signals and systems note3

8/7/2019 Signals and Systems note3

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6.003: Signals and SystemsFeedback, Poles, and Fundamental Modes

February9,2010


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Last Time: Multiple Representations of DT SystemsVerbal descriptions: preserve the rationale.To reduce thenumberofbitsneeded to storea sequenceoflargenumbersthatarenearlyequal, recordthefirstnumber,and then record successivedifferences.

Difference equations: mathematicallycompact.y[n] = x[n] x[n 1]

Block diagrams: illustrate signalflowpaths.1 Delay

+x[n] y[n]

Operator representations: analyze systemsaspolynomials.Y = (1R

) X


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Last Time: Feedback, Cyclic Signal Paths, and ModesSystems with signals that depend on previous values of the samesignalare said tohave feedback.Example: Theaccumulator systemhas feedback.

+X YDelay

Bycontrast, thedifferencemachinedoesnothave feedback.1 Delay

+X Y


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Last Time: Feedback, Cyclic Signal Paths, and ModesThe effect of feedback can be visualized by tracing each cyclethrough thecyclic signalpaths.

Delay

+p0

X Y

x[n] = [n] y[n]

n n1 0 1 2 3 4 1 0 1 2 3 4Eachcyclecreatesanother sample in theoutput.


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Delay

+p0

X Y

x[n] = [n] y[n]



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Delay

+p0

X Y

x[n] = [n] y[n]



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X + YDelayp0

x[n] = [n] y[n]



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X + YDelayp0

x[n] = [n] y[n]



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X + YDelayp0

x[n] = [n] y[n]

n n1 0 1 2 3 4 1 0 1 2 3 4Eachcyclecreatesanother sample in theoutput.The responsewillpersisteven though the input is transient.


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Geometric Growth: PolesTheseunit-sampleresponsescanbecharacterizedbyasinglenumber the polewhich is thebaseof thegeometric sequence.

Delay

+p0

X Y

ny[n] =

p0, if n >= 0;0, otherwise.

y[n] y[n] y[n]

n n n

1 0 1 2 3 4 1 0 1 2 3 4 1 0 1 2 3 4p0 = 0.5 p0 = 1 p0 = 1.2


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Check YourselfHowmanyof the following unit-sample responses can berepresentedbya singlepole?

n n

n n

n


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Check YourselfHowmanyof the following unit-sample responses can berepresentedbya singlepole? 3

n n

n n

n


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Geometric GrowthThevalueof p0 determines the rateofgrowth.

y[n] y[n] y[n] y[n]

1 0 1z

p0 < 1:1< p0 < 0:

0< p0 < 1:p0 > 1:

magnitudediverges,alternating signmagnitudeconverges,alternating signmagnitudeconvergesmonotonicallymagnitudedivergesmonotonically


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Second-Order SystemsThe unit-sample responses ofmore complicated cyclic systems aremorecomplicated.

y[n]

+R

R1.6

0.63

X Y

n

1 0 1 2 3 4 5 6 7 8Notgeometric. This responsegrows thendecays.


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Factoring Second-Order SystemsFactortheoperatorexpressiontobreakthesystem intotwosimplersystems (divideandconquer).

+R

R1.6

0.63

X Y

Y =X+ 1.6RY0.63R2Y(11.6R+ 0.63R2)Y =X(1

0.7

R)(1

0.9

R)Y =X


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Factoring Second-Order SystemsThe

factored

form

corresponds

to

a

cascade

of

simpler

systems.

(10.7R)(10.9R)Y =X

X +0.7 R

+0.9 R

YY2

(10.7R)Y2 =X (10.9R)Y =Y2

+0.9

R+

0.7 RY

Y1X

(10.9R)Y1 =X (10.7R)Y =Y1

Theorderdoesntmatter (if systemsare initiallyat rest).


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Factoring Second-Order SystemsThe

unit-sample

response

of

the

cascaded

system

can

be

found

by

multiplying thepolynomial representationsof the subsystems.Y 1 1 1X

= (1

0.7R

)(1

0.9R

)= (1

0.7R

)

(1

0.9R

)

= (1 + 0.7R+ 0.72

R2+ 0.73

R3+

)

(1 + 0.9

R+ 0.92

R2+ 0.93

R3+

)

Multiply, thencollect termsofequalorder:Y = 1 + (0.7 + 0.9)R+ (0.72+ 0.7 0.9 + 0.92)R2X

+ (0.73+ 0.72 0.9 + 0.7 0.92+ 0.93)R3+


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Multiplying PolynomialGraphical

representation

of

polynomial

multiplication.

Y = (1 + aR+ a2R2+ a3R3+ ) (1 + bR+ b2R2+ b3R3+ )X1

a Ra2 R2

a3 R3

+

1b Rb2 R2

b3 R3

+

... ... ... ...

X Y

Collect termsofequalorder:Y = 1 + (a+ b)R+ (a

2

+ ab+ b2)R2

+ (a3+ a2b+ ab2+ b3)R3

+ X


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Multiplying PolynomialsTabular representationofpolynomialmultiplication.

(1 + aR+ a2R2+ a3R3+ ) (1 + bR+ b2R2+ b3R3+ )1 bR b2R2 b3R3

1 1 bR b2R2 b3R3 aR aR abR2 ab2R3 ab3R4

a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 a3

R3 a3

R3 a3b

R4 a3b2

R5 a3b3

R6

Group samepowersofRby following reversediagonals:

Y = 1 + (a+ b)R+ (a2

+ ab+ b2)R2

+ (a3+ a2b+ ab2+ b3)R3

+ X y[n]

n1 0 1 2 3 4 5 6 7 8


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Partial FractionsUse

partial

fractions

to

rewrite

as

a

sum

of

simpler

parts.

+

R

R1.6

0.63

X Y

Y 1 1 4.5 3.5X = 11.6R+ 0.63R2 = (10.9R)(10.7R) = 10.9R 10.7R

S d O d S t E i l t F


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Second-Order Systems: Equivalent FormsThe

sum

of

simpler

parts

suggests

a

parallel

implementation.

Y 4.5 3.5X = 10.9R 10.7R

+0.9 R

4.5 +

3.5R0.7

+

X YY1

Y2

n nIf x[n] = [n] then y1[n] = 0.9 and y2[n] = 0.7 for n0.n

Thus, y[n] = 4.5(0.9)n3.5(0.7) for n0.

P ti l F ti


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Partial FractionsGraphical

representation

of

the

sum

of

geometric

sequences.

ny1[n] = 0.9 for n0n1 0 1 2 3 4 5 6 7 8

ny2[n] = 0.7 for n0n1 0 1 2 3 4 5 6 7 8

ny[n] = 4.5(0.9)n3.5(0.7)for n0

n1 0 1 2 3 4 5 6 7 8

Partial Fractions


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Partial FractionsPartial fractionsprovidesa remarkableequivalence.

R

R1.6

0.63

+X Y

+0.9

R4.5 +

3.5R0.7

+

X YY1

Y2

follows from thinkingabout systemaspolynomial (factoring).

Poles


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PolesThe

key

to

simplifying

a

higher-order

system

is

identifying

its

poles.

Poles are the roots of the denominator of the system functionalwhenR 1.zStartwith system functional:

Y 1 1 1= = =X 1 1.6R+0.63R2 (1p0R)(1p1R) (10.7R) (10.9R) p0=0.7 p1=0.91SubstituteR andfind rootsofdenominator:zY 1 z2 z2= = =X 1.6 0.63 z21.6z +0.63 (z0.7) (z0.9)1 + 2

z z z0=0.7 z1=0.9

Thepolesareat0.7and0.9.

Check Yourself


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Check YourselfConsider the systemdescribedby

y[n] =14 y[n1]+18

y[n2]+x[n1]12 x[n2]Howmanyof the followingare true?

1. Theunit sample responseconverges to zero.2. Therearepolesat z= 12 and z= 14 .3. There isapoleat z= 12 .4. Thereare twopoles.5. Noneof theabove

Check Yourself


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Check Yourself1 1 1y[n] =4

y[n 1] +8y[n 2]+x[n 1]2

x[n 2](1 +4

1R 81R2)Y = (R 2

1R2)XR R2R

12

1 121 1

2z 14zH(

R) =

YX

2zz

1 += = =11 +4 R218 14 1 18 1 18z2+ 2zz

z12z =

12

14z+

1. Theunit sample responseconverges to zero.12 and z= 14. X2. Therearepolesat z=12. X 3. There isapoleat z=

4. Thereare twopoles.5. Noneof theabove X

Check Yourself


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Check YourselfConsider the systemdescribedby

y[n] =14 y[n1]+18

y[n2]+x[n1]12 x[n2]Howmanyof the followingare true? 2

1. Theunit sample responseconverges to zero.2. Therearepolesat z= 12 and z= 14 .3. There isapoleat z= 12 .4. Thereare twopoles.5. Noneof theabove

Population Growth


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Population Growth

Population Growth


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Population Growth

Population Growth


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Population Growth

Population Growth


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opu at o G owt

Population Growth


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p

Population Growth


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p

Population Growth


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Population Growth


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Population Growth


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Population Growth


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Check Yourself


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Whatare thepole(s)of theFibonacci system?1.

1

2. 1and13. 1and24. 1.618. . . and0.618. . . 5. noneof theabove

Check Yourself


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Whatare thepole(s)of theFibonacci system?Differenceequation forFibonacci system:

y[n] =

x[n] +

y[n

1]

+

y[n

2]

System functional:

Y 1H= =X 1RR2Denominator is secondorder2poles.

Check Yourself


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Find thepolesby substitutingR 1/z in system functional.Y 1 1 z2H= = =X 1RR2 11z 12 z2 z 1z

Polesareat1 5 1z= =,2

where represents thegolden ratio1 + 5= 1.6182

The twopolesareat1z0 = 1.618 and z1 = 0.618

Check Yourself


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Whatare thepole(s)of theFibonacci system? 41. 12. 1and13. 1and24. 1.618. . . and0.618. . . 5. noneof theabove

Example: Fibonaccis Bunnies


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Eachpolecorresponds toa fundamentalmode.1 1.618 and 0.618

1 n

n

n

n

1 0 1 2 3 4 1 0 1 2 3 4Onemodediverges,onemodeoscillates!

Example: Fibonaccis Bunnies


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The unit-sample response of the Fibonacci system can be writtenasaweighted sumof fundamentalmodes.

1Y 1 5 5H= =

= +

X 1RR2 1 R 1 +1Rh[n] = n+ 1 ()n; n 05

5

Butwealreadyknow that h[n] is theFibonacci sequence f:

f : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, . . . Thereforewecancalculate f[n]withoutknowing f[n 1]or f[n 2]!

Complex Poles


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What ifapolehasanon-zero imaginarypart?Example:

Y 1=X 1R+R21 z2= =11z +z12 z2 z+ 1 Polesare z=

1

2 23j= ej/3.Whatare the implicationsofcomplexpoles?

Complex Poles


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Partial fractionsworkevenwhen thepolesarecomplex.

Y 1 1 1 ej/3 ej/3X = 1ej/3R 1ej/3R = j3 1ej/3R 1ej/3R

Thereare two fundamentalmodes (bothgeometric sequences):ejn/3 =cos(n/3) +jsin(n/3) and ejn/3 =cos(n/3)jsin(n/3)

n n

Complex Poles


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Complexmodesareeasier tovisualize in thecomplexplane.ejn/3 =cos(n/3)+jsin(n/3)

j2/3 Im ej1/3ej3/3 j0/3e e Re n

j4/3 j5/3e eejn/3 =cos(n/3)jsin(n/3)

j4/3 Im ej5/3ej3/3 j0/3e e Re n

j2/3 j1/3e e

Complex Poles


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Theoutputofareal systemhas realvalues.y[n] =x[n] +y[n 1] y[n 2]

Y 1H= =X 1R+R21 1= 1 ej/3R 1 ej/3R

1 ej/3 ej/3= j 3 1 ej/3

R 1 ej/3

Rh[n] = 1 ej(n+1)/3 ej(n+1)/3 =2 sin(n+ 1)j 3 3 3

h[n]1

1

n

Check Yourself


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Unit-sample responseofa systemwithpolesat z =rej.

n

Which

of

the

following

is/are

true?

1. r < 0.5and 0.52. 0.5< r < 1and 0.53. r < 0.5and 0.084. 0.5< r < 1and 0.085. noneof theabove

Check Yourself


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Unit-sample responseofa systemwithpolesat z =rej.

n

Whichof the following is/are true? 21. r < 0.5and 0.52. 0.5< r < 1and 0.53. r < 0.5and 0.084. 0.5< r < 1and 0.085. noneof theabove

Check Yourself


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R R R+X Y

Howmanyof the following statementsare true?1.This systemhas3 fundamentalmodes.2.Allof the fundamentalmodescanbewrittenasgeometrics.3.Unit-sample response is y[n] : 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1 . . . 4.Unit-sample response is y[n] : 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1 . . . 5.One of the fundamental modes of this system is the unitstep.

Check Yourself


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R R R+X Y

Howmanyof the following statementsare true? 41.This systemhas3 fundamentalmodes.2.Allof the fundamentalmodescanbewrittenasgeometrics.3.Unit-sample response is y[n] : 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1 . . . 4.Unit-sample response is y[n] : 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1 . . . 5.One of the fundamental modes of this system is the unitstep.

SummaryS d f dd i d d l b h i d


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Systemscomposedofadders,gains,anddelayscanbecharacterizedby theirpoles.Thepolesofa systemdetermine its fundamentalmodes.Theunit-sampleresponseofasystemcanbeexpressedasaweightedsumof fundamentalmodes.Theseproperties follow fromapolynomial interpretationofthesystem functional.

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6.003 Signals and Systems

Spring 2010

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