ecen3714 network analysis lecture #30 30 march 2015 dr. george scheets problems: olde quiz #8...
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ECEN3714 Network AnalysisLecture #30 30 March 2015Dr. George Scheetswww.okstate.edu/elec-eng/scheets/ecen3714
ECEN3714 Network AnalysisLecture #30 30 March 2015Dr. George Scheetswww.okstate.edu/elec-eng/scheets/ecen3714
Problems: Olde Quiz #8 Exam #2 this Friday
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ECEN3714 Network AnalysisLecture #31 1 April 2015Dr. George Scheetswww.okstate.edu/elec-eng/scheets/ecen3714
ECEN3714 Network AnalysisLecture #31 1 April 2015Dr. George Scheetswww.okstate.edu/elec-eng/scheets/ecen3714
Problems: Olde Exams (#2) Exam #2 this Friday Quiz #7 Results
Hi = 10, Low = 2.0, Average = 6.22Standard Deviation = 2.70
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2015 OSU ECE Spring Banquet2015 OSU ECE Spring Banquet Hosted by Student Branch of IEEE Wednesday, 15 April, at Meditations Doors open at 5:30 pm, meal at 6:00 pm Cash Bar Sign up in ES202 to reserve your seat(s)
$5 if pay in advance (< 8 April) & resume submitted to [email protected]
$10 on 9 & 10 April. (A $20 value.)
Speakers: Ron Sinnes (Level3, Director IP) & Eric Miller (VYVX, Sports Manager)
Dress is Business Casual Many door prizes available!
Known best: 8.4" Samsung Galaxy Tab S, two Fluke Multimeters
All are invited!
Sponsored in part by:
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Quiz 7Quiz 7
-
+
10 μF
vout
0.1 H50 Ω
50 Ω
Suppose you needed to find the input current iin(t)
What can be ignored?
iin(t) 10 μF
0.1 H50 Ω
50 Ω
iin(t)
1 MΩ
10 μF
0.1 H50 Ω
50 Ω
iin(t)
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Quiz 7Quiz 7
-
+
10 μF
vout
0.1 H50 Ω
50 Ω
Suppose you needed to find the input impedance Zin
What can be ignored?
Zin
10 μF
0.1 H50 Ω
50 Ω 1 MΩ
10 μF
0.1 H50 Ω
50 ΩZin
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Quiz 7Quiz 7
-
+
10 μF
vout
0.1 H50 Ω
50 Ω
Suppose you needed to find the voltage transfer function H(s) = Vout(s)/Vin(s)
What can be ignored?
vin
-
+
10 μF
vout
0.1 H50 Ωvin
10 μF
vout
0.1 H50 Ω
vin
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H(f) for Quiz #7H(f) for Quiz #7
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H(f) for Quiz #7 w/o InductorH(f) for Quiz #7 w/o Inductor
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H(f) for Quiz #7 w/small InductorH(f) for Quiz #7 w/small Inductor
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The Ideal FilterThe Ideal Filter
Frequency Response H(f) = K1; over some frequency range = 0; elsewhere
Phase Response θ(f) = K2*f; over same frequency range
Will delay all frequencies by same time amount Input with energy 100% in frequency range?
Amplitude may be changed May be shifted in time General shape unchanged
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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)
H(f) = 100 & θ(f) = 0; 0 < f < 50
H(f) = 100 & θ(f) = 18f degrees; 0 < f < 50
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Some 5th Order FiltersSome 5th Order Filters
source: Wikipedia – Alessio Damato
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Single Real Pole, Two Real Poles 1/(jω+3), 1/(jω+3)2
Single Real Pole, Two Real Poles 1/(jω+3), 1/(jω+3)2
0 5 10 15 200
0.2
0.40.333
2.445 103
hi
h2i
200 wi
|H(ω)|
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Single Real Pole, Two Real Poles 1/(jω+3), 3/(jω+3)2
Single Real Pole, Two Real Poles 1/(jω+3), 3/(jω+3)2
0 5 10 15 200
0.2
0.40.333
7.335 103
hi
h2i
200 wi
|H(ω)|
Note: 2nd order system has sharper roll-off.Also, 3 dB break point has moved.
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Complex Conjugate Poles, |real| = 0H(s) = 1/(s2 + 100) = 1/[(s + j10)(s – j10)]
H(jω) = 1/(ω2 – 100)
Complex Conjugate Poles, |real| = 0H(s) = 1/(s2 + 100) = 1/[(s + j10)(s – j10)]
H(jω) = 1/(ω2 – 100)
0 5 10 15 20 250
0.01
0.02
.025
3.333 103
h4i
202 10
8 wi
|H(ω)|
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Complex Conjugate Poles, real < 0H(s) = 1/(s2 + 4s + 104) = 1/[(s + 2 + j10)(s + 2 – j10)]
H(jω) = 1/[104 – ω2 +j4ω]
Complex Conjugate Poles, real < 0H(s) = 1/(s2 + 4s + 104) = 1/[(s + 2 + j10)(s + 2 – j10)]
H(jω) = 1/[104 – ω2 +j4ω]
0 5 10 15 20 250
0.01
0.02
.025
3.261 103
h4i
202 10
8 wi
|H(ω)|
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H(f) for Quiz #7 w/small InductorH(f) for Quiz #7 w/small Inductor
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Complex Conjugate Poles, real < 0H(s) = 1/(s2 + 10s + 125) = 1/[(s + 5 + j10)(s + 5 – j10)]
H(jω) = 1/[125 – ω2 +j10ω]
Complex Conjugate Poles, real < 0H(s) = 1/(s2 + 10s + 125) = 1/[(s + 5 + j10)(s + 5 – j10)]
H(jω) = 1/[125 – ω2 +j10ω]
0 5 10 15 20 250
0.01
0.02
.025
2.941 103
h4i
202 10
8 wi
|H(ω)|
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Let to = α & f( ) = h( )h(t – α)u(t – α); α > 0
h(t – α); α > 0, t > α
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