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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)
5. none of the above
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
5. none of the above
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
5. none of the above
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
5. none of the above
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
5. none of the above
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
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
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
Series Expansion of System Functionals
Determine response directly from system functional.
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
Series Expansion of System Functionals
Determine response directly from system functional.
Y
X= A
1− pA=∞∑k=0pkAk+1
If x(t) = δ(t) then
y(t) =∞∑k=0pkAk+1δ(t)
= 1 + pt+ · · ·
t
y(t)
1
0
Series Expansion of System Functionals
Determine response directly from system functional.
Y
X= A
1− pA=∞∑k=0pkAk+1
If x(t) = δ(t) then
y(t) =∞∑k=0pkAk+1δ(t)
= 1 + pt+ (pt)2
2+ · · ·
t
y(t)
1
0
Series Expansion of System Functionals
Determine response directly from system functional.
Y
X= A
1− pA=∞∑k=0pkAk+1
If x(t) = δ(t) then
y(t) =∞∑k=0pkAk+1δ(t)
= 1 + pt+ (pt)2
2+ (pt)3
3!+ · · ·
t
y(t)
1
0
Series Expansion of System Functionals
Determine response directly from system functional.
Y
X= A
1− pA=∞∑k=0pkAk+1
If x(t) = δ(t) then
y(t) =∞∑k=0pkAk+1δ(t)
= 1 + pt+ (pt)2
2+ (pt)3
3!+ (pt)4
4!+ · · ·
t
y(t)
1
0
Series Expansion of System Functionals
Determine response directly from system functional.
Y
X= A
1− pA=∞∑k=0pkAk+1
If x(t) = δ(t) then
y(t) =∞∑k=0pkAk+1δ(t)
= 1 + pt+ (pt)2
2+ (pt)3
3!+ (pt)4
4!+ (pt)5
5!+ · · ·
t
y(t)
1
0
Series Expansion of System Functionals
Determine response directly from system functional.
Y
X= A
1− pA=∞∑k=0pkAk+1
If x(t) = δ(t) then
y(t) =∞∑k=0pkAk+1δ(t)
= 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
Comparison of CT and DT representations
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
Mass and Spring System
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
Mass and Spring System
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
Mass and Spring System
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
Mass and Spring System
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.
Mass and Spring System
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
Mass and Spring System
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 −+ · · ·
If x(t) = δ(t) then
y(t) = ω20t− ω4
0t3
3!+ ω6
0t5
5!−+ · · · , t ≥ 0
Mass and Spring System
Look at successive approximations to this infinite series.
Y
X=ω2
0A2
1 + ω20A2 = ω2
0A2∞∑l=0
(−ω2
0A2)l
If x(t) = δ(t) then
y(t) =∞∑l=0ω2
0(−ω2
0)lA2l+2δ(t)
= ω20t
t
y(t)
0
Mass and Spring System
Look at successive approximations to this infinite series.
Y
X=ω2
0A2
1 + ω20A2 = ω2
0A2∞∑l=0
(−ω2
0A2)l
If x(t) = δ(t) then
y(t) =∞∑l=0ω2
0(−ω2
0)lA2l+2δ(t)
= ω20t− ω4
0t3
3!
t
y(t)
0
Mass and Spring System
Look at successive approximations to this infinite series.
Y
X=ω2
0A2
1 + ω20A2 = ω2
0A2∞∑l=0
(−ω2
0A2)l
If x(t) = δ(t) then
y(t) =∞∑l=0ω2
0(−ω2
0)lA2l+2δ(t)
= ω20t− ω4
0t3
3!+ ω6
0t5
5!
t
y(t)
0
Mass and Spring System
Look at successive approximations to this infinite series.
Y
X=ω2
0A2
1 + ω20A2 = ω2
0A2∞∑l=0
(−ω2
0A2)l
If x(t) = δ(t) then
y(t) =∞∑l=0ω2
0(−ω2
0)lA2l+2δ(t)
= ω20t− ω4
0t3
3!+ ω6
0t5
5!− ω8
0t7
7!
t
y(t)
0
Mass and Spring System
Look at successive approximations to this infinite series.
Y
X=ω2
0A2
1 + ω20A2 = ω2
0A2∞∑l=0
(−ω2
0A2)l
If x(t) = δ(t) then
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
Mass and Spring System
Look at successive approximations to this infinite series.
Y
X=ω2
0A2
1 + ω20A2 = ω2
0A2∞∑l=0
(−ω2
0A2)l
If x(t) = δ(t) then
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
Mass and Spring System
Look at successive approximations to this infinite series.
Y
X=ω2
0A2
1 + ω20A2 = ω2
0A2∞∑l=0
(−ω2
0A2)l
If x(t) = δ(t) then
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
Mass and Spring System
Look at successive approximations to this infinite series.
Y
X=ω2
0A2
1 + ω20A2 = ω2
0A2∞∑l=0
(−ω2
0A2)l
If x(t) = δ(t) then
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
Mass and Spring System
Look at successive approximations to this infinite series.
Y
X=ω2
0A2
1 + ω20A2 = ω2
0A2∞∑l=0
(−ω2
0A2)l
If x(t) = δ(t) then
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
Mass and Spring System
Look at successive approximations to this infinite series.
Y
X=ω2
0A2
1 + ω20A2 = ω2
0A2∞∑l=0
(−ω2
0A2)l
If x(t) = δ(t) then
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
Mass and Spring System
Look at successive approximations to this infinite series.
Y
X=ω2
0A2
1 + ω20A2 = ω2
0A2∞∑l=0
(−ω2
0A2)l
If x(t) = δ(t) then
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
Mass and Spring System
Look at successive approximations to this infinite series.
Y
X=ω2
0A2
1 + ω20A2 = ω2
0A2∞∑l=0
(−ω2
0A2)l
If x(t) = δ(t) then
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
Mass and Spring System
Look at successive approximations to this infinite series.
Y
X=ω2
0A2
1 + ω20A2 = ω2
0A2∞∑l=0
(−ω2
0A2)l
If x(t) = δ(t) then
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
Mass and Spring System
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:
• homogeneous and particular equations
• singularity functions
• Laplace transform
• 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
)
If x(t) = δ(t) then
y(t) = 1τ1 − τ2
(e−t/τ1 − e−t/τ2
); t ≥ 0
Leaky Tanks System
Impulse response.
If x(t) = δ(t) then
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:
• homogeneous and particular equations
• singularity functions
• Laplace transform
• series expansion of system functional
Comparison of CT and DT representations
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