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Revision Lecture 1: Nodal Analysis & Frequency Responses Revision Lecture 1: Nodal Analysis & Frequency Responses Exam Nodal Analysis Op Amps Block Diagrams Diodes Reactive Components Phasors Plotting Frequency Responses LF and HF Asymptotes Corner frequencies (linear factors) Sketching Magnitude Responses (linear factors) Filters Resonance E1.1 Analysis of Circuits (2018-10453) Revision Lecture 1 – 1 / 14

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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