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

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

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)

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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)

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

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

H(f) for Quiz #7H(f) for Quiz #7

H(f) for Quiz #7 w/o InductorH(f) for Quiz #7 w/o Inductor

H(f) for Quiz #7 w/small InductorH(f) for Quiz #7 w/small Inductor

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

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

Some 5th Order FiltersSome 5th Order Filters

source: Wikipedia – Alessio Damato

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)|

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.

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)|

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)|

H(f) for Quiz #7 w/small InductorH(f) for Quiz #7 w/small Inductor

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)|

Let to = α & f( ) = h( )h(t – α)u(t – α); α > 0

h(t – α); α > 0, t > α