ECEN3714 Network AnalysisLecture #9 2 February 2015Dr. George Scheetswww.okstate.edu/elec-eng/scheets/ecen3714

ECEN3714 Network AnalysisLecture #9 2 February 2015Dr. George Scheetswww.okstate.edu/elec-eng/scheets/ecen3714

Read 13.8 Problems: 13.16a, 19a, & 31c Exam #1, Lecture 11, Friday 6 February

Open Book & Notes, Closed Instructor Chapter 13, Forward & Inverse Laplace Transforms Simple R, L, & C circuits

Quiz 2 ResultsHi = 8.0, Low = 5.6, Average = 5.60Standard Deviation = 1.31

V(s) = 10s/D(s)D(s) = s2 + 10s + 25 = (s + 5)2

V(s) = 10s/D(s)D(s) = s2 + 10s + 25 = (s + 5)2

Re(s)

D(s)

-5 Denominator has two real roots at s = -5 Using Partial Fraction Expansion, can write as

V(s) = -50/(s+5)2 + 10/(s+5) These two terms have transform pairs in the tables

V(s) = -50/(s+5)2 + 10/(s+5)↕

v(t) = -50te-5t + 10e-5t

V(s) = -50/(s+5)2 + 10/(s+5)↕

v(t) = -50te-5t + 10e-5t

t

v(t)

10

V(s) has 2nd order pole at -5Zero at 0

V(s) has 2nd order pole at -5Zero at 0

Re(s)

V(s) = 10s/(s+5)2

-5

Re(s) = σ

Im(s) = jω

xx-5

|V(s)|The figure above is a plot along the real axis.

3D Plotof

|V(s)| =

|10s/(s+5)|

3D Plotof

|V(s)| =

|10s/(s+5)|ω

σ

ω = 0 axis = real axis

-50+50

25

-25

v(t) = -50te-5t + 10e-5tv(t) = -50te-5t + 10e-5t

t

v(t)

10Re(s) = σ

Im(s) = jω

xx-5

|V(s)|

Frequency Content (magnitude) along σ = 0 axis

Fourier TransformFourier Transform

Is built into the Laplace Transform Provided nothing much happens before t = 0 Set σ = 0 and examine jω axis

Got a waveform x(t)? F.T. indicates sinusoids needed to generate x(t)

Got a filter transfer function? F.T. indicates frequencies system will pass or

block.

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

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

Re(s)

D(s)

-1 Roots exist even if equation doesn't hit zero

They'll be complex

V(s) = 10s/(s2 + 2s + 25)↕

v(t) = 10.2e-tcos(4.899t + .0641π)

V(s) = 10s/(s2 + 2s + 25)↕

v(t) = 10.2e-tcos(4.899t + .0641π)

t

v(t)10.2

V(s) has poles at -1 + j4.899Zero at 0

V(s) has poles at -1 + j4.899Zero at 0

s

V(s) = __10s___ (s2+2s+25)

Im(s) = jω

x

x

-1 Re(s) = σ

4.899 The figure above is a plot along the real axis.

3D Plotof

|V(s)| =

10s(s2+2s+25)

3D Plotof

|V(s)| =

10s(s2+2s+25) ω

σ

ω = 0 axis = real axis

-10+10

+10

-10

v(t) = 10.2e-tcos(4.899t + .0641π)v(t) = 10.2e-tcos(4.899t + .0641π)

t

v(t)10.2

Im(s) = jω

x

x

-1 Re(s) = σ

4.899

Frequency Content (magnitude) along σ = 0 axis

V(s) = 10s/(s+5)2V(s) = 10s/(s+5)2

Re

Im

xx-5

t

v(t)

10

v(t) = -50te-5t + 10e-5t

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

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:Location of poles on Real axis sets decay rate.

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:Complex conjugate poles indicate oscillation.Location of poles on Imaginary axis sets oscillation rate.

S Domain Voltage & Currents S Domain Voltage & Currents Resistor (v = iR)

V(s) = I(s) R I(s) = V(s)/R

Capacitor (i = C dv/dt)I(s) = sCV(s) – Cv(0-)V(s) = I(s)/(sC) – v(0-)/s

Inductor (v = L di/dt)V(s) = sLI(s) – Li(0-)I(s) = V(s)/(sL) – i(0-)/s