6.003: signals and systems - massachusetts …web.mit.edu/6.003/f09/www/handouts/lec08.pdf6.003:...

6.003: Signals and Systems

Operator Representations for Continuous-Time Systems

October 6, 2009

Mid-term Examination #1

Tomorrow, October 7, 7:30-9:30pm, Walker Memorial.

No recitations tomorrow.

Coverage: DT Signals and Systems

Lectures 1–5

Homeworks 1–4

Homework 4 includes practice problems for mid-term 1.

It will not collected or graded. Solutions are posted.

Closed book: 1 page of notes (812 × 11 inches; front and back).

Designed as 1-hour exam; two hours to complete.

Analyzing CT Systems

We have used differential equations to represent CT systems.

r0(t)

r1(t)h1(t)

τ r1(t) = r0(t)− r1(t)

Methods to solve differential equations:

• homogeneous and particular equations

• singularity matching

• Laplace transform

Today: new methods based on block diagrams and operators.

Block Diagrams

Block diagrams illustrate signal flow paths.

DT: adders, scalers, and delays – represent systems described by

linear difference equations with constant coefficents.

+ Delay

p

x[n] y[n]

CT: adders, scalers, and integrators – represent systems described

by a linear differential equations with constant coefficients.

+∫ t−∞

( · ) dt

p

x(t) y(t)

Operator Representation

Block diagrams are concisely represented with operators.

We will define the A operator for functional analysis of CT systems.

Applying A to a CT signal generates a new signal that is equal to

the integral of the first signal at all points in time.

Y = AX

is equivalent to

y(t) =∫ t−∞x(τ) dτ

for all time t.

Evaluating Operator Expressions

As with R, A expressions can be manipulated as polynomials.

+ +

A A

X YW

w(t) = x(t) +∫ t−∞x(τ)dτ

y(t) = w(t) +∫ t−∞w(τ)dτ

y(t) = x(t) +∫ t−∞x(τ)dτ +

∫ t−∞x(τ)dτ +

∫ t−∞

∫ τ2−∞x(τ1)dτ1dτ2

W = (1 +A)X

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

Evaluating Operator Expressions

Expressions 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

Continuous-Time Feedback

What is the response of a CT system with feedback?

+∫ t−∞

( · ) dt

p

x(t) y(t)

Check Yourself

+∫ t−∞

( · ) dt

p

x(t) y(t)

Find the corresponding differential equation.

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

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

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

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

5. none of the above

Check Yourself

Find the corresponding differential equation.

+∫ t−∞

( · ) dt

p

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

The input to the integrator is y(t).

Therefore

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

Check Yourself

+∫ t−∞

( · ) dt

p

x(t) y(t)

Find the corresponding differential equation. 1

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

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

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

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


Check Yourself

+∫ t−∞

( · ) dt

p

x(t) y(t)

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

Find the impulse response.

1. e pt, t ≥ 0

2. e−pt, t ≥ 0

3. pe pt, t ≥ 0

4. pe−pt, t ≥ 0


Check Yourself

Find the impulse response of the system: y(t) = x(t) + py(t).

Lots of methods — try singularity matching.

Homogeneous solution: y(t) = Ae pt

Initially at rest: y(t) = Ae ptu(t)Differentiate: y(t) = Aδ(t) + pAe ptu(t)Substitute in diff. eq.: Aδ(t) + pAe ptu(t) = δ(t) + pAe ptu(t)Match singularity functions: A = 1

Answer: y(t) = e ptu(t).

Check Yourself

+∫ t−∞

( · ) dt

p

x(t) y(t)

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

Find the impulse response. 1

1. e pt, t ≥ 0

2. e−pt, t ≥ 0

3. pe pt, t ≥ 0

4. pe−pt, t ≥ 0


Check Yourself

+ A

p

X Y

Which functional represents this system?

1.Y

X= A

1 + pA2.Y

X= 1p+A

3.Y

X= A

1− pA4.Y

X= 1p−A


Check Yourself

Express the block diagram in operator notation.

+ A

p

X YG

Y = AG = A(X + pY )

(1− pA)Y = AX

Y

X= A

1− pA

Black’s equation works just the same in CT as in DT !

Check Yourself

+ A

p

X Y

Which functional represents this system? 3

1.Y

X= A

1 + pA2.Y

X= 1p+A

3.Y

X= A

1− pA4.Y

X= 1p−A


Comparison of CT and DT representations

CT block diagrams, functionals, and solutions are similar but not

identical to DT counterparts.

+ A

p

X Y +

Delayp

X Y

A1− pA

11− pR

e ptu(t) pnu[n]

Continuous-Time Feedback

What is the significance of cyclic signal flow paths in CT systems?

+ A

p

X Y

Y

X= A

1− pA

t

x(t) = δ(t)

1

0t

y(t)

1

0

Series Expansion of System Functionals

Determine response directly from system functional.

+ A

p

X Y

Y

X= A

1− pA= A(

1 + pA+ p2A2 + p3A3 + · · ·)

=∞∑k=0pkAk+1



+ A

p

X Y

Y

X= A

1− pA= A(

1 + pA+ p2A2 + p3A3 + · · ·)

=∞∑k=0pkAk+1

If x(t) = δ(t), then y(t) is a sum of

k = 0 : Aδ(t) = u(t) unit step

k = 1 : pA2δ(t) = p t u(t) unit ramp

k = 2 : p2A3δ(t) = 12 p

2t2u(t) parabola

· · ·k : pkAk+1δ(t) = 1

k!pktku(t) higher-order power



Y

X= A

1− pA=∞∑k=0pkAk+1

If x(t) = δ(t) then

y(t) =∞∑k=0pkAk+1δ(t)

= 1 + · · ·

t

y(t)

1

0



Y

X= A




= 1 + pt+ · · ·

t

y(t)

1

0



Y

X= A




= 1 + pt+ (pt)2

2+ · · ·

t

y(t)

1

0



Y

X= A




= 1 + pt+ (pt)2

2+ (pt)3

3!+ · · ·

t

y(t)

1

0



Y

X= A




= 1 + pt+ (pt)2

2+ (pt)3

3!+ (pt)4

4!+ · · ·

t

y(t)

1

0



Y

X= A




= 1 + pt+ (pt)2

2+ (pt)3

3!+ (pt)4

4!+ (pt)5

5!+ · · ·

t

y(t)

1

0



Y

X= A




= 1 + pt+ (pt)2

2+ (pt)3

3!+ (pt)4

4!+ (pt)5

5!+ · · · = e pt

t

y(t)

1

0

The system has a pole at p.

Convergent and Divergent Poles

The fundamental mode associated with p diverges if p > 0 and con-

verges if p < 0.

+ A

p

X Y

t

y(t)

1

0

p = 1

t

y(t)

1

0

p = −1

Convergent and Divergent Poles

The fundamental mode associated with p diverges if p > 0 and con-

verges if p < 0.

+ A

p

X Y

Re p

Im p

Re p

Convergent Divergent


Shapes of regions of convergence differ for CT and DT systems.

+ A

p

X Y +

Delayp

X Y

A1− pA

11− pR

e ptu(t) pnu[n]

Mass and Spring System

Use the A operator to solve the mass and spring system.

x(t)

y(t)

F = K(x(t)− y(t)

)=My(t)

+ K

MA A

−1

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

Y

X=

KMA

2

1 + KMA2


Factor system functional to find the poles.

Y

X=

KMA

2

1 + KMA2 =KMA

2

(1− p0A)(1− p1A)

1 + KMA2 = 1− (p0 + p1)A+ p0p1A2

The sum of the poles must be zero.

The product of the poles must be K/M .

p0 = j√K

Mp1 = −j

√K

M


Alternatively, find the poles by substituting A → 1s .

The poles are then the roots of the denominator.

Y

X=

KMA

2

1 + KMA2

Substitute A → 1s :

Y

X=

KM

s2 + KM

s = ±j√K

M≡ ±jω0 ; ω2

0 = KM


The poles are complex conjugates.

Re s

Im ss-plane √

KM ≡ ω0

−√KM ≡ −ω0

The corresponding fundamental modes have complex values.

fundamental mode 1: ejω0t = cosω0t+ j sinω0t

fundamental mode 2: e−jω0t = cosω0t− j sinω0t


Real-valued inputs always excite combinations of these modes so

that the imaginary parts cancel.

Example: find the impulse response.

Y

X=

KMA

2

1 + KMA2 =KM

p0 − p1

(A

1− p0A− A

1− p1A

)=ω2

02jω0

(A

1− jω0A− A

1 + jω0A

)= ω0

2j

(A

1− jω0A

)︸︷︷︸makes mode 1

−ω02j

(A

1 + jω0A

)︸︷︷︸makes mode 2

The modes themselves are complex conjugates, and their coefficients

are also complex conjugates. So the sum is a sum of something and

its complex conjugate, which is real.


The impulse response is therefore real.

Y

X= ω0

2j

(A

1− jω0A

)− ω0

2j

(A

1 + jω0A

)

The impulse response is

h(t) = ω02je jω0t − ω0

2je−jω0t = ω0 sinω0t ; t > 0

t

y(t)

0


Alternatively, we can find impulse response by expanding the system

functional.

+ ω20 A A

−1

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

Y

X=ω2

0A2

1 + ω20A2 = ω2

0A2 − ω40A4 + ω6

0A6 −+ · · ·


y(t) = ω20t− ω4

0t3

3!+ ω6

0t5

5!−+ · · · , t ≥ 0


Look at successive approximations to this infinite series.

Y

X=ω2

0A2

1 + ω20A2 = ω2

0A2∞∑l=0

(−ω2

0A2)l


y(t) =∞∑l=0ω2

0(−ω2

0)lA2l+2δ(t)

= ω20t

t

y(t)

0



Y

X=ω2

0A2

1 + ω20A2 = ω2

0A2∞∑l=0

(−ω2

0A2)l


y(t) =∞∑l=0ω2

0(−ω2

0)lA2l+2δ(t)

= ω20t− ω4

0t3

3!

t

y(t)

0



Y

X=ω2

0A2

1 + ω20A2 = ω2

0A2∞∑l=0

(−ω2

0A2)l


y(t) =∞∑l=0ω2

0(−ω2

0)lA2l+2δ(t)

= ω20t− ω4

0t3

3!+ ω6

0t5

5!

t

y(t)

0



Y

X=ω2

0A2

1 + ω20A2 = ω2

0A2∞∑l=0

(−ω2

0A2)l


y(t) =∞∑l=0ω2

0(−ω2

0)lA2l+2δ(t)

= ω20t− ω4

0t3

3!+ ω6

0t5

5!− ω8

0t7

7!

t

y(t)

0



Y

X=ω2

0A2

1 + ω20A2 = ω2

0A2∞∑l=0

(−ω2

0A2)l


y(t) =∞∑l=0ω2

0(−ω2

0)lA2l+2δ(t)

= ω20t− ω4

0t3

3!+ ω6

0t5

5!− ω8

0t7

7!+ ω10

0t9

9!

t

y(t)

0



Y

X=ω2

0A2

1 + ω20A2 = ω2

0A2∞∑l=0

(−ω2

0A2)l


y(t) =∞∑l=0ω2

0(−ω2

0)lA2l+2δ(t)

= ω20t− ω4

0t3

3!+ ω6

0t5

5!− ω8

0t7

7!+ ω10

0t9

9!−+ · · ·

t

y(t)

0



Y

X=ω2

0A2

1 + ω20A2 = ω2

0A2∞∑l=0

(−ω2

0A2)l


y(t) =∞∑l=0ω2

0(−ω2

0)lA2l+2δ(t)

= ω20t− ω4

0t3

3!+ ω6

0t5

5!− ω8

0t7

7!+ ω10

0t9

9!−+ · · · = ω0 sinω0t

t

y(t)

0


Now you know how to find the position directly from the system

functional without having to guess solutions to differential equations.

Y

X=ω2

0A2

1 + ω20A2 = ω2

0t− ω40t3

3!+ ω6

0t5

5!− ω8

0t7

7!+ ω10

0t9

9!−+ · · · = ω0 sinω0t

This is a new method for finding the output of a CT system.

Methods:


• singularity functions


• Series expansion of system functional

Leaky Tanks System

We can similarly use the A operator to solve the leaky tank system.

r0(t)

r1(t)

r2(t)

h1(t)

h2(t)

τ1r1(t) = r0(t)− r1(t)

τ2r2(t) = r1(t)− r2(t)

Leaky Tanks System

Determine the system functional.

τ1r1(t) = r0(t)− r1(t)

τ2r2(t) = r1(t)− r2(t)

+1τ1

A +1τ2

A

−1 −1

r0(t) r2(t)r1(t) r1(t) r2(t)

R2R0

= A/τ11 +A/τ1

× A/τ21 +A/τ2

Leaky Tanks System

Modal decomposition of second-order CT system.

R2R0

= A/τ11 +A/τ1

× A/τ21 +A/τ2

= C1A1− p1A

+ C2A1− p2A

p1 = − 1τ1

p2 = − 1τ2

Re s

Im ss-plane

− 1τ1

− 1τ2

Leaky Tanks System

Modal decomposition of second-order CT system.

R2R0

= A/τ11 +A/τ1

× A/τ21 +A/τ2

= C1A1− p1A

+ C2A1− p2A

p1 = − 1τ1

p2 = − 1τ2

R2R0

= A/τ11 +A/τ1

× A/τ21 +A/τ2

= 1τ1 − τ2

(A

1 +A/τ1− A

1 +A/τ2

)


y(t) = 1τ1 − τ2

(e−t/τ1 − e−t/τ2

); t ≥ 0

Leaky Tanks System

Impulse response.


y(t) = 1τ1 − τ2

(e−t/τ1 − e−t/τ2

); t ≥ 0

t

mode 1

0

1τ1 − τ2

t

mode 2

0

1τ1 − τ2

t

y(t)

0

1τ1 − τ2

Summary

We have a new method to our inventory.

r0(t)

r1(t)h1(t)

τ r1(t) = r0(t)− r1(t)

Methods:


• singularity functions


• series expansion of system functional


Summary.

+ A

s

X Y +

Delayz

X Y

H = YX

= A1− sA

H = YX

= 11− zR

h(t) = e stu(t) h[n] = znu[n]

s-plane s-plane z-plane z-plane

6.003: signals and systems - massachusetts …web.mit.edu/6.003/f09/www/handouts/lec08.pdf6.003:...

Documents