ECEN3714 Network Analysis Lecture #16 18 February 2015 Dr. George Scheets n Read 14.4 n Problems:...

ECEN3714 Network AnalysisECEN3714 Network AnalysisLecture #16 18 February 2015Lecture #16 18 February 2015Dr. George ScheetsDr. George Scheetswww.okstate.edu/elec-eng/scheets/ecen3714www.okstate.edu/elec-eng/scheets/ecen3714


Read 14.4Read 14.4 Problems: Old Quiz #4Problems: Old Quiz #4 Quiz #4 this FridayQuiz #4 this Friday Quiz #3 ResultsQuiz #3 Results

Hi = 9.0, Low = 0.5, Average = 5.50Hi = 9.0, Low = 0.5, Average = 5.50Standard Deviation = 2.69Standard Deviation = 2.69



Read 14.5Read 14.5 Problems: 13.55, 13.57, 14.2Problems: 13.55, 13.57, 14.2 Quiz #5 this FridayQuiz #5 this Friday



Read 14.6Read 14.6 Problems: Old Quiz #5Problems: Old Quiz #5 Quiz #5 this FridayQuiz #5 this Friday Quiz 4 ResultsQuiz 4 Results

Hi = 10, Low = 3, Average = 8.06Hi = 10, Low = 3, Average = 8.06Standard Deviation = 1.97Standard Deviation = 1.97

Gustav KirchoffGustav Kirchoff Born 1824Born 1824 Died 1887Died 1887 German PhysicistGerman Physicist KCL & KVL formulated in 1852KCL & KVL formulated in 1852

Doctoral DissertationDoctoral Dissertation

KVL: KVL: ΣΣ(voltage drops around a closed loop) = 0(voltage drops around a closed loop) = 0 KCL: KCL: ΣΣ(current entering or exiting a node) = 0(current entering or exiting a node) = 0 High side voltage on device end a current entersHigh side voltage on device end a current enters

source: Wikipedia

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

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

Re

Im

xx-5000 t

v(t)

100

.001

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

t

500,000te-5,000t

100

.001

t

100

.001

100e-5,000t

Pulse in: u(t) – u(t-1)... Pulse in: u(t) – u(t-1)...

t

1

10

xin(t)

1

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

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

t

1

10

xin(t)

1

t

0.632

101

xout(t)

V(s) = 4/(s [4Cs2 +(1+4C)s +5])Roots at [ –[1+4C] + {[1+4C]2 – 80C}0.5 ]/2

V(s) = 4/(s [4Cs2 +(1+4C)s +5])Roots at [ –[1+4C] + {[1+4C]2 – 80C}0.5 ]/2

vout

vin 1 Ω

C

1 H

4 Ω

Quiz 4B2005

Will not oscillate when 0 < C < 13 mF

Im

xx-5 Re

C = 0

Im

xx-5.27 Re

C = 3 mF

x-79.04

Im

xx-6.13 Re

x-22.64

C = 9 mFIm

xx-7.04 Re

x-14.79

C = 12 mF

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

vout

vin 1 Ω

C

1 H

4 Ω

Quiz 4B

Oscillates when 14 mF < C < 4.49 F

Im

xx

-.75 Rex

1.39C = .5 F

Im

x

x-4.96

Re

C = 28 mF

x

4.47

Im

xx

-9.43 Re

C = 14 mF

.627x

C = 3 F Im

xx

-.54Rex

.35

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

vout

vin 1 Ω

C

1 H

4 Ω

Quiz 4B

Will not oscillate when C > 4.49 F

Im

xx-.2 Rex

-.83C = 7.5 F

Im

xx-.36

Re

C = 5 F

x-.69

C = 4.48 F Im

xx

-.53Re

x.02

Im

xx-.14 Re

x-.88

C = 10 F

s = σ + jω…s = σ + jω… Pole real (x axis) coordinate Pole real (x axis) coordinate σσ

Provides time constantProvides time constant Time Waveform fades away if < 0Time Waveform fades away if < 0

Pole imaginary (y axis) coordinate Pole imaginary (y axis) coordinate ωω Frequency of oscillationFrequency of oscillation

0 = no oscillation0 = no oscillation

1st Order Equation "s"1st Order Equation "s" Has 1 poleHas 1 pole

2nd Order Equation "s2nd Order Equation "s22"" Has 2 polesHas 2 poles

Etc.Etc.

y(t) = x(t) + y(t-1) → H(s) = 1/[1 – e–s]y(t) = x(t) + y(t-1) → H(s) = 1/[1 – e–s]

σ

Frequency Responseσ = 0 axisThis system, with a

feedback loop, has anInfinite # of poles alongThe σ = 0 (jω) axis.

V(s) = 10s/(s2+2s+25)V(s) = 10s/(s2+2s+25)

Im

x

x

-1 Ret

10.2

v(t) = 10.2e-tcos(4.899t + .0641π)

Stability Issues: UnderdampedLocation of poles on Real axis sets decay rate.Location of poles on Imagninary axis sets oscillation rate.

V(s) = 105/(s2 + 2*105s + 108)V(s) = 105/(s2 + 2*105s + 108)

t

v(t)

.5

.002

v(t) = .5025(e-500t - e-199,500t)

Im

xx-500 Re-199,500

Stability Issues: OverdampedNo complex poles = no oscillation.Location of poles on real axis sets decay rate.

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

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

Re

Im

xx-5000 t

v(t)

100

.001

Stability Issues: Critically dampedLocation of poles on Real axis sets decay rate.

If system is under damped it'seasy to get values of α & β

α = decay rate & β = oscillation frequency

If system is under damped it'seasy to get values of α & β

α = decay rate & β = oscillation frequency

Must have imaginaryroots to oscillate.

ECEN3714 Network Analysis Lecture #16 18 February 2015 Dr. George Scheets n Read 14.4 n Problems:...

Documents

Transcript of ECEN3714 Network Analysis Lecture #16 18 February 2015 Dr. George Scheets n Read 14.4 n Problems:...