Sinusoids and Phasors

Match against the standard form \(v = V_m\cos(\omega t + \phi)\):

The value 377 is worth recognising on sight: it is \(2\pi \times 60\), the mains angular frequency in North America, as 314 is for 50 Hz systems.

Instantaneous values. At \(t = 0\):

Not 12 V — the phase shift means the peak has already passed.

At \(t = 2\ \text{ms}\), converting carefully:

Note that \(\omega t\) arrives in radians and \(\phi\) is quoted in degrees. They cannot be added until one is converted — the most common arithmetic slip in this topic.

When does the peak occur? When the argument is zero:

A negative time, meaning the most recent peak was 1.389 ms before the origin — a positive phase shifts the waveform to the left, i.e. earlier. This is the geometric content of "leading".

The phasor discards everything that is not amplitude or phase:

The frequency is not lost but held aside — it is the same for every quantity in the circuit, so carrying it through the algebra would be redundant. It must be restored when transforming back.

Convert to a common reference. Both must be cosines before any comparison is meaningful. Using \(\sin\theta = \cos(\theta - 90°)\):

Now compare the phases:

\(v_1\) leads \(v_2\) by 10°, or equivalently \(v_2\) lags \(v_1\) by 10°.

What "leads" means physically. \(v_1\) reaches its peak earlier in time. At 60 Hz, 10° corresponds to

A positive phase difference means earlier; the sign convention is fixed by the argument increasing with \(t\).

The conversion identities, all of which follow from shifting the argument:

A negative amplitude is never left standing — it is absorbed as a 180° phase shift, since a phasor's magnitude is by definition positive.

A graphical aid. On a phasor diagram, adding 90° rotates anticlockwise. So \(\sin\) lags \(\cos\) by 90°, \(-\cos\) is \(\cos\) reversed, and the four functions sit at 90° intervals around the diagram. Reading the conversion off a sketch is more reliable than recalling which identity carries which sign.

Why the phase difference is meaningful but the phase is not. Shifting the time origin changes both \(\phi_1\) and \(\phi_2\) equally, leaving their difference untouched. Only relative phase has physical content — which is why one quantity in a circuit is usually chosen as the reference and assigned \(0°\).

The direct route. Assume a steady-state solution of the same frequency — which it must be, since a linear circuit cannot create frequencies:

Substituting and expanding both terms gives, after collecting \(\cos\omega t\) and \(\sin\omega t\) separately, two equations in \(I_m\) and \(\theta\). The work is tedious and the trigonometric identities are easy to mis-apply.

The complex route. Note that \(V_m\cos\omega t = \operatorname{Re}\left(V_me^{j\omega t}\right)\), and solve the complex problem instead:

Because the equation has real coefficients, the real part of its solution solves the real problem. That is the whole trick, and it works only because the circuit is linear.

Try \(\tilde{i} = \mathbf{I}e^{j\omega t}\) with \(\mathbf{I}\) a complex constant. The key property of the exponential:

Differentiation reproduces the function, multiplied by \(j\omega\). No other function has this property — which is exactly why the complex exponential, and not the sinusoid itself, is the right object to work with.

Substituting, the common factor \(e^{j\omega t}\) cancels from every term:

The differential equation has become a division. Every trace of \(t\) has gone.

Recovering the time function:

Identical to what the trigonometric route yields, with a fraction of the effort.

The three conditions this rests on:

Convert the sine to cosine reference first, as Problem 2 requires:

Write both as phasors and convert to rectangular form, because addition is rectangular:

Add:

Note what did not happen. The amplitudes did not add: \(20 + 15 = 35\) but the answer is 34.71. They would add only if the phases were equal; here the 15° difference costs a little. The general result is the cosine rule:

Verification in the time domain. Evaluating both expressions at any instant must agree. At \(\omega t = 0\):

Checking at a second instant — say \(\omega t = 90°\) — confirms the phase as well as the amplitude, and is worth the few seconds it takes.

Why this matters beyond arithmetic. Every KVL and KCL equation in an AC circuit is a sum of sinusoids of the same frequency. Doing it by trigonometry is possible but unpleasant; doing it as complex addition is mechanical. That is where most of the phasor method's practical value lies.

RThe resistor. Ohm's law is already algebraic:

Real and positive, so voltage and current are in phase. Resistance does not depend on frequency.

LThe inductor, using \(d/dt \to j\omega\) from Problem 3:

Multiplying by \(j\) adds 90°, so the voltage leads the current by 90°. Mnemonic: ELI — \(E\) before \(I\) in an \(L\).

CThe capacitor:

The current leads the voltage by 90°. Mnemonic: ICE — \(I\) before \(E\) in a \(C\). Together: ELI the ICE man.

Summary, with the frequency dependence made explicit:

The last two columns recover Set 18's DC rules as the \(\omega \to 0\) limit, and Set 19's \(t=0^+\) rules as \(\omega \to \infty\) — the same facts, now as endpoints of a continuum.

Reactance versus impedance. The imaginary part is the reactance \(X\), so \(\mathbf{Z} = R + jX\) with \(X_L = \omega L > 0\) and \(X_C = -1/\omega C < 0\). Reactance stores and returns energy; only the real part dissipates it — Set 23.

Transform to the frequency domain. The source becomes a phasor and the elements become impedances:

Series impedances add, exactly as resistances did in Set 2:

The magnitude is \(\sqrt{40^2+50^2}\) and the angle \(\tan^{-1}(50/40)\). Note it is not \(40+50 = 90\) — impedances add as complex numbers, so the magnitudes do not.

Ohm's law, unchanged:

The current lags the voltage by 51.34°, as it must in an inductive circuit.

The element voltages:

The inductor's voltage leads the current by exactly 90°, and the resistor's is in phase with it.

Verify KVL. The magnitudes are 62.47 and 78.09, summing to 140.6 — far more than 100 V. But phasors add as vectors:

The imaginary parts cancel exactly, leaving \(100\angle0°\). KVL holds for phasors, not for magnitudes.

Compare with the DC case. At \(\omega = 0\) the inductor is a short, giving \(I = 100/40 = 2.5\ \text{A}\) in phase. The reactance has reduced the current to 1.562 A and shifted it by 51°, without dissipating anything additional — the inductor stores and returns energy each cycle rather than consuming it.

The parallel combination, by the product-over-sum rule of Set 2 — which holds unchanged for impedances:

Evaluate in polar form for the division:

A clean result: 8 Ω of resistance and 4 Ω of capacitive reactance.

Something worth noticing. Combining a pure resistance with a pure reactance in parallel has produced both a resistance and a reactance — and the resistive part, 8 Ω, is less than the 10 Ω resistor that produced it. Parallel combination reduces the real part, just as with resistors.

Add the series impedance, which is rectangular addition:

The inductive and capacitive reactances have largely cancelled: \(+6\) and \(-4\) leave \(+2\). The network is only slightly inductive.

Reactance cancellation is the distinctive AC phenomenon. Resistances can only accumulate; reactances of opposite sign subtract. Had \(\mathbf{Z}_3\) been \(-j15\) instead, the parallel part would have been different and the total could have been made purely resistive — which is what tuning a circuit to resonance means (Set 29).

Practical procedure, to avoid the commonest errors:

Total impedance and current:

Voltage division, the Set 2 formula with impedances:

The surprise: \(|\mathbf{V}_1| = 137.9\ \text{V}\) exceeds the 120 V source. A resistive divider can never do this — Set 2's division always gives a fraction less than one.

How it is possible. KVL is satisfied, as the phasor sum confirms:

The two voltages are nearly 120° apart in phase, so a large part of each cancels the other. Their magnitudes sum to 190.9 V but their phasor sum is 120 V.

The magnification factor is essentially \(Q\) from Set 19. The reactances here are \(+12\) and \(-4\), partially cancelling, while the total resistance is only 8 Ω:

A 15% magnification here. In a circuit tuned so the reactances cancel exactly, the denominator becomes the resistance alone and the factor becomes \(Q\) — which can be hundreds. Set 29 develops this.

Why this matters practically. Voltages inside a reactive network can exceed the supply, and components must be rated for what actually appears across them, not for the source voltage. Series compensation capacitors on transmission lines and the ignition coil in a petrol engine both exploit the effect; insulation failures in resonant circuits are caused by ignoring it.

The transformation formulas from Set 2 apply unchanged, with \(R \to \mathbf{Z}\):

The sum, where the reactances cancel completely:

A purely real denominator, which makes the arithmetic unusually clean.

The three wye arms:

Note \(j \times (-j) = -j^2 = +1\): multiplying an inductive by a capacitive reactance gives a positive real number. That is how a wye arm of pure resistance appears from a delta containing none.

What has happened. The delta had one resistor and two reactances; the wye has one resistor and two reactances too — but they are not the same elements, and the resistance has moved to a different arm:

The equivalence is exact at this frequency and only at this frequency — since each impedance depends on \(\omega\) differently, the wye that matches at 500 rad/s will not match at 1000.

A caution about realisability. The formulas can produce an arm with negative resistance if the delta impedances are chosen adversely. Such a wye cannot be built from passive components, though it remains perfectly valid as an intermediate step in a calculation — the negative resistance always cancels before the final answer.

The definition is chosen so that an AC quantity delivers the same average power to a resistor as a DC quantity of the same value:

Root of the mean of the square — read backwards, the name is the recipe.

For a sinusoid \(v = V_m\cos\omega t\), using \(\overline{\cos^2} = \tfrac12\):

Which is why a "120 V" outlet has a peak of about 170 V, and a "240 V" supply peaks near 340 V — a distinction that matters for insulation and for component voltage ratings.

The factor \(\sqrt2\) is not universal. It applies to sinusoids alone:

A square wave spends all its time at the peak, so its RMS is the peak. Applying \(0.707\) to a non-sinusoid is a common and consequential error — an inexpensive multimeter does exactly this, which is why "true RMS" instruments exist.

Deriving the triangular result as an example. Over a quarter period the wave rises linearly, \(v = 4V_mt/T\):

The DC-offset case, which arises constantly in rectifier and switching circuits. For \(v = V_{DC} + V_m\cos\omega t\) the squares add, not the values:

This generalises: for any sum of a DC term and sinusoids of different frequencies, the total RMS is the square root of the sum of the individual mean squares. Set 33's Fourier series makes this a general theorem.

Definitions, the AC extensions of Set 2's conductance:

\(G\) is the conductance, \(B\) the susceptance. As with impedance, the real part is associated with dissipation and the imaginary part with storage.

Compute the reactance at 60 Hz:

Invert. In polar form this is trivial:

Inverting reciprocates the magnitude and negates the angle — an inductive impedance gives a capacitive-looking admittance angle.

In rectangular form, using the conjugate:

Now the point: \(1/R = 1/30 = 33.3\ \text{mS}\), but \(G = 4.56\ \text{mS}\) — different by more than sevenfold. In general

The reactance limits the current, so the circuit conducts less than its resistance alone suggests. Only in a purely resistive circuit do the two coincide.

Why admittance is worth having. It makes parallel combination additive:

The capacitor's admittance \(j\omega C\) is the tidiest expression in the table, which is why nodal analysis of AC circuits (Set 21) is naturally written in admittances.

Voltage division with the output across \(R\):

Multiply numerator and denominator by \(j\omega C\) to clear the compound fraction:

Magnitude and phase, writing \(x = \omega RC\):

The numerator's \(j\) contributes \(+90°\) and the denominator subtracts \(\tan^{-1}x\).

Tabulating:

At \(\omega RC = 1\), that is \(\omega = 1/RC\), the output is \(1/\sqrt2\) of the input — the half-power or \(-3\ \text{dB}\) point, and the same \(0.707\) that appeared as the settling-optimal \(\zeta\) in Set 19.

The achievable phase range. As \(x\) runs from 0 to \(\infty\) the phase runs from \(+90°\) to \(0°\):

And the two ends are useless: \(90°\) comes with zero output, \(0°\) with no shift. This is the fundamental limitation of a single \(RC\) stage, and the reason phase-shift oscillators use three cascaded stages to obtain 180°.

The complementary circuit. Taking the output across the capacitor instead gives

a low-pass with lagging phase, where the first was a high-pass with leading phase. The two magnitudes satisfy \(|\mathbf{H}_R|^2 + |\mathbf{H}_C|^2 = 1\) at every frequency — the power splits between them.

The impedance as a function of frequency:

The two reactances oppose each other, one rising with frequency and the other falling.

They cancel when \(\omega L = 1/\omega C\):

The same \(\omega_0 = 1/\sqrt{LC}\) as Set 19's undamped natural frequency — not a coincidence, and Problem 19 explains why.

Evaluating at the three frequencies:

The magnitudes:

At resonance the impedance falls to the resistance alone — a factor of fifteen below its value a decade away. The same source voltage would drive fifteen times the current.

The character changes with frequency because the two elements dominate in different regions:

Below resonance the capacitor's large reactance dominates; above it the inductor's does. The circuit is a capacitor at low frequency, a resistor at \(\omega_0\), and an inductor at high frequency — three behaviours from one network.

The characteristic impedance is the common value of the two reactances at resonance:

and \(Q = Z_0/R = 3.16\) — the same quantity Set 19 defined from the damping, now read off the impedances.

Construction rules. Each phasor is drawn as an arrow: length proportional to magnitude, angle equal to phase, measured anticlockwise from the positive real axis.

Choosing the common quantity as reference means every other phasor can be placed relative to it by the element's known 90° rule.

For Problem 6's circuit, taking \(\mathbf{I} = 1.562\angle0°\) as reference:

The three form a right triangle, since \(\mathbf{V}_R\) and \(\mathbf{V}_L\) are perpendicular. Pythagoras gives the source magnitude directly:

The impedance triangle is the same figure divided by \(|\mathbf{I}|\):

Similar triangles, so the angle is the same 51.34°. Set 23 will add a third similar triangle — the power triangle — completing the family.

What the diagram shows that algebra does not.

A worked use. To make Problem 6's circuit unity power factor, a capacitor must be added whose \(\mathbf{V}_C\) cancels \(\mathbf{V}_L\). On the diagram this is obvious — draw an arrow of 78.09 V pointing at \(-90°\) — and it requires \(X_C = 50\ \Omega\), hence \(C = 1/(500 \times 50) = 40\ \mu\text{F}\). Set 23 uses exactly this reasoning for power-factor correction.

The transfer function, output across the capacitor:

The corner frequency, where \(\omega RC = 1\):

Equivalently \(1/2\pi\tau\), with \(\tau = RC = 1\ \text{ms}\) the time constant of Set 18 — the same circuit, described in the frequency domain instead.

Evaluating, with the three frequencies chosen a decade apart around the corner:

The decibel definition for a voltage ratio:

The factor is 20 for amplitude ratios and 10 for power ratios, because power goes as amplitude squared. Confusing them is a factor-of-two error in the exponent, and is common.

Two facts worth memorising from the table:

Twenty decibels per decade is the asymptotic slope of a single pole. Set 28 builds the entire Bode construction on these two landmarks.

The connection to Set 18's rise time. There the 10%–90% rise time was \(t_r = 2.2\tau\), and here \(f_c = 1/2\pi\tau\), so

recovering the rule quoted there. The time and frequency descriptions are the same information, and either can be derived from the other.

The proof for KVL. In the time domain, round any loop,

Each is a sinusoid of the same frequency, so \(v_k = \operatorname{Re}\left(\mathbf{V}_ke^{j\omega t}\right)\). Substituting:

The conclusion. A complex number whose product with \(e^{j\omega t}\) has zero real part at every instant must itself be zero — because \(e^{j\omega t}\) sweeps through all phases, so any non-zero sum would show up at some instant. Hence

The identical argument at a node gives \(\sum\mathbf{I}_k = 0\). This is why Problem 6's element voltages summed to the source as phasors and not as magnitudes.

What transfers, and it is nearly everything:

Only one entry is genuinely altered: maximum power transfer now requires the conjugate match, so the load's reactance must cancel the source's rather than equal it.

What does not transfer:

Why so much survives. Kirchhoff's laws come from charge and energy conservation, which know nothing about frequency; and every theorem in Part 1 was derived from those laws plus linearity. Both survive the transform intact, so the theorems must too. The proofs need no re-derivation — only the substitution \(R \to \mathbf{Z}\).

Method 1 — substitute into the differential equation. The circuit obeys

With \(i = I_m\cos(\omega t+\theta)\), the derivative is \(-\omega I_m\sin(\omega t+\theta)\), so the left side is

Evaluate at a convenient instant. At \(t = 0\), with \(\theta = -51.34°\) and \(I_m = 1.562\):

Matching the source's value of 100 at \(t = 0\). Repeating at \(\omega t = 90°\) should give zero, and does — one instant checks the amplitude, two check the phase as well.

Method 2 — check Kirchhoff's law as a phasor sum. From Problem 6:

The imaginary parts cancel to the last digit. This is the faster check and the one to use routinely.

Why two methods. They fail differently, so agreement is strong evidence:

Three quick sanity checks worth applying before any detailed verification:

The middle one generalises usefully: adding reactance can only reduce the current magnitude below its purely resistive value, never increase it.

The three requirements identified in Problem 3, and what fails without each:

The two-frequency source. There is no single phasor for \(10\cos100t + 6\cos300t\) — writing \(10\angle0° + 6\angle0° = 16\angle0°\) would be meaningless, since the two terms are never in step. Instead, treat them as two separate sources and superpose:

The procedure — the essential point being that the impedances must be recomputed:

Add the two time functions — never the two phasors. That is the whole rule, and it is where superposition across frequencies most often goes wrong.

A consequence for power. Since power is quadratic, cross terms between different frequencies must be considered — but their time average vanishes:

So average powers do add across frequencies, even though instantaneous powers do not. This orthogonality is the foundation of Set 33's Fourier analysis of power.

Non-linearity is the case with no remedy. A diode, a saturating iron core or a transistor produces output frequencies absent from the input — harmonics, and sums and differences of input frequencies. Since superposition itself fails, no decomposition into single-frequency problems is available, and numerical or specialised methods are required.

Why the steady-state restriction is usually harmless. The transient decays as \(e^{-t/\tau}\), so in a circuit with \(\tau = 1\ \text{ms}\) it is gone within 5 ms. For a mains circuit examined over seconds, that is invisible. It matters only when the switching instant matters — inrush current, relay contact bounce, or the first cycle after a fault.

Observe a coincidence. Set 19's series \(RLC\) had

and this set's series impedance is

Substituting \(s\) for \(j\omega\) in the second and multiplying by \(s/L\) reproduces the first exactly. The two are not merely similar — they are the same expression.

The unifying variable. Both arise from assuming a solution \(e^{st}\) and using \(d/dt \to s\):

Phasor analysis is what happens when \(s\) is restricted to the imaginary axis.

What each part of \(s = \sigma + j\omega\) means:

The imaginary axis is precisely the boundary between decay and growth, which is why it describes a steady state that neither dies nor grows.

Why the natural frequency reappeared. Problem 13 found the reactances cancel at \(\omega_0 = 1/\sqrt{LC}\) — the same \(\omega_0\) as Set 19's undamped natural frequency. Now the reason is visible: both come from the same polynomial. Resonance in the frequency domain and natural oscillation in the time domain are one phenomenon, examined from two directions.

What this anticipates. Set 31's Laplace transform takes \(s\) as a genuine complex variable, and Set 32 analyses circuits with \(\mathbf{Z}(s)\) directly. The result contains everything: the transient (from the poles' real parts), the steady state (from the behaviour on \(s = j\omega\)), and the initial conditions, in one calculation.

What it achieves. The transform converts calculus into arithmetic:

The last row is the real prize. Twelve sets of network theory transfer intact under \(R \to \mathbf{Z}\).

What it costs. Three things are given up, of which only the first is serious:

What remains in Part 3:

Only Set 23 introduces genuinely new physics. The rest is either Part 1 re-run or the consequences of examining \(\mathbf{Z}(\omega)\) across a range rather than at a point.

The one thing genuinely new so far is that phase now matters. In a DC circuit a voltage and a current either coexist or they do not; here they can be 90° apart, in which case the product averages to zero and no power flows at all despite both being non-zero. Set 23 develops this into real, reactive and apparent power.

A word on why sinusoids specifically. The method works because the sinusoid is the only waveform a linear circuit passes without changing its shape — only its amplitude and phase. Set 33 shows this makes sinusoids a basis: any periodic waveform can be decomposed into them, analysed frequency by frequency, and reassembled. The restriction to one frequency is therefore far less limiting than it appears.