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ECEN3714 Network AnalysisLecture #21 2 March 2015Dr. George Scheetswww.okstate.edu/elec-eng/scheets/ecen3714
ECEN3714 Network AnalysisLecture #21 2 March 2015Dr. George Scheetswww.okstate.edu/elec-eng/scheets/ecen3714
Read 14.7 Problems: 14.5, 7, & 55 Quiz #6 this Friday Quiz #5 Results
Hi = 10, Low = 4.0, Average = 7.89Standard Deviation = 2.02
ECEN3714 Network AnalysisLecture #24 9 March 2015Dr. George Scheetswww.okstate.edu/elec-eng/scheets/ecen3714
ECEN3714 Network AnalysisLecture #24 9 March 2015Dr. George Scheetswww.okstate.edu/elec-eng/scheets/ecen3714
Problems: 14.11, 22,37 No Quizzes or Labs week prior to Spring Break Exam #2 on 3 April Note: Lab Practicum dropped
10 Labs & 1 Design Problem Still worth 220 points total in the end ≈ 1/3 of grade
ECEN3714 Network AnalysisLecture #25 11 March 2015Dr. George Scheetswww.okstate.edu/elec-eng/scheets/ecen3714
ECEN3714 Network AnalysisLecture #25 11 March 2015Dr. George Scheetswww.okstate.edu/elec-eng/scheets/ecen3714
Problems: 14.51, 52, 57 Next Quiz is week after Spring Break Exam #2 on 3 April
ECEN3714 Network AnalysisLecture #26 13 March 2015Dr. George Scheetswww.okstate.edu/elec-eng/scheets/ecen3714
ECEN3714 Network AnalysisLecture #26 13 March 2015Dr. George Scheetswww.okstate.edu/elec-eng/scheets/ecen3714
Problems: 14.58, 59, 72 Next Quiz is week after Spring Break Exam #2 on 3 April
TransformsTransforms
X(s) = x(t) e-st dt
0-
∞Laplace
s = σ +jω
∞
X(f) = x(t) e-j2πft dt
-∞Fourier
Got a Laplace Transform?Got a Laplace Transform?
Re(s) = σ
Im(s) = jω
|V(s)|
The Fourier Transform of x(t) is on the jω axis.*
*Provided x(t) = 0; t < 0
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 = 0
Frequency Responseσ = 0 axis
Re(s) = σ
Im(s) = jω
|V(s)|
ω = 0
Generating a Square Wave...Generating a Square Wave...
0
1.5
-1.50 1.0
5 Hz+
15 Hz+
25 Hz+
35 Hz
cos2*pi*5t - (1/3)cos2*pi*15t + (1/5)cos2*pi*25t - (1/7)cos2*pi*35t)
5 cycle per second square wave generated using 4 sinusoids
Generating a Square Wave...Generating a Square Wave...
5 cycle per second square wave generated using 100 sinusoids.
Max frequency at (N*10 – 5) Hz = 995 Hz
0
1.5
-1.50 1.0
5 Hz square wave after Single Pole Low Pass Filtering159.2 Hz half power frequency
5 Hz square wave after Single Pole Low Pass Filtering159.2 Hz half power frequency
0
1.5
-1.50 1.0
Not much visible change.Blue = Error = Output(t) – Input(t)
Generating a Square Wave...Generating a Square Wave...
5 cycle per second square wave generated using 100 sinusoids.
Max frequency at (N*10 – 5) Hz = 995 Hz
0
1.5
-1.50 1.0
5 Hz square wave after Single Pole High Pass Filtering159.2 Hz half power frequency
5 Hz square wave after Single Pole High Pass Filtering159.2 Hz half power frequency
0
1.5
-1.50 1.0
FiltersFilters Low Pass, High Pass, & Band Pass
All are LTI if made from R's, C's, and/or L's LTI Systems have an impulse response h(t)
δ(t) in? h(t) is output. If LTI, for any input x(t),
In Time Domain: x(t)☺h(t) = y(t)In Laplace Domain: X(s)H(s) = Y(s)In Frequency Domain: X(jω)H(jω) = Y(jω)
If not LTI, h(t) does not exist We'll hit ☺ in more detail later
OpAmpsOpAmps
High Gain Devices Typically, Voltage Gain > 10,00 Vout(t) = [Vp(t) – Vn(t)]*Voltage Gain
High Input Impedance Zin typically > 1 MΩ
Integrator: H(jω) = -1/jωRCIntegrator: H(jω) = -1/jωRC
ω
|H(ω)|
10
10000 100
5 Hz Square Wave In...5 Hz Square Wave In...
0
1.5
-1.50 1.0
This one made up of 100 sinusoids.Fundamental frequency of 5 Hz &
next 99 harmonics (15, 25, ..., 995 Hz).
5 Hz Triangle Out...5 Hz Triangle Out...
0
1.5
-1.50 1.0
-50
50
Differentiator: H(jω) = -jωRCDifferentiator: H(jω) = -jωRC
ω
|H(ω)|
1
10000 100
5 Hz Square Wave In...5 Hz Square Wave In...
0
1.5
-1.50 1.0
This curve made up of 100 sinusoids.Fundamental frequency of 5 Hz &
next 99 harmonics (15, 25, ..., 995 Hz).
Spikes Out...Spikes Out...
0
1.5
-1.50 1.0