RevisionLecture1: Nodal Analysis & FrequencyResponses · Nodal Analysis RevisionLecture1: Nodal...

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

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

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

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

Nodal Analysis

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

(1) Pick reference node.

Nodal Analysis

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

(2) Label nodes: 8

Nodal Analysis

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

(2) Label nodes: 8, X

Nodal Analysis

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

(2) Label nodes: 8, X and X + 2 sinceit is joined to X via a voltage source.

Nodal Analysis

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

(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

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

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

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

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

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

• 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

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

Non-inverting amplifier

Y =(1 + 3

Op Amps

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

Y =(1 + 3

Inverting amplifier

Y = −81 X1 +

−82 X2 +

−82 X3

Op Amps

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

Y =(1 + 3

Inverting amplifier

Y = −81 X1 +

−82 X2 +

−82 X3

Nodal Analysis: Use two separate KCL equations at V+ and V−.

Op Amps

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

Y =(1 + 3

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

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

Blocks are labelled with their gains and connected using adder/subtractorsshown as circles. Adder inputs are marked + for add or − for subtract.

Block Diagrams

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

To analyse:

1. Label the inputs, the outputs and the output of each adder.

Block Diagrams

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

To analyse:

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

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

To analyse:

• U = X − FGU , Y = FU + FGHU• Follow signals back though the blocks until you meet a labelled node.

Block Diagrams

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

To analyse:

3. Solve the equations (eliminate intermediate node variables):

• U(1 + FG) = X

Block Diagrams

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

To analyse:

• U(1 + FG) = X ⇒ U = X1+FG

Block Diagrams

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

To analyse:

• U(1 + FG) = X ⇒ U = X1+FG

• Y = (1 +GH)FU

Block Diagrams

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

To analyse:

• U(1 + FG) = X ⇒ U = X1+FG

• Y = (1 +GH)FU= (1+GH)F1+FG X

Block Diagrams

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

To analyse:

• 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

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

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

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

• Off: ID = 0, VD < 0.7 ⇒ Diode = open circuit

• On: VD = 0.7, ID > 0

• Assume Diode Off

X = 5 + 2 = 7

Diodes

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

• On: VD = 0.7, ID > 0

X = 5 + 2 = 7VD = 2

Diodes

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

• On: VD = 0.7, ID > 0

X = 5 + 2 = 7VD = 2 Fail: VD > 0.7

Diodes

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

• On: VD = 0.7, ID > 0 ⇒ Diode = 0.7V voltage source

X = 5 + 2 = 7VD = 2 Fail: VD > 0.7

• Assume Diode On

X = 5 + 0.7 = 5.7

Diodes

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

X = 5 + 2 = 7VD = 2 Fail: VD > 0.7

• Assume Diode On

X = 5 + 0.7 = 5.7

ID + 0.71 k = 2mA

Diodes

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

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

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

• Impedances: R, jωL, 1jωC = −j

◦ Admittances: 1R , 1

jωL = −jωL , jωC

Reactive Components

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

• In a capacitor or inductor, the Current and Voltage are 90◦ apart :

◦ CIVIL: Capacitor - I leads V ; Inductor - I lags V

Reactive Components

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

• Average current (or DC current) through a capacitor is always zero

• Average voltage across an inductor is always zero

Reactive Components

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

• 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

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

A phasor represents a time-varying sinusoidal waveform by a fixed complexnumber.

Waveform Phasor

x(t) = F cosωt−G sinωt X = F + jG

Phasors

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

Waveform Phasor

x(t) = F cosωt−G sinωt X = F + jG [Note minus sign]

Phasors

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

Waveform Phasor

x(t) = A cos (ωt+ θ) X = Aejθ = A∠θ

Phasors

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

Waveform Phasor

max (x(t)) = A |X| = A

Phasors

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

Waveform Phasor

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

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

Waveform Phasor

max (x(t)) = A |X| = A

RMS Phasor: V = 1√2V

Phasors

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

Waveform Phasor

max (x(t)) = A |X| = A

RMS Phasor: V = 1√2V ⇒

∣∣∣V∣∣∣2

=⟨x2(t)

Phasors

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

Waveform Phasor

max (x(t)) = A |X| = A

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

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

• 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.

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

• If V2

= A (jω)k = A× jk × ωk (where A is real)◦ magnitude is a straight line with gradient k:

log∣∣∣V2

∣∣∣ = log |A|+ k logω

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

• If V2

log∣∣∣V2

◦ phase is a constant k × π2 (+π if A < 0):

)= ∠A+ k∠j = ∠A+ k π

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

• If V2

log∣∣∣V2

)= ∠A+ k∠j = ∠A+ k π

• Measure magnitude response using decibels = 20 log10|V2||V1| .

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

• If V2

log∣∣∣V2

)= ∠A+ k∠j = ∠A+ k π

A gradient of k on log axes is equivalent to 20k dB/decade(×10 in frequency) or 6k dB/octave ( ×2 in frequency).

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

• If V2

log∣∣∣V2

)= ∠A+ k∠j = ∠A+ k π

= 1jωRC+1

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

• If V2

log∣∣∣V2

)= ∠A+ k∠j = ∠A+ k π

0.1/RC 1/RC 10/RC-30

ω (rad/s)

0.1/RC 1/RC 10/RC

ω (rad/s)

= 1jωRC+1=

1jωωc

+1where ωc =

LF and HF Asymptotes

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

• 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.

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

Example: H(jω) = 60(jω)2+720(jω)

3(jω)3+165(jω)2+762(jω)+600

0.1 1 10 100 1000

ω (rad/s)0.1 1 10 100 1000

ω (rad/s)

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

3(jω)3+165(jω)2+762(jω)+600

0.1 1 10 100 1000

ω (rad/s)0.1 1 10 100 1000

ω (rad/s)

πLF: H(jω) ≃ 1.2jω

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

3(jω)3+165(jω)2+762(jω)+600

0.1 1 10 100 1000

ω (rad/s)0.1 1 10 100 1000

ω (rad/s)

πLF: H(jω) ≃ 1.2jω

HF: H(jω) ≃ 20 (jω)−1

Corner frequencies (linear factors)

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

• We can factorize the numerator and denominator into linear terms of

the form (ajω + b) ≃

{b ω <

∣∣ ba

∣∣ajω ω >

∣∣ ba

∣∣ .

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

{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.

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

{b ω <

∣∣ ba

∣∣ajω ω >

∣∣ ba

∣∣ .

• 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).

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

{b ω <

∣∣ ba

∣∣ajω ω >

∣∣ ba

∣∣ .

• When a and b are real and positive, it is often convenient to write

(ajω + b) = b(

jωωc

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

{b ω <

∣∣ ba

∣∣ajω ω >

∣∣ ba

∣∣ .

• When a and b are real and positive, it is often convenient to write

(ajω + b) = b(

jωωc

• The corner frequencies are the absolute values of the roots of thenumerator and denominator polynomials (values of jω).

Sketching Magnitude Responses (linear factors)

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

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)| =(

|H(jω1)| if the gradient between them is k.

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

4. |H(jω2)| =(

H(jω) = 1.2jω( jω

( jω1+1)( jω

4+1)( jω

0.1 1 10 100 1000

ω (rad/s)

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

4. |H(jω2)| =(

( jω1+1)( jω

4+1)( jω

LF: 1.2jω ⇒|H(j1)| = 1.2 (1.6 dB)

0.1 1 10 100 1000

ω (rad/s)

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

4. |H(jω2)| =(

( jω1+1)( jω

4+1)( jω

LF: 1.2jω ⇒|H(j1)| = 1.2 (1.6 dB)

|H(j4)| =(41

)0× 1.2 = 1.2 0.1 1 10 100 1000

ω (rad/s)

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

4. |H(jω2)| =(

( jω1+1)( jω

4+1)( jω

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

ω (rad/s)

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

4. |H(jω2)| =(

( jω1+1)( jω

4+1)( jω

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

ω (rad/s)

|H(j50)| =(5012

)0× 0.4 = 0.4 (−8 dB).

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

4. |H(jω2)| =(

( jω1+1)( jω

4+1)( jω

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

ω (rad/s)

|H(j50)| =(5012

)0× 0.4 = 0.4 (−8 dB). As a check: HF: 20 (jω)−1

Filters

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

• 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

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

YX = R

R+1/jωC

Filters

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

YX = R

R+1/jωC= jωRC

jωRC+1

Filters

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

YX = R

R+1/jωC= jωRC

jωRC+1 =jωRCjωa

Filters

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

YX = R

R+1/jωC= jωRC

jωRC+1 =jωRCjωa

Filters

• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes

• Filters

• Resonance

YX = R

R+1/jωC= jωRC

jωRC+1 =jωRCjωa

ZX = Z

Y × YX =

(1 + RB

)× jωRC

Resonance

• Resonant circuits have quadratic factors that cannot be factorized

◦ H(jω) = a (jω)2 + bjω + c = c

((jωω0

+ 2ζ(

jωω0

Resonance

◦ H(jω) = a (jω)2 + bjω + c = c

((jωω0

+ 2ζ(

jωω0

◦ Corner frequency: ω0 =√

ca determines the horizontal position

◦ Damping Factor: ζ = bω0

2c = b√4ac

determines the response shape

Resonance

◦ H(jω) = a (jω)2 + bjω + c = c

((jωω0

+ 2ζ(

jωω0

2c = b√4ac

◦ Equivalently Quality Factor: Q ,ω×Average Stored EnergyAverage Power Dissipation ≈ 1

2ζ = cbω0

Resonance

◦ H(jω) = a (jω)2 + bjω + c = c

((jωω0

+ 2ζ(

jωω0

2c = b√4ac

2ζ = cbω0

• At ω = ω0, outer terms cancel (a (jω)2= −c): ⇒ H(jω) = jbω0 = 2jcζ

Resonance

◦ H(jω) = a (jω)2 + bjω + c = c

((jωω0

+ 2ζ(

jωω0

2c = b√4ac

2ζ = cbω0

◦ |H(jω0)| = 2ζ times the straight line approximation at ω0.

Resonance

◦ H(jω) = a (jω)2 + bjω + c = c

((jωω0

+ 2ζ(

jωω0

2c = b√4ac

2ζ = cbω0

R+jωL+ 1

= 1(jω)2LC+jωRC+1

Resonance

◦ H(jω) = a (jω)2 + bjω + c = c

((jωω0

+ 2ζ(

jωω0

2c = b√4ac

2ζ = cbω0

R+jωL+ 1

= 1(jω)2LC+jωRC+1

ω0 =√

1LC , ζ = R

√CL , Q = ω0L

R = 12ζ

Resonance

◦ H(jω) = a (jω)2 + bjω + c = c

((jωω0

+ 2ζ(

jωω0

2c = b√4ac

2ζ = cbω0

R = 5, 20, 60, 120

ζ = 140 ,

100 1k 10k-40

ω (rad/s)

R+jωL+ 1

= 1(jω)2LC+jωRC+1

ω0 =√

1LC , ζ = R

√CL , Q = ω0L

R = 12ζ

Resonance

◦ H(jω) = a (jω)2 + bjω + c = c

((jωω0

+ 2ζ(

jωω0

2c = b√4ac

2ζ = cbω0

R = 5, 20, 60, 120

ζ = 140 ,

Q = |ZC(ω0) or ZL(ω0)|R = 20, 5, 53 ,

100 1k 10k-40

ω (rad/s)

R+jωL+ 1

= 1(jω)2LC+jωRC+1

ω0 =√

1LC , ζ = R

√CL , Q = ω0L

R = 12ζ

Resonance

◦ H(jω) = a (jω)2 + bjω + c = c

((jωω0

+ 2ζ(

jωω0

2c = b√4ac

2ζ = cbω0

R = 5, 20, 60, 120

ζ = 140 ,

Q = |ZC(ω0) or ZL(ω0)|R = 20, 5, 53 ,

Gain@ω0

CornerGain = 12ζ ≈ Q

100 1k 10k-40

ω (rad/s)

R+jωL+ 1

= 1(jω)2LC+jωRC+1

ω0 =√

1LC , ζ = R

√CL , Q = ω0L

R = 12ζ

Resonance

◦ H(jω) = a (jω)2 + bjω + c = c

((jωω0

+ 2ζ(

jωω0

2c = b√4ac

2ζ = cbω0

◦ |H(jω0)| = 2ζ times the straight line approximation at ω0.◦ 3 dB bandwidth of peak ≃ 2ζω0 ≈ ω0

R = 5, 20, 60, 120

ζ = 140 ,

Q = |ZC(ω0) or ZL(ω0)|R = 20, 5, 53 ,

Gain@ω0

100 1k 10k-40

ω (rad/s)

R+jωL+ 1

= 1(jω)2LC+jωRC+1

ω0 =√

1LC , ζ = R

√CL , Q = ω0L

R = 12ζ

Resonance

◦ H(jω) = a (jω)2 + bjω + c = c

((jωω0

+ 2ζ(

jωω0

2c = b√4ac

2ζ = cbω0

◦ |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 ,

Q = |ZC(ω0) or ZL(ω0)|R = 20, 5, 53 ,

Gain@ω0

100 1k 10k-40

ω (rad/s)

R+jωL+ 1

= 1(jω)2LC+jωRC+1

ω0 =√

1LC , ζ = R

√CL , Q = ω0L

R = 12ζ

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