Sinusoids and Phasors

A sinusoid is a signal having the form of the sine or cosine function. Take a sinusoidally varying voltage:

where \(V_{m}\) is the amplitude in volts and \(\omega\) the angular frequency in radians per second.

The waveform repeats every \(2\pi\) radians of argument, which takes \(T\) seconds:

\(T\) is the period in seconds. The number of periods executed each second is the frequency \(f\), measured in hertz:

Keep the two frequencies distinct. \(f\) is in hertz — cycles per second — while \(\omega\) is in radians per second, and they differ by the factor \(2\pi\). The mains supply at 50 Hz corresponds to \(\omega = 314.2~\mathrm{rad/s}\); at 60 Hz, to \(\omega = 377~\mathrm{rad/s}\). That figure of 377 recurs constantly in North American practice and is worth recognising on sight.

The general form adds a phase angle:

Although \(\omega t\) is in radians, \(\theta\) is conventionally quoted in degrees. Writing \(100\sin(2\pi 1000t - 30^{\circ})\) mixes the two units in one expression, which is technically inconsistent but universal — and harmless provided you convert before evaluating anything numerically.

Two sinusoids of the same frequency but different phase are said to be out of phase. If

then \(v_{1}\) leads \(v_{2}\) by \(\theta\) — it reaches its peak earlier. A negative \(\theta\) means \(v_{1}\) lags. If \(\theta = 0\) they are in phase.

Three conditions must hold before phases can be compared. Two sinusoids can be compared only if they are:

• both written as sines, or both as cosines — not one of each;

• both written with positive amplitudes;

• of the same frequency.

Failing to normalise before comparing is the commonest error in this section, and it produces an answer wrong by exactly 90° or 180°.

Phasors are complex numbers, so a brief review is in order. A complex number can be written three equivalent ways:

Electrical engineering writes \(j = \sqrt{-1}\) rather than the mathematician's \(i\), because \(i\) is already taken by current. Conversion between forms:

The link between polar and exponential form is Euler's identity, which underpins the whole phasor method:

from which \(\cos\phi = \mathrm{Re}\left(e^{j\phi}\right)\) and \(\sin\phi = \mathrm{Im}\left(e^{j\phi}\right)\).

A warning about the arctangent. The formula \(\phi = \tan^{-1}(y/x)\) is ambiguous, because it cannot distinguish the second quadrant from the fourth. Always check which quadrant \(x\) and \(y\) place the number in, and add 180° if \(x\) is negative. A calculator's \(\tan^{-1}\) returns only \(-90^{\circ}\) to \(+90^{\circ}\).

A phasor is a complex number that represents the amplitude and phase of a sinusoid.

The idea rests on writing a cosine as the real part of a complex exponential. Given

Euler's identity lets this be written

where the phasor is

Since every voltage and current in a linear circuit shares the same \(\omega\), the factor \(e^{j\omega t}\) is common to all of them and can be suppressed. What remains — the amplitude and the phase — is all the information the problem contains.

The mental picture is a vector of length \(V_{m}\) rotating anticlockwise at \(\omega\) radians per second, starting at angle \(\phi\). Its projection onto the real axis traces out the cosine. Suppressing \(e^{j\omega t}\) amounts to photographing the rotating vector at \(t=0\) and agreeing that everything rotates together thereafter.

With the transform established, each element's \(v\)–\(i\) relation can be rewritten in phasor form. All three follow in a line or two.

Resistor. Ohm's law \(v = iR\) contains no calculus, so it transforms unchanged:

The phasors differ only in magnitude; there is no phase shift. Voltage and current in a resistor are in phase.

Inductor. Starting from \(v = L\,di/dt\) and replacing the derivative by \(j\omega\):

Since \(j = 1\angle 90^{\circ}\), multiplying by \(j\) advances the phase by 90°. The voltage therefore leads the current by 90° — equivalently, the current lags the voltage by 90°.

Capacitor. Starting from \(i = C\,dv/dt\):

Here the current leads the voltage by 90°, exactly reversing the inductor.

The mnemonic used throughout the English-speaking world is CIVIL: in a Capacitor, I leads V; V leads I in an L. Crude, but it has saved many examination answers.

Look at the three relations just derived. Every one has the form "phasor voltage equals something times phasor current". That something is the impedance:

This is Ohm's law in phasor form, and it is the reason every technique of Part 1 survives into AC analysis. Impedance is measured in ohms and represents the opposition a circuit presents to sinusoidal current.

Impedance is not a phasor. It is a complex number, but it does not correspond to any sinusoidally varying quantity — it is a ratio of two phasors. This distinction matters: you never write an impedance as a time function.

Note also the frequency dependence. An inductor's impedance \(j\omega L\) grows without limit as frequency rises, so an inductor blocks high frequencies and passes DC. A capacitor's impedance \(1/j\omega C\) does the opposite. At \(\omega = 0\) the inductor becomes a short and the capacitor an open circuit — recovering the DC results of Section 6.8 as a special case.

Kirchhoff's laws hold instant by instant in the time domain. Since the phasor transform is linear, they hold in the phasor domain too — and this is what licenses everything that follows.

For KVL, suppose \(v_{1}+v_{2}+\cdots+v_{n} = 0\) around a loop, with every term a sinusoid at the same \(\omega\). Writing each as \(\mathrm{Re}\left(\mathbf{V}_{k}e^{j\omega t}\right)\) and factoring out the common exponential gives

For this to hold at every instant the bracket must vanish, so

and by the identical argument applied to currents at a node,

This is the moment the whole of Part 1 becomes available again. With \(\mathbf{V} = \mathbf{Z}\mathbf{I}\) playing the role of Ohm's law and both Kirchhoff laws intact, every analysis technique from Chapters 2 to 4 carries over unchanged — series and parallel reduction, voltage and current division, wye–delta, nodal and mesh analysis, superposition, source transformation, Thévenin and Norton. The only difference is that the arithmetic is complex.

One thing does not survive. Phasors at different frequencies cannot be added, so a circuit driven at two frequencies must be solved once per frequency and the time-domain results superposed at the end. Chapter 10 returns to this.

Series. The same current flows through each element, so the voltages add:

Parallel. The same voltage appears across each, so the currents add:

For two in parallel, the product-over-sum rule applies as before:

Voltage and current division likewise transfer without change:

Note that the crossover in the current divider — the other impedance on top — is exactly as it was for resistors in Section 2.8.

The wye–delta transformation of Section 2.9 also carries over, with \(\mathbf{Z}\) replacing \(R\) throughout:

A phase shifter delivers an output that is a controlled number of degrees ahead of or behind its input. The simplest is a series RC network with the output taken across one element — and because the impedances are complex, a division of two of them naturally produces a phase angle.

Taking the output across the resistor of a series RC pair:

This is a leading network: the output leads the input by an angle between 0° and 90°, depending on how \(\omega RC\) compares with 1. Taking the output across the capacitor instead gives a lagging network. Cascading identical stages multiplies the shift, which is how RC phase-shift oscillators achieve the 180° they require.

• A sinusoid applied to a linear circuit produces a sinusoid of the same frequency; only amplitude and phase change.

• A sinusoid is described by amplitude, angular frequency and phase, with \(T = 2\pi/\omega\) and \(\omega = 2\pi f\).

• Before comparing phases, convert both sinusoids to positive-amplitude cosines of the same frequency.

• A phasor is a complex number carrying the amplitude and phase: \(V_{m}\cos(\omega t + \phi) \leftrightarrow V_{m}\angle\phi\).

• Differentiation becomes multiplication by \(j\omega\); integration becomes division by it.

• Impedances: \(R\) for a resistor, \(j\omega L\) for an inductor, \(1/j\omega C\) for a capacitor.

• Current lags voltage by 90° in an inductor and leads by 90° in a capacitor — CIVIL.

• \(\mathbf{Z} = R + jX\); positive \(X\) is inductive, negative capacitive. \(\mathbf{Y} = 1/\mathbf{Z} = G + jB\), and \(G \neq 1/R\) in general.

• KVL and KCL hold for phasors, so every technique from Part 1 carries over with complex arithmetic.

• Phasors of different frequencies cannot be added; use superposition instead.

• Comparing phases without normalising first. Both sinusoids must be positive-amplitude cosines (or both sines) at the same frequency. Skipping this gives an answer wrong by 90° or 180°.

• Adding complex numbers in polar form. Magnitudes and angles do not add. Convert to rectangular, add, convert back.

• Trusting the calculator's arctangent. It cannot distinguish the second quadrant from the fourth. Check the signs of the real and imaginary parts and add 180° when the real part is negative.

• Writing \(G = 1/R\) for a complex impedance. The correct relation is \(G = R/(R^{2}+X^{2})\). The whole quantity inverts, not the parts.

• Losing the sign of capacitive reactance. Since \(1/j = -j\), the capacitor impedance is \(-j/\omega C\). Reactance is negative for a capacitor and positive for an inductor.

• Adding phasor magnitudes. KVL applies to the phasors themselves. Two series voltages of 80 V and 45 V can perfectly well sum to 100 V.

• Adding phasors of different frequencies. The suppressed \(e^{j\omega t}\) factors differ, so the sum is meaningless. Use superposition, one frequency at a time.

• Treating a phasor as a function of time. Writing \(\mathbf{V}(t)\) confuses the two domains. The whole point is that time has been removed.

• Confusing \(f\) with \(\omega\). A 50 Hz supply has \(\omega = 314.2~\mathrm{rad/s}\). Using 50 for \(\omega\) makes every reactance wrong by a factor of \(2\pi\).

• Forgetting that impedance depends on frequency. A network that is inductive at one frequency may be capacitive at another; there is no single answer independent of \(\omega\).

Work these before opening the answers. Quote phasors in polar form to four significant figures with angles in degrees.

1. P9.1 For \(v = 20\sin\left(4\pi t - 30^{\circ}\right)~\mathrm{V}\), find \(V_{m}\), \(\omega\), \(f\), \(T\) and \(v\) at \(t = 0.1~\mathrm{s}\).

   Show answer

   \[V_{m} = 20~\mathrm{V}, \quad \omega = 4\pi = 12.57~\mathrm{rad/s}, \quad f = \frac{\omega}{2\pi} = 2~\mathrm{Hz}, \quad T = 0.5~\mathrm{s}\]

   At \(t = 0.1\) s the argument is \(4\pi(0.1) = 0.4\pi~\mathrm{rad} = 72^{\circ}\):

   \[v = 20\sin\left(72^{\circ}-30^{\circ}\right) = 20\sin 42^{\circ} = 13.38~\mathrm{V}\]

2. P9.2 Find the phase relationship between \(v_{1} = 10\cos\left(\omega t + 30^{\circ}\right)\) and \(v_{2} = 8\sin\left(\omega t + 10^{\circ}\right)\).

   Show answer

   \[v_{2} = 8\cos\left(\omega t + 10^{\circ} - 90^{\circ}\right) = 8\cos\left(\omega t - 80^{\circ}\right)\]

   \[30^{\circ} - \left(-80^{\circ}\right) = 110^{\circ}\]

   \(v_{1}\) leads \(v_{2}\) by 110°.

3. P9.3 Transform \(v = 15\sin\left(377t + 25^{\circ}\right)~\mathrm{V}\) to a phasor, and state the supply frequency in hertz.

   Show answer

   \[v = 15\cos\left(377t + 25^{\circ} - 90^{\circ}\right) = 15\cos\left(377t - 65^{\circ}\right) \;\Longrightarrow\; \mathbf{V} = 15\angle-65^{\circ}~\mathrm{V}\]

   \[f = \frac{377}{2\pi} = 60.00~\mathrm{Hz}\]

4. P9.4 Given \(\mathbf{V} = 30\angle-90^{\circ}~\mathrm{V}\) at \(\omega = 60~\mathrm{rad/s}\), write \(v(t)\) in both cosine and sine form.

   Show answer

   \[v(t) = 30\cos\left(60t - 90^{\circ}\right)~\mathrm{V} = 30\sin 60t~\mathrm{V}\]

   since \(\cos(x-90^{\circ}) = \sin x\).

5. P9.5 At \(f = 60~\mathrm{Hz}\), find the impedances of a 50 \(\Omega\) resistor, a 0.2 H inductor and a 100 \(\mu\mathrm{F}\) capacitor.

   Show answer

   \[\omega = 2\pi(60) = 377~\mathrm{rad/s}\]

   \[\mathbf{Z}_{R} = 50~\Omega, \qquad \mathbf{Z}_{L} = j(377)(0.2) = j75.40~\Omega\]

   \[\mathbf{Z}_{C} = \frac{-j}{(377)\left(100\times10^{-6}\right)} = -j26.53~\Omega\]

6. P9.6 Find the admittance of \(\mathbf{Z} = 8 - j6~\Omega\), and state whether the load is inductive or capacitive.

   Show answer

   \[\left|\mathbf{Z}\right| = \sqrt{64+36} = 10~\Omega\]

   \[\mathbf{Y} = \frac{8+j6}{100} = 0.0800 + j0.0600~\mathrm{S}\]

   Since \(X = -6 < 0\) the load is capacitive, and the current leads the voltage.

7. P9.7 Find \(\mathbf{Z}_{\mathrm{eq}}\) for \(\left(4+j3\right)~\Omega\) in series with \(\left(2-j5\right)~\Omega\), in polar form.

   Show answer

   \[\mathbf{Z}_{\mathrm{eq}} = \left(4+2\right) + j\left(3-5\right) = 6 - j2~\Omega\]

   \[\left|\mathbf{Z}\right| = \sqrt{36+4} = 6.325~\Omega, \qquad \theta = \tan^{-1}\frac{-2}{6} = -18.43^{\circ}\]

8. P9.8 An inductor of reactance \(j10~\Omega\) is in parallel with a capacitor of reactance \(-j5~\Omega\). Find the equivalent impedance and comment.

   Show answer

   \[\mathbf{Z}_{\mathrm{eq}} = \frac{\left(j10\right)\left(-j5\right)}{j10 - j5} = \frac{50}{j5} = -j10~\Omega\]

   The combination is capacitive, with a reactance larger in magnitude than either element. Parallel reactances of opposite sign can produce this — and if the two were equal in magnitude the denominator would vanish, giving infinite impedance. That is parallel resonance, and Chapter 14 treats it fully.

9. P9.9 A source \(120\angle 0^{\circ}~\mathrm{V}\) drives 30 \(\Omega\) in series with \(j40~\Omega\). Find the voltage across the inductor.

   Show answer

   \[\mathbf{V}_{L} = \frac{j40}{30+j40}\left(120\angle 0^{\circ}\right) = \frac{40\angle 90^{\circ}}{50\angle 53.13^{\circ}}\left(120\right)\]

   \[\mathbf{V}_{L} = 96.00\angle 36.87^{\circ}~\mathrm{V}\]

10. P9.10 For the phase shifter of Worked Example 9.9, what happens to the phase shift as \(\omega\) becomes very large, and very small?

    Show answer

    With the output across \(R\), \(\mathbf{V}_o/\mathbf{V}_i = j\omega RC/(1+j\omega RC)\).
    As \(\omega \to \infty\) the ratio tends to 1, so the shift tends to 0° and the output equals the input — the capacitor becomes a short.
    As \(\omega \to 0\) the ratio tends to \(j\omega RC\), so the shift tends to +90° while the amplitude falls to zero. The full 90° is approached only where there is no output left to shift — a limitation of every single-stage RC network.