revisionlecture1: nodal analysis & frequencyresponses · 2018. 3. 27. · exam...
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Revision Lecture 1: NodalAnalysis & Frequency Responses
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 1 / 14
Exam
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 2 / 14
Exam Format
Question 1 (40%): eight short parts covering the whole syllabus.
Questions 2 and 3: single topic questions (answer both)
Syllabus
Does include: Everything in the notes.
Does not include: Two-port parameters (2008:1j), Gaussian elimination(2007:2c), Application areas (2008:3d), Nullators and Norators (2008:4c),Small-signal component models (2008:4e), Gain-bandwidth product(2006:4c), Zener Diodes (2008/9 syllabus), Non-ideal models of L, C andtransformer (2011/12 syllabus), Transmission lines VSWR and crankdiagram (2011/12 syllabus).
Nodal Analysis
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 3 / 14
Nodal Analysis
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 3 / 14
(1) Pick reference node.
Nodal Analysis
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 3 / 14
(1) Pick reference node.
(2) Label nodes: 8
Nodal Analysis
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 3 / 14
(1) Pick reference node.
(2) Label nodes: 8, X
Nodal Analysis
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 3 / 14
(1) Pick reference node.
(2) Label nodes: 8, X and X + 2 sinceit is joined to X via a voltage source.
Nodal Analysis
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 3 / 14
(1) Pick reference node.
(2) Label nodes: 8, X and X + 2 sinceit is joined to X via a voltage source.
(3) Write KCL equations but count all thenodes connected via floating voltagesources as a single “super-node” givingone equation
X−81 + X
2 + (X+2)−03 = 0
Nodal Analysis
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 3 / 14
(1) Pick reference node.
(2) Label nodes: 8, X and X + 2 sinceit is joined to X via a voltage source.
(3) Write KCL equations but count all thenodes connected via floating voltagesources as a single “super-node” givingone equation
X−81 + X
2 + (X+2)−03 = 0
Ohm’s law always involves the differencebetween the voltages at either end of aresistor. (Obvious but easily forgotten)
Nodal Analysis
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 3 / 14
(1) Pick reference node.
(2) Label nodes: 8, X and X + 2 sinceit is joined to X via a voltage source.
(3) Write KCL equations but count all thenodes connected via floating voltagesources as a single “super-node” givingone equation
X−81 + X
2 + (X+2)−03 = 0
Ohm’s law always involves the differencebetween the voltages at either end of aresistor. (Obvious but easily forgotten)
(4) Solve the equations: X = 4
Op Amps
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 4 / 14
• Ideal Op Amp: (a) Zero input current, (b) Infinite gain(b) ⇒ V+ = V− provided the circuit has negative feedback.
• Negative Feedback: An increase in Vout makes (V+ − V−) decrease.
Op Amps
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 4 / 14
• Ideal Op Amp: (a) Zero input current, (b) Infinite gain(b) ⇒ V+ = V− provided the circuit has negative feedback.
• Negative Feedback: An increase in Vout makes (V+ − V−) decrease.
Non-inverting amplifier
Y =(1 + 3
1
)X
Op Amps
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 4 / 14
• Ideal Op Amp: (a) Zero input current, (b) Infinite gain(b) ⇒ V+ = V− provided the circuit has negative feedback.
• Negative Feedback: An increase in Vout makes (V+ − V−) decrease.
Non-inverting amplifier
Y =(1 + 3
1
)X
Inverting amplifier
Y = −81 X1 +
−82 X2 +
−82 X3
Op Amps
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 4 / 14
• Ideal Op Amp: (a) Zero input current, (b) Infinite gain(b) ⇒ V+ = V− provided the circuit has negative feedback.
• Negative Feedback: An increase in Vout makes (V+ − V−) decrease.
Non-inverting amplifier
Y =(1 + 3
1
)X
Inverting amplifier
Y = −81 X1 +
−82 X2 +
−82 X3
Nodal Analysis: Use two separate KCL equations at V+ and V−.
Op Amps
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 4 / 14
• Ideal Op Amp: (a) Zero input current, (b) Infinite gain(b) ⇒ V+ = V− provided the circuit has negative feedback.
• Negative Feedback: An increase in Vout makes (V+ − V−) decrease.
Non-inverting amplifier
Y =(1 + 3
1
)X
Inverting amplifier
Y = −81 X1 +
−82 X2 +
−82 X3
Nodal Analysis: Use two separate KCL equations at V+ and V−.Do not do KCL at Vout except to find the op-amp output current.
Block Diagrams
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 5 / 14
Blocks are labelled with their gains and connected using adder/subtractorsshown as circles. Adder inputs are marked + for add or − for subtract.
Block Diagrams
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 5 / 14
Blocks are labelled with their gains and connected using adder/subtractorsshown as circles. Adder inputs are marked + for add or − for subtract.
To analyse:
1. Label the inputs, the outputs and the output of each adder.
Block Diagrams
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 5 / 14
Blocks are labelled with their gains and connected using adder/subtractorsshown as circles. Adder inputs are marked + for add or − for subtract.
To analyse:
1. Label the inputs, the outputs and the output of each adder.
2. Write down an equation for each variable:
• U = X − FGU• Follow signals back though the blocks until you meet a labelled node.
Block Diagrams
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 5 / 14
Blocks are labelled with their gains and connected using adder/subtractorsshown as circles. Adder inputs are marked + for add or − for subtract.
To analyse:
1. Label the inputs, the outputs and the output of each adder.
2. Write down an equation for each variable:
• U = X − FGU , Y = FU + FGHU• Follow signals back though the blocks until you meet a labelled node.
Block Diagrams
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 5 / 14
Blocks are labelled with their gains and connected using adder/subtractorsshown as circles. Adder inputs are marked + for add or − for subtract.
To analyse:
1. Label the inputs, the outputs and the output of each adder.
2. Write down an equation for each variable:
• U = X − FGU , Y = FU + FGHU• Follow signals back though the blocks until you meet a labelled node.
3. Solve the equations (eliminate intermediate node variables):
• U(1 + FG) = X
Block Diagrams
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 5 / 14
Blocks are labelled with their gains and connected using adder/subtractorsshown as circles. Adder inputs are marked + for add or − for subtract.
To analyse:
1. Label the inputs, the outputs and the output of each adder.
2. Write down an equation for each variable:
• U = X − FGU , Y = FU + FGHU• Follow signals back though the blocks until you meet a labelled node.
3. Solve the equations (eliminate intermediate node variables):
• U(1 + FG) = X ⇒ U = X1+FG
Block Diagrams
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 5 / 14
Blocks are labelled with their gains and connected using adder/subtractorsshown as circles. Adder inputs are marked + for add or − for subtract.
To analyse:
1. Label the inputs, the outputs and the output of each adder.
2. Write down an equation for each variable:
• U = X − FGU , Y = FU + FGHU• Follow signals back though the blocks until you meet a labelled node.
3. Solve the equations (eliminate intermediate node variables):
• U(1 + FG) = X ⇒ U = X1+FG
• Y = (1 +GH)FU
Block Diagrams
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 5 / 14
Blocks are labelled with their gains and connected using adder/subtractorsshown as circles. Adder inputs are marked + for add or − for subtract.
To analyse:
1. Label the inputs, the outputs and the output of each adder.
2. Write down an equation for each variable:
• U = X − FGU , Y = FU + FGHU• Follow signals back though the blocks until you meet a labelled node.
3. Solve the equations (eliminate intermediate node variables):
• U(1 + FG) = X ⇒ U = X1+FG
• Y = (1 +GH)FU= (1+GH)F1+FG X
Block Diagrams
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 5 / 14
Blocks are labelled with their gains and connected using adder/subtractorsshown as circles. Adder inputs are marked + for add or − for subtract.
To analyse:
1. Label the inputs, the outputs and the output of each adder.
2. Write down an equation for each variable:
• U = X − FGU , Y = FU + FGHU• Follow signals back though the blocks until you meet a labelled node.
3. Solve the equations (eliminate intermediate node variables):
• U(1 + FG) = X ⇒ U = X1+FG
• Y = (1 +GH)FU= (1+GH)F1+FG X
[Note: “Wires” carry information not current: KCL not valid]
Diodes
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 6 / 14
Each diode in a circuit is in one of two modes; each has an equalitycondition and an inequality condition:
• Off: ID = 0, VD < 0.7
• On: VD = 0.7, ID > 0
(a) Guess the mode(b) Do nodal analysis assuming the equality condition(c) Check the inequality condition. If the inequality condition fails, you
made the wrong guess.
Diodes
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 6 / 14
Each diode in a circuit is in one of two modes; each has an equalitycondition and an inequality condition:
• Off: ID = 0, VD < 0.7 ⇒ Diode = open circuit
• On: VD = 0.7, ID > 0
(a) Guess the mode(b) Do nodal analysis assuming the equality condition(c) Check the inequality condition. If the inequality condition fails, you
made the wrong guess.
• Assume Diode Off
X = 5 + 2 = 7
Diodes
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 6 / 14
Each diode in a circuit is in one of two modes; each has an equalitycondition and an inequality condition:
• Off: ID = 0, VD < 0.7 ⇒ Diode = open circuit
• On: VD = 0.7, ID > 0
(a) Guess the mode(b) Do nodal analysis assuming the equality condition(c) Check the inequality condition. If the inequality condition fails, you
made the wrong guess.
• Assume Diode Off
X = 5 + 2 = 7VD = 2
Diodes
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 6 / 14
Each diode in a circuit is in one of two modes; each has an equalitycondition and an inequality condition:
• Off: ID = 0, VD < 0.7 ⇒ Diode = open circuit
• On: VD = 0.7, ID > 0
(a) Guess the mode(b) Do nodal analysis assuming the equality condition(c) Check the inequality condition. If the inequality condition fails, you
made the wrong guess.
• Assume Diode Off
X = 5 + 2 = 7VD = 2 Fail: VD > 0.7
Diodes
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 6 / 14
Each diode in a circuit is in one of two modes; each has an equalitycondition and an inequality condition:
• Off: ID = 0, VD < 0.7 ⇒ Diode = open circuit
• On: VD = 0.7, ID > 0 ⇒ Diode = 0.7V voltage source
(a) Guess the mode(b) Do nodal analysis assuming the equality condition(c) Check the inequality condition. If the inequality condition fails, you
made the wrong guess.
• Assume Diode Off
X = 5 + 2 = 7VD = 2 Fail: VD > 0.7
• Assume Diode On
X = 5 + 0.7 = 5.7
Diodes
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 6 / 14
Each diode in a circuit is in one of two modes; each has an equalitycondition and an inequality condition:
• Off: ID = 0, VD < 0.7 ⇒ Diode = open circuit
• On: VD = 0.7, ID > 0 ⇒ Diode = 0.7V voltage source
(a) Guess the mode(b) Do nodal analysis assuming the equality condition(c) Check the inequality condition. If the inequality condition fails, you
made the wrong guess.
• Assume Diode Off
X = 5 + 2 = 7VD = 2 Fail: VD > 0.7
• Assume Diode On
X = 5 + 0.7 = 5.7
ID + 0.71 k = 2mA
Diodes
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 6 / 14
Each diode in a circuit is in one of two modes; each has an equalitycondition and an inequality condition:
• Off: ID = 0, VD < 0.7 ⇒ Diode = open circuit
• On: VD = 0.7, ID > 0 ⇒ Diode = 0.7V voltage source
(a) Guess the mode(b) Do nodal analysis assuming the equality condition(c) Check the inequality condition. If the inequality condition fails, you
made the wrong guess.
• Assume Diode Off
X = 5 + 2 = 7VD = 2 Fail: VD > 0.7
• Assume Diode On
X = 5 + 0.7 = 5.7
ID + 0.71 k = 2mA OK: ID > 0
Reactive Components
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 7 / 14
• Impedances: R, jωL, 1jωC = −j
ωC .
◦ Admittances: 1R , 1
jωL = −jωL , jωC
Reactive Components
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 7 / 14
• Impedances: R, jωL, 1jωC = −j
ωC .
◦ Admittances: 1R , 1
jωL = −jωL , jωC
• In a capacitor or inductor, the Current and Voltage are 90◦ apart :
◦ CIVIL: Capacitor - I leads V ; Inductor - I lags V
Reactive Components
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 7 / 14
• Impedances: R, jωL, 1jωC = −j
ωC .
◦ Admittances: 1R , 1
jωL = −jωL , jωC
• In a capacitor or inductor, the Current and Voltage are 90◦ apart :
◦ CIVIL: Capacitor - I leads V ; Inductor - I lags V
• Average current (or DC current) through a capacitor is always zero
• Average voltage across an inductor is always zero
Reactive Components
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 7 / 14
• Impedances: R, jωL, 1jωC = −j
ωC .
◦ Admittances: 1R , 1
jωL = −jωL , jωC
• In a capacitor or inductor, the Current and Voltage are 90◦ apart :
◦ CIVIL: Capacitor - I leads V ; Inductor - I lags V
• Average current (or DC current) through a capacitor is always zero
• Average voltage across an inductor is always zero
• Average power absorbed by a capacitor or inductor is always zero
Phasors
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 8 / 14
A phasor represents a time-varying sinusoidal waveform by a fixed complexnumber.
Waveform Phasor
x(t) = F cosωt−G sinωt X = F + jG
Phasors
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 8 / 14
A phasor represents a time-varying sinusoidal waveform by a fixed complexnumber.
Waveform Phasor
x(t) = F cosωt−G sinωt X = F + jG [Note minus sign]
Phasors
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 8 / 14
A phasor represents a time-varying sinusoidal waveform by a fixed complexnumber.
Waveform Phasor
x(t) = F cosωt−G sinωt X = F + jG [Note minus sign]
x(t) = A cos (ωt+ θ) X = Aejθ = A∠θ
Phasors
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 8 / 14
A phasor represents a time-varying sinusoidal waveform by a fixed complexnumber.
Waveform Phasor
x(t) = F cosωt−G sinωt X = F + jG [Note minus sign]
x(t) = A cos (ωt+ θ) X = Aejθ = A∠θ
max (x(t)) = A |X| = A
Phasors
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 8 / 14
A phasor represents a time-varying sinusoidal waveform by a fixed complexnumber.
Waveform Phasor
x(t) = F cosωt−G sinωt X = F + jG [Note minus sign]
x(t) = A cos (ωt+ θ) X = Aejθ = A∠θ
max (x(t)) = A |X| = A
x(t) is the projection of a rotating rod onto thereal (horizontal) axis.
X = F + jG is its starting position at t = 0.
Phasors
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 8 / 14
A phasor represents a time-varying sinusoidal waveform by a fixed complexnumber.
Waveform Phasor
x(t) = F cosωt−G sinωt X = F + jG [Note minus sign]
x(t) = A cos (ωt+ θ) X = Aejθ = A∠θ
max (x(t)) = A |X| = A
x(t) is the projection of a rotating rod onto thereal (horizontal) axis.
X = F + jG is its starting position at t = 0.
RMS Phasor: V = 1√2V
Phasors
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 8 / 14
A phasor represents a time-varying sinusoidal waveform by a fixed complexnumber.
Waveform Phasor
x(t) = F cosωt−G sinωt X = F + jG [Note minus sign]
x(t) = A cos (ωt+ θ) X = Aejθ = A∠θ
max (x(t)) = A |X| = A
x(t) is the projection of a rotating rod onto thereal (horizontal) axis.
X = F + jG is its starting position at t = 0.
RMS Phasor: V = 1√2V ⇒
∣∣∣V∣∣∣2
=⟨x2(t)
⟩
Phasors
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 8 / 14
A phasor represents a time-varying sinusoidal waveform by a fixed complexnumber.
Waveform Phasor
x(t) = F cosωt−G sinωt X = F + jG [Note minus sign]
x(t) = A cos (ωt+ θ) X = Aejθ = A∠θ
max (x(t)) = A |X| = A
x(t) is the projection of a rotating rod onto thereal (horizontal) axis.
X = F + jG is its starting position at t = 0.
RMS Phasor: V = 1√2V ⇒
∣∣∣V∣∣∣2
=⟨x2(t)
⟩
Complex Power: V I∗ = |I |2Z = |V |2Z∗
= P + jQ
P is average power (Watts), Q is reactive power (VARs)
Plotting Frequency Responses
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 9 / 14
• Plot the magnitude response and phase response as separate graphs.Use log scale for frequency and magnitude and linear scale for phase:this gives graphs that can be approximated by straight line segments.
Plotting Frequency Responses
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 9 / 14
• Plot the magnitude response and phase response as separate graphs.Use log scale for frequency and magnitude and linear scale for phase:this gives graphs that can be approximated by straight line segments.
• If V2
V1
= A (jω)k = A× jk × ωk (where A is real)◦ magnitude is a straight line with gradient k:
log∣∣∣V2
V1
∣∣∣ = log |A|+ k logω
Plotting Frequency Responses
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 9 / 14
• Plot the magnitude response and phase response as separate graphs.Use log scale for frequency and magnitude and linear scale for phase:this gives graphs that can be approximated by straight line segments.
• If V2
V1
= A (jω)k = A× jk × ωk (where A is real)◦ magnitude is a straight line with gradient k:
log∣∣∣V2
V1
∣∣∣ = log |A|+ k logω
◦ phase is a constant k × π2 (+π if A < 0):
∠
(V2
V1
)= ∠A+ k∠j = ∠A+ k π
2
Plotting Frequency Responses
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 9 / 14
• Plot the magnitude response and phase response as separate graphs.Use log scale for frequency and magnitude and linear scale for phase:this gives graphs that can be approximated by straight line segments.
• If V2
V1
= A (jω)k = A× jk × ωk (where A is real)◦ magnitude is a straight line with gradient k:
log∣∣∣V2
V1
∣∣∣ = log |A|+ k logω
◦ phase is a constant k × π2 (+π if A < 0):
∠
(V2
V1
)= ∠A+ k∠j = ∠A+ k π
2
• Measure magnitude response using decibels = 20 log10|V2||V1| .
Plotting Frequency Responses
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 9 / 14
• Plot the magnitude response and phase response as separate graphs.Use log scale for frequency and magnitude and linear scale for phase:this gives graphs that can be approximated by straight line segments.
• If V2
V1
= A (jω)k = A× jk × ωk (where A is real)◦ magnitude is a straight line with gradient k:
log∣∣∣V2
V1
∣∣∣ = log |A|+ k logω
◦ phase is a constant k × π2 (+π if A < 0):
∠
(V2
V1
)= ∠A+ k∠j = ∠A+ k π
2
• Measure magnitude response using decibels = 20 log10|V2||V1| .
A gradient of k on log axes is equivalent to 20k dB/decade(×10 in frequency) or 6k dB/octave ( ×2 in frequency).
Plotting Frequency Responses
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 9 / 14
• Plot the magnitude response and phase response as separate graphs.Use log scale for frequency and magnitude and linear scale for phase:this gives graphs that can be approximated by straight line segments.
• If V2
V1
= A (jω)k = A× jk × ωk (where A is real)◦ magnitude is a straight line with gradient k:
log∣∣∣V2
V1
∣∣∣ = log |A|+ k logω
◦ phase is a constant k × π2 (+π if A < 0):
∠
(V2
V1
)= ∠A+ k∠j = ∠A+ k π
2
• Measure magnitude response using decibels = 20 log10|V2||V1| .
A gradient of k on log axes is equivalent to 20k dB/decade(×10 in frequency) or 6k dB/octave ( ×2 in frequency).
YX =
1
jωC
R+ 1
jωC
= 1jωRC+1
Plotting Frequency Responses
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 9 / 14
• Plot the magnitude response and phase response as separate graphs.Use log scale for frequency and magnitude and linear scale for phase:this gives graphs that can be approximated by straight line segments.
• If V2
V1
= A (jω)k = A× jk × ωk (where A is real)◦ magnitude is a straight line with gradient k:
log∣∣∣V2
V1
∣∣∣ = log |A|+ k logω
◦ phase is a constant k × π2 (+π if A < 0):
∠
(V2
V1
)= ∠A+ k∠j = ∠A+ k π
2
• Measure magnitude response using decibels = 20 log10|V2||V1| .
A gradient of k on log axes is equivalent to 20k dB/decade(×10 in frequency) or 6k dB/octave ( ×2 in frequency).
0.1/RC 1/RC 10/RC-30
-20
-10
0
ω (rad/s)
|Gai
n| (
dB)
0.1/RC 1/RC 10/RC
-0.4
-0.2
0
ω (rad/s)
Pha
se/π
YX =
1
jωC
R+ 1
jωC
= 1jωRC+1=
1jωωc
+1where ωc =
1RC
LF and HF Asymptotes
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 10 / 14
• Frequency response is always a ratio of two polynomials in jω withreal coefficients that depend on the component values.
◦ The terms with the lowest power of jω on top and bottom givesthe low-frequency asymptote
◦ The terms with the highest power of jω on top and bottomgives the high-frequency asymptote.
LF and HF Asymptotes
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 10 / 14
• Frequency response is always a ratio of two polynomials in jω withreal coefficients that depend on the component values.
◦ The terms with the lowest power of jω on top and bottom givesthe low-frequency asymptote
◦ The terms with the highest power of jω on top and bottomgives the high-frequency asymptote.
Example: H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600
0.1 1 10 100 1000
-40
-20
0
ω (rad/s)0.1 1 10 100 1000
-0.5
0
0.5
ω (rad/s)
π
LF and HF Asymptotes
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 10 / 14
• Frequency response is always a ratio of two polynomials in jω withreal coefficients that depend on the component values.
◦ The terms with the lowest power of jω on top and bottom givesthe low-frequency asymptote
◦ The terms with the highest power of jω on top and bottomgives the high-frequency asymptote.
Example: H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600
0.1 1 10 100 1000
-40
-20
0
ω (rad/s)0.1 1 10 100 1000
-0.5
0
0.5
ω (rad/s)
πLF: H(jω) ≃ 1.2jω
LF and HF Asymptotes
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 10 / 14
• Frequency response is always a ratio of two polynomials in jω withreal coefficients that depend on the component values.
◦ The terms with the lowest power of jω on top and bottom givesthe low-frequency asymptote
◦ The terms with the highest power of jω on top and bottomgives the high-frequency asymptote.
Example: H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600
0.1 1 10 100 1000
-40
-20
0
ω (rad/s)0.1 1 10 100 1000
-0.5
0
0.5
ω (rad/s)
πLF: H(jω) ≃ 1.2jω
HF: H(jω) ≃ 20 (jω)−1
Corner frequencies (linear factors)
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 11 / 14
• We can factorize the numerator and denominator into linear terms of
the form (ajω + b) ≃
{b ω <
∣∣ ba
∣∣ajω ω >
∣∣ ba
∣∣ .
Corner frequencies (linear factors)
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 11 / 14
• We can factorize the numerator and denominator into linear terms of
the form (ajω + b) ≃
{b ω <
∣∣ ba
∣∣ajω ω >
∣∣ ba
∣∣ .
• At the corner frequency, ωc =∣∣ ba
∣∣, the slope of the magnituderesponse changes by ±1 ( ±20 dB/decade) because the linear termintroduces another factor of ω into the numerator or denominator forω > ωc.
Corner frequencies (linear factors)
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 11 / 14
• We can factorize the numerator and denominator into linear terms of
the form (ajω + b) ≃
{b ω <
∣∣ ba
∣∣ajω ω >
∣∣ ba
∣∣ .
• At the corner frequency, ωc =∣∣ ba
∣∣, the slope of the magnituderesponse changes by ±1 ( ±20 dB/decade) because the linear termintroduces another factor of ω into the numerator or denominator forω > ωc.
• The phase changes by ±π2 because the linear term introduces
another factor of j into the numerator or denominator for ω > ωc.
◦ The phase change is gradual and takes place over the range0.1ωc to 10ωc (±π
2 spread over two decades in ω).
Corner frequencies (linear factors)
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 11 / 14
• We can factorize the numerator and denominator into linear terms of
the form (ajω + b) ≃
{b ω <
∣∣ ba
∣∣ajω ω >
∣∣ ba
∣∣ .
• At the corner frequency, ωc =∣∣ ba
∣∣, the slope of the magnituderesponse changes by ±1 ( ±20 dB/decade) because the linear termintroduces another factor of ω into the numerator or denominator forω > ωc.
• The phase changes by ±π2 because the linear term introduces
another factor of j into the numerator or denominator for ω > ωc.
◦ The phase change is gradual and takes place over the range0.1ωc to 10ωc (±π
2 spread over two decades in ω).
• When a and b are real and positive, it is often convenient to write
(ajω + b) = b(
jωωc
+ 1)
.
Corner frequencies (linear factors)
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 11 / 14
• We can factorize the numerator and denominator into linear terms of
the form (ajω + b) ≃
{b ω <
∣∣ ba
∣∣ajω ω >
∣∣ ba
∣∣ .
• At the corner frequency, ωc =∣∣ ba
∣∣, the slope of the magnituderesponse changes by ±1 ( ±20 dB/decade) because the linear termintroduces another factor of ω into the numerator or denominator forω > ωc.
• The phase changes by ±π2 because the linear term introduces
another factor of j into the numerator or denominator for ω > ωc.
◦ The phase change is gradual and takes place over the range0.1ωc to 10ωc (±π
2 spread over two decades in ω).
• When a and b are real and positive, it is often convenient to write
(ajω + b) = b(
jωωc
+ 1)
.
• The corner frequencies are the absolute values of the roots of thenumerator and denominator polynomials (values of jω).
Sketching Magnitude Responses (linear factors)
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 12 / 14
1. Find corner frequencies: (a) factorize the numerator/denominatorpolynomials or (b) find their roots
2. Find LF and HF asymptotes. A (jω)k
has a slope of k; substituteω = ωc to get the response at first/last corner frequency.
3. At a corner frequency, the gradient of the magnitude responsechanges by ±1 (±20 dB/decade). + for numerator (top line) and −for denominator (bottom line).
4. |H(jω2)| =(
ω2
ω1
)k
|H(jω1)| if the gradient between them is k.
Sketching Magnitude Responses (linear factors)
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 12 / 14
1. Find corner frequencies: (a) factorize the numerator/denominatorpolynomials or (b) find their roots
2. Find LF and HF asymptotes. A (jω)k
has a slope of k; substituteω = ωc to get the response at first/last corner frequency.
3. At a corner frequency, the gradient of the magnitude responsechanges by ±1 (±20 dB/decade). + for numerator (top line) and −for denominator (bottom line).
4. |H(jω2)| =(
ω2
ω1
)k
|H(jω1)| if the gradient between them is k.
H(jω) = 1.2jω( jω
12+1)
( jω1+1)( jω
4+1)( jω
50+1)
0.1 1 10 100 1000
-40
-20
0
ω (rad/s)
Sketching Magnitude Responses (linear factors)
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 12 / 14
1. Find corner frequencies: (a) factorize the numerator/denominatorpolynomials or (b) find their roots
2. Find LF and HF asymptotes. A (jω)k
has a slope of k; substituteω = ωc to get the response at first/last corner frequency.
3. At a corner frequency, the gradient of the magnitude responsechanges by ±1 (±20 dB/decade). + for numerator (top line) and −for denominator (bottom line).
4. |H(jω2)| =(
ω2
ω1
)k
|H(jω1)| if the gradient between them is k.
H(jω) = 1.2jω( jω
12+1)
( jω1+1)( jω
4+1)( jω
50+1)
LF: 1.2jω ⇒|H(j1)| = 1.2 (1.6 dB)
0.1 1 10 100 1000
-40
-20
0
ω (rad/s)
Sketching Magnitude Responses (linear factors)
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 12 / 14
1. Find corner frequencies: (a) factorize the numerator/denominatorpolynomials or (b) find their roots
2. Find LF and HF asymptotes. A (jω)k
has a slope of k; substituteω = ωc to get the response at first/last corner frequency.
3. At a corner frequency, the gradient of the magnitude responsechanges by ±1 (±20 dB/decade). + for numerator (top line) and −for denominator (bottom line).
4. |H(jω2)| =(
ω2
ω1
)k
|H(jω1)| if the gradient between them is k.
H(jω) = 1.2jω( jω
12+1)
( jω1+1)( jω
4+1)( jω
50+1)
LF: 1.2jω ⇒|H(j1)| = 1.2 (1.6 dB)
|H(j4)| =(41
)0× 1.2 = 1.2 0.1 1 10 100 1000
-40
-20
0
ω (rad/s)
Sketching Magnitude Responses (linear factors)
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 12 / 14
1. Find corner frequencies: (a) factorize the numerator/denominatorpolynomials or (b) find their roots
2. Find LF and HF asymptotes. A (jω)k
has a slope of k; substituteω = ωc to get the response at first/last corner frequency.
3. At a corner frequency, the gradient of the magnitude responsechanges by ±1 (±20 dB/decade). + for numerator (top line) and −for denominator (bottom line).
4. |H(jω2)| =(
ω2
ω1
)k
|H(jω1)| if the gradient between them is k.
H(jω) = 1.2jω( jω
12+1)
( jω1+1)( jω
4+1)( jω
50+1)
LF: 1.2jω ⇒|H(j1)| = 1.2 (1.6 dB)
|H(j4)| =(41
)0× 1.2 = 1.2
|H(j12)| =(124
)−1× 1.2 = 0.4
0.1 1 10 100 1000
-40
-20
0
ω (rad/s)
Sketching Magnitude Responses (linear factors)
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 12 / 14
1. Find corner frequencies: (a) factorize the numerator/denominatorpolynomials or (b) find their roots
2. Find LF and HF asymptotes. A (jω)k
has a slope of k; substituteω = ωc to get the response at first/last corner frequency.
3. At a corner frequency, the gradient of the magnitude responsechanges by ±1 (±20 dB/decade). + for numerator (top line) and −for denominator (bottom line).
4. |H(jω2)| =(
ω2
ω1
)k
|H(jω1)| if the gradient between them is k.
H(jω) = 1.2jω( jω
12+1)
( jω1+1)( jω
4+1)( jω
50+1)
LF: 1.2jω ⇒|H(j1)| = 1.2 (1.6 dB)
|H(j4)| =(41
)0× 1.2 = 1.2
|H(j12)| =(124
)−1× 1.2 = 0.4
0.1 1 10 100 1000
-40
-20
0
ω (rad/s)
|H(j50)| =(5012
)0× 0.4 = 0.4 (−8 dB).
Sketching Magnitude Responses (linear factors)
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 12 / 14
1. Find corner frequencies: (a) factorize the numerator/denominatorpolynomials or (b) find their roots
2. Find LF and HF asymptotes. A (jω)k
has a slope of k; substituteω = ωc to get the response at first/last corner frequency.
3. At a corner frequency, the gradient of the magnitude responsechanges by ±1 (±20 dB/decade). + for numerator (top line) and −for denominator (bottom line).
4. |H(jω2)| =(
ω2
ω1
)k
|H(jω1)| if the gradient between them is k.
H(jω) = 1.2jω( jω
12+1)
( jω1+1)( jω
4+1)( jω
50+1)
LF: 1.2jω ⇒|H(j1)| = 1.2 (1.6 dB)
|H(j4)| =(41
)0× 1.2 = 1.2
|H(j12)| =(124
)−1× 1.2 = 0.4
0.1 1 10 100 1000
-40
-20
0
ω (rad/s)
|H(j50)| =(5012
)0× 0.4 = 0.4 (−8 dB). As a check: HF: 20 (jω)−1
Filters
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 13 / 14
• Filter: a circuit designed to amplify some frequencies and/orattenuate others. Very widely used.
• The order of the filter is the highest power of jω in the denominatorof the frequency response.
• Often use op-amps to provide a low impedance output.
Filters
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 13 / 14
• Filter: a circuit designed to amplify some frequencies and/orattenuate others. Very widely used.
• The order of the filter is the highest power of jω in the denominatorof the frequency response.
• Often use op-amps to provide a low impedance output.
YX = R
R+1/jωC
Filters
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 13 / 14
• Filter: a circuit designed to amplify some frequencies and/orattenuate others. Very widely used.
• The order of the filter is the highest power of jω in the denominatorof the frequency response.
• Often use op-amps to provide a low impedance output.
YX = R
R+1/jωC= jωRC
jωRC+1
Filters
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 13 / 14
• Filter: a circuit designed to amplify some frequencies and/orattenuate others. Very widely used.
• The order of the filter is the highest power of jω in the denominatorof the frequency response.
• Often use op-amps to provide a low impedance output.
YX = R
R+1/jωC= jωRC
jωRC+1 =jωRCjωa
+1
Filters
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 13 / 14
• Filter: a circuit designed to amplify some frequencies and/orattenuate others. Very widely used.
• The order of the filter is the highest power of jω in the denominatorof the frequency response.
• Often use op-amps to provide a low impedance output.
YX = R
R+1/jωC= jωRC
jωRC+1 =jωRCjωa
+1
Filters
Revision Lecture 1: NodalAnalysis &FrequencyResponses
• Exam
• Nodal Analysis
• Op Amps
• Block Diagrams
• Diodes
• Reactive Components
• Phasors• Plotting FrequencyResponses
• LF and HF Asymptotes
• Corner frequencies (linearfactors)
• Sketching MagnitudeResponses (linear factors)
• Filters
• Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 13 / 14
• Filter: a circuit designed to amplify some frequencies and/orattenuate others. Very widely used.
• The order of the filter is the highest power of jω in the denominatorof the frequency response.
• Often use op-amps to provide a low impedance output.
YX = R
R+1/jωC= jωRC
jωRC+1 =jωRCjωa
+1
ZX = Z
Y × YX =
(1 + RB
RA
)× jωRC
jωa
+1
Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 14 / 14
• Resonant circuits have quadratic factors that cannot be factorized
◦ H(jω) = a (jω)2 + bjω + c = c
((jωω0
)2
+ 2ζ(
jωω0
)+ 1
)
Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 14 / 14
• Resonant circuits have quadratic factors that cannot be factorized
◦ H(jω) = a (jω)2 + bjω + c = c
((jωω0
)2
+ 2ζ(
jωω0
)+ 1
)
◦ Corner frequency: ω0 =√
ca determines the horizontal position
◦ Damping Factor: ζ = bω0
2c = b√4ac
determines the response shape
Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 14 / 14
• Resonant circuits have quadratic factors that cannot be factorized
◦ H(jω) = a (jω)2 + bjω + c = c
((jωω0
)2
+ 2ζ(
jωω0
)+ 1
)
◦ Corner frequency: ω0 =√
ca determines the horizontal position
◦ Damping Factor: ζ = bω0
2c = b√4ac
determines the response shape
◦ Equivalently Quality Factor: Q ,ω×Average Stored EnergyAverage Power Dissipation ≈ 1
2ζ = cbω0
Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 14 / 14
• Resonant circuits have quadratic factors that cannot be factorized
◦ H(jω) = a (jω)2 + bjω + c = c
((jωω0
)2
+ 2ζ(
jωω0
)+ 1
)
◦ Corner frequency: ω0 =√
ca determines the horizontal position
◦ Damping Factor: ζ = bω0
2c = b√4ac
determines the response shape
◦ Equivalently Quality Factor: Q ,ω×Average Stored EnergyAverage Power Dissipation ≈ 1
2ζ = cbω0
• At ω = ω0, outer terms cancel (a (jω)2= −c): ⇒ H(jω) = jbω0 = 2jcζ
Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 14 / 14
• Resonant circuits have quadratic factors that cannot be factorized
◦ H(jω) = a (jω)2 + bjω + c = c
((jωω0
)2
+ 2ζ(
jωω0
)+ 1
)
◦ Corner frequency: ω0 =√
ca determines the horizontal position
◦ Damping Factor: ζ = bω0
2c = b√4ac
determines the response shape
◦ Equivalently Quality Factor: Q ,ω×Average Stored EnergyAverage Power Dissipation ≈ 1
2ζ = cbω0
• At ω = ω0, outer terms cancel (a (jω)2= −c): ⇒ H(jω) = jbω0 = 2jcζ
◦ |H(jω0)| = 2ζ times the straight line approximation at ω0.
Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 14 / 14
• Resonant circuits have quadratic factors that cannot be factorized
◦ H(jω) = a (jω)2 + bjω + c = c
((jωω0
)2
+ 2ζ(
jωω0
)+ 1
)
◦ Corner frequency: ω0 =√
ca determines the horizontal position
◦ Damping Factor: ζ = bω0
2c = b√4ac
determines the response shape
◦ Equivalently Quality Factor: Q ,ω×Average Stored EnergyAverage Power Dissipation ≈ 1
2ζ = cbω0
• At ω = ω0, outer terms cancel (a (jω)2= −c): ⇒ H(jω) = jbω0 = 2jcζ
◦ |H(jω0)| = 2ζ times the straight line approximation at ω0.
XU =
1
jωC
R+jωL+ 1
jωC
= 1(jω)2LC+jωRC+1
Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 14 / 14
• Resonant circuits have quadratic factors that cannot be factorized
◦ H(jω) = a (jω)2 + bjω + c = c
((jωω0
)2
+ 2ζ(
jωω0
)+ 1
)
◦ Corner frequency: ω0 =√
ca determines the horizontal position
◦ Damping Factor: ζ = bω0
2c = b√4ac
determines the response shape
◦ Equivalently Quality Factor: Q ,ω×Average Stored EnergyAverage Power Dissipation ≈ 1
2ζ = cbω0
• At ω = ω0, outer terms cancel (a (jω)2= −c): ⇒ H(jω) = jbω0 = 2jcζ
◦ |H(jω0)| = 2ζ times the straight line approximation at ω0.
XU =
1
jωC
R+jωL+ 1
jωC
= 1(jω)2LC+jωRC+1
ω0 =√
1LC , ζ = R
2
√CL , Q = ω0L
R = 12ζ
Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 14 / 14
• Resonant circuits have quadratic factors that cannot be factorized
◦ H(jω) = a (jω)2 + bjω + c = c
((jωω0
)2
+ 2ζ(
jωω0
)+ 1
)
◦ Corner frequency: ω0 =√
ca determines the horizontal position
◦ Damping Factor: ζ = bω0
2c = b√4ac
determines the response shape
◦ Equivalently Quality Factor: Q ,ω×Average Stored EnergyAverage Power Dissipation ≈ 1
2ζ = cbω0
• At ω = ω0, outer terms cancel (a (jω)2= −c): ⇒ H(jω) = jbω0 = 2jcζ
◦ |H(jω0)| = 2ζ times the straight line approximation at ω0.
R = 5, 20, 60, 120
ζ = 140 ,
110 ,
310 ,
610
100 1k 10k-40
-20
0
20
ω (rad/s)
R=5
R=120
XU =
1
jωC
R+jωL+ 1
jωC
= 1(jω)2LC+jωRC+1
ω0 =√
1LC , ζ = R
2
√CL , Q = ω0L
R = 12ζ
Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 14 / 14
• Resonant circuits have quadratic factors that cannot be factorized
◦ H(jω) = a (jω)2 + bjω + c = c
((jωω0
)2
+ 2ζ(
jωω0
)+ 1
)
◦ Corner frequency: ω0 =√
ca determines the horizontal position
◦ Damping Factor: ζ = bω0
2c = b√4ac
determines the response shape
◦ Equivalently Quality Factor: Q ,ω×Average Stored EnergyAverage Power Dissipation ≈ 1
2ζ = cbω0
• At ω = ω0, outer terms cancel (a (jω)2= −c): ⇒ H(jω) = jbω0 = 2jcζ
◦ |H(jω0)| = 2ζ times the straight line approximation at ω0.
R = 5, 20, 60, 120
ζ = 140 ,
110 ,
310 ,
610
Q = |ZC(ω0) or ZL(ω0)|R = 20, 5, 53 ,
56
100 1k 10k-40
-20
0
20
ω (rad/s)
R=5
R=120
XU =
1
jωC
R+jωL+ 1
jωC
= 1(jω)2LC+jωRC+1
ω0 =√
1LC , ζ = R
2
√CL , Q = ω0L
R = 12ζ
Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 14 / 14
• Resonant circuits have quadratic factors that cannot be factorized
◦ H(jω) = a (jω)2 + bjω + c = c
((jωω0
)2
+ 2ζ(
jωω0
)+ 1
)
◦ Corner frequency: ω0 =√
ca determines the horizontal position
◦ Damping Factor: ζ = bω0
2c = b√4ac
determines the response shape
◦ Equivalently Quality Factor: Q ,ω×Average Stored EnergyAverage Power Dissipation ≈ 1
2ζ = cbω0
• At ω = ω0, outer terms cancel (a (jω)2= −c): ⇒ H(jω) = jbω0 = 2jcζ
◦ |H(jω0)| = 2ζ times the straight line approximation at ω0.
R = 5, 20, 60, 120
ζ = 140 ,
110 ,
310 ,
610
Q = |ZC(ω0) or ZL(ω0)|R = 20, 5, 53 ,
56
Gain@ω0
CornerGain = 12ζ ≈ Q
100 1k 10k-40
-20
0
20
ω (rad/s)
R=5
R=120
XU =
1
jωC
R+jωL+ 1
jωC
= 1(jω)2LC+jωRC+1
ω0 =√
1LC , ζ = R
2
√CL , Q = ω0L
R = 12ζ
Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 14 / 14
• Resonant circuits have quadratic factors that cannot be factorized
◦ H(jω) = a (jω)2 + bjω + c = c
((jωω0
)2
+ 2ζ(
jωω0
)+ 1
)
◦ Corner frequency: ω0 =√
ca determines the horizontal position
◦ Damping Factor: ζ = bω0
2c = b√4ac
determines the response shape
◦ Equivalently Quality Factor: Q ,ω×Average Stored EnergyAverage Power Dissipation ≈ 1
2ζ = cbω0
• At ω = ω0, outer terms cancel (a (jω)2= −c): ⇒ H(jω) = jbω0 = 2jcζ
◦ |H(jω0)| = 2ζ times the straight line approximation at ω0.◦ 3 dB bandwidth of peak ≃ 2ζω0 ≈ ω0
Q .
R = 5, 20, 60, 120
ζ = 140 ,
110 ,
310 ,
610
Q = |ZC(ω0) or ZL(ω0)|R = 20, 5, 53 ,
56
Gain@ω0
CornerGain = 12ζ ≈ Q
100 1k 10k-40
-20
0
20
ω (rad/s)
R=5
R=120
XU =
1
jωC
R+jωL+ 1
jωC
= 1(jω)2LC+jωRC+1
ω0 =√
1LC , ζ = R
2
√CL , Q = ω0L
R = 12ζ
Resonance
E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 14 / 14
• Resonant circuits have quadratic factors that cannot be factorized
◦ H(jω) = a (jω)2 + bjω + c = c
((jωω0
)2
+ 2ζ(
jωω0
)+ 1
)
◦ Corner frequency: ω0 =√
ca determines the horizontal position
◦ Damping Factor: ζ = bω0
2c = b√4ac
determines the response shape
◦ Equivalently Quality Factor: Q ,ω×Average Stored EnergyAverage Power Dissipation ≈ 1
2ζ = cbω0
• At ω = ω0, outer terms cancel (a (jω)2= −c): ⇒ H(jω) = jbω0 = 2jcζ
◦ |H(jω0)| = 2ζ times the straight line approximation at ω0.◦ 3 dB bandwidth of peak ≃ 2ζω0 ≈ ω0
Q . ∆phase = ±π over 2ζ decades
R = 5, 20, 60, 120
ζ = 140 ,
110 ,
310 ,
610
Q = |ZC(ω0) or ZL(ω0)|R = 20, 5, 53 ,
56
Gain@ω0
CornerGain = 12ζ ≈ Q
100 1k 10k-40
-20
0
20
ω (rad/s)
R=5
R=120
XU =
1
jωC
R+jωL+ 1
jωC
= 1(jω)2LC+jωRC+1
ω0 =√
1LC , ζ = R
2
√CL , Q = ω0L
R = 12ζ
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