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# ECEN3713 Network Analysis Lecture #11 14 February 2006 Dr. George Scheets - PowerPoint PPT Presentation

ECEN3713 Network Analysis Lecture #11 14 February 2006 Dr. George Scheets. Read 13.4 – 13.5, 13.7 Problems: 13.9 – 13.12 Exam #1, Thursday 16 February Chapter 12 & 13.1 - 13.5, no initial conditions. v(t). 100. Im. t. .001. x. x. -5000. Re.

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ECEN3713 Network AnalysisLecture #11 14 February 2006Dr. George Scheets

• Read 13.4 – 13.5, 13.7

• Problems: 13.9 – 13.12

• Exam #1, Thursday 16 February

• Chapter 12 & 13.1 - 13.5, no initial conditions

100

Im

t

.001

x

x

-5000

Re

V(s) = (100s + 106)/(s + 5000)2

v(t) = 500,000te-5000t + 100e-5000t

Stability Issues:

Location of poles on Real axis sets decay rate.

Shape of curve indicates

not pure exponential.

v(t) = 500,000te-5,000t + 100e-5,000t)

500,000te-5,000t

100

t

.001

100e-5,000t

100

t

.001

xin(t)

1

t

1

10

Smeared pulse out: u(t)(1–e-0.5t) - u(t-1)(1+e-0.5(t-1))

xin(t)

1

t

1

10

xout(t)

0.632

t

10

1

Im

C = 3 mF

C = 0

-79.04

x

x

x

x

x

Re

Re

-5

-5.27

Im

Im

C = 12 mF

C = 9 mF

-14.79

-22.64

x

x

x

x

x

x

Re

Re

-7.04

-6.13

V(s) = 4/(s(4Cs2 +(1+4C)s +5)

Will not oscillate when 0 < C < 13 mF

1 H

vin

1 Ω

C

vout

4 Ω

Quiz 4B

2005

Im

C = 28 mF

C = 14 mF

x

4.47

.627

-4.96

x

x

x

x

Re

Re

x

-9.43

Im

Im

C = 3 F

C = .5 F

x

1.39

.35

x

x

x

x

x

-.75

Re

Re

-.54

V(s) = 4/(s(4Cs2 +(1+4C)s +5)

Oscillate when 14 mF < C < 4.49 F

1 H

vin

1 Ω

C

vout

4 Ω

Quiz 4B

Im

C = 4.48 F

C = 5 F

.02

x

x

-.36

x

x

x

x

Re

-.69

-.53

Re

Im

Im

C = 10 F

C = 7.5 F

-.88

-.83

x

x

x

x

x

x

Re

-.14

Re

-.2

V(s) = 4/(s(4Cs2 +(1+4C)s +5)

Will not oscillate when C > 4.49 F

1 H

vin

1 Ω

C

vout

4 Ω

Quiz 4B