1. ECEN3713 Network Analysis Lecture #15 15 February 2016 Dr
ECEN3713 Network Analysis Lecture # February 2016 Dr. George Scheets
Read 13.7
Problems: 13.8, 10, & 11
Friday, Quiz #4
Initial Conditions (i.e. everything ≠ 0 when t < 0)

2. ECEN3713 Network Analysis Lecture #16 17 February 2016 Dr
ECEN3713 Network Analysis Lecture # February 2016 Dr. George Scheets
Problems: & 13.78; Old Quiz #5
Friday, Quiz #4
Initial Conditions (i.e. everything ≠ 0 when t < 0)
Exam 1 Results
Hi = 100, Low = 53, Average = 80.73, Standard Deviation = 10.25
A > 89, B > 79, C > 68, D > 57

3. OSU IEEE February General Meeting 5:30-6:30 pm, Wednesday, 17 February
ES201b
Reps from Zeeco will
present
Dinner will be served + 3 points extra credit
All are invited

4. V(s) = (100s + 106)/(s )2
v(t) = 500,000te-5000t + 100e-5000t
Stability Issues:
Location of poles on Real axis sets decay rate. t
v(t)
100
.001
Shape of curve indicates
not pure exponential. Re Im x
-5000

5. v(t) = 500,000te-5,000t + 100e-5,000t) 500,000te-5,000t 100e-5,000t
.001
100e-5,000t
100 t
.001

6. s = σ + jω… Pole real (x axis) coordinate σ Provides time constant
Time Waveform fades away if < 0
Pole imaginary (y axis) coordinate ω
Frequency of oscillation
0 = no oscillation
1st Order Equation "s"
Has 1 pole
2nd Order Equation "s2"
Has 2 poles
Etc.

7. Pulse in: u(t) – u(t-1)...
xin(t) 1 t 1 10

8. Pulse & RC Low Pass Filter y(t) = u(t)(1–e-t/[RC]) - u(t-1)(1-e-(t-1)/[RC])10
xin(t)

9. e-αs term in Laplace Domain
In numerator?
Indicates a delay in the time domain
Ignore it for Partial Fraction Expansion Find undelayed time domain equation
Then delay it by α seconds
In denominator?
Indicates a math error

10. V(s) = 4/(s(4Cs2 +(1+4C)s +5)
Will not oscillate when 0 < C < 13 mF
1 H
vin Im x -5 Re
C = 0 Im x
-5.27 Re
C = 3 mF
-79.04
1 Ω C
vout
4 Ω
Quiz 4B
2005 Im x
-7.04 Re
-14.79
C = 12 mF Im x
-6.13 Re
-22.64
C = 9 mF

11. V(s) = 4/(s(4Cs2 +(1+4C)s +5) Oscillate when 14 mF < C < 4.49 F
vin Im x
-9.43 Re
C = 14 mF
.627 Im x
-4.96 Re
C = 28 mF
4.47
1 Ω C
vout
4 Ω
Quiz 4B
C = 3 F Im x
-.54 Re
.35 Im x
-.75 Re
1.39
C = .5 F

12. V(s) = 4/(s(4Cs2 +(1+4C)s +5) Will not oscillate when C > 4.49 F Im
vin Im x
-.36 Re
C = 5 F
-.69
1 Ω
C = 4.48 F Im x
-.53 Re
.02 C
vout
4 Ω
Quiz 4B Im x
-.14 Re
-.88
C = 10 F Im x
-.2 Re
-.83
C = 7.5 F

13. H(s) = 1/[1 – e–Δs]; s = σ + jω
Δ = 1 second
Frequency Response
H(jω) = 1/[1 – e–jω]
Plot of jω axis; σ = 0
1st & 2nd
Quadrant
Blows Up at Integer
Multiples of
1 Hz (2π rads/sec) jω σ

14. Need Voltage Gain at ω = 5.5 rad/sec?
Plot of jω axis; σ = 0
Frequency Response
H(j5.5) = 1/[1 – e–j5.5]
[1 - e–j5.5] = [1 - e–j2π0.8754] = [1 – e–j315.1◦]
= [1 – ej44.9◦] = [1 – ( j0.7059)]
= [ j0.7059] = ∟-67.55◦
H(j5.5) = 1.309∟67.55◦
Voltage Gain = 1.309, degree phase shift