Difference between revisions of "EGR 224/Concept List/S22"
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* Several ways to solve: | * Several ways to solve: | ||
** If there are only independent sources, turn independent sources off and find $$R_{eq}$$ between terminals of interest to get $$R_{T}$$. Then find $$v_{oc}=v_{T}$$ and recall that $$v_T=R_Ti_N$$ | ** If there are only independent sources, turn independent sources off and find $$R_{eq}$$ between terminals of interest to get $$R_{T}$$. Then find $$v_{oc}=v_{T}$$ and recall that $$v_T=R_Ti_N$$ | ||
− | ** If there are both independent sources and dependent sources, solve for $$v_{oc}=v_T$$ first, then put a | + | ** If there are both independent sources and dependent sources, solve for $$v_{oc}=v_T$$ first, then put a short circuit between the terminals and solve for $$i_{sc}=i_N$$. Recall that $$v_T=R_Ti_N$$ |
** If there are '''only''' dependent sources, you have to activate the circuit with an external source. | ** If there are '''only''' dependent sources, you have to activate the circuit with an external source. | ||
Line 99: | Line 99: | ||
== Lecture 17 - 3/4 - == | == Lecture 17 - 3/4 - == | ||
+ | * Transfer function: ratio of an output phasor (for a voltage or current) to a input phasor (for a voltage or current) | ||
+ | ** A transfer function is not a phasor - a transfer function does not represent a magnitude and phase of a single frequency sinusoid; instead, the magnitude represents a ratio of amplitudes for the output and the input while the angle represents the phase difference between the output and the input. | ||
+ | * Filters | ||
+ | ** Low-pass, High-pass, Band-pass, Band-reject; names based on which frequencies of the transfer function have magnitudes that are 70.71% ($$1/\sqrt{2}$$) or more of the maximum possible magnitude | ||
+ | ** Cutoff frequency is where the magnitude is at 70.71% of the maximum | ||
+ | * Decibel - based on the base-10 logarithm of a power ratio, multiplied by 10 (to get the deci- part) | ||
+ | ** We are assuming power is related to voltage or current squared, and the log of a square is twice the log, so $$H_{dB}=20\,\log_{10}|\mathbb{H}(j\omega)|$$ | ||
+ | * Bode plots | ||
+ | ** Magnitude plot is $$20\,\log_{10}|\mathbb{H}(j\omega)|$$ versus $$\omega$$ with $$\omega$$ on a log scale | ||
+ | ** Angle plot is $$\angle \mathbb{H}(j\omega)$$ versus $$\omega$$ with $$\omega$$ on a log scale | ||
+ | * Translation between decibels and magnitudes | ||
+ | ** 1 $$\leftrightarrow$$ 0 dB | ||
+ | ** 10 $$\leftrightarrow$$ 20 dB | ||
+ | ** 100 $$\leftrightarrow$$ 40 dB | ||
+ | ** 0.1 $$\leftrightarrow$$ -20 dB | ||
+ | ** 2 $$\leftrightarrow$$ $$\approx$$ 6 dB | ||
+ | ** 5 $$\leftrightarrow$$ $$\approx$$ 14 dB | ||
+ | ** 0.25=1/2/2 $$\leftrightarrow$$ 0-6-6 dB=-12 dB | ||
+ | ** 8=2*2*2=$$2^3$$ $$\leftrightarrow$$ 6+6+6 dB = 3 * 5 dB = 18 dB | ||
+ | ** $$\sqrt{2}$$=$$2^{1/2}$$ $$\leftrightarrow$$ $$\frac{1}{2}$$ 6 dB = 3 dB | ||
+ | * 3 dB below maximum for a transfer function amplitude is critical because that indicates where the half-power frequency is. | ||
+ | * Four main terms in transfer functions: | ||
+ | ** $$\mathbb{H}(j\omega)=K\,\color{red}{(j\omega)^2}\,\color{green}{\Pi_k(j\omega+a_k)^{n_k}}\,\color{blue}{\Pi_l((j\omega)^2+2\zeta_l\omega_{n_l}+(\omega_{n_l})^2)^{m_l}}$$ |
Latest revision as of 19:46, 26 September 2022
The notes below are not meant to be comprehensive but rather to capture the general topics of covered during lectures in EGR 224 for Spring 2022. These notes are in no way a replacement for actively attending class.
Contents
- 1 Lecture 1 - 1/5 - No Class
- 2 Lecture 2 - 1/7 - Course Introduction, Nomenclature
- 3 Lecture 3 - 1/10- Voltage and Current; Power and Energy
- 4 Lecture 4 - 1/14 - Equivalents
- 5 Lecture 5 - 1/21 - Voltage Division and Current Division
- 6 Lecture 6 - 1/24 - Node Voltage Method
- 7 Lecture 7 - 1/28 - Mesh and Branch Current Method
- 8 Lecture 8 - 1/31 - Linearity and Superposition
- 9 Lecture 9 - 2/4 - Thévenin and Norton Equivalent Circuits
- 10 Lecture 10 - 2/7 - Capacitors and Inductors
- 11 Lecture 11 - 2/11 - First-Order Circuits (constant forcing functions)
- 12 Lecture 12 - 2/14 - More First-Order Circuits
- 13 Lecture 13 - 2/18 - AC Steady State
- 14 Lecture 14 - 2/21 - Test 1
- 15 Lecture 15 - 2/25 - Phasors 1
- 16 Lecture 16 - 2/28 - Phasors 2
- 17 Lecture 17 - 3/4 -
Lecture 1 - 1/5 - No Class
Lecture 2 - 1/7 - Course Introduction, Nomenclature
- Circuit terms (Element, Circuit, Path, Branch and Essential Branch, Node and Essential Node, Loop and Mesh).
- Circuit topology (parallel, series)
- Electrical quantities (charge, current, voltage, power)
Lecture 3 - 1/10- Voltage and Current; Power and Energy
- Power redux
- Passive Sign Convention and Active Sign Convention and relation to calculating power absorbed and/or power delivered.
- Example of how to find $$i$$, $$v$$, and $$p_{\mathrm{abs}}$$
- $$i$$-$$v$$ characteristics of various elements (ideal independent voltage source, ideal independent current source, short circuit, open circuit, switch, resistor)
- Kirchhoff's Laws
Lecture 4 - 1/14 - Equivalents
- Combining voltage sources in series; ability to move series items
- Combining current sources in parallel; ability to move parallel items
- Equivalent resistances
- series, parallel, and other (Delta-Wye)
- Examples/Req
Lecture 5 - 1/21 - Voltage Division and Current Division
- Voltage division (actually covered during Lab 2)
- Current Division
- Beginning of Node Voltage Method and label techniques
Lecture 6 - 1/24 - Node Voltage Method
- More NVM
- Start of Mesh Current Method
Lecture 7 - 1/28 - Mesh and Branch Current Method
- More MCM
- Branch Current Method
Lecture 8 - 1/31 - Linearity and Superposition
- Definition of a linear system
- Examples of nonlinear systems and linear systems
- Nonlinear system examples (additive constants, powers other than 1, trig):
- $$\begin{align*} y(t)&=x(t)+1\\ y(t)&=(x(t))^n, n\neq 1\\ y(t)&=\cos(x(t)) \end{align*} $$
- Linear system examples (multiplicative constants, derivatives, integrals):
- $$\begin{align*} y(t)&=ax(t)\\ y(t)&=\frac{d^nx(t)}{dt^n}\\ y(t)&=\int x(\tau)~d\tau \end{align*} $$
- Superposition
- Redraw the circuit as many times as needed to focus on each independent source individually
- If there are dependent sources, you must keep them activated and solve for measurements each time
Lecture 9 - 2/4 - Thévenin and Norton Equivalent Circuits
- Thévenin and Norton Equivalents
- Circuits with independent sources, dependent sources, and resistances can be reduced to a single source and resistance from the perspective of any two nodes
- Equivalents are electrically indistinguishable from one another
- Several ways to solve:
- If there are only independent sources, turn independent sources off and find $$R_{eq}$$ between terminals of interest to get $$R_{T}$$. Then find $$v_{oc}=v_{T}$$ and recall that $$v_T=R_Ti_N$$
- If there are both independent sources and dependent sources, solve for $$v_{oc}=v_T$$ first, then put a short circuit between the terminals and solve for $$i_{sc}=i_N$$. Recall that $$v_T=R_Ti_N$$
- If there are only dependent sources, you have to activate the circuit with an external source.
Lecture 10 - 2/7 - Capacitors and Inductors
- Intro to capacitors and inductors
- Basic physical models
- Basic electrical models
- Energy storage
- Continuity requirements
- DCSS equivalents
Lecture 11 - 2/11 - First-Order Circuits (constant forcing functions)
- First-order switched circuits with constant forcing functions
- Sketching basic exponential decays
- Using Node Voltage Method to get model equation
Lecture 12 - 2/14 - More First-Order Circuits
Lecture 13 - 2/18 - AC Steady State
- Solving ACSS using just trig gets complex very quickly - we will use complex analysis to simplify the process
- At the heart of complex analysis is an understanding of Complex Numbers
Lecture 14 - 2/21 - Test 1
Lecture 15 - 2/25 - Phasors 1
Lecture 16 - 2/28 - Phasors 2
- EGR_224/Spring_2022/Sandbox
- Motivation for phasors
- Reminder: a phasor is a complex number whose magnitude represents the amplitude of a single frequency sinusoid and whose angle represents the phase of a single frequency sinusoid
- Impedance: a ratio of phasors (though not a phasor itself)
- $$\mathbb{Z}_R=R$$
- $$\mathbb{Z}_L=j\omega L$$
- $$\mathbb{Z}_R=\frac{1}{j\omega C}$$
- Conservation laws (KCL, KVL), methods derived from conservation laws (NVM, MCM, BCM), and methods derived from Ohm's Law (voltage division, current division) apply in the phasor domain
Lecture 17 - 3/4 -
- Transfer function: ratio of an output phasor (for a voltage or current) to a input phasor (for a voltage or current)
- A transfer function is not a phasor - a transfer function does not represent a magnitude and phase of a single frequency sinusoid; instead, the magnitude represents a ratio of amplitudes for the output and the input while the angle represents the phase difference between the output and the input.
- Filters
- Low-pass, High-pass, Band-pass, Band-reject; names based on which frequencies of the transfer function have magnitudes that are 70.71% ($$1/\sqrt{2}$$) or more of the maximum possible magnitude
- Cutoff frequency is where the magnitude is at 70.71% of the maximum
- Decibel - based on the base-10 logarithm of a power ratio, multiplied by 10 (to get the deci- part)
- We are assuming power is related to voltage or current squared, and the log of a square is twice the log, so $$H_{dB}=20\,\log_{10}|\mathbb{H}(j\omega)|$$
- Bode plots
- Magnitude plot is $$20\,\log_{10}|\mathbb{H}(j\omega)|$$ versus $$\omega$$ with $$\omega$$ on a log scale
- Angle plot is $$\angle \mathbb{H}(j\omega)$$ versus $$\omega$$ with $$\omega$$ on a log scale
- Translation between decibels and magnitudes
- 1 $$\leftrightarrow$$ 0 dB
- 10 $$\leftrightarrow$$ 20 dB
- 100 $$\leftrightarrow$$ 40 dB
- 0.1 $$\leftrightarrow$$ -20 dB
- 2 $$\leftrightarrow$$ $$\approx$$ 6 dB
- 5 $$\leftrightarrow$$ $$\approx$$ 14 dB
- 0.25=1/2/2 $$\leftrightarrow$$ 0-6-6 dB=-12 dB
- 8=2*2*2=$$2^3$$ $$\leftrightarrow$$ 6+6+6 dB = 3 * 5 dB = 18 dB
- $$\sqrt{2}$$=$$2^{1/2}$$ $$\leftrightarrow$$ $$\frac{1}{2}$$ 6 dB = 3 dB
- 3 dB below maximum for a transfer function amplitude is critical because that indicates where the half-power frequency is.
- Four main terms in transfer functions:
- $$\mathbb{H}(j\omega)=K\,\color{red}{(j\omega)^2}\,\color{green}{\Pi_k(j\omega+a_k)^{n_k}}\,\color{blue}{\Pi_l((j\omega)^2+2\zeta_l\omega_{n_l}+(\omega_{n_l})^2)^{m_l}}$$