The course web site is: ...trueman/elec273files/ELEC273_11... · Final Exam (confirmed): ... Find the Thevenin Equivalent Circuit at a frequency of 𝜔𝜔=3 rad/sec. Use phasors

ELEC273 Lecture Notes Set 11 AC Circuit Theorems

Homework on phasors and impedance.

The course web site is: http://users.encs.concordia.ca/~trueman/web_page_273.htmFinal Exam (confirmed): Friday December 15, 2017 from 9:00 to 12:00 (confirmed)

Network Theorems:

Chapter 10 Sinusoidal Steady State Analysis• Superposition• Thevenin’s Theorem• Norton’s Theorem

Chapter 11 AC Power Analysis• The Maximum Power Transfer Theorem

Thevenin’ s Theorem for AC Circuits

Thevenin’s TheoremAny two-terminal network consisting of voltage sources, current sources, dependent sources, inductors, capacitors and resistors is equivalent to a voltage source 𝑉𝑉𝑇𝑇 in series with an impedance 𝑍𝑍𝑇𝑇.

Procedure to find the Thevenin Equivalent Circuit:1.Open-circuit test: terminate the circuit with an open circuit and find the open-circuit voltage, 𝑉𝑉𝑜𝑜𝑜𝑜.2.Short-circuit test: terminate the circuit with a short circuit and find the short-circuit current, 𝐼𝐼𝑠𝑠𝑜𝑜. 3.The Thevenin equivalent voltage source is 𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜.4.The Thevenin equivalent impedance is 𝑍𝑍𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜

𝐼𝐼𝑠𝑠𝑜𝑜.

Note that 𝑍𝑍𝑇𝑇 can also be found as the impedance of the “dead circuit” with the independent sources set equal to zero.

TwoTerminal Circuit

𝑉𝑉𝑇𝑇

𝑍𝑍𝑇𝑇

Example: Thevenin’ s Theorem

1)Find the Thevenin Equivalent Circuit at a frequency of 𝜔𝜔=3 rad/sec. Use phasors relative to “sine”.

2)Find the voltage across a load consisting of a 1 ohm resistor in series with a capacitor of value C=0.16667 F, using: • The Thevenin equivalent circuit• The original circuitThe answer must be the same in both cases!

2 Ω

1 Ω

12

H10 sin 3𝑡𝑡

A

B

Find the Thevenin Equivalent Circuit.

1) Find the open-circuit voltage.2) Find the short-circuit current.

Convert the circuit to phasors and impedance.The impedance of the inductance is 𝑗𝑗𝜔𝜔𝜔𝜔 = j3x 1

2=

𝑗𝑗𝑗.5 ohms

In this example I have written my phasors relative to “sine”.

Hence 10 sin 3t becomes phasor 10 angle zero.

However, we must remember to convert phasors back to “sine”.

Thus phasor 𝑉𝑉 = 𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃 converts to time function𝑣𝑣 𝑡𝑡 = 𝐴𝐴 sin(𝜔𝜔𝑡𝑡 + 𝜃𝜃)

2 Ω

1 Ω

12

H10 sin 3𝑡𝑡

A

B

2 Ω

1 Ω𝑗𝑗𝑗.5 Ω10

A

B

+𝑉𝑉𝑜𝑜𝑜𝑜−

Find the open-circuit voltage

015.12

10=−−

− ocococ VjVV

01

65.1

62

106 =−−− ocococ Vj

jVjVj

064)10(3 =−−− ocococ jVVVj

064330 =−−− ocococ jVVjVj

jjVV ococ 3094 =+

°∠=∠=∠

∠=

+= 24046.34182.0046.3

153.1849.9571.130

9430

jjVoc

°∠== 24046.3ocT VV

2 Ω

1 Ω𝑗𝑗𝑗.5 Ω10

A

B

+𝑉𝑉𝑜𝑜𝑜𝑜−

2)Find the short-circuit current

°∠== 052

10scI

2 Ω

1 Ω𝑗𝑗𝑗.5 Ω10

A

B

𝐼𝐼𝑠𝑠𝑜𝑜

3)Find the impedance

°∠= 05scI°∠= 24046.3ocV

°∠== 24046.3ocT VV

2477.05565.00.246092.005

0.24046.3 jIVZ

sc

ocT +=∠=

∠∠

==

The reactance of 0.2477 ohms is equivalent to an inductor of value 𝜔𝜔𝜔𝜔 = 0.2477 so 𝜔𝜔 = 0.2477𝜔𝜔

= 0.24773

= 0.08257 H

This phasor is written relative to sine so the time function is 3.046 sin(𝜔𝜔𝑡𝑡 + 24°)

𝑉𝑉𝑇𝑇 = 3.046∠24°

𝑍𝑍𝑇𝑇 = 0.5565 + 𝑗𝑗𝑗.2477 Ω

𝑉𝑉𝑇𝑇 = 3.046∠24°

0.5565 Ω 0.08257 H

The impedance of the capacitor is 𝑍𝑍𝐶𝐶 = 1𝐽𝐽𝜔𝜔𝐶𝐶

= 1𝑗𝑗3𝑗𝑗0.16667

= −𝑗𝑗2 ohms

Draw the circuit in the frequency domain using phasors and impedances.

2)Find the voltage across a load consisting of a 1 ohm resistor in series with a capacitor of value C=0.16667 F, using: • The original circuit• The Thevenin equivalent circuit

Using the original circuit:

𝑗𝑗𝑗.5 Ω

2 Ω

1 Ω

12

H10 sin 3𝑡𝑡

A

B

1 Ω

0.16667 F

2 Ω

1 Ω10

B

1 Ω

−𝑗𝑗2 Ω

Using the original circuit, find 𝑉𝑉𝐿𝐿:

02115.12

10=

−−−−

−j

VVjVV LLLL

Write a node equation:

210

2115.12=

−+++

jVV

jVV LLLL

( ) ( ) ( )( ) 5

21)5.1(221)5.1(2212215.1=

−+−+−+−

jjjjjjjVL

536

5.311=

++

jjVL

°∠=++

= 9.8906.25.311

365jjVL

Do we get the same answer using the Thevenin Equivalent Circuit?

𝑗𝑗𝑗.5 Ω2 Ω

1 Ω10

B

1 Ω

−𝑗𝑗2 Ω

𝑉𝑉𝐿𝐿

2477.05565.00.246092.0 jZT +=∠=

Find the load voltage using the Thevenin Equivalent Circuit. Define 𝑍𝑍𝐿𝐿 = 1 − 𝑗𝑗2 Ω

°∠== 24046.3ocT VV

0=−−

L

L

T

LT

ZV

ZVV

T

T

LTL Z

VZZ

V =

+

11

T

T

LT

LTL Z

VZZ

ZZV =

+

LT

TLL ZZ

VZV+

=

21 jZL −=

)21(2477.05565.0)0.24046.3)(21(jj

jVL −++∠−

=

7523.15565.1)0.24046.3)(4.63236.2(

jVL −

∠−∠=

8.48343.2)0.24046.3)(4.63236.2(

−∠∠−∠

=LV

°∠= 915.8906.2LV

°∠= 9.8906.2LVThis agrees with the answer obtained above of

Note that this equation is simply a voltage divider between 𝑍𝑍𝑇𝑇 and 𝑍𝑍𝐿𝐿.

𝑉𝑉𝑇𝑇𝑍𝑍𝑇𝑇

𝑉𝑉𝐿𝐿

−𝑗𝑗2 Ω

1 Ω

Norton’s Theorem for AC Circuits.

Norton’s TheoremAny two-terminal network consisting of voltage sources, current sources, dependent sources, capacitors, inductors and resistors is equivalent to a current source 𝐼𝐼𝑁𝑁 in parallel with an admittance 𝑌𝑌𝑁𝑁.

Procedure to find the Norton Equivalent Circuit:1.Open-circuit test: terminate the circuit with an open circuit and find the open-circuit voltage, 𝑉𝑉𝑜𝑜𝑜𝑜.2.Short-circuit test: terminate the circuit with a short circuit and find the short-circuit current, 𝐼𝐼𝑠𝑠𝑜𝑜. 3.The Norton equivalent current source is 𝐼𝐼𝑁𝑁 = 𝐼𝐼𝑠𝑠𝑜𝑜.4.The Norton equivalent susceptance is 𝑌𝑌𝑁𝑁 = 𝐼𝐼𝑠𝑠𝑜𝑜

𝑉𝑉𝑜𝑜𝑜𝑜.

Note that 𝑌𝑌𝑁𝑁 can also be found as the admittance of the circuit with the independent sources set equal to zero.

Two TerminalCircuit

A

B

A

B

𝐼𝐼𝑁𝑁 𝑌𝑌𝑁𝑁

Norton Equivalent from the Thevenin EquivalentFind the open-circuit voltage 𝑉𝑉𝑜𝑜𝑜𝑜 and the short-circuit current 𝐼𝐼𝑠𝑠𝑜𝑜.

Thevenin: 𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜𝑍𝑍𝑇𝑇 =

𝑉𝑉𝑜𝑜𝑜𝑜𝐼𝐼𝑠𝑠𝑜𝑜

Norton:𝐼𝐼𝑁𝑁 = 𝐼𝐼𝑆𝑆𝐶𝐶𝑌𝑌𝑁𝑁 = 𝐼𝐼𝑠𝑠𝑜𝑜

𝑉𝑉𝑜𝑜𝑜𝑜

We can change The Thevenin Equivalent Circuit to the Norton Equivalent Circuits using:𝑌𝑌𝑁𝑁 = 1

𝑍𝑍𝑇𝑇and𝐼𝐼𝑁𝑁 = 𝐼𝐼𝑆𝑆𝐶𝐶 = 𝑉𝑉𝑇𝑇

𝑍𝑍𝑇𝑇

𝑉𝑉𝑇𝑇

𝑍𝑍𝑇𝑇

𝐼𝐼𝑁𝑁 𝑌𝑌𝑁𝑁

𝑉𝑉𝑇𝑇

𝑍𝑍𝑇𝑇

𝐼𝐼𝑆𝑆𝐶𝐶 =𝑉𝑉𝑇𝑇𝑍𝑍𝑇𝑇

Example: Find the Norton Equivalent Circuit

Find the Norton Equivalent Circuit at a frequency of 𝜔𝜔=3 rad/sec.

Method:1.Find the open-circuit voltage.

2.Find the short-circuit current.

3.Find the Norton Equivalent Circuit:𝐼𝐼𝑁𝑁 = 𝐼𝐼𝑆𝑆𝐶𝐶𝑌𝑌𝑁𝑁 = 𝐼𝐼𝑠𝑠𝑜𝑜

𝑉𝑉𝑜𝑜𝑜𝑜

We found the Thevenin Equivalent Circuit earlier in this set of notes.

Open-circuit “load”.

Short-circuit “load”.

2 Ω

1 Ω12

H10 sin 3𝑡𝑡

A

B

2 Ω

1 Ω12

H10 sin 3𝑡𝑡

A

B

+𝑉𝑉𝑜𝑜𝑜𝑜−

𝐼𝐼𝑠𝑠𝑜𝑜10 sin 3𝑡𝑡

2 Ω12

H1 Ω

A

B

Find the open-circuit voltage:

1) Find the open-circuit voltage

015.12

10=−−

− ocococ VjVV

Convert the circuit to phasors and impedance.The impedance of the inductance is 𝑗𝑗𝜔𝜔𝜔𝜔 = j3x 1

2= 𝑗𝑗𝑗.5

°∠== 24046.3ocT VV

We solved for the open-circuit voltage and the short-circuit current when we found the Thevenin equivalent earlier in these notes.

2 Ω

1 Ω

12

H10 sin 3𝑡𝑡

A

B

2 Ω

1 Ω𝑗𝑗𝑗.5 Ω10

A

B

+𝑉𝑉𝑜𝑜𝑜𝑜−

Find the short-circuit current:

°∠== 052

10scI

2 Ω

1 Ω𝑗𝑗𝑗.5 Ω10

A

B

𝐼𝐼𝑠𝑠𝑜𝑜

3)Find the admittance:

°∠= 05scI°∠= 24046.3ocV

°∠== 05scN II

°−∠=∠∠

== 0.24641.10.24046.3

05

oc

scN V

IY

2 Ω

1 Ω𝑗𝑗𝑗.5 Ω10

A

B

A

B

𝐼𝐼𝑁𝑁 𝑌𝑌𝑁𝑁

Norton Equivalent, example #2:

• Find the Norton Equivalent Circuit at 60 Hz.• Convert the Norton Equivalent to the Thevenin Equivalent Circuit.

Procedure to find the Norton Equivalent Circuit:1.Open-circuit test: terminate the circuit with an open circuit and find the open-circuit voltage, 𝑉𝑉𝑜𝑜𝑜𝑜.2.Short-circuit test: terminate the circuit with a short circuit and find the short-circuit current, 𝐼𝐼𝑠𝑠𝑜𝑜. 3.The Norton equivalent current source is 𝐼𝐼𝑁𝑁 = 𝐼𝐼𝑠𝑠𝑜𝑜.4.The Norton equivalent susceptance is 𝑌𝑌𝑁𝑁 = 𝐼𝐼𝑠𝑠𝑜𝑜

𝑉𝑉𝑜𝑜𝑜𝑜.

Convert to phasors and impedances:

Voltage source:𝑣𝑣𝑠𝑠 𝑡𝑡 = 110 cos𝜔𝜔𝑡𝑡 becomes phasor 𝑉𝑉𝑠𝑠 = 110 volts

At 𝑓𝑓 =60 Hz the radian frequency is 𝜔𝜔 = 2𝜋𝜋𝑓𝑓 = 377 r/s

𝑗𝑗𝜔𝜔𝜔𝜔 = 𝑗𝑗𝑗𝑗3𝑗𝑗𝑗𝑗𝑗.3𝑗𝑗𝑗10−3 = 𝑗𝑗2 ohms

𝑗𝑗𝜔𝜔𝜔𝜔 = 𝑗𝑗𝑗𝑗3𝑗𝑗𝑗𝑗10.62𝑗𝑗10−3 = 𝑗𝑗4 ohms

1𝑗𝑗𝜔𝜔𝐶𝐶

= 1𝑗𝑗𝑗𝑗377𝑗𝑗133𝑗𝑗10−6

= −𝑗𝑗2 ohms

1.Find the open-circuit voltage:

𝑉𝑉𝑜𝑜𝑜𝑜5 + 𝑗𝑗28 − 𝑗𝑗2 = 110

𝑉𝑉𝑜𝑜𝑜𝑜 = 1108 − 𝑗𝑗25 + 2𝑗𝑗

𝑉𝑉𝑜𝑜𝑜𝑜 = 1108.25∠ − 14.0°5.39∠ − 21.8°

𝑉𝑉𝑜𝑜𝑜𝑜 = 168.4∠ − 35.8°

110 − 𝑉𝑉𝑜𝑜𝑜𝑜1 + 𝑗𝑗2 −

𝑉𝑉𝑜𝑜𝑜𝑜1 + 𝑗𝑗4 −

𝑉𝑉𝑜𝑜𝑜𝑜−𝑗𝑗2 = 0

𝑉𝑉𝑜𝑜𝑜𝑜1

1 + 𝑗𝑗2 +1

1 + 𝑗𝑗4 +1−𝑗𝑗2 =

1101 + 𝑗𝑗2

𝑉𝑉𝑜𝑜𝑜𝑜1 + 𝑗𝑗4 −𝑗𝑗2 + 1 + 𝑗𝑗2 −𝑗𝑗2 + (1 + 𝑗𝑗2)(1 + 𝑗𝑗4)

(1 + 𝑗𝑗2)(1 + 𝑗𝑗4)(−𝑗𝑗2) =110

1 + 𝑗𝑗2

𝑉𝑉𝑜𝑜𝑜𝑜1 + 𝑗𝑗4 −𝑗𝑗2 + 1 + 𝑗𝑗2 −𝑗𝑗2 + (1 + 𝑗𝑗2)(1 + 𝑗𝑗4)

(1 + 𝑗𝑗4)(−𝑗𝑗2) = 110

𝑉𝑉𝑜𝑜𝑜𝑜−2𝑗𝑗 + 8 − 2𝑗𝑗 + 4 + 1 + 𝑗𝑗2 + 𝑗𝑗4 − 8

−𝑗𝑗2 + 8 = 110

𝑉𝑉𝑜𝑜𝑜𝑜5 + 𝑗𝑗28 − 𝑗𝑗2 = 110

𝑉𝑉𝑜𝑜𝑜𝑜

+

-

𝑉𝑉𝑜𝑜𝑜𝑜

2.Find the short-circuit current:

𝐼𝐼𝑠𝑠𝑜𝑜 =110

1 + 𝑗𝑗2 = 22 − j44 = 49.19∠ − 63.4°

𝐼𝐼𝑠𝑠𝑜𝑜

3 and 4.Find the Norton Equivalent Circuit:

°−∠== 4.6319.49scN II

6.272921.08.354.1684.6319.49

−∠=°−∠°−∠

==oc

scN V

IY

𝐼𝐼𝑠𝑠𝑜𝑜 =110

1 + 𝑗𝑗2= 22 − j44 = 49.19∠ − 63.4°

𝑉𝑉𝑜𝑜𝑜𝑜 = 168.4∠ − 35.8°

Siemens

Amps

A

B

𝐼𝐼𝑁𝑁 𝑌𝑌𝑁𝑁

Check by finding the dead-circuit impedance:

𝑍𝑍𝑇𝑇

𝑍𝑍𝑇𝑇

1 + 𝑗𝑗4 ∥ −𝑗𝑗2 = 1+𝑗𝑗4 −𝑗𝑗21+𝑗𝑗4−𝑗𝑗2

= 0.8 − 𝑗𝑗3.6 ohms

𝑍𝑍𝑇𝑇 = 1 + 𝑗𝑗2 ∥ 0.8 − 𝑗𝑗3.6 =1 + 𝑗𝑗2 0.8 − 𝑗𝑗3.61 + 𝑗𝑗2 + 0.8 − 𝑗𝑗3.6

𝑍𝑍𝑇𝑇 = 3.034 + 𝑗𝑗1.586 = 3.423∠27.6° ohms

𝑌𝑌𝑁𝑁 = 1𝑍𝑍𝑇𝑇

= 13.423∠27.6°

= 0.2920∠ − 27.6° Siemens

This agrees with the value found on the previous slide:

6.272921.0 −∠==oc

scN V

IY

Convert the Norton Equivalent to the Thevenin Equivalent Circuit:

°−∠= 4.6319.49NI°−∠= 6.272921.0NY

𝑉𝑉𝑜𝑜𝑜𝑜 = 𝐼𝐼𝑁𝑁𝑌𝑌𝑁𝑁

= 49.19∠−63.4°0.2921∠−27.6°

= 136.6 − 𝑗𝑗𝑗𝑗.51 = 168.4∠ − 35.8° volts

𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜 = 98.51 = 168.4∠ − 35.8°𝑍𝑍𝑇𝑇 = 1

𝑌𝑌𝑁𝑁= 1

0.2921∠−27.6°= 3.423∠27.6° ohms

+𝑉𝑉𝑜𝑜𝑜𝑜-

The Superposition Theorem for AC Circuits

The Superposition Theorem is exactly the same for AC circuits as for DC circuits.

The response of a circuit 𝑉𝑉0 with two sources 𝑉𝑉𝑠𝑠1 and 𝑉𝑉𝑠𝑠2 acting together is equal to the sum of:• the response 𝑉𝑉01 with source 𝑉𝑉𝑠𝑠1 acting alone and 𝑉𝑉𝑠𝑠2=0, • plus the response 𝑉𝑉02 with source 𝑉𝑉𝑠𝑠2 acting alone and 𝑉𝑉𝑠𝑠1=0,𝑉𝑉0 = 𝑉𝑉01 + 𝑉𝑉02.

Two sources acting together. Source #1 acting alone. Source #2 acting alone. Two sources acting together. Source #1 acting alone. Source #2 acting alone.

𝑉𝑉𝑠𝑠1 𝑉𝑉𝑠𝑠2

+𝑉𝑉𝑜𝑜 −

Linear Circuit 𝑉𝑉𝑠𝑠2

+𝑉𝑉𝑜𝑜2 −

Linear Circuit𝑉𝑉𝑠𝑠1

+𝑉𝑉𝑜𝑜1 −

Linear Circuit

Superposition: DC Sources and AC Source

Use Superposition to separate this problem into:1)The DC response to the DC sources 𝑉𝑉2 = 3 volts and 𝐼𝐼1 = 1 amp2)The AC response to the AC source 𝑣𝑣1 𝑡𝑡 = 0.1 cos(𝑗𝑗𝑗𝑗𝑡𝑡) volts at 𝜔𝜔=1000 r/s

In the circuit below there are two DC sources, 𝑉𝑉2 = 3 volts and 𝐼𝐼1 = 1 amp. There is also an AC source, 𝑣𝑣1 𝑡𝑡 = 0.1 cos 𝑗𝑗𝑗𝑗𝑡𝑡 volts.

Use Superposition to find the voltage across the current generator 𝑣𝑣 𝑡𝑡 .

The component values are 𝑅𝑅1 = 1 Ω, 𝑅𝑅2 = 2 Ω, 𝐶𝐶1 = 500 microFarads, 𝐶𝐶2 = 40 microFarads, 𝜔𝜔 = 1 mH.

0.1 cos 𝑗𝑗𝑗𝑗𝑡𝑡

𝐶𝐶1 = 500 𝜇𝜇F

𝑅𝑅1 = 1 Ω

𝐶𝐶2 = 40 𝜇𝜇F 𝑅𝑅2 = 2 Ω 𝜔𝜔 = 1 mH

5𝑣𝑣𝑗𝑗+𝑣𝑣𝑗𝑗−

𝑉𝑉2 = 3 volts𝐼𝐼1 = 1 amp+𝑣𝑣−

𝑣𝑣(𝑡𝑡)

Solve with the DC sources:

• The AC source is set to zero.• The capacitors become open circuits.• The inductor becomes a short circuit. • Since 𝐶𝐶1 and 𝐶𝐶2 are open-circuits at DC, there is zero DC current in 𝑅𝑅1 and

so 𝑣𝑣𝑗𝑗 = 0• Then the dependent source is zero: 5𝑣𝑣𝑗𝑗 = 0

Node equation:

−𝑣𝑣 − 3

2 − 1 = 03 − 𝑣𝑣 − 2 = 0

𝑣𝑣 = 1 volt at DC

Draw the DC circuit:• AC source becomes zero volts = short circuit• Capacitor: 𝑖𝑖 = 𝐶𝐶 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑= 0, C’s become open

circuits. • Inductor: 𝑣𝑣 = 𝜔𝜔 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑= 0, L’s become short

circuits.

+𝑣𝑣𝑗𝑗−

5𝑣𝑣𝑗𝑗 12 Ω

3+𝑣𝑣−

3 volts0.1 cos 𝑗𝑗𝑗𝑗𝑡𝑡

𝐶𝐶1

𝑅𝑅1

𝐶𝐶2 𝑅𝑅2 𝜔𝜔

5𝑣𝑣𝑗𝑗+𝑣𝑣𝑗𝑗−

1 amp+𝑣𝑣−

𝑣𝑣(𝑡𝑡)

Solve with the AC source: use phasors and impedance.

Draw the AC circuit.

0.1 cos 𝑗𝑗𝑗𝑗𝑡𝑡

𝐶𝐶1 = 500 𝜇𝜇F

1 Ω

𝐶𝐶2 = 40 𝜇𝜇F 2 Ω 1 mH

5𝑣𝑣𝑗𝑗+𝑣𝑣𝑗𝑗−

Use Superposition to separate this problem into:1)The DC response to the DC sources 𝑉𝑉2 = 3 volts and 𝐼𝐼1 = 1 amp2)The AC response to the AC source 𝑣𝑣1 𝑡𝑡 = 0.1 cos(𝑗𝑗𝑗𝑗𝑡𝑡) volts at 𝜔𝜔=1000 r/s

+𝑣𝑣−

0.1 cos 𝑗𝑗𝑗𝑗𝑡𝑡

𝐶𝐶1 = 500 𝜇𝜇F

𝑅𝑅1 = 1 Ω

𝐶𝐶2 = 40 𝜇𝜇F 𝑅𝑅2 = 2 Ω 𝜔𝜔 = 1 mH

5𝑣𝑣𝑗𝑗+𝑣𝑣𝑗𝑗−

𝑉𝑉2 = 3 volts𝐼𝐼1 = 1 amp+𝑣𝑣−

𝑣𝑣(𝑡𝑡)

Use phasors and impedance:Draw the AC circuit: • The DC sources are set to zero.• 𝑣𝑣1 𝑡𝑡 = 0.1 cos(𝑗𝑗𝑗𝑗𝑡𝑡)

becomes phasor 𝑉𝑉1 = 0.1• The frequency is 𝜔𝜔 = 1000 r/s.• 1

𝑗𝑗𝜔𝜔𝐶𝐶1= 1

𝑗𝑗𝑗𝑗1000𝑗𝑗𝑗00𝑗𝑗10−6= −𝑗𝑗2

• 1𝑗𝑗𝜔𝜔𝐶𝐶2

= 1𝑗𝑗𝑗𝑗1000𝑗𝑗40𝑗𝑗10−6

= −𝑗𝑗2𝑗• 𝑗𝑗𝜔𝜔𝜔𝜔 = 𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗10−3 = 𝑗𝑗𝑗

Node equations:

0.1 − 𝑉𝑉𝑗𝑗−𝑗𝑗2 −

𝑉𝑉𝑗𝑗1 −

𝑉𝑉𝑗𝑗 − 𝑉𝑉−𝑗𝑗2𝑗 = 0

𝑉𝑉𝑗𝑗 − 𝑉𝑉−𝑗𝑗2𝑗 − 5𝑉𝑉𝑗𝑗 −

𝑉𝑉2 + 𝑗𝑗 = 0

0.1 cos 𝑗𝑗𝑗𝑗𝑡𝑡

𝐶𝐶1 = 500 𝜇𝜇F

1 Ω

𝐶𝐶2 = 40 𝜇𝜇F 2 Ω 1 mH

5𝑣𝑣𝑗𝑗+𝑣𝑣𝑗𝑗−

0.1

−𝑗𝑗2

1 Ω

−𝑗𝑗2𝑗 2 Ω 𝑗𝑗𝑗

5𝑉𝑉𝑗𝑗+𝑉𝑉𝑗𝑗−

𝑉𝑉𝑗𝑗 𝑉𝑉

+𝑉𝑉−

Solve the equations for V:Node equations:0.1 − 𝑉𝑉𝑗𝑗−𝑗𝑗2

−𝑉𝑉𝑗𝑗1−𝑉𝑉𝑗𝑗 − 𝑉𝑉−𝑗𝑗2𝑗

= 0

𝑉𝑉𝑗𝑗 − 𝑉𝑉−𝑗𝑗2𝑗

− 5𝑉𝑉𝑗𝑗 −𝑉𝑉

2 + 𝑗𝑗= 0

Multiply the 1st equation by (-j2)(-j25):−𝑗𝑗2𝑗 0.1 − 𝑉𝑉𝑗𝑗 − −𝑗𝑗2 −𝑗𝑗2𝑗 𝑉𝑉𝑗𝑗 − (−𝑗𝑗2)(𝑉𝑉𝑗𝑗 − 𝑉𝑉) = 0−𝑗𝑗2.5 + 𝑗𝑗2𝑗𝑉𝑉𝑗𝑗 + 50𝑉𝑉𝑗𝑗 + 𝑗𝑗2𝑉𝑉𝑗𝑗 − 𝑗𝑗2𝑉𝑉 = 0+50 + 𝑗𝑗2𝑗 𝑉𝑉𝑗𝑗 − 𝑗𝑗2𝑉𝑉 = 𝑗𝑗2.5

Multiply the 2st equation by (-j25)(2+j):2 + 𝑗𝑗 𝑉𝑉𝑗𝑗 − 𝑉𝑉 − −𝑗𝑗2𝑗 2 + 𝑗𝑗 5𝑉𝑉𝑗𝑗 − −𝑗𝑗2𝑗 𝑉𝑉 = 02 + 𝑗𝑗 𝑉𝑉𝑗𝑗 − 2 + j 𝑉𝑉 − −j50 + 25 5𝑉𝑉𝑗𝑗 + 𝑗𝑗2𝑗𝑉𝑉 = 02 + 𝑗𝑗 + 𝑗𝑗2𝑗𝑗 − 125 𝑉𝑉𝑗𝑗 + (−2 − 𝑗𝑗 + 𝑗𝑗25)𝑉𝑉 = 0−123 + 𝑗𝑗2𝑗𝑗 𝑉𝑉𝑗𝑗 + (−2 + 𝑗𝑗24)𝑉𝑉 = 0

Solve the equations for V:First Equation:+50 + 𝑗𝑗2𝑗 𝑉𝑉𝑗𝑗 − 𝑗𝑗2𝑉𝑉 = 𝑗𝑗2.5

𝑉𝑉𝑗𝑗 =𝑗𝑗2𝑉𝑉 + 𝑗𝑗2.550 + 𝑗𝑗2𝑗

Second equation:−123 + 𝑗𝑗2𝑗𝑗 𝑉𝑉𝑗𝑗 + (−2 + 𝑗𝑗24)𝑉𝑉 = 0

Substitute 𝑉𝑉𝑗𝑗 = 𝑗𝑗2𝑉𝑉+𝑗𝑗2.𝑗𝑗0+𝑗𝑗27

to get:

−123 + 𝑗𝑗2𝑗𝑗𝑗𝑗2𝑉𝑉 + 𝑗𝑗2.550 + 𝑗𝑗2𝑗 + (−2 + 𝑗𝑗24)𝑉𝑉 = 0

−123 + 𝑗𝑗2𝑗𝑗 (𝑗𝑗2𝑉𝑉 + 𝑗𝑗2.5) + (50 + 𝑗𝑗2𝑗)(−2 + 𝑗𝑗24)𝑉𝑉 = 0−123 + 𝑗𝑗2𝑗𝑗 𝑗𝑗2𝑉𝑉 + 50 + 𝑗𝑗2𝑗 −2 + 𝑗𝑗24 𝑉𝑉 = − −123 + 𝑗𝑗2𝑗𝑗 𝑗𝑗2.5−𝑗𝑗24𝑗 − 502 𝑉𝑉 + −100 + 𝑗𝑗𝑗2𝑗𝑗 − 𝑗𝑗𝑗4 − 648 𝑉𝑉 = 𝑗𝑗307.5 + 627.5−502 − 𝑗𝑗24𝑗 𝑉𝑉 + −748 + 𝑗𝑗𝑗𝑗4𝑗 𝑉𝑉 = 627.5 + 𝑗𝑗3𝑗𝑗.5−1250 + 𝑗𝑗𝑗𝑗𝑗 𝑉𝑉 = 627.5 + 𝑗𝑗3𝑗𝑗.5

𝑉𝑉 =627.5 + 𝑗𝑗3𝑗𝑗.5−1250 + 𝑗𝑗𝑗𝑗𝑗 = −0.2139 − 𝑗𝑗𝑗.4001 = 0.4𝑗3𝑗∠ − 118.1°

Superposition: add the DC solution to the AC solution.

DC Solution:

𝑣𝑣 = 1 volt

AC Solution:𝑉𝑉 = 0.4𝑗3𝑗∠ − 118.1°𝑣𝑣 𝑡𝑡 = 0.4536 cos 𝑗𝑗𝑗𝑗𝑡𝑡 − 118.1° volts

With both the DC sources and the AC source active: 𝑣𝑣 𝑡𝑡 = 1 + 0.4536 cos 𝑗𝑗𝑗𝑗𝑡𝑡 − 118.1° volts

+𝑣𝑣𝑗𝑗−

5𝑣𝑣𝑗𝑗 12 Ω

3+𝑣𝑣−

0.1−𝑗𝑗2

1 Ω

−𝑗𝑗2𝑗 2 Ω 𝑗𝑗𝑗5𝑉𝑉𝑗𝑗

+𝑉𝑉𝑗𝑗−

𝑉𝑉𝑗𝑗 𝑉𝑉

+𝑉𝑉−

3 volts0.1 cos 𝑗𝑗𝑗𝑗𝑡𝑡

𝐶𝐶1

𝑅𝑅1

𝐶𝐶2 𝑅𝑅2 𝜔𝜔

5𝑣𝑣𝑗𝑗+𝑣𝑣𝑗𝑗−

1 amp+𝑣𝑣−

𝑣𝑣(𝑡𝑡)

Superposition with two frequencies:

An antenna receives two signals at the same time, one at frequency , 𝑓𝑓1 = 1 GHz and the other at frequency 𝑓𝑓2 = 2 GHz. The antenna is modelled with two voltage sources in series. The circuit uses two ideal op-amps. The component values are 𝑅𝑅 = 1 ohm, 𝑅𝑅1 = 1ohm, 𝐶𝐶 = 1 pF and 𝜔𝜔 = 6.333 nH.

Use Superposition to find the output voltage 𝑣𝑣4(𝑡𝑡).

𝑣𝑣𝑠𝑠1 𝑡𝑡 = 0.1 cos𝜔𝜔1𝑡𝑡

𝑣𝑣𝑠𝑠2 𝑡𝑡 = 0.1 cos𝜔𝜔2𝑡𝑡

50 Ω50 Ω

𝑅𝑅

𝑅𝑅1

𝑣𝑣4𝑣𝑣3

+𝑣𝑣4(𝑡𝑡)−

𝜔𝜔𝐶𝐶𝑣𝑣2𝑣𝑣1

Analysis of the first stage:𝑣𝑣𝑠𝑠1 + 𝑣𝑣𝑠𝑠2 − 𝑣𝑣1

50−𝑣𝑣1 − 𝑣𝑣2

50= 0

The op-amps are ideal with infinite gain and very high input impedance. Consider 𝑣𝑣1 to be a virtual ground, so 𝑣𝑣1 = 0. Then

𝑣𝑣𝑠𝑠1 + 𝑣𝑣𝑠𝑠250

+𝑣𝑣250

= 0

𝑣𝑣2 = − 𝑣𝑣𝑠𝑠1 + 𝑣𝑣𝑠𝑠2

The output of the first stage is the sum of the two voltages.

𝑣𝑣𝑠𝑠1 𝑡𝑡

𝑣𝑣𝑠𝑠2 𝑡𝑡

50 Ω50 Ω

𝑣𝑣2

𝑣𝑣1 ≈ 0

≈ 0

𝑣𝑣𝑠𝑠1 + 𝑣𝑣𝑠𝑠2

Use phasors to analyze the second stage at some frequency 𝜔𝜔:

Write a node equation at 𝑉𝑉3:

𝑉𝑉2 − 𝑉𝑉3

𝑅𝑅 + 𝑗𝑗 𝜔𝜔𝜔𝜔 − 1𝜔𝜔𝐶𝐶

−𝑉𝑉3 − 𝑉𝑉4𝑅𝑅1

= 0

Amplifier input 𝑉𝑉3 is a virtual ground, 𝑉𝑉3 = 0

𝑉𝑉2

𝑅𝑅 + 𝑗𝑗 𝜔𝜔𝜔𝜔 − 1𝜔𝜔𝐶𝐶

+𝑉𝑉4𝑅𝑅1

= 0

phasor

Drive the circuit with a generator at frequency 𝜔𝜔:𝑣𝑣𝑠𝑠 𝑡𝑡 = 𝐴𝐴 cos𝜔𝜔𝑡𝑡

The phasor is𝑉𝑉𝑠𝑠 = 𝐴𝐴𝑒𝑒𝑗𝑗0

The output of the first stage is 𝑉𝑉2 = −𝑉𝑉𝑠𝑠

𝑉𝑉4 = 𝑅𝑅1𝑉𝑉𝑠𝑠

𝑅𝑅 + 𝑗𝑗 𝜔𝜔𝜔𝜔 − 1𝜔𝜔𝐶𝐶

𝑣𝑣𝑠𝑠 𝑡𝑡 = 𝐴𝐴 cos𝜔𝜔𝑡𝑡

50 Ω50 Ω

𝑅𝑅

𝑅𝑅1

𝑉𝑉4𝑉𝑉3 ≈ 0

+𝑉𝑉4−

𝑗𝑗𝜔𝜔𝜔𝜔1𝑗𝑗𝜔𝜔𝐶𝐶

𝑉𝑉2𝑉𝑉1 ≈ 0

𝑉𝑉𝑠𝑠 = 𝐴𝐴

≈ 0 ≈ 0

At frequency 𝑓𝑓1 = 1 GHz, 𝜔𝜔1 = 2𝜋𝜋𝑓𝑓1 = 6.2𝑗3𝑗𝑗109:

𝑅𝑅 = 1 ohm, 𝑅𝑅1 = 1 ohm, 𝐶𝐶 = 1 pF and 𝜔𝜔 = 6.333 nH, 𝑉𝑉𝑠𝑠 = 0.1

𝜔𝜔1𝜔𝜔 −1

𝜔𝜔1𝐶𝐶= 6.2𝑗3𝑗𝑗109𝑗𝑗𝑗.333𝑗𝑗10−9 −

16.2𝑗3𝑗𝑗109𝑗𝑗𝑗𝑗𝑗10−12 = 39.79 − 159.15 = −119.36

𝑉𝑉4 = 𝑅𝑅1𝑉𝑉𝑠𝑠

𝑅𝑅+𝑗𝑗 𝜔𝜔1𝐿𝐿−1

𝜔𝜔1𝐶𝐶

= 0.11−𝑗𝑗119.36

= 8.3𝑗𝑗𝑗𝑗10−4∠𝑗𝑗.5 degrees

𝑣𝑣4 𝑡𝑡 = 8.3𝑗𝑗𝑗𝑗10−4 cos 𝜔𝜔1𝑡𝑡 + 89.5°

𝑣𝑣𝑠𝑠 𝑡𝑡 = 0.1 cos𝜔𝜔1𝑡𝑡

50 Ω50 Ω

𝑅𝑅

𝑅𝑅1

𝑉𝑉4𝑉𝑉3 ≈ 0

+

𝑉𝑉4 = 𝑅𝑅1𝑉𝑉𝑠𝑠

𝑅𝑅 + 𝑗𝑗 𝜔𝜔1𝜔𝜔 −1

𝜔𝜔1𝐶𝐶−

𝑗𝑗𝜔𝜔1𝜔𝜔1𝑗𝑗𝜔𝜔1𝐶𝐶

𝑉𝑉2𝑉𝑉1 ≈ 0

𝑉𝑉𝑠𝑠

≈ 0 ≈ 0

At frequency 𝑓𝑓2 = 2 GHz, 𝜔𝜔2 = 2𝜋𝜋𝑓𝑓2 = 1.2𝑗𝑗𝑗𝑗1010:

𝑅𝑅 = 1 ohm, 𝑅𝑅1 = 1 ohm, 𝐶𝐶 = 1 pF and 𝜔𝜔 = 6.333 nH, 𝑉𝑉𝑠𝑠 = 0.1

𝜔𝜔2𝜔𝜔 −1

𝜔𝜔2𝐶𝐶= 1.2𝑗𝑗𝑗𝑗1010𝑗𝑗𝑗.333𝑗𝑗10−9 −

11.2𝑗𝑗𝑗𝑗1010𝑗𝑗𝑗𝑗𝑗10−12 = 79.54 − 79.55 = −0.01 ≈ 0

𝑉𝑉4 = 𝑅𝑅1𝑉𝑉𝑠𝑠

𝑅𝑅 + 𝑗𝑗 𝜔𝜔2𝜔𝜔 −1

𝜔𝜔2𝐶𝐶=

0.11 + 𝑗𝑗𝑗 = 0.𝑗∠𝑗 degrees

𝑣𝑣4 𝑡𝑡 = 0.1 cos 𝜔𝜔2𝑡𝑡

𝑣𝑣𝑠𝑠 𝑡𝑡 = 0.1 cos𝜔𝜔2𝑡𝑡

50 Ω50 Ω

𝑅𝑅

𝑅𝑅1

𝑉𝑉4𝑉𝑉3 ≈ 0

+

𝑉𝑉4 = 𝑅𝑅1𝑉𝑉𝑠𝑠

𝑅𝑅 + 𝑗𝑗 𝜔𝜔2𝜔𝜔 −1

𝜔𝜔2𝐶𝐶−

𝑗𝑗𝜔𝜔2𝜔𝜔1𝑗𝑗𝜔𝜔2𝐶𝐶

𝑉𝑉2𝑉𝑉1 ≈ 0

𝑉𝑉𝑠𝑠

≈ 0 ≈ 0

Superposition with two frequencies:

By superposition, with both sources active:𝑣𝑣4 𝑡𝑡 = 8.3𝑗𝑗𝑗𝑗10−4 cos 𝜔𝜔1𝑡𝑡 + 89.5° + 0.1 cos 𝜔𝜔2𝑡𝑡

The design of the circuit makes 𝜔𝜔𝜔𝜔 − 1𝜔𝜔𝐶𝐶

=0 at 2 GHz.Then the circuit “passes” 2 GHz but “rejects” 1 GHz.The circuit behaves as a ‘filter’ to recover the 2 GHz signal from the two signals received by the antenna.

𝑣𝑣𝑠𝑠1 𝑡𝑡 = 0.1 cos𝜔𝜔1𝑡𝑡

𝑣𝑣𝑠𝑠2 𝑡𝑡 = 0.1 cos𝜔𝜔2𝑡𝑡

50 Ω50 Ω

𝑅𝑅

𝑅𝑅1

𝑣𝑣4𝑣𝑣3

+𝑣𝑣4(𝑡𝑡)−

𝜔𝜔𝐶𝐶𝑣𝑣2𝑣𝑣1