EEM 205 CIRCUIT ANALYSISeem.eskisehir.edu.tr/aaybar/EEM 209/icerik/EEM209_WEEK1... ·...

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EEM 209 CIRCUIT ANALYSIS

Transcript of EEM 205 CIRCUIT ANALYSISeem.eskisehir.edu.tr/aaybar/EEM 209/icerik/EEM209_WEEK1... ·...

Page 1: EEM 205 CIRCUIT ANALYSISeem.eskisehir.edu.tr/aaybar/EEM 209/icerik/EEM209_WEEK1... · 2018-10-03 · William H. Hayt, Jack E. Kemmerly and Steven M. Durbin “Engineering Circuit

EEM 209CIRCUIT ANALYSIS

Page 2: EEM 205 CIRCUIT ANALYSISeem.eskisehir.edu.tr/aaybar/EEM 209/icerik/EEM209_WEEK1... · 2018-10-03 · William H. Hayt, Jack E. Kemmerly and Steven M. Durbin “Engineering Circuit

Text Books:

William H. Hayt, Jack E. Kemmerly and Steven M. Durbin

“Engineering Circuit Analysis”, Mc Graw Hill, 6th Ed. 2002.

James W.Nilsson and Susan A. Riedel “Electric Circuits”,

Prentice Hall, 6th Ed. 2002.

Grading:

First Midterm : 20%

Second Midterm: 25%

Quiz: 15% (3 - 4)

Final: 40%

Page 3: EEM 205 CIRCUIT ANALYSISeem.eskisehir.edu.tr/aaybar/EEM 209/icerik/EEM209_WEEK1... · 2018-10-03 · William H. Hayt, Jack E. Kemmerly and Steven M. Durbin “Engineering Circuit

• Transient Response

• The Simple RL Circuit

• The Simple RC Circuit

• Unit-Step Function

• Response of an RL Circuit to a Step Function

• RLC Circuits : Passive Series RLC Circuit

• Complex Frequency

• Frequency Response

• Poles and the Natural Response

Page 4: EEM 205 CIRCUIT ANALYSISeem.eskisehir.edu.tr/aaybar/EEM 209/icerik/EEM209_WEEK1... · 2018-10-03 · William H. Hayt, Jack E. Kemmerly and Steven M. Durbin “Engineering Circuit

• Complete Response

• Resonance:The Simple Series RLC Circuit

• Scaling

• Bode Diagrams

• The Operational Amplifier

• Two Port Networks

• The Fourier Transform

• The Laplace Transform

Page 5: EEM 205 CIRCUIT ANALYSISeem.eskisehir.edu.tr/aaybar/EEM 209/icerik/EEM209_WEEK1... · 2018-10-03 · William H. Hayt, Jack E. Kemmerly and Steven M. Durbin “Engineering Circuit

Transient ResponseA change in the source output, or a change in the circuit or

element values will cause the currents and voltages in the circuit tochange.The question is how these changes will take place.

Ex:

AIt

AIt

34

12:0

242

12:0

1

1

Page 6: EEM 205 CIRCUIT ANALYSISeem.eskisehir.edu.tr/aaybar/EEM 209/icerik/EEM209_WEEK1... · 2018-10-03 · William H. Hayt, Jack E. Kemmerly and Steven M. Durbin “Engineering Circuit
Page 7: EEM 205 CIRCUIT ANALYSISeem.eskisehir.edu.tr/aaybar/EEM 209/icerik/EEM209_WEEK1... · 2018-10-03 · William H. Hayt, Jack E. Kemmerly and Steven M. Durbin “Engineering Circuit
Page 8: EEM 205 CIRCUIT ANALYSISeem.eskisehir.edu.tr/aaybar/EEM 209/icerik/EEM209_WEEK1... · 2018-10-03 · William H. Hayt, Jack E. Kemmerly and Steven M. Durbin “Engineering Circuit

Ex:

A

t

AIt

34

12I :t

jumps)sudden shownot do currents (The ?:0

242

12:0

1

1

Page 9: EEM 205 CIRCUIT ANALYSISeem.eskisehir.edu.tr/aaybar/EEM 209/icerik/EEM209_WEEK1... · 2018-10-03 · William H. Hayt, Jack E. Kemmerly and Steven M. Durbin “Engineering Circuit

The Simple RL Circuit

0tfor circuit following the have we

osuddenly.S change not cani current the butsuddenly, opens switch the0,t At

i current the of because L in stored isEnergy :0

L

L

t

Page 10: EEM 205 CIRCUIT ANALYSISeem.eskisehir.edu.tr/aaybar/EEM 209/icerik/EEM209_WEEK1... · 2018-10-03 · William H. Hayt, Jack E. Kemmerly and Steven M. Durbin “Engineering Circuit

tL

R

L

tti

I

L

L

L

eItitL

R

t

tL

Ri

tI

I

dtL

Rdt

L

R

ii

L

R

Ri

L

L

L

0

0

L

0L

0

)(

LL

L

RL

0L

)(I

(t)i ln

)0(L

R-)ln(I-(t)i ln

ln

)(di

dt

di

0dt

diL

0VV

I(0)i current the Let

0

0

Page 11: EEM 205 CIRCUIT ANALYSISeem.eskisehir.edu.tr/aaybar/EEM 209/icerik/EEM209_WEEK1... · 2018-10-03 · William H. Hayt, Jack E. Kemmerly and Steven M. Durbin “Engineering Circuit

Note that, the power delivered to the resistor is

inductor the in storedenergy initial the to equalexactly is which

2

11

2

1)(W

is heat to convertedenergy total The

12

11

2

1

2)()(W

isresistor the in heat to convertedenergy the and

Re)()(

2

0

2

0R

22

0

22

0

0

22

0

0

22

0

0

R

22

0

2

LIeLIt

eLIeLI

eR

LRIdteRIdttPt

ItRitP

tL

Rt

L

R

tt

L

Rtt

L

Rt

R

tL

R

LR

Page 12: EEM 205 CIRCUIT ANALYSISeem.eskisehir.edu.tr/aaybar/EEM 209/icerik/EEM209_WEEK1... · 2018-10-03 · William H. Hayt, Jack E. Kemmerly and Steven M. Durbin “Engineering Circuit

Ex:

• The switch has been connected

• to the terminal a for very long

• time and then switched to b at

• t=0.

• a) Find i(t)

b) Plot i(t)

c) In how many seconds after t=0 does the current i drop down to 50% of its initialvalue?

d) In how many seconds does i drop down to 10% of its inital value

e)In how many seconds does the power in become 75%of its initial value

f) In how many seconds does the energy in the L become 60% of its initial value?

Page 13: EEM 205 CIRCUIT ANALYSISeem.eskisehir.edu.tr/aaybar/EEM 209/icerik/EEM209_WEEK1... · 2018-10-03 · William H. Hayt, Jack E. Kemmerly and Steven M. Durbin “Engineering Circuit

Aeti

eti

t

AiiiAi

ta

t

t

5

2

10

3)(

3)(

0

3)0()0()0(3105

45

0 )

stee

LIeLItWf

stetP

WP

eeRitPe

sttetid

st

te

etic

dtt

tL

R

dR

ct

cR

R

ttR

bb

b

a

at

ta

dd

d

c

a

a

051.06.040.01

2

140.01

2

1)()

029.010

75.0ln)90(75.090

90)0(

90310)()

41.05

10.0ln)3(10.03)()

139.0

693.05.0ln550.0

)3(50.03)( )

1010

2

0

22

0

10

10252

5

5

5

Page 14: EEM 205 CIRCUIT ANALYSISeem.eskisehir.edu.tr/aaybar/EEM 209/icerik/EEM209_WEEK1... · 2018-10-03 · William H. Hayt, Jack E. Kemmerly and Steven M. Durbin “Engineering Circuit

Properties of the Exponential Response

Normalized current expression :

When would the current become zero if it had continued to decrease at its initial rate? The line tangent to the curve at (0,1) point will give as the answer.

tL

R

eI

ti

0

)(

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The slope of the line :

τ

τττ

of multiples at current normalized the of Values

Constant) Time:( 1L

R- 0 : zero be to y(t)for Time

1L

R-y(t) : line the of equation The

)(

000

R

L

t

L

Re

L

R

I

ti

dt

d

t

tL

R

t

Page 16: EEM 205 CIRCUIT ANALYSISeem.eskisehir.edu.tr/aaybar/EEM 209/icerik/EEM209_WEEK1... · 2018-10-03 · William H. Hayt, Jack E. Kemmerly and Steven M. Durbin “Engineering Circuit

t İ(t)/I0 Appoximate Value

τ 𝑖(τ)

𝐼0=𝑒−

𝑅

𝐿 𝑒𝐿

𝑅 = 𝑒−1=0.3679 ~1

3

2τ 𝑒−2=0.1353 <5%

3τ 𝑒−3=0.0498 <5%

4τ 𝑒−4 =0.0183 <1%

5τ 𝑒−5=0.0067 <1%

desired.) is

precision more when5(or 3after value statesteady its reached it that assumed

bemay it nscalculatio mustfor but , t at 0 to goes response The: Note

ττ

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