ECE 211 · Lecture 12: Sinusoids, Phasors, and Complex Impedance

§1 · The sinusoidal steady state

The AC steady state is a sinusoid at the source frequency

A switched DC circuit produces a transient that decays and a steady state that survives. An AC circuit behaves the same way, except that the surviving part is itself a sinusoid at the source frequency. That part is the sinusoidal steady state, and it is the subject of this lecture; the transient is set aside.

DC switched circuit → transient + steady state AC circuit → transient + steady state the AC steady state is a sinusoid at the source frequency

Only two numbers per signal are therefore unknown: amplitude and phase. A method that carries exactly those two, and discards the rest, is sufficient. That method is the phasor.

What the phasor method replaces

A series RL circuit driven by $v_s = V_m\cos\omega t$ obeys

L\frac{di}{dt} + R\,i = V_m\cos\omega t

Solving this directly requires a trial solution $i = A\cos\omega t + B\sin\omega t$ , substitution, matching of the cosine and sine coefficients, and a two-equation solve for $A$ and $B$ , followed by recombination into a single cosine. The phasor method reduces the same problem to

\tilde{I} = \frac{\tilde{V}_s}{R + j\omega L}

which is one division of complex numbers. The saving grows with circuit size, because the differential equation grows in order while the phasor equation remains algebraic.

The three parameters of a sinusoid

The general sinusoid is written in cosine form, $v(t) = V_m \cos(\omega t + \theta)$ , with amplitude $V_m$ , angular frequency $\omega = 2\pi f$ in rad/s, and phase angle $\theta$ . The period follows as $T = 2\pi/\omega = 1/f$ . The first three sliders set the waveform; the fourth scrubs the observation time and is not a property of it.

The waveform in the time domain. The teal marker rides the curve at the scrubbed instant.

The same signal as a vector of length V_m turning at ω. Its shadow on the real axis is v(t).

V_m (V) 5

f (Hz) 100

θ (degrees) 30

observation time t (ms) 2

v(t) = 5.0 cos(628t + 30°) V T = 10.00 ms ωt + θ = 102.0° v(t) = −1.04 V

The right-hand figure states the whole idea. A vector of fixed length $V_m$ , started at angle $\theta$ , turns counterclockwise at $\omega$ rad/s, and $v(t)$ is its projection on the real axis. The turning is identical for every signal in a circuit driven at one frequency, so it can be factored out. What remains is the starting vector.

Check yourself

Raising $f$ from 100 Hz to 200 Hz halves the period. Does it change the amplitude, the phase angle, or the value of $v$ at a fixed instant?

Answer. Neither the amplitude nor the phase angle changes, because both are set by their own sliders. The value at a fixed instant does change, since the total angle $\omega t + \theta$ has doubled its rate of advance. Frequency and phase are independent parameters, which is exactly why the phasor can carry the phase while $\omega$ is recorded separately.

§2 · Complex numbers, in three forms

Rectangular, polar, and exponential

A phasor is a complex number, so the three forms and their conversions are required equipment. The imaginary unit is written $j$ , because $i$ is reserved for current: $j = \sqrt{-1}$ .

rectangular: x + jy polar: r∠θ exponential: r e^jθ e^jθ = cos θ + j sin θ

r = \sqrt{x^2 + y^2} \qquad \theta = \tan^{-1}\!\left[\frac{y}{x}\right] \qquad x = r\cos\theta \qquad y = r\sin\theta

Euler's identity ties the three forms into one statement, on which the entire phasor method rests:

x + jy = r e^{j\theta} = r\cos\theta + j\,r\sin\theta \qquad\Longrightarrow\qquad \boxed{\,e^{j\theta} = \cos\theta + j\sin\theta\,}

The point is set in rectangular form by the sliders; the polar form is read back from the triangle. The angle is measured counterclockwise from the positive real axis.

x (real part) 3

y (imaginary part) 4

3.0 + j4.0 = 5.00∠53.1° check: r cos θ = 3.00, r sin θ = 4.00

The arctangent needs the quadrant

The expression $\theta = \tan^{-1}(y/x)$ is correct only for $x > 0$ , because $\tan^{-1}$ returns an angle in $(-90°, 90°)$ and cannot distinguish $-3 + j4$ from $3 - j4$ . For $x < 0$ , $180°$ must be added or subtracted. Sketching the point before converting is the reliable habit.

Arithmetic: which form suits which operation

Addition and subtraction are performed in rectangular form, because real and imaginary parts add separately. Multiplication and division are performed in polar form, because magnitudes multiply and angles add:

r_1 e^{j\theta_1} \cdot r_2 e^{j\theta_2} = r_1 r_2\, e^{j(\theta_1 + \theta_2)} \qquad \frac{r_1 e^{j\theta_1}}{r_2 e^{j\theta_2}} = \frac{r_1}{r_2}\, e^{j(\theta_1 - \theta_2)}

A network reduction therefore alternates between the forms: series impedances are added in rectangular form, while the product over sum of a parallel pair is evaluated in polar form for the product and in rectangular form for the sum.

Check yourself

Express $5\angle 143.13°$ in rectangular form.

Answer. $x = 5\cos 143.13° = -4$ and $y = 5\sin 143.13° = 3$ , so the number is $-4 + j3$ . Converting back with $\tan^{-1}(3/{-4})$ returns $-36.87°$ on a calculator; the point lies in the second quadrant, so $180°$ must be added.

§3 · The phasor transform

A sinusoid becomes a complex number

The transform is obtained by writing the cosine as the real part of a complex exponential, then separating the time-dependent factor from the constant one:

v(t) = V_m\cos(\omega t + \theta) = \mathrm{Re}\left\{V_m e^{j(\omega t + \theta)}\right\} = \mathrm{Re}\left\{\underbrace{V_m e^{j\theta}}_{\textstyle \tilde{V}}\, e^{j\omega t}\right\}

The factor $e^{j\omega t}$ is common to every signal in a circuit driven at one frequency and is therefore suppressed. What survives is the complex constant $\tilde{V} = V_m e^{j\theta}$ , the phasor.

time domain: v(t) = V_m cos(ωt + θ) phasor domain: Ṽ = V_m∠θ = V_m e^jθ

The cosine reference is part of the definition

A phasor angle is meaningful only against a stated reference, and this course uses the cosine reference. A source given as a sine must be converted first, using $\sin x = \cos(x - 90°)$ : the phasor of $4\sin(\omega t + 20°) = 4\cos(\omega t - 70°)$ is $4\angle -70°$ , not $4\angle 20°$ . The frequency is never carried in the phasor and must be recorded separately.

Degrees and radians appear in the same expression

An answer such as $0.0745\cos(10t + 36.30°)$ mixes units deliberately: $10t$ is in radians and $36.30°$ is in degrees. The convention is universal in circuit analysis, because phasor angles are read in degrees while $\omega t$ comes from a rate in rad/s. Before evaluating such an expression numerically, convert one of the two so that both terms share a unit.

Differentiation becomes multiplication by $j\omega$

The derivative rule is the result that makes the transform worth performing. Differentiating a cosine shifts it by $+90°$ and scales it by $\omega$ , and a $+90°$ shift is multiplication by $e^{j90°} = j$ :

\frac{d}{dt}\Big[V_m\cos(\omega t + \theta)\Big] = -\omega V_m \sin(\omega t + \theta) = \omega V_m \cos(\omega t + \theta + 90°)

\frac{dv(t)}{dt} = \mathrm{Re}\left\{V_m e^{j\theta}\cdot \omega e^{j90°} \cdot e^{j\omega t}\right\} = \mathrm{Re}\left\{\tilde{V}\cdot j\omega\cdot e^{j\omega t}\right\}

The differential equations of a circuit therefore become algebraic equations.

\frac{dv(t)}{dt} \;\Longrightarrow\; j\omega\,\tilde{V}

Differentiation multiplies the magnitude by $\omega$ and advances the phase by $90°$ . The inductor, whose voltage is $L\,di/dt$ , inherits this rule directly.

\int v(t)\,dt \;\Longrightarrow\; \frac{\tilde{V}}{j\omega}

Integration divides the magnitude by $\omega$ and retards the phase by $90°$ . The capacitor, whose voltage is the integral of its current, inherits this rule.

v(t) \;\Longrightarrow\; \tilde{V}

The signal passes through unchanged. Kirchhoff's laws, which contain only sums of signals, therefore hold in the phasor domain in the form they hold in the time domain.

Check yourself

The current through a 5 mH inductor is $i(t) = 2\cos(1000t - 40°)$ A. Find the voltage across it.

Answer. $\tilde{I} = 2\angle -40°$ A and $\tilde{V} = j\omega L\,\tilde{I} = (j5)(2\angle-40°) = 10\angle 50°$ V, so $v(t) = 10\cos(1000t + 50°)$ V. The voltage leads the current by $90°$ , as $j$ requires.

§4 · Complex impedance

Ohm's law, restored

Impedance is defined as the ratio of the voltage phasor to the current phasor at a pair of terminals, which restores Ohm's law to its familiar shape:

Z = \frac{\text{phasor } V}{\text{phasor } I} = \frac{\tilde{V}}{\tilde{I}} \qquad\Longrightarrow\qquad \boxed{\;\tilde{V} = Z\,\tilde{I}\;}

Each element is converted by applying the transform rules of Section 3 to its defining relation.

v_R = R\,i_R \;\Longrightarrow\; \tilde{V}_R = R\,\tilde{I}_R \;\Longrightarrow\; \boxed{Z_R = R}

The resistor introduces no phase shift, so its impedance is real and independent of frequency.

v_L = L\frac{di_L}{dt} \;\Longrightarrow\; \tilde{V}_L = L\,(j\omega)\,\tilde{I}_L \;\Longrightarrow\; \boxed{Z_L = j\omega L}

The inductor voltage leads its current by $90°$ . Its impedance grows with frequency: an inductor is a short at DC and an open at very high frequency.

i_C = C\frac{dv_C}{dt} \;\Longrightarrow\; \tilde{I}_C = C\,(j\omega)\,\tilde{V}_C \;\Longrightarrow\; \boxed{Z_C = \frac{1}{j\omega C} = \frac{-j}{\omega C}}

The capacitor voltage lags its current by $90°$ . Its impedance falls with frequency: a capacitor is an open at DC and a short at very high frequency.

The three elements with their impedances evaluated at the selected ω.

Impedance magnitude against ω on logarithmic axes. The resistor is flat, the inductor rises, and the capacitor falls.

ω (rad/s) 1000

R = 50 Ω |Z_L| = 20.00 Ω |Z_C| = 100.00 Ω

ω = 1000 rad/s, f = 159.2 Hz the series R, L, C combination is capacitive

Resistance and reactance

Any impedance separates into a real part and an imaginary part, and the sign of the imaginary part names the character of the load:

Z = R + jX \;\ \Omega \qquad\text{impedance} = \text{resistance} + j\,\text{reactance}

X > 0 ⇒ inductive X < 0 ⇒ capacitive X = 0 ⇒ resistive

The procedure, in four steps

Because Kirchhoff's laws survive the transform, so does every technique built on them. Nodal analysis, mesh analysis, superposition, source transformation, Thevenin and Norton equivalents, and ideal operational amplifiers all apply unchanged, with resistances replaced by impedances and real arithmetic replaced by complex arithmetic.

1 · read ω from the source 2 · convert every element to an impedance 3 · solve by complex algebra 4 · convert the answer back to a cosine at that ω

This strip is repeated at the head of each worked example, because the four steps never change.

§5 · Warm-up: one loop, one impedance

The four steps on the smallest possible circuit

1 · ω 2 · impedances 3 · complex algebra 4 · back to a cosine

The problem

Find $i(t)$ for the source $v_s(t) = 12\cos 500t$ V driving 8 Ω in series with 12 mH.

Step 1 and 2: the frequency, then the impedances

$\omega = 500$ rad/s is read from the source argument. Only the inductor changes:

Z_L = j\omega L = j(500)(0.012) = j6\ \Omega

Step 3: series impedances add

Z = \tm{wu.r8}{8} + \tm{wu.l12}{j6} = 10\angle 36.87°\ \Omega

The reactance is positive, so the branch is inductive and the current is expected to lag.

Step 4: Ohm's law, then back to the time domain

\tilde{I} = \frac{\tilde{V}_s}{Z} = \frac{12\angle 0°}{10\angle 36.87°} = 1.2\angle -36.87°\ \text{A} \;\Longrightarrow\; \boxed{i(t) = 1.2\cos(500t - 36.87°)\ \text{A}}

The current lags by $36.87°$ , as predicted. Note that $\omega = 500$ rad/s was never carried by a phasor; it was restored from step 1.

Check yourself

What happens to $|\tilde{I}|$ if the frequency is doubled to 1000 rad/s?

Answer. The inductive reactance doubles to $j12$ Ω, so $Z = 8 + j12 = 14.42\angle 56.31°$ Ω and $|\tilde{I}| = 12/14.42 = 0.832$ A. The current falls and lags further, because the inductor opposes change more strongly at higher frequency.

§6 · Example A: the impedance looking in at a port

Find $Z_{ab}$ and the source current

1 · ω 2 · impedances 3 · complex algebra 4 · back to a cosine

The problem

The source is $v_s(t) = 6\cos(10t - 30°)$ V, so $\omega = 10$ rad/s and the phasor is $\tilde{V}_s = 6\angle-30°$ V. Find the complex impedance $Z_{ab}$ looking in at a and b, then the source current $i_s(t)$ .

Convert every element to an impedance at ω = 10 rad/s

Z_C = \frac{-j}{\omega C}:\quad \tm{xa.c1}{\frac{-j}{10(0.002)} = -j50\ \Omega} \qquad \tm{xa.c2}{\frac{-j}{10(0.004)} = -j25\ \Omega}

Z_L = j\omega L:\quad \tm{xa.l1}{j(10)(2) = j20\ \Omega}

The 20 Ω and 50 Ω resistors are unchanged.

Combine the series branch on the right

The inductor and the 50 Ω resistor carry the same current, so their impedances add:

Z_1 = 50 + j20\ \Omega

The parallel pair, by product over sum

Z_2 = \tm{xa.ser}{(50 + j20)} \parallel \tm{xa.c2}{(-j25)} = \frac{(-j25)(50 + j20)}{50 + j20 - j25} = \frac{500 - j1250}{50 - j5}

Multiplying numerator and denominator by the conjugate $50 + j5$ gives $(31250 - j60000)/2525$ , so $Z_2 = 12.38 - j23.76$ Ω.

Add the series input branch

Z_{ab} = Z_2 + \tm{xa.r20}{20} + \tm{xa.c1}{(-j50)} = 12.38 - j23.76 + 20 - j50

\boxed{Z_{ab} = 32.38 - j73.76\ \Omega = 80.56\angle -66.30°\ \Omega}

The reactance is negative, so the port presents a capacitive load and the current leads the voltage.

Ohm's law, then back to the time domain

\tilde{I}_s = \frac{\tilde{V}_s}{Z_{ab}} = \frac{6\angle -30°}{80.56\angle -66.30°} = 0.0745\angle 36.30°\ \text{A}

\boxed{i_s(t) = 0.0745\cos(10t + 36.30°)\ \text{A}}

The answer as a phasor diagram

A phasor diagram states the relationship between two phasors at a glance, and it is the fastest check on any AC answer. The source voltage and the source current are drawn from a common origin, with the angle between them read directly.

The source voltage sits at $-30°$ and the source current at $+36.30°$ , so the current leads by $66.30°$ , which equals the negative of the port angle. Voltage and current are drawn to separate scales, as is standard, because their magnitudes differ by three orders.

Check: does the answer pass inspection?

Three tests are available without redoing the algebra. First, the port angle is $-66.30°$ , so the current must lead by $66.30°$ , and it does. Second, the input branch alone contributes $|20 - j50| = 53.9$ Ω, so a port magnitude near 80 Ω is the right order. Third, a current of 74.5 mA against a 6 V source is consistent with an 80 Ω port. An answer that fails any of the three should be recomputed rather than reported.

§7 · Example B: two sources at two frequencies

Superposition across the frequency boundary

An AC source and a DC source drive the same node. The phasor method handles one frequency at a time, so the two contributions are computed in separate circuits and added in the time domain.

Phasors at different frequencies cannot be added

A phasor is defined relative to a specific $e^{j\omega t}$ . Two phasors belonging to different frequencies describe vectors turning at different rates, so their sum has no fixed length or angle. Superposition remains valid, but the summation must be performed on the time functions.

The problem

Find $v(t)$ at the marked node. An AC source $5\cos 50t$ V and a DC source of $\tfrac12$ A act together, at $\omega = 50$ rad/s and $\omega = 0$ respectively, so two separate circuits are required.

Impedances at ω = 50 rad/s

\tm{xb.ind}{Z_L = j(50)\left(\tfrac{1}{10}\right) = j5\ \Omega} \qquad \tm{xb.cap}{Z_C = \frac{-j}{50 \cdot \frac{1}{100}} = -j2\ \Omega}

These values apply to the AC circuit only.

Circuit 1: the AC source acting alone

The DC current source is turned off, which makes it an open circuit. Two branches remain at the node: the series $10 + j5$ path to the source and the series $5 - j2$ path to ground.

KCL at the node solves the AC part

\tm{xb.r10 xb.ind}{\frac{\tilde{V} - 5\angle 0°}{10 + j5}} + \tm{xb.r5 xb.cap}{\frac{\tilde{V} - 0}{5 - j2}} = 0

Collecting terms gives $\tilde{V}(0.2524 + j0.0290) = 0.4 - j0.2$ , hence

\boxed{\tilde{V}\big|_{\omega = 50} = 1.76\angle -33.11°\ \text{V}}

Circuit 2: the DC source acting alone

The AC voltage source is turned off, which makes it a short circuit. At $\omega = 0$ the reactive elements degenerate: the inductor becomes a short and the capacitor an open.

The DC part is a one-line calculation

With the capacitor open, the 5 Ω branch carries nothing, so only the 10 Ω path to ground remains:

\tm{xb.r10}{\frac{\tilde{V} - 0}{10}} - \tm{xb.idc}{\frac12} = 0 \;\Longrightarrow\; \boxed{\tilde{V}\big|_{\omega = 0} = 5\ \text{V}}

Superpose the time functions

\boxed{v(t) = \underbrace{1.76\cos(50t - 33.11°)}_{\text{AC part}} + \underbrace{5}_{\text{DC part}}\ \ \text{V}}

The result is a sinusoid riding on a 5 V offset. It is not a single sinusoid, and it has no single phasor.

§8 · Your turn

Practice: a series RLC branch

The circuit exercises the four steps once more, and adds one observation that surprises most students on first encounter.

The source $v_s(t) = 20\cos(1000t + 30°)$ V drives 40 Ω, 30 mH, and 25 μF in series. Find (a) the three impedances at this frequency, (b) $Z$ and whether the branch is inductive or capacitive, (c) $i(t)$ , and (d) the capacitor voltage $v_C(t)$ .

Hint 1: the two conversions

$Z_L = j\omega L$ with $\omega = 1000$ rad/s and $L = 0.03$ H. $Z_C = -j/(\omega C)$ with $C = 25 \times 10^{-6}$ F. Note that the reactances partly cancel, since one is positive and the other negative.

Hint 2: the capacitor voltage

Once $\tilde{I}$ is known, the capacitor voltage follows from Ohm's law applied to that element alone: $\tilde{V}_C = \tilde{I}\,Z_C$ . Multiplying by $-j40 = 40\angle -90°$ scales the magnitude by 40 and subtracts $90°$ from the angle.

Solution

(a) $Z_R = 40$ Ω, $Z_L = j(1000)(0.03) = j30$ Ω, and $Z_C = -j/(1000 \cdot 25\times10^{-6}) = -j40$ Ω.

(b) Series impedances add:

Z = 40 + j30 - j40 = 40 - j10 = 41.23\angle -14.04°\ \Omega

The net reactance is negative, so the branch is capacitive: the capacitor outweighs the inductor at this frequency.

(c) Ohm's law and the return to the time domain:

\tilde{I} = \frac{20\angle 30°}{41.23\angle -14.04°} = 0.485\angle 44.04°\ \text{A} \;\Longrightarrow\; \boxed{i(t) = 0.485\cos(1000t + 44.04°)\ \text{A}}

(d) Applying Ohm's law to the capacitor alone:

\tilde{V}_C = \tilde{I}\,Z_C = (0.485\angle 44.04°)(40\angle -90°) = 19.40\angle -45.96°\ \text{V}

\boxed{v_C(t) = 19.40\cos(1000t - 45.96°)\ \text{V}}

The observation. The capacitor holds 19.40 V and the inductor holds $0.485 \times 30 = 14.55$ V, yet the source supplies only 20 V. No law is broken: the two reactive voltages are $180°$ apart and cancel, so KVL is satisfied by the phasor sum and not by the magnitudes. Adding $19.40\angle-45.96°$ , $14.55\angle 134.04°$ , and the resistor voltage $19.40\angle 44.04°$ returns exactly $20\angle 30°$ V.

Lecture 13 keeps this machinery and asks what it costs to run: real, reactive, and complex power, the power factor, and the load that extracts the most power from a source.

The AC steady state is a sinusoid at the source frequency

The three parameters of a sinusoid

Rectangular, polar, and exponential

A sinusoid becomes a complex number

Differentiation becomes multiplication by jωj\omegajω

Ohm's law, restored

Resistance and reactance

The procedure, in four steps

The four steps on the smallest possible circuit

The problem

Step 1 and 2: the frequency, then the impedances

Step 3: series impedances add

Step 4: Ohm's law, then back to the time domain

Find ZabZ_{ab}Zab​ and the source current

The problem

Convert every element to an impedance at ω = 10 rad/s

Combine the series branch on the right

The parallel pair, by product over sum

Add the series input branch

Ohm's law, then back to the time domain

The answer as a phasor diagram

Superposition across the frequency boundary

The problem

Impedances at ω = 50 rad/s

Circuit 1: the AC source acting alone

KCL at the node solves the AC part

Circuit 2: the DC source acting alone

The DC part is a one-line calculation

Superpose the time functions

Practice: a series RLC branch

Differentiation becomes multiplication by $j\omega$

Find $Z_{ab}$ and the source current