Sinusoidal Steady-State Analysis

Every problem in this chapter follows the same three steps.

1. Transform the circuit to the phasor domain. Replace every source by its phasor, and every element by its impedance at the source frequency \(\omega\). Resistors keep their value; inductors become \(j\omega L\); capacitors become \(1/j\omega C\).

2. Solve the problem using any circuit technique — nodal, mesh, superposition, Thévenin. The arithmetic is complex; the method is unchanged.

3. Transform the answer back to the time domain, if a time function was asked for. A phasor \(\mathbf{I} = I_{m}\angle\theta\) at frequency \(\omega\) becomes \(i(t) = I_{m}\cos\left(\omega t + \theta\right)\).

Two points of practice. Step 1 requires \(\omega\), so read it from the source before doing anything else — an impedance is meaningless without a frequency. And step 3 is often forgotten: if the question asks for \(i(t)\) rather than \(\mathbf{I}\), a phasor is not an answer.

Since this chapter's methods produce simultaneous complex equations, a calculator or software able to solve them directly is a considerable help. Solving \(2\times2\) complex systems by hand using Cramer's rule, as in Section 3.6, remains entirely practical and is what the worked examples do.

Since KCL holds for phasors, nodal analysis proceeds exactly as in Section 3.2. Choose a reference node, assign phasor node voltages, and write KCL at each non-reference node with every branch current expressed as a potential difference divided by an impedance.

Dependent sources are handled as in Section 3.8: treat them as ordinary sources when writing the equations, then add a constraint expressing the controlling variable in terms of the node voltages.

A voltage source between two non-reference nodes requires a supernode, exactly as in Section 3.3. Enclose the source and both nodes, write KCL for the branches crossing the boundary, and add the source constraint.

KVL holds for phasors, so mesh analysis transfers likewise. Assign mesh currents, conventionally all clockwise, and write KVL around each mesh with element voltages expressed as \(\mathbf{Z}\mathbf{I}\). For an impedance shared between meshes \(j\) and \(k\), the drop seen from mesh \(j\) is \(\mathbf{Z}\left(\mathbf{I}_{j}-\mathbf{I}_{k}\right)\) — the own mesh current first, exactly as in Section 3.4.

The by-inspection method of Section 3.6 also survives, with the resistance matrix becoming an impedance matrix. With only independent sources present it remains symmetric, which is as useful an error check here as it was there.

A current source shared by two meshes requires a supermesh, as in Section 3.5: take KVL around the combined loop excluding the branch containing the source, and supply the constraint \(\mathbf{I}_{j}-\mathbf{I}_{k} = \mathbf{I}_{s}\).

A source belonging to one mesh alone is easier still: that mesh current is simply known, and no KVL equation is written for it. In the four-mesh circuit below, two of the four unknowns are fixed this way:

Four equations, of which two are constraints rather than KVL statements. Solving gives \(\mathbf{I}_{1} = 3.618\angle 274.5^{\circ}~\mathrm{A}\), and the voltage across the \(-j2~\Omega\) element follows:

Superposition applies to phasor circuits for the same reason it applied to resistive ones: linearity. Deactivate all independent sources but one, solve, and repeat; then sum the partial responses. The deactivation rules are unchanged — voltage sources become short circuits, current sources become open circuits, and dependent sources are never deactivated.

For a circuit whose sources all share one frequency, superposition is legitimate but rarely economical: it means solving several circuits instead of one. Its indispensable role appears in the next section, where the sources have different frequencies and no other method will do.

Here superposition is not merely convenient but necessary. Recall from Section 9.5 that phasors suppress the factor \(e^{j\omega t}\). If two sources have different \(\omega\), the suppressed factors differ, and their phasors cannot be added — the sum would be meaningless.

Worse, the impedances differ too. An inductor presents \(j\omega L\), which is not the same element at 50 rad/s as at 500 rad/s. There is no single frequency-domain circuit that represents the problem.

The procedure:

1. Take one frequency at a time, deactivating sources at all other frequencies.

2. Draw the frequency-domain circuit for that frequency, recomputing every impedance.

3. Solve, and immediately transform the answer back to the time domain.

4. Repeat for each frequency, and add the time-domain results.

The addition happens in step 4, in the time domain, where sinusoids of different frequencies coexist perfectly happily. A DC source counts as a separate "frequency" with \(\omega = 0\), at which inductors are shorts and capacitors open circuits.

One consequence worth noting for Chapter 11: powers from different frequencies do add directly, because the cross terms average to zero over time. But voltages and currents must be combined as time functions.

The transformation of Section 4.4 carries over directly, with impedance replacing resistance:

A voltage source \(\mathbf{V}_{s}\) in series with \(\mathbf{Z}_{s}\) is indistinguishable, at its terminals, from a current source \(\mathbf{V}_{s}/\mathbf{Z}_{s}\) in parallel with the same \(\mathbf{Z}_{s}\). The current-source arrow points towards what was the positive terminal, and an ideal source with no series or parallel impedance cannot be transformed.

The two most useful theorems of Chapter 4 transfer intact. Any linear two-terminal AC network can be replaced by

• a voltage source \(\mathbf{V}_{Th}\) in series with an impedance \(\mathbf{Z}_{Th}\), or

• a current source \(\mathbf{I}_{N}\) in parallel with the same impedance,

related by the same conversion as before:

The methods for finding \(\mathbf{Z}_{Th}\) are those of Sections 4.5 and 4.7: deactivate the independent sources and reduce, when only independent sources are present; otherwise use \(\mathbf{V}_{oc}/\mathbf{I}_{sc}\) or apply a test source.

One difference matters. \(\mathbf{Z}_{Th}\) is complex and frequency-dependent. A network's Thévenin equivalent is valid only at the frequency for which it was computed — change \(\omega\) and both \(\mathbf{V}_{Th}\) and \(\mathbf{Z}_{Th}\) change with it.

The two golden rules of Chapter 5 depended only on the ideal op-amp assumptions and negative feedback, neither of which involves frequency. They therefore apply unchanged to phasor circuits:

The inverting amplifier's gain becomes a ratio of impedances rather than resistances:

Since the ratio is now complex, the circuit provides both gain and phase shift, and both vary with frequency. This single observation is the foundation of the active filter, and Chapter 14 develops it fully.

The integrator and differentiator of Section 6.9 are special cases. With \(\mathbf{Z}_{f} = 1/j\omega C\) and \(\mathbf{Z}_{1} = R\) the gain is \(-1/j\omega RC\), which falls as frequency rises — the frequency-domain signature of integration.

• The methodology is: transform to the frequency domain, solve by any method, transform back if a time function is required.

• Nodal analysis, mesh analysis, supernodes and supermeshes work exactly as in Chapter 3, with impedances and complex arithmetic.

• Dependent sources require a constraint equation and are never deactivated.

• Superposition applies, with voltage sources shorted and current sources opened.

• For sources at different frequencies, superposition is compulsory: solve at each frequency separately, recomputing every impedance, and add the results in the time domain.

• Source transformation uses \(\mathbf{V}_{s} = \mathbf{Z}_{s}\mathbf{I}_{s}\).

• Thévenin and Norton equivalents exist for any linear AC network, with \(\mathbf{Z}_{Th} = \mathbf{V}_{oc}/\mathbf{I}_{sc}\) and both parameters valid only at the frequency for which they were found.

• Op-amp golden rules hold unchanged, so the inverting gain becomes \(-\mathbf{Z}_{f}/\mathbf{Z}_{1}\) — complex, and frequency-dependent.

• Adding phasors from sources at different frequencies. The suppressed \(e^{j\omega t}\) factors differ, and so do all the impedances. Solve separately and add the time functions.

• Leaving the answer as a phasor when a time function was asked for. Step 3 of the methodology exists for a reason. \(\mathbf{I} = 5\angle 30^{\circ}\) is not an answer to "find \(i(t)\)".

• Computing impedances without checking \(\omega\). An impedance is meaningless without a frequency, and a circuit with two sources has two sets of impedances.

• Adding complex quantities in polar form. Convert to rectangular first. This costs more keystrokes in AC analysis than any other single habit.

• Deactivating a dependent source. Only independent sources are ever deactivated — in superposition, and when finding \(\mathbf{Z}_{Th}\).

• Adding magnitudes instead of phasors. In Example 10.5 two contributions of 14.14 V sum to 20 V, not 28.28 V.

• Forgetting that \(1/j = -j\). This sign error propagates silently through an entire solution.

• Assuming a Thévenin equivalent holds at all frequencies. It holds at one frequency only.

• Reversing the order in a shared mesh branch. From mesh \(j\) the drop is \(\mathbf{Z}\left(\mathbf{I}_{j}-\mathbf{I}_{k}\right)\), exactly as in Chapter 3.

• Losing track of which angle unit is in use. Phasor angles are conventionally in degrees; the argument \(\omega t\) is in radians. Convert deliberately, not accidentally.

Work these before opening the answers. Quote phasors in polar form to four significant figures, and give time functions where asked.

1. P10.1 A \(4\angle 0^{\circ}~\mathrm{A}\) source feeds a node connected to ground through a 5 \(\Omega\) resistor and a \(j5~\Omega\) inductor in parallel. Find the node voltage.

   Show answer

   \[\mathbf{Z}_{p} = \frac{5\left(j5\right)}{5+j5} = 2.5 + j2.5~\Omega\]

   \[\mathbf{V} = \left(4\right)\left(2.5+j2.5\right) = 10 + j10~\mathrm{V} = 14.14\angle 45^{\circ}~\mathrm{V}\]

2. P10.2 A \(100\angle 0^{\circ}~\mathrm{V}\) source drives a single loop containing 6 \(\Omega\) and \(j8~\Omega\). Find the current, and state whether it leads or lags.

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   \[\mathbf{I} = \frac{100}{6+j8} = \frac{100\left(6-j8\right)}{100} = 6 - j8~\mathrm{A} = 10\angle-53.13^{\circ}~\mathrm{A}\]

   The current lags by 53.13°, as it must in an inductive circuit.

3. P10.3 A series circuit of 4 \(\Omega\) and 1 H is driven by a 12 V DC source in series with \(6\cos 4t~\mathrm{V}\). Find \(i(t)\).

   Show answer

   DC: the inductor is a short, so \(I_{1} = 12/4 = 3~\mathrm{A}\).
   AC at \(\omega = 4\):

   \[\mathbf{Z} = 4 + j4 = 5.657\angle 45^{\circ}~\Omega, \qquad \mathbf{I}_{2} = \frac{6}{5.657\angle 45^{\circ}} = 1.061\angle-45^{\circ}~\mathrm{A}\]

   \[i(t) = 3 + 1.061\cos\left(4t - 45^{\circ}\right)~\mathrm{A}\]

4. P10.4 Convert a \(20\angle 0^{\circ}~\mathrm{V}\) source in series with \(\left(4+j3\right)~\Omega\) into its Norton form.

   Show answer

   \[\mathbf{I}_{s} = \frac{20}{4+j3} = \frac{20\left(4-j3\right)}{25} = 3.2 - j2.4~\mathrm{A} = 4\angle-36.87^{\circ}~\mathrm{A}\]

   in parallel with \(\left(4+j3\right)~\Omega\).

5. P10.5 A \(50\angle 0^{\circ}~\mathrm{V}\) source in series with 10 \(\Omega\) feeds terminals \(a\)–\(b\), across which a \(-j10~\Omega\) capacitor is connected. Find the Thévenin equivalent.

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   \[\mathbf{V}_{Th} = \frac{-j10}{10-j10}\left(50\right) = 25 - j25~\mathrm{V} = 35.36\angle-45^{\circ}~\mathrm{V}\]

   \[\mathbf{Z}_{Th} = 10\parallel\left(-j10\right) = \frac{10\left(-j10\right)}{10-j10} = 5 - j5~\Omega\]

6. P10.6 Convert the result of P10.5 into a Norton equivalent.

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   \[\mathbf{I}_{N} = \frac{\mathbf{V}_{Th}}{\mathbf{Z}_{Th}} = \frac{25-j25}{5-j5} = 5\angle 0^{\circ}~\mathrm{A}\]

   with \(\mathbf{Z}_{N} = 5 - j5~\Omega\). The ratio is purely real because both quantities share the phase angle \(-45^{\circ}\).

7. P10.7 Repeat Worked Example 10.5 by nodal analysis instead of superposition, and confirm the answer.

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   KCL at the node, with both sources present:

   \[\frac{20-\mathbf{V}}{5} + j4 = \frac{\mathbf{V}}{-j5}\]

   Multiplying through by 5: \(20 - \mathbf{V} + j20 = \frac{5\mathbf{V}}{-j5} = j\mathbf{V}\), so

   \[20 + j20 = \mathbf{V}\left(1+j\right) \quad\Longrightarrow\quad \mathbf{V} = \frac{20\left(1+j\right)}{1+j} = 20\angle 0^{\circ}~\mathrm{V} \;\checkmark\]

   One equation instead of two circuits — nodal analysis was the better choice here.

8. P10.8 An inverting op-amp stage has \(\mathbf{Z}_{1} = 10~\mathrm{k}\Omega\) and \(\mathbf{Z}_{f} = -j20~\mathrm{k}\Omega\). Find the gain and describe what the circuit does.

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   \[\frac{\mathbf{V}_{o}}{\mathbf{V}_{i}} = -\frac{-j20}{10} = j2 = 2\angle 90^{\circ}\]

   Gain 2 with a 90° phase lead. Since \(\mathbf{Z}_{f}\) is a pure capacitance, this is an integrator, and its gain falls as frequency rises.

9. P10.9 A circuit is driven by \(10\cos 2t~\mathrm{V}\) and \(8\cos 5t~\mathrm{V}\) in series. Explain, without solving, why a single phasor circuit cannot be drawn.

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   The phasor transform suppresses \(e^{j\omega t}\), and the two sources have different \(\omega\), so their phasors are not comparable. More concretely, every inductor and capacitor has a different impedance at 2 rad/s than at 5 rad/s, so no single frequency-domain circuit represents both. Two separate circuits are required, and the results are added as time functions.

10. P10.10 A planar AC circuit has 5 nodes and 7 branches, driven entirely by voltage sources. Which analysis method would you choose?

    Show answer

    Nodal needs \(n-1 = 4\) equations; mesh needs \(b-n+1 = 7-5+1 = 3\). Mesh analysis is more economical on count, and with only voltage sources present it avoids supernodes entirely — voltage sources are handled trivially in mesh analysis. Mesh analysis on both grounds.