RevisionLecture1: Nodal Analysis & FrequencyResponses · 2018. 3. 27. · Exam RevisionLecture1: Nodal Analysis & FrequencyResponses •Exam •Nodal Analysis •Op Amps •Block

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


• 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

1

)X

Op Amps


• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes






• Filters

• Resonance





Y =(1 + 3

1

)X

Inverting amplifier

Y = −81 X1 +

−82 X2 +

−82 X3

Op Amps


• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes






• Filters

• Resonance





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


• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes






• Filters

• Resonance





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


• 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

ωC .

◦ Admittances: 1R , 1

jωL = −jωL , jωC

Reactive Components


• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes






• Filters

• Resonance



ωC .



• 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



ωC .





• 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



ωC .





• 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

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



• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes






• Filters

• Resonance



• If V2

V1


log∣∣∣V2

V1


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

∠

(V2

V1

)= ∠A+ k∠j = ∠A+ k π

2



• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes






• Filters

• Resonance



• If V2

V1


log∣∣∣V2

V1



∠

(V2

V1

)= ∠A+ k∠j = ∠A+ k π

2

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



• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes






• Filters

• Resonance



• If V2

V1


log∣∣∣V2

V1



∠

(V2

V1

)= ∠A+ k∠j = ∠A+ k π

2


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

V1


log∣∣∣V2

V1



∠

(V2

V1

)= ∠A+ k∠j = ∠A+ k π

2



YX =

1

jωC

R+ 1

jωC

= 1jωRC+1



• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes






• Filters

• Resonance



• If V2

V1


log∣∣∣V2

V1



∠

(V2

V1

)= ∠A+ k∠j = ∠A+ k π

2



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


• 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

-40

-20

0

ω (rad/s)0.1 1 10 100 1000

-0.5

0

0.5

ω (rad/s)

π



• 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

-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



• 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

-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)


• 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

+ 1)

.



• 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

+ 1)

.

• 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)| =(

ω2

ω1

)k

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



• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes






• Filters

• Resonance






4. |H(jω2)| =(

ω2

ω1

)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)



• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes






• Filters

• Resonance






4. |H(jω2)| =(

ω2

ω1

)k



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)



• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes






• Filters

• Resonance






4. |H(jω2)| =(

ω2

ω1

)k



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)



• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes






• Filters

• Resonance






4. |H(jω2)| =(

ω2

ω1

)k



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)



• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes






• Filters

• Resonance






4. |H(jω2)| =(

ω2

ω1

)k



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).



• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes






• Filters

• Resonance






4. |H(jω2)| =(

ω2

ω1

)k



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


• 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

+1

Filters


• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes






• Filters

• Resonance





YX = R

R+1/jωC= jωRC

jωRC+1 =jωRCjωa

+1

Filters


• Exam

• Nodal Analysis

• Op Amps

• Block Diagrams

• Diodes






• Filters

• Resonance





YX = R

R+1/jωC= jωRC

jωRC+1 =jωRCjωa

+1

ZX = Z

Y × YX =

(1 + RB

RA

)× jωRC

jωa

+1

Resonance


• Resonant circuits have quadratic factors that cannot be factorized

◦ H(jω) = a (jω)2 + bjω + c = c

((jωω0

)2

+ 2ζ(

jωω0

)+ 1

)

Resonance



◦ 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



◦ H(jω) = a (jω)2 + bjω + c = c

((jωω0

)2

+ 2ζ(

jωω0

)+ 1

)




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

)2

+ 2ζ(

jωω0

)+ 1

)




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

)2

+ 2ζ(

jωω0

)+ 1

)




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

)2

+ 2ζ(

jωω0

)+ 1

)




2c = b√4ac



2ζ = cbω0



XU =

1

jωC

R+jωL+ 1

jωC

= 1(jω)2LC+jωRC+1

Resonance



◦ H(jω) = a (jω)2 + bjω + c = c

((jωω0

)2

+ 2ζ(

jωω0

)+ 1

)




2c = b√4ac



2ζ = cbω0



XU =

1

jωC

R+jωL+ 1

jωC

= 1(jω)2LC+jωRC+1

ω0 =√

1LC , ζ = R

2

√CL , Q = ω0L

R = 12ζ

Resonance



◦ H(jω) = a (jω)2 + bjω + c = c

((jωω0

)2

+ 2ζ(

jωω0

)+ 1

)




2c = b√4ac



2ζ = cbω0



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



◦ H(jω) = a (jω)2 + bjω + c = c

((jωω0

)2

+ 2ζ(

jωω0

)+ 1

)




2c = b√4ac



2ζ = cbω0



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



◦ H(jω) = a (jω)2 + bjω + c = c

((jωω0

)2

+ 2ζ(

jωω0

)+ 1

)




2c = b√4ac



2ζ = cbω0



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



◦ H(jω) = a (jω)2 + bjω + c = c

((jωω0

)2

+ 2ζ(

jωω0

)+ 1

)




2c = b√4ac



2ζ = cbω0


◦ |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


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



◦ H(jω) = a (jω)2 + bjω + c = c

((jωω0

)2

+ 2ζ(

jωω0

)+ 1

)




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 ,

110 ,

310 ,

610

Q = |ZC(ω0) or ZL(ω0)|R = 20, 5, 53 ,

56

Gain@ω0


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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RevisionLecture1: Nodal Analysis & FrequencyResponses · 2018. 3. 27. · Exam RevisionLecture1: Nodal Analysis & FrequencyResponses •Exam •Nodal Analysis •Op Amps •Block