Circuit Analysisjoelsd/electronics/Homework/... · Homework Answers #3 • Calculate the Currents...

BME 372 Electronics I –J.Schesser

21

Circuit Analysis

Lesson #2


22

Homework

• Voltage and Current division– How does the voltage divide across two capacitors in series? Show

your results.– How does the current divide among two capacitors in parallel?

Show your results.

• Calculate the Currents and Voltages for the following circuits:

1

10Adc

2

1

25Vdc

2Vdc

1

1 10Adc2

1

10Adc

2

1

2Vdc


23

Homework Answers #1

• Voltage and Current division– How does the voltage divide across two

capacitors in series? Show your results.

– How does the current divide among two capacitors in parallel? Show your results.

21

1

21

11

21

11

212121

2211

21

1

21

1

21

1

1

2121

21

1

21

1

/1/1/1

][

;

)(11

1

)(11

1

)()(

11

1

)()(

1)(

]11[11)(

CCC

CjCjCj

ZZZ

II

OrCC

CII

dtdVCC

dtdVC

dtdVCIII

dtdVCI

dtdVCI

tV

CC

CtV

CjCj

CjtVZZ

ZtV

OrCC

CtVtV

idtC

tV

idtCC

idtC

idtC

tV

CC

C

ab

ab

abababab

abab

acacac

CC

C

ab

ac

ab

ab

ac

a b c

C1 C2

a

b

C1 C2Iab

I1I2


24

Homework Answers #2


I1=2/2=1A

1

10A

21

2VI2=10A I3=9A

- 1V +

+

1V

-

-20V ++

22V -


25

Homework Answers #2


I1=2/2=1A

1

10A

21

2VI2=-10A I3=9A

- 1V +

+

1V

-

-20V ++

22V -

1 2 3

2

3

3 3

0Note:

10

10 02Note:

21 10 0 9

NodalI I I

IV I

VI I


26

Homework Answers #2


I1=25/2=12.5A

1

10A

21

25VI2=4A I3=2.5A

- 12.5V +

+

12.5V

-

-20V ++

45V -


27

Homework Answers #2


I1=25/2=12.5A

1

10A

21

25VI2=-10A

I3=-2.5A- 12.5V +

+

12.5V

-

-20V ++

45V -

1 2 3

2

3

3 3

2

0Note:

10

10 02Note:

2512.5 10 0 2.5

25 2 25 ( 10) 2 45CS

NodalI I I

IV I

VI I

V I


28

Homework Answers #3• Calculate the Currents and Voltages for the following circuits:

2Vdc

1

1 10Adc2

I1 I2

I3

V

1 2 3

1 3

Using Nodal Analysis:0

2( 10) 02 22 11 112

11 5.5 ; 10 5.5 4.52

I I IV V

V V

I A I

2Vdc

1

1 10Adc2 I3=4.5

I1=5.5

+

5.5v

-

- 5.5v +

+ 9v -


29

HomeworkCalculate the current labeled i and the voltage labeled v in the following circuit R1 = 1Ω, R2 = 2Ω, R3 = 1Ω,

R4 = 1Ω, R5 = 2Ω, R6 = 2Ω,

R7 = 2Ω, Vcc = 4v6 7

1 6 76 7

6 72 4 1 4

6 7

6 74 5

6 7 6 7 4 5 6 4 5 7 5 6 73 2 5 4 5

6 76 7 4 6 4 7 5 6 5 7 6 74 5

6 7

4 5 6 4 5 7 5 6 74 3 3

4 6 4 7 5 6 5

' ||

' '

( )' ' || ( ) ||

( )

' '

R RR R RR R

R RR R R RR R

R RR RR R R R R R R R R R R R RR R R R R R RR R R R R R R R R R R RR R

R RR R R R R R R R RR R R

R R R R R R R

37 6 7

3 4 6 3 4 7 3 5 6 3 5 7 3 6 7 4 5 6 4 5 7 5 6 7

4 6 4 7 5 6 5 7 6 7

5 2 4

6 1 5

' || '' '

RR R R

R R R R R R R R R R R R R R R R R R R R R R R RR R R R R R R R R R

R R RR R R

R1

R7

R2

R3

R4

R6

R5+VCC

--i

+v--

iin i2

v1 v2


30

HomeworkCalculate the current labeled i and the voltage labeled v in the following circuit R1 = 3Ω, R2 = 6Ω, R3 = 12Ω,

R4 = 4Ω, R5 = 2Ω, R6 = 2Ω,

R7 = 4Ω, Vcc = 4v

R1

R7

R2

R3

R4

R6

R5+VCC

--i

+v--

3 4 6 3 4 7 3 5 6 3 5 7 3 6 7 4 5 6 4 5 7 5 6 75 2 4 2

4 6 4 7 5 6 5 7 6 7

3 4 6 3 4 7 3 5 6 3 5 7 3 6 7 4 5 6 4 5 7 5 6 72

4 6 4 7 5 6 5 7 6 7

3 4 6 3 4 7 3 52

' || ' || R R R R R R R R R R R R R R R R R R R R R R R RR R R RR R R R R R R R R R

R R R R R R R R R R R R R R R R R R R R R R R RRR R R R R R R R R R

R R R R R R R R RR

6 3 5 7 3 6 7 4 5 6 4 5 7 5 6 7

4 6 4 7 5 6 5 7 6 7

2 3 4 6 2 3 4 7 2 3 5 6 2 3 5 7 2 3 6 7 2 4 5 6 2 4 5 7 2 5 6 7

2 4 6 2 4 7 2 5 6 2 5 7 2 6 7 3 4 6 3 4 7 3 5 6 3 5 7 3 6

R R R R R R R R R R R R R R RR R R R R R R R R R

R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R RR R R R R R R R R R R R R R R R R R R R R R R R R R R R R R

7 4 5 6 4 5 7 5 6 7

2 3 4 6 2 3 4 7 2 3 5 6 2 3 5 7 2 3 6 7 2 4 5 6 2 4 5 7 2 5 6 76 1 5 1

2 4 6 2 4 7 2 5 6 2 5 7 2 6 7 3 4 6 3 4 7 3 5 6 3 5 7 3 6 7 4 5 6 4 5 7 5 6 7

1

' '

R R R R R R R R RR R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R RR R R R

R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R RR

2 4 6 1 2 4 7 1 2 5 6 1 2 5 7 1 2 6 7 1 3 4 6 1 3 4 7 1 3 5 6 1 3 5 7 1 3 6 7 1 4 5 6 1 4 5 7 1 5 6 7

2 4 6 2 4 7 2 5 6 2 5 7 2 6 7 3 4 6 3 4 7 3 5 6 3 5 7 3 6 7 4 5 6 4 5 7 5

R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R RR R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R

6 7

2 3 4 6 2 3 4 7 2 3 5 6 2 3 5 7 2 3 6 7 2 4 5 6 2 4 5 7 2 5 6 7

2 4 6 2 4 7 2 5 6 2 5 7 2 6 7 3 4 6 3 4 7 3 5 6 3 5 7 3 6 7 4 5 6 4 5 7 5 6 7

R RR R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R

R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R

i2

v1 v2


31

Homework

1 6 7

2 4 1

3 2 5

4 3 3

5 2 4

6 1 5

' || 2 || 2 1' ' 1 1 2' ' || 2 || 2 1' ' 1 1 2' || ' 2 || 2 1' ' 1 1 2

R R RR R RR R RR R RR R RR R R

R1

R7

R2

R3

R4

R6

R5+VCC

--i

+v--

R’1

R1 R3

R2 R4 R5+VCC

--i R’2

R’1

R2 R’2

R5

+VCC

--i R’3

R3R1

+v--

R2 R’3+VCC

--i

R’4

R3R1

iiniin

iin iin

i2 i2

i2 i2

v1v1

v1 v1

v2 v2

v2 v2

4 22

6 4 2 4 2

3 11 2 4 2 1 2

3 3 1 4

'; ; ;' ' '

' '' ; ;' '

in in inR RVcci i i i i

R R R R RR Rv i R v v v v

R R R R


32

Homework

R2 R’4+VCC

--i R’5

R1

iin i2

v1

6

4

4 2

22

4 2

1 2 4

32 1

3 3

12

1 4

4 2;' 2

' 22 1' 2 2

21 0.5' 2 2

' 0.5 2 1' 11 0.5

' 1 1' 10.5 0.25

' 1 1

in

in

in

VcciR

Ri iR R

Ri iR R

v i RRv v

R RRv v

R R

R’5+VCC

--

R1

iin

v1

R’6

1 6 7

2 4 1

3 2 5

4 3 3

5 2 4

6 1 5

' || 2 || 2 1' ' 1 1 2' ' || 2 || 2 1' ' 1 1 2' || ' 2 || 2 1' ' 1 1 2



33

Homework

6

4

4 2

22

4 2

1 2 4

32 1

3 3

12

1 4

4 2;' 2

' 22 1' 2 2

22 1' 2 2

' 1 2 2' 12 1

' 1 1' 11 0.5

' 1 1

in

in

in

VcciR

Ri iR R

Ri iR R

v i RRv v

R RRv v

R R

Calculate the current labeled i and the voltage labeled v in the following circuit R1 = 1Ω, R2 = 2Ω, R3 = 1Ω,

R4 = 1Ω, R5 = 2Ω, R6 = 2Ω,

R7 = 2Ω, Vcc = 4v

1

2

2

1

1

2

2+4--

1a+0.5v--

2a

1av1=2v

+ 2v -

v2=1v

+ 1v -

1 6 7

2 4 1

3 2 5

4 3 3

5 2 4

6 1 5

' || 2 || 2 1' ' 1 1 2' ' || 2 || 2 1' ' 1 1 2' || ' 2 || 2 1' ' 1 1 2


CHECK YOUR ANSWERS!


34

Homework

1 21 1 2

1 2

1 22 4 1 4

1 2

1 24 3

1 2 1 23 2 3 4 3

1 21 24 3

1 2

1 3 4 2 3 4 1 2 3

1 4 2 4 1 3 2 3 1 2

' || 2 || 2 1

' ' 3 1 4

( )' ' || ( ) ||

( )

8 44 || 26 3

R RR R RR R

R RR R R RR R

R RR RR R R RR R R R R R RR R R R

R RR R R R R R R R R

R R R R R R R R R R

Calculate the current labeled, i.

R1 = 2Ω, R2 = 2Ω, R3 = 2Ω, R4 = 3Ω, Vcc = 2vi4=0.5

R1R2

R3

R4

+VCC

--

i1=0.25

iin= 1.5

i3=1v1

v2

i2=0.25

i=1.25

3

23 1

2 3

34 1

2 3

21 4

2 1

12 4

2 1

1

2 3 ;4' 23' 3 4 3 4 2 1; 2

' 2 4 2 2 6 23 2 1 3;

' 2 6 2 21 2 12 4 41 2 12 4 4

3 1 52 4 4

in

in

in

in

VcciR

Ri i v vR R

Ri i v vR R

Ri iR R

Ri iR R

i i i


35

Homework

An electrode is connected to an oscilloscope which has a purely capacitance input impedance, CIN. Find and plot the output voltage Vab(jω) as function of ω.

Use Matlab to perform the plot.

RS 500RD 29.5k

CD 53n

a

b

1Vac CIN 5.3n


Electrode Impedance

ZE RD RS jRDRSCD

1 jRDCD

Vout

Vin

1jCin

1jCin

ZE

1

1 jCinZE

1

1 jCinRD RS jRDRSCD

1 jRDCD

1 jRDCD

1 jRDCD jCin (RD RS jRDRSCD )

1 jRDCD

1 2RDRSCDCin j (RDCD Cin (RD RS ))

Vout

Vin 0

1 jRDCD

1 2RDRSCDCin j (RDCD Cin (RD RS ))0

1

Vout

Vin

1 jRDCD


jRDCD

2RDRSCDCin

1

RSCin

2 0

2

Vout

Vin 1

RDRSCDCin

1 jRDCD


1RDRSCDCin

1 j 1

RDRSCDCin

RDCD

j 1RDRSCDCin

(RDCD Cin (RD RS ))

RDRSCDCin jRDCD

j(RDCD Cin (RD RS ))

RDRSCDCin (RDCD )2

(RDCD Cin (RD RS )) tan1( RDCD

RDRSCDCin

) 2

Vout

Vin

1 j f

fp

1 ( ffR

)2 jf ( 1fp

1fin

)

1 ( ffp

)2

(1 ( ffR

)2 )2 ( f ( 1fp

1fin

))2

tan1( ffp

) tan1(f ( 1

fp

1fin

)

1 ( ffR

)2)

f 2

, fp 1

2RDCD

, fR 1

2 RDRSCDCin

, fin 1

2Cin (RD RS )

0.5

0.6

0.7

0.8

0.9

1.0

1.1

1.E+00 1.E+01 1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07HZ

RD=29.5k CD=53nf RS=500 Cin= 5.3nf


Electrode Impedance

Vout

Vin

( j ) 1 jRDCD

1 2RDRSCDCin j (RDCD Cin(RD RS ))

Vout

Vin

(s) 1 sRDCD

1 s2RDRSCDCin s(RDCD Cin (RD RS ))

sRDCD 1s2RDRSCDCin s(RDCD Cin(RD RS ))1

RDCD

RDRSCDCin

s 1RDCD

s2 s(RDCD Cin (RD RS )RDRSCDCin

) 1RDRSCDCin

1

RSCin

s 1RDCD

s2 s( RDCD Cin (RD RS )RDRSCDCin

) 1RDRSCDCin

1

RSCin

s sz

(s sp1)(s sp2

)

sz 1

RDCD

;sp1, p2

RDCD Cin (RD RS )RDRSCDCin

( RDCD Cin (RD RS )RDRSCDCin

)2 4 1RDRSCDCin

2RDCD Cin (RD RS ) (RDCD Cin (RD RS ))2 4RDRSCDCin

2RDRSCDCin

sz 640;sp1, p2 581, 4.2 105

fz 102Hz; fp1, p2 92.5Hz, 6.6 104 Hz


38

Homework

• Repeat the analysis of this circuit using Mesh and Nodal Analysis. That is find Vabas a function of frequency

a

b

R2

5k

R1

1k

1VacC1

1n

C2

5n


39

Homework Answers #7

• Repeat the analysis of this circuit using Mesh and Nodal Analysis

11221

12211122

11221

11221

1221

1

1

111

111

321

))(

)()((

)111(

0

0

RCCRR

RCRCRCCR

RCCRR

RCCRR

CCRR

ZVV

ZZZZZZZZZZZZ

ZVV

ZZZZ

ZV

ZV

ZZV

ZV

ZV

ZZV

ZVV

III

a

b

V1

I1 I3

I2

V

R2

5k

R1

1k

1VacC1

1n

C2

5n


40

Homework Answers #8


a

b

V1

I1 I3

I2

V

VZZZZZZZZZZ

ZZ

VZZZZZZZZZZ

ZZZZZ

ZV

ZZZ

V

VZZZZZZZZZZ

ZZZV

RCRRCRCCCR

CC

RCRRCRCCCR

CCR

CR

C

CR

C

ab

RCRRCRCCCR

CCR

1212111212

21

1212111212

122

22

2

22

2

1212111212

122

)()()(

)(

1

1

R2

5k

R1

1k

1VacC1

1n

C2

5n


41

Homework Answers #8• Repeat the analysis of this circuit using Mesh and Nodal Analysis

a

b

V1

I1 I3

I2

V

1 2

2 1 2 1 1 1 2 1 2 1

1 2

2 1 2 1 11 1 2 1 2

22 2 2 1 1 2 2 1 2 1

21 2 2 1 2 2 2 1 2 1

1 1

1 1 1 1 1

11

11 ( )

C Cab

R C C C R C R R C R

Z ZV V

Z Z Z Z Z Z Z Z Z Z

j C j C

R R R R Rj C j C j C j C j C

j C R j C R C C R R j C R

C C R R j C R C R C R

R2

5k

R1

1k

1VacC1

1n

C2

5n


42

Homework Answers #6

22 1 1 2 2 1 2 1 1

6 2 6

61

6 26 2 2 2 12

61

0 6 26 2 2 12

1

1(1 ) [ ( ) ]

11 ( 5 10 ) 31 10

1 31 10tan [ ]1 ( 5 10 )[1 ( 5 10 ) ] 961 10

1 0 31 10| tan [ ]1 (0 5 10 )[1 (0 5 10 ) ] 0 961 10

1 tan1

ab

ab

VV C C R R j C R R C R

j

VV

0[ ] 1 01

6

6 66 6

1 1 11 2 26 25 10 6 2 2 12 2 126

6 6 6

6

1 131 10 31 101 1 1 6.25 10 5 10| tan [ ] tan [ ] tan [ ]1 0 038.441 1 11 ( 5 10 )[1 ( 5 10 ) ] 961 10 [0] 961 105 105 10 5 10 5 101

6.2 21|

1 ( 5 10 )

ab

ab

VV

VV

2 6 6 2

1| 031 10 ( 5 10 )j

a

b

R2

5k

R1

1k

1VacC1

1n

C2

5n


43

Homework Answers #6

2

' 2 22

2

' '

' 1

2 22

2 1 2 1'

2 22

2 1 2 1

2 2 2 22

1 2 2 2 2 2 1 1 2

' '

' 1

11

1 1

11 1 1( ) ( )

1 1 1 1( )

1 1(1 ) ( )

1

ab

a b

a b a b

a b

a b

a b a b

a b

V j CV j R CR

j CV ZV Z R

j C RRj C j C j C j CZ j C RR

j C j C j C j Cj C R j C R

j C j C R j C C R C j C Cj

V ZV Z R

2 22

2 2 1 1 2

2 212

2 2 1 1 2

2 22

2 2 1 2 2 1 1 2

2 22

2 2 1 1 1 1 1 2 2 2

' 2 22

' 2 2 2 2 1 1 1 1 1 2 2 2

( )1

( )1

1 ( ( ))1

1 ( )11

1 1 ( )ab ab a b

a b

C RC R C j C C

j C R RC R C j C C

j C Rj C R R C R C j C C

j C RC R C R j R C R C C R

V V V j C RV V V j R C C R C R j R C R C C R

22 2 1 1 1 1 1 2 2 2

11 ( )C R C R j R C R C C R

a

b

R2

5k

R1

1k

1VacC1

1n

C2

5n

a’

22 2 1 1 1 1 1 2 2 2

1 1 1 1 2 2 222 2 2

2 2 1 12 2 1 1 1 1 1 2 2 2

11 ( )

( )1 tan ( )1(1 ) ( ( ))

abVV C R C R j R C R C C R

R C R C C RC R C RC R C R R C R C C R


44

Matlab Codeclear all;R1=1e3;C1=1e-9;R2=5e3;C2=5e-9;omega=(100:1000:10^7);maxomega=length(omega); z1=R1;GIN=[1/R1];for i=1:maxomega

zC1=1/complex(0,omega(i)*C1);zC2=1/complex(0,omega(i)*C2);z3=R2+zC2;G=1/R1+1/zC1+1/z3;V1=G\GIN;VOUT(i)=V1*(zC2/z3);

endf=omega/(2*pi);subplot(2,1,1);semilogx(f,abs(VOUT));title('RCLadder Nodal Magnitude');xlabel('Hz');axis([f(1) f(maxomega) 0 1]);subplot(2,1,2);semilogx(f,atan2(imag(VOUT),real(VOUT)));title('RCLadder Nodal Phase Angle');xlabel('Hz');axis([f(1) f(maxomega) -2 2]);

102 103 104 105 1060

0.5

1RCLadder Nodal Magnitude

Hz

102 103 104 105 106-2

-1

0

1

2RCLadder Nodal Phase Angle

Hz


45

Homework Answers #6

2

' 2 22

2

' '

' 1

2 22

2 1 2 1'

2 22

2 1 2 1

2 2 2 22

1 2 2 2 2 2 1 1 2

' '

' 1

11

1 1

11 1 1( ) ( )

1 1 1 1( )

1 1(1 ) ( )

1

ab

a b

a b a b

a b

a b

a b a b

a b

V j CV j R CR

j CV ZV Z R

j C RRj C j C j C j CZ j C RR

j C j C j C j Cj C R j C R

j C j C R j C C R C j C Cj

V ZV Z R

2 22

2 2 1 1 2

2 212

2 2 1 1 2

2 22

2 2 1 2 2 1 1 2

2 22

2 2 1 1 1 1 1 2 2 2

' 2 22

' 2 2 2 2 1 1 1 1 1 2 2 2

( )1

( )1

1 ( ( ))1

1 ( )11

1 1 ( )ab ab a b

a b

C RC R C j C C

j C R RC R C j C C

j C Rj C R R C R C j C C

j C RC R C R j R C R C C R

V V V j C RV V V j R C C R C R j R C R C C R

22 2 1 1 1 1 1 2 2 2

11 ( )C R C R j R C R C C R

a

b

R2

1k

R1

1k

1VacC1

1n

C2

1n

a’

22 2 1 1 1 1 1 2 2 2

1 1 1 1 2 2 222 2 2

2 2 1 12 2 1 1 1 1 1 2 2 2

11 ( )

( )1 tan ( )1(1 ) ( ( ))

abVV C R C R j R C R C C R



46

Homework Answers #6

2 1 1 2

22 2 1 1 1 1 1 2 2 2

1 1 1 1 2 2 222 2 2

2 2 1 12 2 1 1 1 1 1 2 2 2

0

22 1 1 2

1

2 1 1

11 ( )

( )1 tan ( )1(1 ) ( ( ))

1| 1 0(1 0) 0

1| | 0 0

1| 1(0)

ab

ab

ab

ab

C C R R

VV C R C R j R C R C C R


VV j

VV C C R R

VV j

C C R

2 1 2 1 1

2

2 1 1 2

2 1 2 1 1

[ ( ) ]

0( ) 2

C R R C RR

C C R RC R R C R

a

b

R2

1k

R1

1k

1VacC1

1n

C2

1n

a’

‐3.5

‐3

‐2.5

‐2

‐1.5

‐1

‐0.5

0

0.00E+00

2.00E‐01

4.00E‐01

6.00E‐01

8.00E‐01

1.00E+00

1.20E+00

1.E+00 1.E+02 1.E+04 1.E+06 1.E+08 1.E+10 1.E+12

Frequency Response

Magnitude Ladder Phase Ladder


47

Homework Answers #6

1

2

0

1

11 1 tan ( )1 1 1 ( )

1| 1 01 0

1| 02

1 1|1 1 42

ab

ab

ab

ab

RC

V j C RCV j RC RCR

j C

VV j

VV j RC

VV j

a

b

R

1k

1VacC

1n

a’

‐1.8

‐1.6

‐1.4

‐1.2

‐1

‐0.8

‐0.6

‐0.4

‐0.2

0

0.00E+00

2.00E‐01

4.00E‐01

6.00E‐01

8.00E‐01

1.00E+00

1.20E+00

1.E+00 1.E+02 1.E+04 1.E+06 1.E+08 1.E+10 1.E+12

Frequency Response

magnitude Single Phase Single


48

Homework Answers #6a

b

R

1k

1VacC

1n

a’a

b

R2

1k

R1

1k

1VacC1

1n

C2

1n

a’

0.00E+00

2.00E‐01

4.00E‐01

6.00E‐01

8.00E‐01

1.00E+00

1.20E+00

1.E+00 1.E+02 1.E+04 1.E+06 1.E+08 1.E+10 1.E+12

Frequency Response

Magnitude Ladder magnitude Single


49

Homework Answers #4• Repeat the analysis of this circuit

using Mesh and Nodal Analysis

)()()(02Mesh

)()(1Mesh

1221122

11111

21122

21211

CCRCCCR

CCRCR

ZZZIIZIIZZZI

ZIZZIIIZZIV

a

b

I1 I2V=

1V ac

R2

5k

R1

1k

C1

1n

C2

5n


50


I1 ZR2

ZC2 ZC1

ZC1

I2;from Mesh 2

V {ZR2

ZC2 ZC1

ZC1

I2}(ZR1 ZC1

) I2ZC1; Substituting this into Mesh 1

(ZR2

ZC2 ZC1

)(ZR1 ZC1

) ZC1

2

ZC1

I2

ZR2

ZR1 ZC2

ZR1 ZC1

ZR1 ZR2

ZC1 ZC2

ZC1 ZC1

ZC1 ZC1

2

ZC1

I2

ZR2

ZR1 ZC2

ZR1 ZC1

ZR1 ZR2

ZC1 ZC2

ZC1

ZC1

I2

a

b

I1 I2V=

1V ac

R2

5k

R1

1k

C1

1n

C2

5n


51


VZZZZZZZZZZ

ZZZIV

VZZZZZZZZZZ

ZI

CCCRRCRCRR

CC

Cab

CCCRRCRCRR

C

1212111212

21

2

1212111212

1

2

2

a

b

I1 I2V=

1V ac

R2

5k

R1

1k

C1

1n

C2

5n


52

Homework Answers #6


2 2

12

1 2 1 1 22 1 1 1 2

11 1 2 1 1 2

2 2

2

2 12 1 1 2 2 1 2 1 2 1 1

2

22 2

2 1 1 2 2 1 2 1 1

21

22 1 1

1

1 1 1 1 1

11

(1 ) [ ( ) ]

(1 ) [ ( ) ]1

(1

ab C C

j CI VR R R R R

j C j C j C j C j C

VCj C R R R R RC j C

j C VCC C R R j C R j C R R C RC

j CV I Z VZC C R R j C R R C R

j Cj C

C C R

2 2 1 2 1 1

22 1 1 2 2 1 2 1 1

) [ ( ) ]1

(1 ) [ ( ) ]

VR j C R R C R

C C R R j C R R C R

a

b

I1 I2V=

1V ac

R2

5k

R1

1k

C1

1n

C2

5n

22 1 1 2 2 1 2 1 1

6 2 6

61

6 26 2 2 2 12

1(1 ) [ ( ) ]

11 ( 5 10 ) 31 10

1 31 10tan [ ]}1 ( 5 10 )[1 ( 5 10 ) ] 961 10

abVC C R R j C R R C R

j


53

Homework Answers #4• Repeat the analysis of this circuit

using Mesh and Nodal Analysis

Mesh 1

V I1R1 1

jC1

(I1 I2 ) I1(R1 1

jC1

) I2

1jC1

Mesh 2

0 I2 (R2 1

jC2

) 1jC1

(I2 I1) 1

jC1

I1 I2 (R2 1

jC2

1

jC1

)

a

b

I1 I2V=

1V ac

R2

5k

R1

1k

C1

1n

C2

5n


54


I1 R2

1jC2

1

jC1

1jC1

I2;from Mesh 2

V {R2

1jC2

1

jC1

1jC1

I2}(R1 1

jC1

) I2

1jC1

; Substituting this into Mesh 1

(R2

1jC2

1

jC1

)(R1 1

jC1

) ( 1jC1

)2

1jC1

I2

R2R1

1jC2

R1 1

jC1

R1 R2

1jC1

1

jC1

1jC2

1

jC1

1jC1

( 1jC1

)22

1jC1

I2

R2R1

1jC2

R1 1

jC1

R1 R2

1jC1

1

jC1

1jC2

1jC1

I2

a

b

I1 I2V=

1V ac

R2

5k

R1

1k

C1

1n

C2

5n


55


I2

1jC1

R2R1 1

jC2

R1 1

jC1

R1 R2

1jC1

1

jC1

1jC2

V

Vab I2

1jC2

1jC1

1jC2

R2R1 1

jC2

R1 1

jC1

R1 R2

1jC1

1

jC1

1jC2

V

a

b

I1 I2V=

1V ac

R2

5k

R1

1k

C1

1n

C2

5n


56


Vab

V

1jC1

1jC2

R2R1 1

jC2

R1 1

jC1

R1 R2

1jC1

1

jC1

1jC2

1

( jC1)( jC2 )R2R1 ( jC1)( jC2 ) 1jC2

R1 ( jC1)( jC2 ) 1jC1

R1 ( jC1)( jC2 )R2

1jC1

( jC1)( jC2 ) 1jC1

1jC2

1

( 2C1C2R2R1) jC1R1 jC2R1 jC2R2 1

1

1 2C1C2R2R1) j (C1R1 C2R1 C2R2 )

a

b

I1 I2V=

1V ac

R2

5k

R1

1k

C1

1n

C2

5n


57

Homework Answers #6


a

b

I1 I2V=

1V ac

R2

5k

R1

1k

C1

1n

C2

5n

22 1 1 2 2 1 2 1 1

6 2 6

61

6 26 2 2 2 12

1(1 ) [ ( ) ]

11 ( 5 10 ) 31 10

1 31 10tan [ ]}1 ( 5 10 )[1 ( 5 10 ) ] 961 10

abVC C R R j C R R C R

j


58

Homework Answers #6


Zab

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.E+00 1.E+02 1.E+04 1.E+06 1.E+08 1.E+10 1.E+12

Frequency (Hz)

Ohm

s

-200

-180

-160

-140

-120

-100

-80

-60

-40

-20

0

Deg

rees

MagnitudePhase

22 1 1 2 2 1 2 1 1

6 2 6

61

6 26 2 2 2 12

61

0 6 26 2 2 12

1

1(1 ) [ ( ) ]

11 ( 5 10 ) 31 10

1 31 10tan [ ]1 ( 5 10 )[1 ( 5 10 ) ] 961 10

1 0 31 10| tan [ ]1 (0 5 10 )[1 (0 5 10 ) ] 0 961 10

1 0tan [1

ab

ab

VC C R R j C R R C R

j

V

] 1 01

6

6 66 6

1 1 11 2 26 25 10 6 2 2 12 2 126

6 6 6

6 2

1 131 10 31 101 1 1 6.25 10 5 10| tan [ ] tan [ ] tan [ ]1 0 038.441 1 11 ( 5 10 )[1 ( 5 10 ) ] 961 10 [0] 961 105 105 10 5 10 5 101

6.2 21|

1 ( 5 10 )

ab

ab

V

V

6 6 2

1| 031 10 ( 5 10 )j


59

Matlab Code

clear all;R1=1e3;C1=1e-9;R2=5e3;C2=5e-9;omega=(100:1000:10^7);maxomega=length(omega);VIN=[1; 0];for i=1:maxomega

zC1=1/complex(0,omega(i)*C1);zC2=1/complex(0,omega(i)*C2);Z=[R1+zC1 -zC1;-zC1 R2+zC1+zC2]; IC=Z\VIN;VOUT(i)=IC(2)*zC2;

endf=omega/(2*pi);subplot(2,1,1);semilogx(f,abs(VOUT));title('RCLadder Mesh Magnitude');xlabel('Hz');axis([f(1) f(maxomega) 0 1]);subplot(2,1,2);semilogx(f,atan2(imag(VOUT),real(VOUT)));title('RCLadder Mesh Phase Angle');xlabel('Hz');axis([f(1) f(maxomega) -2 2]);

102 103 104 105 1060

0.5

1RCLadder Mesh Magnitude

Hz

102 103 104 105 106-2

-1

0

1

2RCLadder Mesh Phase Angle

Hz


60

Homework• Repeat the analysis of this circuit. That is find and

plot Vout / Vin as a function of frequency. Assume Va = 0; C=10nF,R1=2.2MΩ, R2 =330kΩ, R3=2.7MΩ. Perform the plot using Matlab

VinVoutR1

R2 R3

C

C

Va


61

Homework

• Repeat the analysis of this circuit. That is find and plot Vout / Vin as a function of frequency

2 1

2 1 2

3

3 3

Node

01 12 1

1 1 1 1[ ]1 1 1 12 1 2 1

Node

0;if 0, then 11

101 11

b in b b out b a

in out ab

b a out aa

b outb out

bV V V V V V V

R Rj C j C

V V VVR R R

j C j C j C j Ca

V V V V VR

j CV V V V

R j C Rj C

VinVoutR1

R2 R3

C2

C1

Va

Vb

3 2 1 2

3 2 3 1 3 2

3 3 2 3 1 2

1 2

3 3 2 1 2

1 1 1[ ( 1 2)] 112

1 1 ( 1 2)[ ] 21 1 1

( 1 2) 1 1[ ( 2 ]1 1 1

( 1 2)[ ( 2 ]1 1

in outout

inout out

inout

inout

ou

V VV j C Cj C R R R R

j CVj C CV j C V

j C R R j C R R j C R RVC CV j C

C R C R R C R R RVR RC CV j C

C R C R R R RV

1 2 2 1 222

3 3 2 1 3 3 1

3 1

2 1 1 22 3 1

3 11 2

1 22 1 2 3 1

1 1 1( 1 2)( 1 2) ( 2 ] ( 2 )

1 1 1 1

( 1 2) ( 2 )1 1

|2 ( )

t

in

C C C

R R C C R R RC CV R j C j C RC R C R R R C R C R R

R RC C R R R Rj C R R R

C CR R

R RR R j CR R RC


62

Homework


2 1

2 1 2

3

3 3

Node

01 12 1

1 1 1 1[ ]1 1 1 12 1 2 1

Node

0;if 0, then 12

10; 0 (11 2 22

b in b b out b a

in out ab

b out a outa

b out out outb out b out

bV V V V V V V

R Rj C j C

V V VVR R R

j C j C j C j Cc

V V V V VR

j CV V V VV V V V

R j C R j C Rj C

3

3

3

)

2 1( )2b out

j C RV Vj C R

VinVoutR1

R2 R3

C2

C1

Va

Vb

c


63

Homework


VinVoutR1

R2 R3

C2

C1

Va

Vb

c

3

3 2 1 2

3

3 2 1 2

3

3 2 1 2

3 32 1

2 1 1 1( ) [ ( 1 2)] 122

2 1 1 1( )[ ( 1 2)] 222 1 1 1{( )[ ( 1 2)] 2}

21 1{[ ( 1 2)]( 2 1) 2

in outout

inout out

inout

j C R V VV j C Cj C R R R R

j Cj C R Vj C C V j C V

j C R R R Rj C R Vj C C j C V

j C R R R R

j C C j C R j C R jR R

3

2

3 33

2 1 3 2 3

3

3 2 1 2

22}

( 2 ) 2 21 1{( 2 1)[ ( 1 2)] }2 1 2 1

2 1 1 1( ) [ ( 1 2)] 22

out in

inout

inout out

j C RC V VR

j C R j C V j C Rj C R j C C VR R j C R R j C R

j C R VV j C C j C Vj C R R R R

3

3 2 1 2

3 3 3

3 2 3 1 3 2

3 3

3 2 3 1 3 2

2 1 1 1{( )[ ( 1 2)] 2}2

2 1 2 1 2 1{ ( 1 2) ) 2}2 2 22 1 2 1 1{ ( 1 2)(1 ) 2}2 2 2

{[

inout

inout

inout

j C R Vj C C j C Vj C R R R R

j C R j C R j C R Vj C C j C Vj C R R j C R R j C R Rj C R j C R Vj C C j C Vj C R R j C R R j C R R

3 1 1 3 2 2

3 2 1 3 2 1 3 2

3 1 2 1 2 3 2 1 2 1

3 2 1 22

1 2 3 2 1 3 1 2 2

2 2 ( 1 2)( 1 2) )] 2}2 2 2

2 ( ) 2 1 ( 1 2){[ )]}2

[( 2 1 ) [ 2 ( ){

inout

inout

j C R R R j C R R R Vj C Cj C C j C Vj C R R R j C R R R j C R R

j C R R R R R j C R R R j C j R R C C VVj C R R R R

R R C C R R R j C R R R R

1

3 2 1 22

1 2 3 2 1 3 1 2 2 1

3 2 1 3 2 1 22

1 2 3 2 1 3 1 2 2 1

3 2 1 3 2 1 2

( 1 2)]}2

( 2 1 ) [ 2 ( ) ( 1 2)]{ }2 2

( 2 1 ) [ 2 ( ) ( 1 2)]{ }2 2

inout

inout

inout

R C C VVj C R R R R

R R C C R R R j C R R R R R C C VVj C R R R j C R R R R

R R C C R R R C R R R R R C C VVj C R R R C R R R R


64

Homework


VinVoutR1

R2 R3

C2

C1

Va

Vb

c

3 1 3 2

3 2 1 3 2 1 3 2

3 2 3 1 3 2

3 3 2 3 1 2

3

2 2 ( 1 2){[ ) ) 1 )]}2 2 2

1 1 ( 1 2)[ ] 22 2 2

( 1 2) 1 1[ ( 2 ]2 2 2

( 1 2)[ (2

inout

inout out

inout

out

j C R R j C R R VC Cj C Vj C R R R j C R R R C R R

Vj C CV j C Vj C R R j C R R j C R R

VC CV j CC R C R R C R R R

C CV j CC R

1 2

3 2 1 2

1 2 2 1 222

3 3 2 1 3 3 1

3 1

2 1 1 22 3 1

3 11 2

1 22 1 2 3 1

2 ]2

1 1 1( 1 2)( 1 2) ( 2 ] ( 2 )

2 2 2 2

( 1 2) ( 2 )2 2

|2 ( )

in

out

in

C C C

VR RC R R R R

VR R C C R R RC CV R j C j C R

C R C R R R C R C R RR R

C C R R R Rj C R R RC C

R RR RR R j CR R R

C


65

Homework• Repeat the analysis of this circuit. That is

find and plot Vout / Vin as a function of frequency

3 1

1 22 1 2 3 1

1 22 3 1

13 1

2 2 2 11 22 1 2 3 1

3 1 3 1 3 10 0 0

1 2 1 2 1 22 1 2 3 1

2 ( )

( )tan (

2(2 ) ( )

| | |22 ( ) ( )

30 02 2

OR

out

in

out

in

V R RR RV R R j CR R R

CR RCR R RR R C

R RR RR R CR R RC

V R R R R CR RR R R RV R RR R j CR R R j

C C

1 22 3 1

13 10 0 0

2 2 2 11 22 1 2 3 1

1 2

13 10 0 0

1 2 2 1

3 1 3

1 22 1 2 3 1

( )| | { tan ( } |

2(2 ) ( )

3| { tan ( } | 0 { } | 02 2 2

| |2 ( )

out

in

out

in

R RCR R RV R R CV R RR RR R CR R R

CR R

CR R CR R R R

V R R R RR RV R R j CR R R

C

1

2 3 1

1 22 3 1

13 1

2 2 2 11 22 1 2 3 1

13 1 2 3 10

2 3 1 2 1

| 0( ) 2

OR

( )| | { tan ( } |

2(2 ) ( )

| { tan ( } | 0 { } | 0 {2 2 2

out

in

j CR R R

R RCR R RV R R CV R RR RR R CR R R

CR R CR R R

CR R R R R

VinVoutR1

R2 R3

C

C

Va

Vb

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

0.01 0.1 1 10 100 1000 100000

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

MagnitudeAngle


66

Homework• Repeat the analysis of this circuit. That is

find and plot Vout / Vin as a function of frequency

1 22

2 3 1

1 22

2 3 1

3 1 3 3

2 1 2 2

13 1 32

22 1

| |2 0 2 2

OR

| { tan (0)2(2 )

outR R

in C R R R

outR R

in C R R R

V R R R RV R R j R R

V R R RV RR R

VinVoutR1

R2 R3

C

C

Va

Vb

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

0.01 0.1 1 10 100 1000 100000

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

MagnitudeAngle


67

Matlab Code

clear all;R1=2.2e6;R2=330e3;R3=2.7e6;C=10e-9;omega=(.1:1:10^4);maxomega=length(omega);VIN=[1/R2; 0];for i=1:maxomega

zC=1/complex(0,omega(i)*C);G=[1/R1+1/R2+2/zC -1/zC;1/zC 1/R3]; V=G\VIN;VOUT(i)=V(2);

endf=omega/(2*pi);subplot(2,1,1);semilogx(f,abs(VOUT));title('R-wave Filter Nodal Magnitude');xlabel('Hz');axis([f(1) f(maxomega) 0 5]);subplot(2,1,2);semilogx(f,atan2(imag(VOUT),real(VOUT)));title('R-wave Filter Nodal Phase Angle');xlabel('Hz');axis([f(1) f(maxomega) -4 4]);

10-1 100 101 102 1030

1

2

3

4

5R-wave Filter Nodal Magnitude

Hz

10-1 100 101 102 103-4

-2

0

2

4R-wave Filter Nodal Phase Angle

Hz

Date post:	01-Aug-2020
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