5640)=arctan⁡(0.7390)≈36.46∘\theta = \arctan\left(\frac{29.56}{40}\right) = \arctan(0.7390) \approx 36.46^\circ

The impedance in polar notation is Z≈49.74 Ω∠36.46∘Z \approx 49.74\ \Omega \angle 36.46^\circ.

Step 5: Determine Circuit Current and Individual Component Voltages:

I=V∣Z∣=50.0 V49.74 Ω≈1.005 A RMSI = \frac{V}{|Z|} = \frac{50.0\text{ V}}{49.74\ \Omega} \approx 1.005\text{ A RMS}

The voltage drops across each element are:

  • VR=I×R=1.005 A×40 Ω=40.20 VV_R = I \times R = 1.005\text{ A} \times 40\ \Omega = 40.20\text{ V}
  • VL=I×XL=1.005 A×66.92 Ω=67.25 VV_L = I \times X_L = 1.005\text{ A} \times 66.92\ \Omega = 67.25\text{ V}
  • VC=I×XC=1.005 A×37.36 Ω=37.55 VV_C = I \times X_C = 1.005\text{ A} \times 37.36\ \Omega = 37.55\text{ V}

Verifying vector addition of voltages:

Vtotal=VR2+(VL−VC)2=40.202+(67.25−37.55)2=1616.04+882.09≈50.0 VV_{\text{total}} = \sqrt{V_R^2 + (V_L - V_C)^2} = \sqrt{40.20^2 + (67.25 - 37.55)^2} = \sqrt{1616.04 + 882.09} \approx 50.0\text{ V}

Parallel AC Circuits, Admittance, Conductance, and Susceptance

Analyzing parallel AC circuits using impedance requires reciprocal summation (1/Ztotal=1/Z1+1/Z2+…1/Z_{\text{total}} = 1/Z_1 + 1/Z_2 + \dots), which becomes unwieldy when working with complex numbers. To streamline calculations, electrical engineers define Admittance (YY), the reciprocal of impedance:

Y=1Z=G+jBY = \frac{1}{Z} = G + jB

Admittance is measured in siemens (S\text{S}, historically mhos, ℧\mho). In rectangular form:

  • Conductance (GG): The real part of admittance, representing the ease with which in-phase current flows.
  • Susceptance (BB): The imaginary part of admittance, representing the ease with which alternating reactive current flows.

Converting Between Impedance and Admittance

Given an arbitrary branch impedance Z=R+jXZ = R + jX, its admittance is:

Y=1R+jX=R−jX(R+jX)(R−jX)=R−jXR2+X2=RR2+X2−jXR2+X2Y = \frac{1}{R + jX} = \frac{R - jX}{(R + jX)(R - jX)} = \frac{R - jX}{R^2 + X^2} = \frac{R}{R^2 + X^2} - j\frac{X}{R^2 + X^2}

Equating real and imaginary parts:

G=RR2+X2G = \frac{R}{R^2 + X^2} B=−XR2+X2B = \frac{-X}{R^2 + X^2}

Important

Note the crucial sign inversion in susceptance:

  • A pure inductor has B=−1/XLB=-1/X_L and Y=−j/XL=−jBLY=-j/X_L=-jB_L, where BL=1/XLB_L=1/X_L is a positive magnitude.
  • A pure capacitor has B=+1/XCB=+1/X_C and Y=+j/XC=+jBCY=+j/X_C=+jB_C, where BC=1/XCB_C=1/X_C is a positive magnitude. In parallel circuits, total admittance is obtained by simple algebraic addition of branch conductances and susceptances: Ytotal=∑G+j∑BY_{\text{total}} = \sum G + j \sum B.

Parallel-to-Series Equivalent Transformations

In RF filter design and impedance matching networks (such as Antenna Tuning Units and Pi-networks), engineers frequently transform a physical series combination (Rs+jXsR_s + jX_s) into an equivalent parallel combination (Rp∥jXpR_p \parallel jX_p), or vice versa, at a single specified frequency.

Defining the circuit quality factor QQ:

Q=∣Xs∣Rs=Rp∣Xp∣Q = \frac{|X_s|}{R_s} = \frac{R_p}{|X_p|}

The exact conversion relationships are:

Rp=Rs(1+Q2)R_p = R_s \left(1 + Q^2\right) Xp=Xs(1+Q2Q2)=Xs(1+1Q2)X_p = X_s \left(\frac{1 + Q^2}{Q^2}\right) = X_s \left(1 + \frac{1}{Q^2}\right)

Conversely, converting from parallel to series:

Rs=Rp1+Q2R_s = \frac{R_p}{1 + Q^2} Xs=Xp(Q21+Q2)X_s = X_p \left(\frac{Q^2}{1 + Q^2}\right)

When Q≥10Q \ge 10 (a standard high-Q circuit approximation):

  • Rp≈Q2RsR_p \approx Q^2 R_s
  • Xp≈XsX_p \approx X_s

This approximation proves invaluable when determining the loaded impedance presented by tuned LC tanks to RF power amplifier tubes or transistors.


Practical RF Engineering: Complex Antenna Feedpoint Impedance

In amateur radio transmission systems, the load presented to a coaxial transmission line by an antenna is rarely a pure 50 Ω50\ \Omega resistance across an entire operating band. The antenna feedpoint exhibits complex impedance:

Zant=Rant+jXant=(Rr+Rloss)+jXantZ_{\text{ant}} = R_{\text{ant}} + jX_{\text{ant}} = (R_r + R_{\text{loss}}) + jX_{\text{ant}}

where:

  • RrR_r is radiation resistance, representing the equivalent resistance that accounts for power radiated into space as electromagnetic waves.
  • RlossR_{\text{loss}} represents ohmic heat loss in wire conductors, dielectric losses in antenna insulators, and earth ground resistance.
  • XantX_{\text{ant}} is the residual reactive component.
Measured Feedpoint ImpedanceElectrical CharacteristicPhysical Meaning on a DipoleCorrective Tuning Action
Z=50+j0 ΩZ = 50 + j0\ \OmegaPurely resistive, matchedPerfectly resonant, correct physical lengthNone; system is matched to 50 Ω50\ \Omega line
Z=52+j30 ΩZ = 52 + j30\ \OmegaInductive (+j+j)Antenna is cut physically too longPrune wire length, or insert series capacitance (XC=30 ΩX_C = 30\ \Omega)
Z=46−j25 ΩZ = 46 - j25\ \OmegaCapacitive (−j-j)Antenna is cut physically too shortLengthen wire, or insert series loading inductance (XL=25 ΩX_L = 25\ \Omega)
Z=15+j0 ΩZ = 15 + j0\ \OmegaPurely resistive, low RRShort loaded vertical or mobile whipAdd impedance matching transformer/Unun to step 15 Ω15\ \Omega up to 50 Ω50\ \Omega

When an antenna analyser reads Z=50+j25 ΩZ = 50 + j25\ \Omega, the real part (50 Ω50\ \Omega) matches standard coaxial characteristic impedance (Z0=50 ΩZ_0 = 50\ \Omega), but the +j25 Ω+j25\ \Omega inductive component causes standing waves on the line, producing an SWR of approximately 1.6:11.6:1. Inserting a series capacitor having XC=25 ΩX_C = 25\ \Omega (where C=1/(2πf×25)C = 1 / (2\pi f \times 25)) adds −j25 Ω-j25\ \Omega, yielding Z=50+j25−j25=50+j0 ΩZ = 50 + j25 - j25 = 50 + j0\ \Omega. This cancellation establishes a conjugate match, eliminating reflected power at the feedline junction.


Comparison Table: Series vs. Parallel AC Circuit Formulations

Parameter / LawSeries AC ConfigurationParallel AC Configuration
Common ReferenceCurrent II is uniform through all elementsVoltage VV is uniform across all branches
Primary VariableImpedance Z=R+jXZ = R + jX (Ohms, Ω\Omega)Admittance Y=G+jBY = G + jB (Siemens, S\text{S})
Real ComponentResistance R=∣Z∣cos⁡(θ)R = \lvert Z\rvert \cos(\theta)Conductance G=R/(R2+X2)=∣Y∣cos⁡(θ)G = R / (R^2 + X^2) = \lvert Y\rvert \cos(\theta)
Imaginary ComponentReactance X=XL−XC=∣Z∣sin⁡(θ)X = X_L - X_C = \lvert Z\rvert \sin(\theta)Susceptance B=BC−BL=−X/(R2+X2)B = B_C - B_L = -X / (R^2 + X^2)
Inductive SignInductive reactance is positive (+jXL+jX_L)Inductive susceptance is negative (−jBL-jB_L)
Capacitive SignCapacitive reactance is negative (−jXC-jX_C)Capacitive susceptance is positive (+jBC+jB_C)
Magnitude∣Z∣=R2+X2\lvert Z\rvert = \sqrt{R^2 + X^2}∣Y∣=G2+B2=1/∣Ztotal∣\lvert Y\rvert = \sqrt{G^2 + B^2} = 1 / \lvert Z_{\text{total}}\rvert
Phase Angle θ\thetaθ=arctan⁡(X/R)\theta = \arctan(X / R)θ=−arctan⁡(B/G)\theta = -\arctan(B / G)
Total CombinationZtotal=Z1+Z2+⋯+ZnZ_{\text{total}} = Z_1 + Z_2 + \dots + Z_nYtotal=Y1+Y2+⋯+YnY_{\text{total}} = Y_1 + Y_2 + \dots + Y_n
Test Your Knowledge

An alternating current circuit operating at 14.1 MHz consists of a 30 Ω non-inductive resistor, an inductor with inductive reactance of 70 Ω, and a capacitor with capacitive reactance of 30 Ω connected in series. What is the total complex impedance in polar form, and what is the phase relationship between the generator voltage and circuit current?

A

50 Ω ∠ 36.9°, with generator voltage lagging the circuit current

B

50 Ω ∠ 53.1°, with generator voltage leading the circuit current

C

70 Ω ∠ 45.0°, with generator voltage in phase with circuit current

D

30 Ω ∠ 90.0°, with generator voltage leading the circuit current

Test Your Knowledge

When converting a branch with an impedance of Z = 20 + j40 Ω into its equivalent admittance Y = G + jB, what are the resulting conductance (G) and susceptance (B) values?

A

G = 0.050 S and B = +0.025 S

B

G = 0.020 S and B = +0.040 S

C

G = 0.010 S and B = -0.020 S

D

G = 0.040 S and B = -0.010 S

Test Your Knowledge

An antenna analyser connected to a high-frequency dipole feedpoint measures a complex impedance of Z = 50 + j35 Ω at the target operating frequency. What physical condition does this reading indicate, and how can the feedpoint be brought to a purely resistive 50 Ω resonance?

A

The dipole is slightly too long, creating an inductive reactive component that can be cancelled by inserting a series capacitor with X_C = 35 Ω

B

The dipole is slightly too short, creating a capacitive reactive component that can be cancelled by inserting a series inductor with X_L = 35 Ω

C

The transmission line has a short circuit, creating pure radiation loss that requires shortening the dipole legs by half

D

The dipole radiation resistance is too high, requiring an auto-transformer to step down the real part to 25 Ω

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