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6.003: Signals and SystemsContinuous-Time Systems

February11,2010

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Previously: DT SystemsVerbal descriptions: preserve the rationale.

Next year, your account will contain p times your balancefrom this year plus the money that you added this year.

Difference

equations:

mathematically

compact.

y[n+ 1] = x[n] +py[n]Block diagrams: illustrate signal flow paths.

+ Delayp

x[n] y[n]

Operator representations: analyze systems as polynomials.(1

p

R) Y

=

RX

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Analyzing CT SystemsVerbal descriptions: preserve the rationale.

Your account will grow in proportion to the current interestrate plus the rate at which you deposit.

Differential equations:mathematically

compact.

dy(t)

=x(t) + py(t)dt

Block diagrams: illustrate signal flow paths.+ t

()dt

p

x(t) y(t)

Operator representations: analyze systems as polynomials.(1pA)Y =AX

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Differential EquationsDifferential equations are mathematically precise and compact.

r0(t)

r1(t)h1(t)

dr1(t) = r0(t)r1(t)dt Solution methodologies: general methods (separation of variables; integrating factors) homogeneous and particular solutions inspectionToday: new methods based on block diagrams and operators,which provide new ways to think about systems behaviors.

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Block DiagramsBlock diagrams illustrate signal flow paths.DT: adders, scalers, and delays represent systems described bylinear difference equations with constant coefficents.

+ Delayp

x[n] y[n]

CT: adders, scalers, and integrators represent systems describedby a linear differential equations with constant coefficients.

+ t()dt

p

x(t) y(t)

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Operator RepresentationCT Block diagrams are concisely represented with the A operator.

Applying A to a CT signal generates a new signal that is equal tothe integral of the first signal at all points in time.

Y = AXis equivalent to

ty(t) = x() d

for all time t.

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Evaluating Operator ExpressionsAs with R, A expressions can be manipulated as polynomials.

+ +A A

X YW

tw(t) = x(t) + x()d

ty(t) = w(t) + w()d

t t t 2 y(t) = x(t) + x()d+ x()d+ x(1)d1 d2

W = (1 + A) XY = (1 + A) W = (1 + A)(1 + A) X= (1 + 2A+ A2) X

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Evaluating Operator ExpressionsExpressions in A can be manipulated using rules for polynomials. Commutativity: A(1 A)X= (1 A)AX

Distributivity: A(1 A)X= (A A2)X Associativity: (1 A)A (2 A)X= (1 A) A(2 A) X

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Check Yourself

Ap

+X Y

A p+X Y

Ap+X

Y

y(t) = x(t) +py(t)

y(t) = x(t) +py(t)

y(t) = px(t) +py(t)

Which best illustrates the left-right correspondences?

1. 2. 3. 4. 5. none

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Check Yourself

Ap

+X Y

A p+X Y

Ap+X

Y

y(t) = x(t) +py(t)

y(t) = x(t) +py(t)

y(t) = px(t) +py(t)

Which best illustrates the left-right correspondences? 4

1. 2. 3. 4. 5. none

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Elementary Building-Block SignalsElementary DT signal: [n].

1, if n= 0;[n] =

0, otherwise[n]

1n

0 shortest possible duration (most transient) useful for constructing more complex signals

What CT signal serves the same purpose?

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Elementary CT Building-Block SignalConsider the analogous CT signal.

0 t < 0

w(t) = 1 t = 0

0 t > 0

w(t)

t1

0

Is

this

a

good

choice

as

a

building-block

signal?

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Elementary CT Building-Block SignalConsider the analogous CT signal.

0 t < 0

w(t) = 1 t= 0

0 t > 0

tw(t) ( ) dt 0

Is this a good choice as a building-block signal? Not

w(t) ( ) dt 0

The integral of w(t) is zero!

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Unit-Impulse SignalTheunit-impulsesignalactsasapulsewithunitareabutzerowidth.

p(t)

(t)= limp(t)0

p1/2(t) p1/4(t) p1/8(t)

t

12 unit area

42

11 1 t 1 1 t 1 1 t 2 2 4 4 8 8

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Unit-Impulse SignalThe unit-impulse function is represented by an arrow with the number 1, which represents its area or weight.

(t)1

tIt has two seemingly contradictory properties: it is nonzero only at t= 0, and its definite integral (,) is one !Both of these properties follow from thinking about (t) as a limit:

p(t)

(t)=

lim

p(t)0

t

12 unit area

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Unit-Impulse and Unit-Step SignalsThe indefinite integral of the unit-impulse is the unit-step.

t t1; 0u(t) =

()d= 0; otherwise

Equivalentlyt

u(t)1

A(t) u(t)

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Impulse Response of Acyclic CT SystemIftheblockdiagramofaCTsystemhasnofeedback(i.e.,nocycles),then the corresponding operator expression is imperative.

+ +A A

X Y

Y

= (1 + A)(1 + A) X= (1 + 2A

+ A

2

) X

If x(t) = (t) theny(t) = (1 + 2A+ A

2

) (t) = (t) + 2u(t) + tu(t)

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CT FeedbackFind the impulse response of this CT system with feedback.

+ Ap

x(t) y(t)

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+ Ap

x(t) y(t)

Method 1: find differential equation and solve it.y(t) = x(t) +py(t)

Linear, first-order difference equation with constant coefficients.Try y(t) = Cetu(t).Then y(t) = Cetu(t) + Cet(t) = Cetu(t) + C(t).Substituting, we find that Cetu(t) + C(t) = (t) +pCetu(t).Therefore = p and C= 1 y(t) = eptu(t).

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+ Ap

x(t) y(t)

Method 2: use operators.Y = A (X+pY)Y A=X 1 pA

Now expand in ascending series in A:Y

= A(1 +pA +p2A2+p3A3+ )X

If x(t) = (t) theny(t) = A(1 +pA +p2A2+p3A3+ ) (t)

= (1 +pt+21p2t2+

61p3t3+ ) u(t) = eptu(t) .

CT F db k

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CT FeedbackWe can visualize the feedback by tracing each cycle through thecyclic signal path.

+ Ap

x(t) y(t)

y(t) = (A +pA2

+p2

A

3+p

3

A

4+ ) (t)

CT F db k

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+ Ap

x(t) y(t)

y(t) = (A +pA2

+p2

A

3+p

3

A

4+ ) (t)

CT Feedback

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+ Ap

x(t) y(t)

y(t) = (A +pA2

+p2

A

3+p

3

A

4+ ) (t)

CT Feedback

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+ Ap

x(t) y(t)

y(t) = (A +pA2

+p2

A

3+p

3A

4+ ) (t)

CT Feedback

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+ Ap

x(t) y(t)

y(t) = (A +pA2+p2A3+p3A4+ ) (t)= (1 +pt+

2

1p2t2+

6

1p3t3+ ) u(t)

y(t)

t1

0

CT Feedback

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+ Ap

x(t) y(t)

y(t) = (A +pA2+p2A3+p3A4+ ) (t)= (1 +pt+

2

1p2t2+

6

1p3t3+ ) u(t)

y(t)

t1

0

CT Feedback

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+ Ap

x(t) y(t)

y(t) = (A +pA2+p2A3+p3A4+ ) (t)= (1 +pt+

2

1p2t2+

6

1p3t3+ ) u(t)

y(t)

t1

0

CT Feedback

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+ Ap

x(t) y(t)

y(t) = (A +pA2+p2A3+p3A4+ ) (t)= (1 +pt+

2

1p2t2+

6

1p3t3+ ) u(t)

y(t)

t1

0

CT Feedback

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+ Ap

x(t) y(t)

y(t) = (A +pA2+p2A3+p3A4+ ) (t)= (1 +pt+

2

1p2t2+

6

1p3t3+ ) u(t) = eptu(t)

y(t)

t1

0

CT Feedback

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CT FeedbackMakingp negative makes the output converge (instead of diverge).

+ Ax(t) y(t)p

y(t) = (A pA2+p2A3p3A4+ ) (t)= (1 pt+ 2

1p2t26

1p3t3+ ) u(t)

CT Feedback

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C eedback

Makingp negative makes the output converge.+ A

px(t) y(t)

y(t) = (A pA2+p2A3p3A4+ ) (t)= (1 pt+ 2

1p2t26

1p3t3+ ) u(t)

y(t)

0 t

1

CT Feedback

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px(t) y(t)

y(t) = (A pA2+p2A3p3A4+ ) (t)= (1 pt+ 2

1p2t26

1p3t3+ ) u(t)

y(t)

0 t

1

CT Feedback

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px(t) y(t)

y(t) = (A pA2+p2A3p3A4+ ) (t)= (1 pt+ 2

1p2t26

1p3t3+ ) u(t)

y(t)

0 t

1

CT Feedback

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px(t) y(t)

y(t) = (A pA2+p2A3p3A4+ ) (t)= (1 pt+ 2

1p2t26

1p3t3+ ) u(t)

y(t)

0 t

1

CT Feedback

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px(t) y(t)

y(t) = (A pA2+p2A3p3A4+ ) (t)= (1 pt+ 2

1p2t26

1p3t3+ ) u(t) = eptu(t)

y(t)

0 t

1

Convergent and Divergent Poles

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The fundamental mode associated withp diverges ifp > 0 and converges ifp < 0.

+ A

p

X Y

p = 1 p = 1y(t) y(t)1 1

0 t 0 t

Convergent and Divergent Poles

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The fundamental mode associated withp diverges ifp > 0 and converges ifp < 0.

Re p

+

Imp

A

p

X Y

RepConvergent Divergent

CT Feedback

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In CT, each cycle adds a new integration.+ A

px(t) y(t)

y(t) = (A +pA2+p2A3+p3A4+ ) (t)= (1 +pt+ 2

1p

2

t2

+ 6

1p

3

t3

+ ) u(t) = ept

u(t)y(t)

t1

0

Feedback in DT Systems

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In DT, each cycle creates another sample in the output.X + Y

Delayp0

y[n] = (1 +pR +p2R2+p3R3+p4R4+ ) [n]= [n] +p[n 1] +p2[n 2] +p3[n 3] +p4[n 4] +

y[n]

n1 0 1 2 3 4

Comparison of CT and DT representations

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Locations of convergent poles differ for CT and DT systems.

X Y X+ A

p

Y+

DelaypA 1

1pA 1pReptu(t) pnu[n]

Mass and Spring System

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Use the A operator to solve the mass and spring system.x(t)

F = K x(t) y(t) = M y(t)y(t)

+ KM A A

1

x(t) y(t)y(t)y(t)

KMKM

1 + A2Y

=

X A2


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Factor system functional to find the poles.KA2 KA2Y M M

KX = 1 + A2 = (1p0A)(1p1A)M1 + KA2 = 1 (p0+p1)A+p0p1A2

MThe sum of the poles must be zero.The product of the poles must be K/M.

K Kp0 =j p1 =j

M M


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.Alternatively, find the poles by substituting A 1sThe poles are then the roots of the denominator.

Y= KMA2

X

1 +

Substitute A 1:

KMA

2

Y

=s

KM

X s2+ KMK

s= j M


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The poles are complex conjugates.Ims

s-planeK 0M

ResK M 0

The corresponding fundamental modes have complex values.fundamental mode 1: ej0t =cos0t+jsin0tfundamental mode 2: ej0t =cos0tjsin0t


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Real-valued inputs always excite combinations of these modes sothat the imaginary parts cancel.Example: find the impulse response.

KA

2 K Y M M A AKX = 1 + A2 =p0p1 1 p0A1 p1AM0

2 A A=

2j0 1 j0A 1 +j0A 0 A 0 A= 2j 1 j0A 2j 1 +j0A

makesmode1 makesmode2

Themodesthemselvesarecomplexconjugates,andtheircoefficientsare also complex conjugates. Sothe sum is asum of somethingandits complex conjugate, which is real.


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The impulse response is therefore real.

Y 0 A 0 AX = 2j 1j0A 2j 1 + j0A

The impulse response ish(t) =

2j0ej0t2j0ej0t =0sin0t ; t > 0

y(t)

t0


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Alternatively, find impulse response by expanding system functional.

x )(t + A Ay(t)y(t)20

1

y(t)

A21 + A2

If x(t) = (t) then20

20Y = A2 4020 A4+ 60 A6 + =

X

3 55! +

t t(t) = 20 40t 6+ 0 , t 0y 3!


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Look at successive approximations to this infinite series.

Y = 02A2

2A2 = 0

2A202A2lX 1 + 0 l=0

If x(t) = (t) theny(t) = 0202lA2l+2(t)

l=02= 0t

y(t)

t0


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Y = 02A2

2A2 = 0

2A202A2lX 1 + 0 l=0

If x(t) = (t) then

y(t) = 0202lA2l+2(t)

l=03

= 02t04t

3!y(t)

t0


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Y = 02A2

2

A2 = 02A202A2lX 1 +

0 l=0If x(t) = (t) then

y(t) = 0202lA2l+2(t)

l=03 5

2 4t 6t= 0t03!

+ 05!y(t)

t0


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Y = 02A2

2

A2 =02A2

02A2

lX 1 +


y(t) = 0202lA2l+2(t)

l=03 5 7

2 4t 6t 8t=0t03!

+05!

07!

y(t)

t0

Mass and Spring SystemL k i i i hi i fi i i

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Y = 02A2

2

A2 =02A2

02A2

lX 1 +


y(t) = 0202lA2l+2(t)

l=03 5 7 9

2 4t 6t 8t 10t=0t03!

+05!

07!

+09!

y(t)

t0

Mass and Spring SystemL k t i i ti t thi i fi it i

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Y = 02A2

2

A2 =02A2

02A2

lX 1 +


y(t) = 0202lA2l+2(t)

l=03 5 7 9

2 4t 6t 8t 10t=0t 03!

+05!

07!

+09!

+ y(t)

t0

Mass and Spring SystemL k t i i ti t thi i fi it i

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Y = 02A2

2

A2 =02A2

02A2

lX 1 +


y(t) = 0202lA2l+2(t)

l=03 5 7 9

2 4t 6t 8t 10t=0t 03!

+05!

07!

+09!

+ y(t)

t0

Mass and Spring SystemLook at successive approximations to this infinite series

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Y = 02A2

2

A2 =02A2

02A2

lX 1 +


y(t) = 0202lA2l+2(t)

l=03 5 7 9

2 4t 6t 8t 10t=0t 03!

+05!

07!

+09!

+ y(t)

t0


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Look at successive approximations to this infinite series.2 2

Y l= 0 A2A2 =02A2

0

2A2

X 1 + 0 l=0

If x(t) = (t) then

y(t) = 0202lA2l+2(t)l=0

3 5 7 92 4t 6t 8t 10t=0t 0

3!+0

5! 0

7!+0

9! +

y(t)

t0


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Y = 02A2

2A2 =02A2

0

2A2l

X 1 + 0 l=0

If x(t) = (t) then

y(t) = 0202lA2l+2(t)l=0

3 5 7 92 4t 6t 8t 10t=0t 0

3!+0

5! 0

7!+0

9! +

y(t)

t0


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Y = 02A2

2A2 =02A2

0

2A2l

X 1 + 0 l=0

If x(t) = (t) then

y(t) = 0202lA2l+2(t)l=0

3 5 7 92 4t 6t 8t 10t=0t 0

3!+0

5! 0

7!+0

9! +

y(t)

t0


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Y = 02A2

2

A2 =02A2

02A2

lX 1 +


y(t) = 0202lA2l+2(t)

l=03 5 7 9

2 4t 6t 8t 10t=0t 03!

+05!

07!

+09!

+ y(t)

t0


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Look at successive approximations to this infinite series.2 2

l

Y = 0 A2A2 =02A2

0

2A2

X 1 + 0 l=0

If x(t) = (t) then

y(t) = 020

2lA2l+2(t)

l=03 5 7 9

=02t 04t3!

+06t5!

08t

7!+010t

9! + =0sin0t

y(t)

t0

Comparison of CT and DT representationsImportant similarities and important differences.

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Important similarities and important differences.

X Y X+ A

p

Y+

DelaypA 1

1

pA

1

pR

eptu(t) pnu[n]

MIT OpenCourseWarehttp://ocw.mit.edu
http://ocw.mit.edu/http://ocw.mit.edu/

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6.003 Signals and Systems

Spring 2010

For information about citing these materials or our Terms of Use, visit:http://ocw.mit.edu/terms.
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