6.003: Signals and Systems
Continuous-Time Systems
September 20, 2011
1
Multiple Representations of Discrete-Time Systems
Discrete-Time (DT) systems can be represented in different ways
to more easily address different types of issues.
Verbal descriptions: preserve the rationale.
“Next year, your account will contain p times your balance
from 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.
x[n]
+
Delay
y[n]
p
Operator representations: analyze systems as polynomials.
(1 − pR) Y = RX
2
Multiple Representations of Continuous-Time Systems
Similar representations for Continuous-Time (CT) systems.
Verbal descriptions: preserve the rationale.
“Your account will grow in proportion to your balance plus the
rate at which you deposit.”
Differential equations: mathematically compact.
dy(t)
= x(t) + py(t)
dt
Block diagrams: illustrate signal flow paths.
Z t
x(t)
( · ) dt
+
y(t)
−∞
p
Operator representations: analyze systems as polynomials.
(1 − pA)Y = AX
3
Differential Equations
Differential equations are mathematically precise and compact.
r0 (t)
h1 (t)
r1 (t)
We can represent the tank system with a differential equation.
dr1 (t)
r0 (t) − r1 (t)
=
dt
τ
You already know lots of methods to solve differential equations:
• general methods (separation of variables; integrating factors)
• homogeneous and particular solutions
• inspection
Today: new methods based on block diagrams and operators,
which provide new ways to think about systems’ behaviors.
4
Block Diagrams
Block diagrams illustrate signal flow paths.
DT: adders, scalers, and delays – represent systems described by
linear difference equations with constant coefficents.
x[n]
+
Delay
y[n]
p
CT: adders, scalers, and integrators – represent systems described
by a linear differential equations with constant coefficients.
Z t
x(t)
( · ) dt
+
−∞
p
Delays in DT are replaced by integrators in CT.
5
y(t)
Operator Representation
CT Block diagrams are concisely represented with the A operator.
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
t
y(t) =
x(τ ) dτ
−∞
for all time t.
6
Check Yourself
+
X
A
ẏ(t) = ẋ(t) + py(t)
Y
p
X
X
+
p
A
ẏ(t) = x(t) + py(t)
Y
+
ẏ(t) = px(t) + py(t)
Y
p
A
Which block diagrams correspond to which equations?
1.
2.
3.
4.
7
1
5. none
Check Yourself
+
X
A
ẏ(t) = ẋ(t) + py(t)
Y
p
X
X
+
p
A
ẏ(t) = x(t) + py(t)
Y
+
ẏ(t) = px(t) + py(t)
Y
p
A
Which block diagrams correspond to which equations?
1.
2.
3.
4.
8
1
5. none
Evaluating Operator Expressions
As with R, A expressions can be manipulated as polynomials.
Example:
W
+
X
A
+
Y
A
Z t
w(t) = x(t) +
x(τ )dτ
−∞
Z t
y(t) = w(t) +
w(τ )dτ
−∞
Z t
y(t) = x(t) +
Z t
x(τ )dτ +
−∞
Z t
Z τ2
x(τ )dτ +
−∞
−∞
−∞
W = (1 + A) X
Y = (1 + A) W = (1 + A)(1 + A) X = (1 + 2A + A2 ) X
9
x(τ1 )dτ1 dτ2
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
10
Check Yourself
Determine k1 so that these systems are “equivalent.”
+
X
+
A
A
−0.7
X
+
Y
−0.9
A
A
Y
k1
k2
1. 0.7
2. 0.9
3. 1.6
4. 0.63
11
5. none of these
Check Yourself
Write operator expressions for each system.
+
X
W
A
+
A
−0.7
Y
−0.9
(1+0.7A)(1+0.9A)Y = A2 X
(1+0.7A)W = AX
W = A(X −0.7W )
→
→
(1+0.9A)Y = AW
Y = A(W −0.9Y )
(1+1.6A+0.63A2 )Y = A2 X
X
+
W
A
A
Y
k1
k2
W = A(X +k1 W +k2 Y )
Y = AW
k1 = −1.6
→
12
Y = A2 X +k1 AY +k2 A2 Y
(1−k1 A−k2 A2 )Y = A2 X
Check Yourself
Determine k1 so that these systems are “equivalent.”
+
X
+
A
A
−0.7
X
+
Y
−0.9
A
A
Y
k1
k2
1. 0.7
2. 0.9
3. 1.6
4. 0.63
13
5. none of these
Elementary Building-Block Signals
Elementary DT signal: δ[n].
δ[n] =
1,
0,
if n = 0;
otherwise
δ[n]
1
n
0
•
•
•
simplest non-trivial signal (only one non-zero value)
shortest possible duration (most “transient”)
useful for constructing more complex signals
What CT signal serves the same purpose?
14
Elementary CT Building-Block Signal
Consider the analogous CT signal: w(t) is non-zero only at t = 0.
⎧
⎨0
w(t) = 1
⎩
0
t<0
t=0
t>0
w(t)
1
t
0
Is this a good choice as a building-block signal?
Z t
( · ) dt
w(t)
−∞
The integral of w(t) is zero!
15
0
No
Unit-Impulse Signal
The unit-impulse signal acts as a pulse with unit area but zero width.
p (t)
1
2
δ(t) = lim p (t)
unit area
→0
−
p1/4 (t)
p1/2 (t)
t
p1/8 (t)
4
2
1
1
−
2
1
2
t
1
−
4
1
4
16
t
1 1
−
8 8
t
Unit-Impulse Signal
The unit-impulse function is represented by an arrow with the num­
ber 1, which represents its area or “weight.”
δ(t)
1
t
It 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 δ(t) as a limit:
p (t)
1
2
δ(t) = lim p (t)
unit area
→0
17
−
t
Unit-Impulse and Unit-Step Signals
The indefinite integral of the unit-impulse is the unit-step.
Z t
u(t) =
δ(λ) dλ =
−∞
1;
0;
t≥0
otherwise
u(t)
1
t
Equivalently
δ(t)
A
18
u(t)
Impulse Response of Acyclic CT System
If the block diagram of a CT system has no feedback (i.e., no cycles),
then the corresponding operator expression is “imperative.”
+
X
+
A
A
Y = (1 + A)(1 + A) X = (1 + 2A + A2 ) X
If x(t) = δ(t) then
y(t) = (1 + 2A + A2 ) δ(t) = δ(t) + 2u(t) + tu(t)
19
Y
CT Feedback
Find the impulse response of this CT system with feedback.
x(t)
+
A
y(t)
p
Method 1: find differential equation and solve it.
ẏ(t) = x(t) + py(t)
Linear, first-order difference equation with constant coefficients.
Try y(t) = Ceαt u(t).
Then ẏ(t) = αCeαt u(t) + Ceαt δ(t) = αCeαt u(t) + Cδ(t).
Substituting, we find that αCeαt u(t) + Cδ(t) = δ(t) + pCeαt u(t).
Therefore α = p and C = 1
→
y(t) = ept u(t).
20
CT Feedback
Find the impulse response of this CT system with feedback.
x(t)
+
A
y(t)
p
Method 2: use operators.
Y = A (X + pY )
Y
A
=
X
1 − pA
Now expand in ascending series in A:
Y
= A(1 + pA + p2 A2 + p3 A3 + · · ·)
X
If x(t) = δ(t) then
y(t) = A(1 + pA + p2 A2 + p3 A3 + · · ·) δ(t)
1
1
= (1 + pt + p2 t2 + p3 t3 + · · ·) u(t) = ept u(t) .
2
6
21
CT Feedback
We can visualize the feedback by tracing each cycle through the
cyclic signal path.
x(t)
+
y(t)
A
p
y(t) = (A + pA2 + p2 A3 + p3 A4 + · · ·) δ(t)
1
1
= (1 + pt + p2 t2 + p3 t3 + · · ·) u(t)
2
6
y(t)
1
t
0
22
CT Feedback
We can visualize the feedback by tracing each cycle through the
cyclic signal path.
x(t)
+
y(t)
A
p
y(t) = (A + pA2 + p2 A3 + p3 A4 + · · ·) δ(t)
1
1
= (1 + pt + p2 t2 + p3 t3 + · · ·) u(t)
2
6
y(t)
1
t
0
23
CT Feedback
We can visualize the feedback by tracing each cycle through the
cyclic signal path.
x(t)
+
y(t)
A
p
y(t) = (A + pA2 + p2 A3 + p3 A4 + · · ·) δ(t)
1
1
= (1 + pt + p2 t2 + p3 t3 + · · ·) u(t)
2
6
y(t)
1
t
0
24
CT Feedback
We can visualize the feedback by tracing each cycle through the
cyclic signal path.
x(t)
+
y(t)
A
p
y(t) = (A + pA2 + p2 A3 + p3 A4 + · · ·) δ(t)
1
1
= (1 + pt + p2 t2 + p3 t3 + · · ·) u(t)
2
6
y(t)
1
t
0
25
CT Feedback
We can visualize the feedback by tracing each cycle through the
cyclic signal path.
x(t)
+
y(t)
A
p
y(t) = (A + pA2 + p2 A3 + p3 A4 + · · ·) δ(t)
1
1
= (1 + pt + p2 t2 + p3 t3 + · · ·) u(t) = ept u(t)
2
6
y(t)
1
t
0
26
CT Feedback
Making p negative makes the output converge (instead of diverge).
x(t)
+
A
−p
y(t) = (A − pA2 + p2 A3 − p3 A4 + · · ·) δ(t)
1
1
= (1 − pt + p2 t2 − p3 t3 + ·
· ·) u(t)
2
6
27
y(t)
CT Feedback
Making p negative makes the output converge (instead of diverge).
x(t)
+
y(t)
A
−p
y(t) = (A − pA2 + p2 A3 − p3 A4 + · · ·) δ(t)
1
1
= (1 − pt + p2
t2 − p3
t3 + · · ·) u(t)
2
6
y(t)
1
t
0
28
CT Feedback
Making p negative makes the output converge (instead of diverge).
x(t)
+
y(t)
A
−p
y(t) = (A − pA2 + p2 A3 − p3 A4 + · · ·) δ(t)
1
1
= (1 − pt + p2
t2 − p3
t3 + · · ·) u(t)
2
6
y(t)
1
t
0
29
CT Feedback
Making p negative makes the output converge (instead of diverge).
x(t)
+
y(t)
A
−p
y(t) = (A − pA2 + p2 A3 − p3 A4 + · · ·) δ(t)
1
1
= (1 − pt + p2
t2 − p3
t3 + · · ·) u(t)
2
6
y(t)
1
t
0
30
CT Feedback
Making p negative makes the output converge (instead of diverge).
x(t)
+
y(t)
A
−p
y(t) = (A − pA2 + p2 A3 − p3 A4 + · · ·) δ(t)
1
1
= (1 − pt + p2
t2 − p3
t3 + · · ·) u(t)
2
6
y(t)
1
t
0
31
CT Feedback
Making p negative makes the output converge (instead of diverge).
x(t)
+
y(t)
A
−p
y(t) = (A − pA2 + p2 A3 − p3 A4 + · · ·) δ(t)
1
1
= (1 − pt + p2
t2 − p3 t3 + · · ·) u(t) = e−pt u(t)
6
2
y(t)
1
t
0
32
Convergent and Divergent Poles
The fundamental mode associated with p converges if p < 0 and
diverges if p > 0.
X
+
A
Y
p
p<0
p>0
y(t)
y(t)
1
1
0
t
0
33
t
Convergent and Divergent Poles
The fundamental mode associated with p converges if p < 0 and
diverges if p > 0.
X
+
A
Y
p
Im(p)
Convergent
Divergent
Re(p)
Re(p)
34
CT Feedback
In CT, each cycle adds a new integration.
x(t)
+
y(t)
A
p
y(t) = (A + pA2 + p2 A3 + p3 A4 + · · ·) δ(t)
1
1
= (1 + pt + p2
t2 + p3
t3 + · · ·) u(t) = ept u(t)
2
6
y(t)
1
t
0
35
DT Feedback
In DT, each cycle creates another sample in the output.
X
+
Y
p0
Delay
y[n] = (1 + pR + p2 R2 + p3 R3 + p4 R4 + · · ·) δ[n]
= δ[n] + pδ[n − 1] + p2 δ[n − 2] + p3 δ[n − 3] + p4 δ[n − 4] + · · ·
y[n]
n
−1 0 1 2 3 4
36
Summary: CT and DT representations
Many similarities and important differences.
y[n] = x[n] + py[n − 1]
ẏ(t) = x(t) + py(t)
X
+
A
Y
X
+
Y
p
p
Delay
A
1 − pA
1
1 − pR
e pt u(t)
pn u[n]
37
Check Yourself
Which functionals represent convergent systems?
1
1−
1
1 R2
4
1 − 14 A2
1
1 + 2R + 43 R2
√
x
√
1.
x
√ √
2. x x
1
1 + 2A + 43 A2
√ √
3. √ √
√
x
4. x √
38
5. none of these
Check Yourself
1
1−
1 R2
4
1
=
1
(1 −
1 R)(1 + 1 R)
2
2
(1 −
1 A)(1 + 1 A)
2
2
1
√
both inside unit circle
left & right half-planes
X
1
1
=
3
1
2
(1 + 2 R)(1 + 32 R)
1 + 2R + 4 R
inside & outside unit circle
X
1
1
=
3
1
2
(1 + 2 A)(1 + 32 A)
1 + 2A + 4 A
both left half plane
1−
1 A2
4
=
√
39
Check Yourself
Which functionals represent convergent systems?
1
1−
1
1 R2
4
1 − 14 A2
1
1 + 2R + 43 R2
√
x
√
1.
x
√ √
2. x x
4
1
1 + 2A + 43 A2
√ √
3. √ √
√
x
4. x √
40
5. none of these
Mass and Spring System
Use the A operator to solve the mass and spring system.
x(t)
F = K x(t) − y(t) = M ÿ(t)
y(t)
x(t)
+
K
M
ÿ(t)
A
−1
K A2
Y
M
=
K A2
X
1 +
M
41
ẏ(t)
A
y(t)
Mass and Spring System
Factor system functional to find the poles.
K A2
K 2
Y
MA
=
M K
=
(1 − p0 A)(1 − p1 A)
X
1 + M A2
1+
K 2
A = 1 − (p0 + p1 )A + p0 p1 A2
M
The sum of the poles must be zero.
The product of the poles must be K/M .
p0 = j
K
M
p1 = −j
K
M
42
Mass and Spring System
Alternatively, find the poles by substituting A → 1
s .
The poles are then the roots of the denominator.
K A2
Y
= MK
X
1 +
M A2
Substitute A → 1s :
K
Y
M
=
K
X
s2 + M
r
K
s = ±j
M
43
Mass and Spring System
The poles are complex conjugates.
Im s
s-plane
q
K
M
≡ ω0
Re s
q
K ≡ −ω
− M
0
The corresponding fundamental modes have complex values.
fundamental mode 1: ejω0 t = cos ω0 t + j sin ω0 t
fundamental mode 2: e−jω0 t = cos ω0 t − j sin ω0 t
44
Mass and Spring System
Real-valued inputs always excite combinations of these modes so
that the imaginary parts cancel.
Example: find the impulse response.
K A2
K
Y
A
A
M
M
=
=
−
X
1 +
K A2
p 0 − p1 1 − p 0 A 1 − p 1 A
M
ω02
A
A
−
=
2jω0 1 − jω0 A 1 + jω0 A
A
ω0
A
ω0
−
=
2j 1 + jω0 A
2j 1 − jω0 A
'
v
"
'
v
"
makes mode 1
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.
45
Mass and Spring System
The impulse response is therefore real.
Y
ω0
A
ω0
A
=
−
X
2j 1 − jω0 A
2j 1 + jω0 A
The impulse response is
ω0 jω0 t ω0 −jω0 t
e
= ω0 sin ω0 t ;
h(t) =
e
−
2j
2j
t>0
y(t)
t
0
46
Mass and Spring System
Alternatively, find impulse response by expanding system functional.
x(t)
+
ω02
ÿ(t)
A
ẏ(t)
−1
ω02 A2
Y
=
= ω02 A2 − ω04 A4 + ω06 A6 − + · · ·
X
1 + ω02 A2
If x(t) = δ(t) then
3
5
2
4t
6t
− + · · · ,
t ≥ 0
y(t) = ω
0
t − ω
0
+ ω0
5!
3!
47
A
y(t)
Mass and Spring System
Look at successive approximations to this infinite series.
∞ �
�l
0
ω02 A2
Y
2 2
2 2
=
=
ω
A
−ω
A
0
0
X
1 + ω02 A2
l=0
If x(t) = δ(t) then
∞
�
�l
0
y(t) =
ω02 −ω02 A2l+2 δ(t)
l=0
= ω02 t − ω04
t5
t7
t9
t3
+ ω06 − ω08 + ω010 − + · · · = ω0 sin ω0 t
3!
5!
7!
9!
y(t)
t
0
48
Summary: CT and DT representations
Many similarities and important differences.
y[n] = x[n] + py[n − 1]
ẏ(t) = x(t) + py(t)
X
+
A
Y
X
+
Y
p
p
Delay
A
1 − pA
1
1 − pR
e pt u(t)
pn u[n]
49
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6.003 Signals and Systems
Fall 2011
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