ECEN3713 Network AnalysisECEN3713 Network AnalysisLecture #21 28 March 2006Lecture #21 28 March 2006Dr. George ScheetsDr. George Scheets

ECEN3713 Network AnalysisECEN3713 Network AnalysisLecture #21 28 March 2006Lecture #21 28 March 2006Dr. George ScheetsDr. George Scheets

Read Chapter 15.4Read Chapter 15.4 Problems: 13.78, 15.5 – 15.7Problems: 13.78, 15.5 – 15.7 Thursday's QuizThursday's Quiz

Series or Parallel Combinations of Active FiltersSeries or Parallel Combinations of Active Filters

Thursday's AssignmentThursday's AssignmentProblems 15.8, 15.10, 15.11, 15.20Problems 15.8, 15.10, 15.11, 15.20

Quiz 7 kicked back for a regrade.Quiz 7 kicked back for a regrade.

ECEN3713 Network AnalysisECEN3713 Network AnalysisLecture #23 4 April 2006Lecture #23 4 April 2006Dr. George ScheetsDr. George Scheets

ECEN3713 Network AnalysisECEN3713 Network AnalysisLecture #23 4 April 2006Lecture #23 4 April 2006Dr. George ScheetsDr. George Scheets

Read Chapter 15.5 - End of ChapterRead Chapter 15.5 - End of Chapter Problems: 15.21, 15.22, 15.24, 15.25Problems: 15.21, 15.22, 15.24, 15.25 Test on Thursday!Test on Thursday!

Focus is on Material after last testFocus is on Material after last test Chapters 14 & 15Chapters 14 & 15 Initial ConditionsInitial Conditions Chapters 12 & 13 may also show upChapters 12 & 13 may also show up

Op Amp CharacteristicsOp Amp Characteristics

A v

vp(t)

vn(t)

+

-

vout(t) = Av(vp(t)-vn(t))

Zin?Zin? In M ohmsIn M ohms

HHopampopamp(f) f(f) f3dB3dB??

In XX or XXX MHzIn XX or XXX MHz Voltage gain Av?Voltage gain Av?

On order of 10On order of 1044 - 10 - 1066

Zin

+Vcc

-Vcc

Op Amps: No FeedbackOp Amps: No Feedback

A v

+

-

vout(t) = Av(vp(t)-vn(t))

Output likely to hit railsOutput likely to hit rails Unless tiny voltagesUnless tiny voltages

+Vcc

-Vcc

vin(t)

Op Amps: Positive FeedbackOp Amps: Positive Feedback

A vvin(t)+

-

vout(t)

Output likely to hit rails Output likely to hit rails May get stuck thereMay get stuck there

Op Amps: Negative FeedbackOp Amps: Negative Feedback

A vvin(t)+

- vout(t)

Safe to assume vSafe to assume vpp(t) = v(t) = vnn(t)(t)

Safe to assume no current enters Op AmpSafe to assume no current enters Op Amp If low Z outside paths existIf low Z outside paths exist

0 v

Op Amps: Output LoadOp Amps: Output Load

A vvin(t)+

- vout(t)

Ideally, load does not effect characteristicsIdeally, load does not effect characteristics Practically, load Practically, load maymay effect characteristics effect characteristics

If Op Amp output can't source or sink enough currentIf Op Amp output can't source or sink enough current

Zload

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

1st Order RC Low Pass Filter1st Order RC Low Pass Filter

|H(ω)|

ω

1

1

0.707

2nd Order Low Pass Filter(Two back-to-back 1st order active

filters)

2nd Order Low Pass Filter(Two back-to-back 1st order active

filters)

|H(ω)|

ω

1

1

0.707

3dB break point changes.

1st order2nd order

Scaled 2nd Order Low Pass Filter(Two back-to-back 1st order active filters)Scaled 2nd Order Low Pass Filter(Two back-to-back 1st order active filters)

|H(ω)|

ω

1

1

0.707

2nd order filter has faster roll-off.

1st order2nd order

2nd Order Butterworth Filter2nd Order Butterworth Filter

|H(ω)|

ω

1

1

0.707

Butterworth has flatter passband.

2nd order Butterworth2nd order Standard

1 & 2 Hz sinusoids1000 samples = 1 second

1 & 2 Hz sinusoids1000 samples = 1 second

0 200 400 600 800 1000

0

1.5

1.5

i

x1i

1 103

0 i

Suppose we need to maintain a phase relationship(low frequency sinusoid positive slope zero

crossing same as the high frequency sinusoid's).

samples

1 & 2 Hz sinusoids1000 samples = 1 second

Both delayed by 30 degrees

1 & 2 Hz sinusoids1000 samples = 1 second

Both delayed by 30 degrees

0 200 400 600 800 1000

0

1.5

1.5

1 103

0 i

Delaying the two curves by the same phaseangle loses the relationship.

1 & 2 Hz sinusoids1000 samples = 1 second

1 Hz delayed by 30, 2 Hz by 60 degrees

1 & 2 Hz sinusoids1000 samples = 1 second

1 Hz delayed by 30, 2 Hz by 60 degrees

0 200 400 600 800 1000

0

1.5

1.5

1 103

0 i

Delaying the two curves by the same time keeps the relationship. θlow/freqlow needs to = θhi/freqhi.

A transfer function with a linear phase plot θout(f) = Kθin(f)

will maintain the proper relationship.