ECEN3513 Signal Analysis Lecture #26 23 October 2006

ECEN3513 Signal AnalysisECEN3513 Signal AnalysisLecture #26 23 October 2006Lecture #26 23 October 2006ECEN3513 Signal AnalysisECEN3513 Signal AnalysisLecture #26 23 October 2006Lecture #26 23 October 2006

Read 6.8, 6.9Read 6.8, 6.9 Problems 6.5-5, 6.5-6, 6.8-2Problems 6.5-5, 6.5-6, 6.8-2 Test #2 on 27 OctoberTest #2 on 27 October

Chapters 4 & 6Chapters 4 & 6

Read 6.12Read 6.12 Problems: 6.9-1, 6.9-2, 6.12-1Problems: 6.9-1, 6.9-2, 6.12-1 Test #2 next time!! Chapters 4-6Test #2 next time!! Chapters 4-6 Quiz 7 results: Hi =7.6, Low = 0.5, Ave = 4.56Quiz 7 results: Hi =7.6, Low = 0.5, Ave = 4.56

Standard Deviation = 1.70Standard Deviation = 1.70

after filteringafter filteringi 0 200 t

i50 100

i

200 time varies from -50 to +50 seconds

j 0 400 fj

4.00000 8j

400 Frequencies vary from -4 to +4 Hz

in 0.02 Hertz incrementsTs 50

envelope f( )0.5j f 0.099( ) f .101( ) 0.5j f .101( ) f .099( ) f .001( ) f .001( )

1 2j f 0.550 .000001

xi

0

400

j

Re envelope fj cos 2 f

j t

i

0

400

j

Im envelope fj sin 2 f

j t

i

xi

xi

Ts

60 40 20 0 20 40 600

1

2

xi

timax x( ) 1.954

δ(f-0.1) δ(f)δ(f+0.1)

y(t)

output magnitude & phaseoutput magnitude & phase

4 3 2 1 0 1 2 3 40

20

40

60

envelope f j

f j

phasej

atan2 Re envelope fj Im envelope f

j f205

0.1 phase205

180

107.441

4 3 2 1 0 1 2 3 42

0

2

phase j

f j

|Y(f)|

filter impulse responsefilter impulse responsei 0 200 t

i5 10

i


j 0 400 fj

4.00000 8j



envelope f( )1

1 2j f 0.5

xi

0

400

j


j t

i

0

400

j


j t

i

xi

xi

Ts

6 4 2 0 2 4 61

0

1

2

xi

ti

H(f)

h(t)

filter magnitude & phase responsefilter magnitude & phase response

4 3 2 1 0 1 2 3 40

0.5

1

envelope f j

f j

phasej


j f205

0.1 phase205

180

17.441

4 3 2 1 0 1 2 3 42

0

2

phase j

f j

|H(f)|

filter magnitude & phase responsefilter magnitude & phase responsephase

jatan2 Re envelope f

j Im envelope fj

f205

0.1 phase205

180

17.441

4 3 2 1 0 1 2 3 42

0

2

phase j

f jTo avoid distortionAll frequencies should be delayed same time

Higher frequencies should see larger phase shifts(1 second delay = 360 degrees for 1 Hz, 720 degrees for 2 Hz, etc.)

i.e. phase delay should = fFilter phase response should be a straight line

Square wave made up of 100 cosinesEvery other cosine has phase = 180○

Square wave made up of 100 cosinesEvery other cosine has phase = 180○

yi

0

99

j

1( )j cos 2 5 1 2 j( ) i

points

1 2 j=

0 200 400 600 800 10001

0

1

yi

i

Improperly aligned phasesImproperly aligned phases

yi

0

99

j

1( )j cos 2 5 1 2 j( ) i

points

1 2 j=

0 200 400 600 800 10005

0

5

yi

i

filter magnitude & phase responsefilter magnitude & phase responsephase

jatan2 Re envelope f

j Im envelope fj

f205

0.1 phase205

180

17.441

4 3 2 1 0 1 2 3 42

0

2

phase j

f j

To avoid phase distortionPhase delay should = f

Filter phase response should be a straight line

2vp, 1 second pulse in2vp, 1 second pulse ini 0 200 t

i5 10

i


j 0 500 fj

5.00001 10j



envelope f( )2sin f

f

xi

0

400

j


j t

i

0

400

j


j t

i

xi

xi

Ts

6 4 2 0 2 4 61

0

1

2

3

xi

ti

x(t)

pulse magnitude & phasepulse magnitude & phase

6 4 2 0 2 4 60

1

2

envelope f j

f j

phasej


j f255

0.1 phase255

180

0

6 4 2 0 2 4 60

2

4

phase j

f j

|X(f)|

pulse outpulse outi 0 200 t

i5 10

i


j 0 500 fj

5.00001 10j



envelope f( )2sin f

f 1 2j f 0.5

xi

0

400

j


j t

i

0

400

j


j t

i

xi

xi

Ts

6 4 2 0 2 4 6

0

2

xi

ti

y(t)

output magnitude & phaseoutput magnitude & phase

6 4 2 0 2 4 60

1

2

envelope f j

f j

phasej


j f255

0.1 phase255

180

17.439

6 4 2 0 2 4 62

0

2

phase j

f j

|Y(f)|

Quiz 7 - Filter Impulse ResponseQuiz 7 - Filter Impulse Responsei 0 200 t

i5 10

i


j 0 500 fj

5.000001 10j



X f( )1

1 .8 e 4j f

xi

0

500

j

Re X fj

cos 2 fj

ti

= 0

500

j

Im X fj

sin 2 fj

ti

=

xi

xi

Ts

6 4 2 0 2 4 65

0

5

10

15

xi

ti

0.02

delay = 2

∑

y(t)

.8

x(t)

.8y(t-2)

+

+← H(f)

h(t)

Quiz 7: Transfer Function H(f)Quiz 7: Transfer Function H(f)

6 4 2 0 2 4 60

2

4

6

X fj

fj

phasej

atan2 Re X fj

Im X fj

6 4 2 0 2 4 61

0

1

phasej

fj

Little or nophase distortion.

Quiz 7 - Energy SpectrumQuiz 7 - Energy Spectrumi 0 200 t

i5 10

i


j 0 500 fj

5.000001 10j



X f( )1

1 .8 e 4j f

1

1 .8 e4j f

xi

0

500

j

Re X fj

cos 2 fj

ti

= 0

500

j

Im X fj

sin 2 fj

ti

=

xi

xi

Ts

6 4 2 0 2 4 620

0

20

40

xi

ti

0.02

← |H(f)|2

(1|H(f)|2 = SYY(f) if delta function input)

↕RYY(τ) for output if delta function input.

Energy SpectrumEnergy Spectrum

6 4 2 0 2 4 60

10

20

30

X fj

fj

phasej

atan2 Re X fj

Im X fj

6 4 2 0 2 4 61

0

1

phasej

fj

Energy Transfer Response|H(f)|2

Output y(t) when input = u(t)Output y(t) when input = u(t)i 0 200 t

i5 50

i


j 0 500 fj

5.000001 10j



X f( )1

1 .8 e 4j f0.5 50 f .001( ) .001 f( ) 1

2j f

xi

0

500

j

Re X fj

cos 2 fj

ti

= 0

500

j

Im X fj

sin 2 fj

ti

=

xi

xi

Ts

10 0 10 20 30 40 50

0

5

xi

ti

max x( ) 4.975

δ(f)H(f)U(f) = Y(f)

y(t)

|Y(f)| and phase plot|Y(f)| and phase plot

6 4 2 0 2 4 60

5

10

X fj

fj

phasej

atan2 Re X fj

Im X fj

6 4 2 0 2 4 64

2

0

2

4

phasej

fj

Need to find power?Need to find power?

Evaluate time average.Evaluate time average. Limit Limit 11 ∫∫ x(t) x(t)22dtdt

TT→∞ T →∞ T TT

Evaluate power REvaluate power RXXXX((ττ) ) at at ττ = 0. = 0.

Find area under power spectrum SFind area under power spectrum SXXXX(f).(f). Units are W/Hz. Units are W/Hz.

Integrate out Hz & left with WattsIntegrate out Hz & left with Watts

Convert X(f) to SConvert X(f) to SXXXX(f) via Lim (f) via Lim 11 |X(f)| |X(f)|22

Find area under curve T→∞ TFind area under curve T→∞ T Limit = |X(f)|Limit = |X(f)|22 if line spectra if line spectra

Need to find energy?Need to find energy?

Evaluate time average.Evaluate time average. Limit Limit ∫∫ x(t) x(t)22dtdt

TT→∞ →∞ TT

Evaluate energy REvaluate energy RXXXX((ττ) ) at at ττ = 0. = 0.

Find area under energy spectrum GFind area under energy spectrum GXXXX(f).(f). Units are J/Hz. Units are J/Hz.

Integrate out Hz & left with JoulesIntegrate out Hz & left with Joules

Convert X(f) to GConvert X(f) to GXXXX(f) = |X(f)|(f) = |X(f)|22

Find area under curve Find area under curve

ECEN3513 Signal Analysis Lecture #26 23 October 2006

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