ECE 1230 F26 U2 L8 First Order Sinusoid Complex Math Examples
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Overview
Professor Angela Rasmussen reviews sinusoid descriptions, phasor conversions, complex-number operations, and first-order RC/RL circuit responses in preparation for an open-notes exam. She works through a graph with a -2 V DC offset, 5 V amplitude, and approximately 42 rad/s frequency, then demonstrates phasor arithmetic and circuit solutions using continuity, steady-state models, Thevenin resistance, and time constants.
Key takeaways
- A graph with a -2 V center, 10 V peak-to-peak range, and four cycles in about 0.6 seconds corresponds to -2 + 5 cos(42t).
- A sinusoid that lags in time uses a negative phase angle, while a leading sinusoid uses a positive phase angle.
- For phasor arithmetic, rectangular form simplifies addition and subtraction, while polar form simplifies multiplication and division; the original angular frequency is restored when converting back to time domain.
- Capacitor voltage and inductor current are continuous across a switching instant, which sets the initial conditions for transient analysis.
- First-order responses use final + (initial - final)e^(-t/τ), with τ = RTHC for RC circuits and τ = L/RTH for RL circuits.
- In the worked RL example, 1 mH and 1.3 kΩ produce a 75 μs time constant; in the RC example, 1 mF and 250 Ω produce a 0.25 s time constant.
Chapters
- The class begins with reports that the campus Wi-Fi is slow or not connecting, delaying an online-tool demonstration.
- Professor Angela Rasmussen says the upcoming exam will take place during class on Wednesday.
- The exam covers an RC circuit and an RL circuit, including initial and final conditions that may require circuit analysis.
- Sinusoid questions include deriving an expression from a graph, converting to and from phasor form, and complex-number arithmetic.
- The exam is open-notes; students should bring a handheld calculator and may bring colored markers or highlighters.
- Students revisit a graph-based sinusoid exercise from Lecture 7 and identify its full mathematical expression.
- The signal is separated into a DC component and an AC component, a superposition approach useful in later circuit analysis.
- The graph oscillates around -2 V, so its center value is the DC component.
- The signal is written in the form DC plus AC, making the offset explicit rather than treating the waveform as centered at zero.
- The graph's peak-to-peak range is 10 V, so its cosine amplitude is half that value, or 5 V.
- Four cycles span approximately 0.6 seconds, giving a period of about 0.15 seconds per cycle.
- Using ω = 2π/T gives approximately 42 rad/s, producing the expression -2 + 5 cos(42t).
- The blue sinusoid is compared with the red sinusoid along the time axis; a waveform shifted right occurs later and lags.
- A lagging waveform has a negative phase angle, while a leading waveform has a positive phase angle.
- The goal is to recognize lead, lag, and phase-sign conventions from a plot, not to calculate a specific phase shift.
- For a sinusoid with amplitude 10, angular frequency 20, and phase 90°, the phasor uses the amplitude and phase: 10∠90°.
- The frequency is omitted during phasor-domain algebra because it remains unchanged when the result is converted back to the time domain.
- The frequency 20 is restored in the cosine expression after completing the complex-domain operation.
- Students apply both division and multiplication by the complex number 10 - 2j to the input phasor.
- The results are then converted back into time-domain sinusoidal expressions.
- Rectangular form is preferred for addition and subtraction, while polar form makes multiplication and division easier.
- Wolfram Alpha is recommended as a calculator-based way to check numerical work, including systems of equations.
- When entering phasor expressions, users should specify degrees and may need to replace engineering notation j with calculator notation i.
- Electrical engineering uses j for the imaginary unit to avoid confusion with i, which commonly denotes current.
- Dividing 10∠90° by 10 - 2j gives approximately 0.98∠101.31°, corresponding to 0.98 cos(20t + 101.31°).
- Multiplying 10∠90° by 10 - 2j gives approximately 102∠78.7°, corresponding to 102 cos(20t + 78.7°).
- The original angular frequency, 20, is retained in both output expressions.
- The magnitude of 10 - 2j is approximately 10.2, so dividing an amplitude of 10 should produce a result near 1.
- The angle of 10 - 2j lies in the fourth quadrant; subtracting that angle during division places the result near 101°, in the second quadrant.
- For multiplication, adding the angles produces a result near 79° in the first quadrant and a magnitude near 100.
- A capacitor's voltage cannot change instantaneously because its stored charge cannot change instantaneously.
- At the switching instant, capacitor voltage is therefore continuous, and the transient-state capacitor can be modeled as a voltage source equal to its initial voltage.
- This continuity rule identifies which capacitor quantity must be carried from the pre-switch circuit into the post-switch analysis.
- An inductor's current cannot change instantaneously, so its transient model is a current source set by the initial current.
- At steady state, a capacitor behaves as an open circuit and an inductor as a wire; immediately after switching, continuity determines the transient replacement.
- The first-order response is final value plus (initial value minus final value) multiplied by e^(-t/τ).
- For RC circuits, τ = RTHC; for RL circuits, τ = L/RTH.
- The review checklist includes component continuity, redrawing circuits at each stage, finding RTH, and evaluating the first-order response.
- Sinusoid checks include identifying the DC offset, amplitude, angular frequency, phase reference, and the sign associated with lead or lag.
- Phasor checks include choosing rectangular or polar form appropriately, verifying the result's quadrant, and restoring angular frequency in the time-domain expression.
- The class begins a switched RL example, seeking inductor current and a node voltage.
- At t = 0−, the inductor is a steady-state wire; current summation gives an initial inductor current of 5 mA.
- Continuity carries that current into the transient state, where node-voltage analysis gives an initial Vx of 17/3 V.
- In the final steady state, the inductor is again a wire; the solution gives a final Vx of 11/2 V and final inductor current of 5.5 mA.
- The resistance seen by the inductor is 1.3 kΩ, so with L = 1 mH the time constant is 75 μs; the initial and final values are inserted into the exponential response.
- Before the switch closes, the capacitor is an open circuit; node-voltage analysis gives an initial capacitor voltage of 3 V.
- Capacitor-voltage continuity makes the voltage at 0+ equal to the 3 V initial value.
- In the final steady state, the capacitor is open again, and node-voltage analysis gives a final voltage of 4 V.
- Four 1 kΩ resistors in parallel give RTH = 250 Ω; with C = 1 mF, τ = 0.25 s and vC(t) = 4 + (3 - 4)e^(-t/0.25).
Transcript excerpt
0:07 If anyone notices um that it's back on and connected while I'm lecturing, let me know. I wanted to just show you um another tool that is useful. So, but I need to connect to the internet. So, >> what do you mean Wi-Fi? >> Yeah, you're connected on Wi-Fi. Sounds like it's happening more than this building. >> Okay. But several of you are not connecting, right? >> It's just super slow >> or just slow. Yeah, mine's not even connecting. Okay. Um,
1:09 okay. Well, we'll start here. Give it a little bit. If it Let me get back on. It will show you because it's not even loading here. Okay. Yeah. All right. Um, so exam will be on Wednesday. It'll be just in this class and you'll have an RC circuit, RL circuit that you'll have to solve. I had mentioned before that for both of these um initial and final um may need
2:08 analysis. So they may not be easy to see that they're exactly zero. So they will need analysis for both of those stages. You're going to have sinosoid a math description from from the graph and then you're going to have to do a phaser transform of sinosoid. Oops. and then an inverse phaser
2:54 transform. So this is basically given the sinosoid um be able to write it in polar form and then given polar form write it into the time domain. So those conversions and then manipulation and complex domain. So you may need to add, subtract, multiply, divide. Remember adding subtracting you would want rectangular form and for multiplication and division you
3:43 want polar form. Right? Uh open notes. Make sure you bring a handheld calculator. Also bring colored bring colors if you want. Some of you are still finding that to be helpful. So
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Related notes
Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, University of Utah: Professor Angela Rasmussen.