lecture_EE1C2_lec1_sinusoidal-voltages-and-currents_2025-11-09

📌 Overview

Introduction to sinusoidal steady-state analysis in linear circuits. Covers sinusoidal signals, phasors for simplifying AC analysis, impedance/admittance concepts, and circuit analysis techniques. Builds on Q1 transients to handle periodic steady-states. Enables solving AC circuits via algebraic equations instead of differentials.

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🎯Learning Objectives

• Define sinusoidal functions, including amplitude, frequency, phase; explain lead/lag.
• Apply phasor transform: direct (time-to-phasor) and inverse (phasor-to-time).
• Derive phasor relations and impedances for R, L, C; understand phase shifts.
• Use nodal/mesh analysis with phasors; compute impedances for series/parallel.
• Draw phasor diagrams; identify resonance conditions.
• Solve example circuits: find currents/voltages in AC networks.
• Recognize method limits: steady-state only, sinusoidal sources, linear time-invariant circuits.

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💡Key Concepts & Definitions

➗ Formulas

Sinusoidal Function:

$$x(t)=X_M\sin (\omega t+\theta )$$

• $X_M$: amplitude
• $\omega =2\pi f=2\pi /T$: angular frequency (rad/s)
• $\theta$: phase (rad)
• Period $T=1/f$; leads/lags if ${\theta }_1>{\theta }_2$ ($x_1$ leads $x_2$ by ${\theta }_1-{\theta }_2$).

Phasor Transform (Direct):

$$\mathbf{V}=V_M\angle \theta \text{from}v(t)=V_M\cos (\omega t+\theta )=\Re \{V_M{e}^{j(\omega t+\theta )}\}$$

• Ignore $e^{j\omega t}$; work in complex plane.
• Derivative: $\frac{d}{dt}\to j\omega$
• Integral: $\int \to \frac{1}{j\omega }$

Inverse Phasor Transform:

$$v(t)=\Re \{\mathbf{V}{e}^{j\omega t}\}=V_M\cos (\omega t+\theta )$$

Impedances ($Z(\omega )$):

Element	Time-Domain	Phasor Eq.	Impedance $Z$	Phase Shift
Resistor $R$	$v(t)=Ri(t)$	$\mathbf{V}=R\mathbf{I}$	$Z_R=R$	0° (in phase)
Inductor $L$	$v(t)=L\frac{di}{dt}$	$\mathbf{V}=j\omega L\mathbf{I}$	$Z_L=j\omega L$	+90° (V leads I)
Capacitor $C$	$i(t)=C\frac{dv}{dt}$	$\mathbf{I}=j\omega C\mathbf{V}$	$Z_C=\frac{1}{j\omega C}=-j\frac{1}{\omega C}$	-90° (I leads V)

• $Z=R+jX$ ($R$: resistance, $X$: reactance)
• Admittance $Y=1/Z=G+jB$ (G: conductance, B: susceptance)
• Series: $Z_{eq}=\sum Z_i$
• Parallel: $\frac{1}{Z_{eq}}=\sum \frac{1}{Z_i}$

Magnitude/Phase:

$$|Z|=\sqrt{R^2+X^2},\angle Z={\tan }^{-1}(X/R)$$

Resonance (Series RLC): ${\omega }_0=1/\sqrt{LC}$ (I in phase with V; $X_L=-X_C$).

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✍️ Notes

1. Recapitulation of Q1

• Mastered DC/transient analysis: KVL/KCL, sources, R/L/C/op-amps.
• Transients: 1st/2nd-order responses to step changes.
• Now: Sinusoidal steady-state (periodic “smooth” changes) – responses are also sinusoidal (linearity).

When Allowed: Linear time-invariant (LTI) circuits with sinusoidal sources. Ignores transients; assumes steady-state reached. Rare: Non-sinusoidal (use Fourier); nonlinear (distortion).

2. Sinusoidal Voltages/Currents

Sinusoids model AC power: $v(t)=V_M\sin (\omega t+\theta )$.

Properties:

• Periodic: $x(t+T)=x(t)$.
• Phase: $x_1(t)=X_M\sin (\omega t+\theta )$, $x_2(t)=X_M\sin (\omega t+\varphi )$; if $\theta >\varphi$, $x_1$ leads $x_2$ by $\theta -\varphi$.

Example: RL Circuit Steady-State Circuit: Voltage $v(t)=V_M\cos (\omega t)$ across R-L series.

• Diff. Eq.: $L\frac{di}{dt}+Ri(t)=v(t)$.
• Solution: $i(t)=I_M\cos (\omega t-\varphi )$, $\varphi ={\tan }^{-1}(\omega L/R)$ (I lags V by $\varphi \in [0^{\circ },90^{\circ }]$).
• Limits: $\omega =0$ or $L=0$ → DC, in phase; $R=0$ → pure inductive, 90° lag.

Stepwise Solve (Time-Domain, Manual):

1. Write diff. eq. from KVL.
2. Assume $i(t)=I_M\cos (\omega t+\psi )$.
3. Differentiate: $\frac{di}{dt}=-\omega I_M\sin (\omega t+\psi )$.
4. Substitute; equate coeffs. for cos/sin terms.
5. Solve: $I_M=V_M/\sqrt{R^2+(\omega L{)}^2}$, $\psi =-{\tan }^{-1}(\omega L/R)$.

Visual: RL Phasor Diagram

\begin{tikzpicture}[circuit ee IEC, scale=1.5]
  \draw (0,0) to[R, l=$R$] (2,0) to[L, l=$L$] (4,0);
  \draw (2,-0.5) node[ground] {};
  \source (0,0) to[V, l=v(t)] (0,1);
  \draw (0,1) -- (4,1);
\end{tikzpicture}

Phasors: $\mathbf{V}={\mathbf{V}}_R+{\mathbf{V}}_L$; ${\mathbf{V}}_L$ leads $\mathbf{I}$ by 90°.

3. Complex Sources & Euler’s Relation

Euler: $e^{j\varphi }=\cos \varphi +j\sin \varphi$; $\cos \varphi =\Re \{e^{j\varphi }\}$.

• Sinusoid: $v(t)=\Re \{V_M{e}^{j(\omega t+\theta )}\}$.
• Complex source: $\tilde{v}(t)=V_M{e}^{j\omega t}$ (non-physical; response $\tilde{i}(t)=I_M{e}^{j(\omega t+\varphi )}$).
• Real part gives physical $i(t)$.

Advantage: Turns diff. eq. into algebraic: $j\omega LI+RI=V$.

Rare: Assumes steady-state; initial conditions ignored (transients decay).

4. Phasors

Phasors: Complex reps. ignoring $e^{j\omega t}$ (fixed $\omega$ known).

Algorithm (Phasor Method):

1. Transform sources/elements to phasors (extract $\omega$; use $Z$).
2. Apply KVL/KCL/nodal/mesh in phasor domain (algebraic).
3. Solve for phasor unknowns.
4. Inverse transform: $v(t)=\Re \{\mathbf{V}{e}^{j\omega t}\}$.

When Allowed: Sinusoidal steady-state, LTI circuits. Not for transients/nonlinear. $\omega$ fixed per analysis.

Example Question: Find $i(t)$ in RL (V=10 cos(100t) V, R=5Ω, L=0.1H).

1. $\omega =100$ rad/s; $\mathbf{V}=10\angle 0^{\circ }$.
2. $Z=R+j\omega L=5+j10=\sqrt{125}\angle {\tan }^{-1}(2)\approx 11.18\angle 63.43^{\circ }$.
3. $\mathbf{I}=\mathbf{V}/Z\approx 0.894\angle -63.43^{\circ }$ A.
4. $i(t)=0.894\cos (100t-63.43^{\circ })$ A.

Visual: Phasor Calc

$$\mathbf{I}=\frac{10}{5+j10}=\frac{10(5-j10)}{25+100}=\frac{50-j100}{125}=0.4-j0.8$$

Magnitude: $\sqrt{0.4^2+0.8^2}=0.894$; Phase: ${\tan }^{-1}(-0.8/0.4)=-63.43^{\circ }$.

5. Phasor Relations & Impedances

• R: In phase; $Z=R$.
• L: V leads I by 90°; $Z=j\omega L$ (inductive reactance $X_L=\omega L>0$).
• C: I leads V by 90°; $Z=1/(j\omega C)$ ($X_C=-1/(\omega C)<0$).

Admittance: $Y=1/Z$; parallel easier with Y.

Phasor Diagram (Series RLC):

• $\mathbf{V}=\mathbf{I}(R+j(\omega L-1/(\omega C)))$.
• At resonance (${\omega }_0=1/\sqrt{LC}$): $X=0$, $\mathbf{V}=IR$ (in phase, max I).

Mermaid Diagram: Phase Behavior

graph TD
    A[Low ω] --> B[I lags V, capacitive dominant]
    C[Resonance ω=1/√(LC)] --> D[I in phase with V]
    E[High ω] --> F[I lags V, inductive dominant]

Rare: Negative frequencies (not physical); $\omega =0$ → DC ($Z_L=0$, $Z_C=\infty$).

6. Analysis Examples

Ex1: Nodal Analysis (Circuit: Vs=6∠0° V, Is=2∠0° A, R=1Ω, L=1H, C=1F, ω=1). Find Io.

1. Impedances: $Z_R=1$, $Z_L=j1$, $Z_C=-j1$.
2. Super-node: $V_1-V_2=6\angle 0°$.
3. KCL at V2: $(V_2-6\angle 0°)/j1+V_2/(-j1)+2\angle 0°=0$ → Solve: $V_2=3-j3$.
4. Io = (V2)/1 = 3 - j3 A (mag 4.24∠-45°).

Stepwise Plan:

• Transform sources to phasors.
• Label nodes; write KCL (admittances for parallels).
• Solve system (2x2 for super-node).
• Back-substitute for currents.

Ex2: Thevenin (Find Vx; Vs=10V, Rs=2Ω, R=4Ω, C=1F, ω=1).

1. Source transform: Is=5∠0° A, Zs=2Ω.
2. Parallel: Y_eq = 1/2 + j1 = 0.5 + j1; Z_eq=1/Y_eq ≈ 0.894 ∠-63.43°.
3. Vx = Is * (4 || Z_eq) = … (compute: Vx ≈ 2.68 ∠-26.57° V).

Limits: Dependent sources need care; mutual inductances (advanced, not here).

7. Exam Example

Circuit: Series R1-C parallel with L-R2; Vin, find Vout across R2. a) Transfer: $H(j\omega )=\frac{R_2}{R_1+R_2+j\omega L+1/(j\omega C)}$ (simplify parallels). b) Freq-independent if $L/C=R_1{R}_2$ (cancels $\omega$ terms).

Stepwise:

1. Equivalent Z: Parallel (R1 + 1/(jωC)) || (jωL + R2)? Wait, derive from KVL.
2. Vout/Vin = Z_load / Z_total.
3. Set imag. parts to cancel for independence.

When: Steady-state AC; no DC bias assumed.

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🔗 Resources

• Presentation:

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❓ Post lecture

• Why phasors simplify AC vs. time-domain? (Algebraic vs. diff. eq.)
• Draw phasor for pure C: How does I relate to V?
• Resonance: Why max power transfer?

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📖 Homework

• Do SGH1 (sinusoidal analysis exercises).
• Seminars: Tue/Fri practice phasor circuits.
• Prep: Next week – transfer functions.