6.003: Signals and Systems
Discrete Approximation of Continuous-Time Systems
September 29, 2011
1
Mid-term Examination #1
Wednesday, October 5, 7:30-9:30pm,
No recitations on the day of the exam.
Coverage:
CT and DT Systems, Z and Laplace Transforms
Lectures 1–7
Recitations 1–7
Homeworks 1–4
Homework 4 will not collected or graded. Solutions will be posted.
Closed book: 1 page of notes (8 12 × 11 inches; front and back).
No calculators, computers, cell phones, music players, or other aids.
Designed as 1-hour exam; two hours to complete.
Review sessions during open office hours.
Conflict? Contact before Friday, Sept. 30, 5pm.
Prior term midterm exams have been posted on the 6.003 website.
2
Concept Map
Today we will look at relations between CT and DT representations.
Delay → R
Block Diagram
X
+
+
Delay
DT
System Functional
Y
Y
1
= H(R) =
X
1 − R − R2
Delay
Unit-Sample Response
index shift
h[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .
Difference Equation
CT
System Function
Y (z)
z2
= 2
X(z)
z −z−1
y[n] = x[n] + y[n−1] + y[n−2]
H(z) =
Delay → R
Block Diagram
+
X
R
−
+
R
−
System Functional
1
2
1
T
D
Y
2A2
=
X
2 + 3A + A2
Y
Impulse Response
ẋ(t)
R
x(t)
h(t) = 2(e−t/2 − e−t ) u(t)
CT
Differential Equation
2ÿ(t) + 3ẏ(t) + y(t) = 2x(t)
System Function
Y (s)
2
= 2
X(s)
2s + 3s + 1
3
Discrete Approximation of CT Systems
Example: leaky tank
r0 (t)
h1 (t)
r1 (t)
R
X
AX
Block Diagram
+
X
−
1
τ
System Functional
R
A
Y
=
X
A+τ
Y
Impulse Response
ẋ(t)
R
x(t)
h(t) = τ1 e−t/τ u(t)
Differential Equation
System Function
Y (s)
1
=
X(s)
1 + τs
τ ṙ1 (t) = r0 (t) − r1 (t)
H(s) =
Today: compare step responses of leaky tank and DT approximation.
4
Check Yourself (Practice for Exam)
What is the “step response” of the leaky tank system?
u(t)
s(t) =?
Leaky Tank
1
1
1.
2.
τ
t
τ
1
t
1
3.
4.
τ
t
τ
5. none of the above
5
t
Check Yourself
What is the “step response” of the leaky tank system?
de:
τ ṙ1 (t) = u(t) − r1 (t)
t < 0: r1 (t) = 0
t > 0: r1 (t) = c1 + c2 e−t/τ
ṙ1 (t) = − cτ2 e−t/τ
Substitute into de: τ − cτ2 e−t/τ = 1 − c1 − c2 e−t/τ
→
c1 = 1
Combine t < 0 and t > 0:
r1 (t) = u(t) + c2 e−t/τ u(t)
c2
ṙ1 (t) = δ(t) + c2 δ(t) − e−t/τ u(t)
τ
Substitute into de:
c2
τ (1 + c2 )δ(t) − τ e−t/τ u(t) = u(t) − u(t) − c2 e−t/τ u(t)
τ
r1 (t) = (1 − e−t/τ )u(t)
6
→
c2 = −1
Check Yourself
Alternatively, reason with systems!
δ(t)
δ(t)
δ(t)
A
A+τ
h(t) = τ1 e−t/τ u(t)
u(t)
A
A+τ
s(t) =?
A
A+τ
s(t) =?
u(t)
A
A
A+τ
Z t
h(t)
A
s(t) =
−∞
t 1
I
1 −tI /τ ' '
u(t )dt =
e
e−t /τ dt' = (1 − e−t/τ ) u(t)
−∞ τ
0 τ
t
s(t) =
h(t0 )dt0
7
Check Yourself
What is the “step response” of the leaky tank system?
u(t)
s(t) =?
Leaky Tank
1
2
1
1.
2.
τ
t
τ
1
t
1
3.
4.
τ
t
τ
5. none of the above
8
t
Forward Euler Approximation
Approximate leaky-tank system using forward Euler approach.
Approximate continuous signals by discrete signals:
xd [n] = xc (nT )
yd [n] = yc (nT )
Approximate derivative at t = nT by looking forward in time:
y [n+1] − yd [n]
ẏc (nT ) = d
T
yc (t)
yd [n+1]
yd [n]
nT
(n+1) T
9
t
Forward Euler Approximation
Approximate leaky-tank system using forward Euler approach.
Substitute
xd [n] = xc (nT )
yd [n] = yc (nT )
yc (n + 1)T − yc nT
y [n + 1] − yd [n]
ẏc (nT ) ≈
= d
T
T
into the differential equation
τ ẏc (t) = xc (t) − yc (t)
to obtain
τ
y [n + 1] − yd [n] = xd [n] − yd [n] .
T d
Solve:
T
T
yd [n + 1] − 1 −
yd [n] = xd [n]
τ
τ
10
Forward Euler Approximation
Plot.
1
T
τ
= 0.1
T
τ
= 0.3
t
1
t
1
T
τ
=1
t
1
T
τ
= 1.5
t
1
T
τ
τ
=2
t
Why is this approximation badly behaved for large Tτ ?
11
Check Yourself
DT approximation:
T
T
yd [n] = xd [n]
yd [n + 1] − 1 −
τ
τ
Find the DT pole.
T
τ
τ
3. z =
T
2. z = 1 −
1. z =
4. z = −
5. z =
1
1 + Tτ
12
τ
T
T
τ
Check Yourself
DT approximation:
T
T
yd [n] = xd [n]
yd [n + 1] − 1 −
τ
τ
Take the Z transform:
T
T
Yd (z) = Xd (z)
zYd (z) − 1 −
τ
τ
Solve for the system function:
T
Y (z)
τ
H(z) = d
=
Xd (z)
z− 1− T
τ
Pole at z = 1 −
T
.
τ
13
Check Yourself
DT approximation:
T
T
yd [n] = xd [n]
yd [n + 1] − 1 −
τ
τ
Find the DT pole.
2
T
τ
τ
3. z =
T
2. z = 1 −
1. z =
4. z = −
5. z =
1
1 + Tτ
14
τ
T
T
τ
Dependence of DT pole on Stepsize
z
1
t
T
τ
= 0.1
T
τ
= 0.3
z
1
t
z
1
T
τ
t
=1
z
1
T
τ
t
= 1.5
z
1
τ
T
τ
t
=2
The CT pole was fixed (s = − τ1 ). Why is the DT pole changing?
15
Dependence of DT pole on Stepsize
Dependence of DT pole on T is generic property of forward Euler.
Approach: make a systems model of forward Euler method.
CT block diagrams: adders, gains, and integrators:
X
A
Y
ẏ(t) = x(t)
Forward Euler approximation:
y[n + 1] − y[n]
= x[n]
T
Equivalent system:
X
T
+
R
Y
Forward Euler: substitute equivalent system for all integrators.
16
Example: leaky tank system
Started with leaky tank system:
+
X
−
1
τ
R
Y
Replace integrator with forward Euler rule:
X
+
−
1
τ
+
T
R
Write system functional:
T R
TR
TR
Y
τ
τ
= τ 1−R
=
=
R
X
1 − R + Tτ R
1 + Tτ 1−R
1 − 1 − Tτ R
Equivalent to system we previously developed:
T
T
yd [n] = xd [n]
yd [n + 1] − 1 −
τ
τ
17
Y
Model of Forward Euler Method
Replace every integrator in the CT system
X
A
Y
with the forward Euler model:
X
T
+
Substitute the DT operator for A:
T
1
TR
T
A= →
= z 1 =
z−1
s
1−R
1− z
z−1
.
T
Or equivalently: z = 1 + sT .
Forward Euler maps s →
18
R
Y
Dependence of DT pole on Stepsize
Pole at z = 1 − Tτ = 1 + sT .
z
1
t
T
τ
= 0.1
T
τ
= 0.3
z
1
t
z
1
T
τ
t
=1
z
1
T
τ
t
= 1.5
z
1
τ
T
τ
t
19
=2
Forward Euler: Mapping CT poles to DT poles
Forward Euler Map:
→
s
0
1
− T1
0
− T2
−1
s
1
T
− T1
− T2
z = 1 + sT
z
z → 1 + sT
−1
1
DT stability: CT pole must be inside circle of radius T1 at s = − T1 .
−
2
1
<− <0
τ
T
→
T
<2
τ
20
Backward Euler Approximation
We can do a similar analysis of the backward Euler method.
Approximate continuous signals by discrete signals:
xd [n] = xc (nT )
yd [n] = yc (nT )
Approximate derivative at t = nT by looking backward in time:
y [n] − yd [n−1]
ẏc (nT ) = d
T
yc (t)
yd [n]
yd [n−1]
(n−1)T
nT
21
t
Backward Euler Approximation
We can do a similar analysis of the backward Euler method.
Substitute
xd [n] = xc (nT )
yd [n] = yc (nT )
yc nT − yc (n − 1)T
y [n] − yd [n − 1]
= d
T
T
into the differential equation
ẏc (nT ) ≈
τ ẏc (t) = xc (t) − yc (t)
to obtain
τ
yd [n] − yd [n − 1] = xd [n] − yd [n] .
T
Solve:
T
T
1+
yd [n] − yd [n − 1] = xd [n]
τ
τ
22
Backward Euler Approximation
Plot.
1
T
τ
= 0.1
T
τ
= 0.3
t
1
t
1
T
τ
=1
t
1
T
τ
= 1.5
t
1
T
τ
=2
τ
This approximation is better behaved. Why?
23
t
Check Yourself
DT approximation:
T
T
yd [n] − yd [n − 1] = xd [n]
1+
τ
τ
Find the DT pole.
T
τ
τ
3. z =
T
2. z = 1 −
1. z =
4. z = −
5. z =
1
1 + Tτ
24
τ
T
T
τ
Check Yourself
DT approximation:
T
T
1+
yd [n] − yd [n − 1] = xd [n]
τ
τ
Take
the Z
transform:
T
T
1+
Yd (z) − z −1 Yd (z) = Xd (z)
τ
τ
Find the system function:
Tz
Y (z)
τ
H(z) = d
=
T
Xd (z)
1+
z−1
τ
Pole at z =
1
.
1 + Tτ
25
Check Yourself
DT approximation:
T
T
yd [n] = xd [n]
yd [n + 1] − 1 −
τ
τ
Find the DT pole.
5
T
τ
τ
3. z =
T
2. z = 1 −
1. z =
4. z = −
5. z =
1
1 + Tτ
26
τ
T
T
τ
Dependence of DT pole on Stepsize
z
1
t
T
τ
= 0.1
T
τ
= 0.3
z
1
t
z
1
T
τ
t
=1
z
1
T
τ
t
= 1.5
z
1
τ
T
τ
t
=2
Why is this approximation better behaved?
27
Dependence of DT pole on Stepsize
Make a systems model of backward Euler method.
CT block diagrams: adders, gains, and integrators:
X
A
Y
ẏ(t) = x(t)
Backward Euler approximation:
y[n] − y[n − 1]
= x[n]
T
Equivalent system:
X
T
+
Y
R
Backward Euler: substitute equivalent system for all integrators.
28
Model of Backward Euler Method
Replace every integrator in the CT system
X
A
Y
with the backward Euler model:
X
T
+
Y
R
Substitute the DT operator for A:
1
T
T
A= →
=
s
1−R
1 − z1
Backward Euler maps z →
1
.
1 − sT
29
Dependence of DT pole on Stepsize
Pole at z =
1
1+ Tτ
1 .
= 1−sT
z
1
t
T
τ
= 0.1
T
τ
= 0.3
z
1
t
z
1
T
τ
t
=1
z
1
T
τ
t
= 1.5
z
1
τ
T
τ
t
30
=2
Backward Euler: Mapping CT poles to DT poles
Backward Euler Map:
s
→
1
z = 1−sT
0
1
− T1
1
2
1
3
− T2
s
0
z
1
z → 1−sT
−1
1
The entire left half-plane maps inside a circle with radius 12 at z = 12 .
If CT system is stable, then DT system is also stable.
31
Masses and Springs, Forwards and Backwards
In Homework 2, you investigated three numerical approximations to
a mass and spring system:
• forward Euler
• backward Euler
• centered method
x(t)
y(t)
32
Trapezoidal Rule
The trapezoidal rule uses centered differences.
Approximate CT signals at points between samples:
1 y [n] + yd [n − 1]
yc (n− )T = d
2
2
Approximate derivatives at points between samples:
1 y [n] − yd [n − 1]
ẏc (n− )T = d
2
T
1 yd [n] + yd [n−1]
T =
2 2
1
y [n] − yd [n−1]
ẏc n− T = d
2
T
yc
n−
yc (t)
yd [n]
yd [n−1]
(n−1)T
33
nT
t
Trapezoidal Rule
The trapezoidal rule uses centered differences.
ẏ(t) = x(t)
Trapezoidal rule:
y[n] − y[n − 1]
x[n] + x[n − 1]
=
T
2
Z transform:
Y (s)
T 1 + z −1
T z+1
H(z) =
=
=
2 z−1
X(s)
2 1 − z −1
Map:
A=
1
T
→
s
2
z+1
z−1
Trapezoidal rule maps z →
1 + sT
2
.
1 − sT
2
34
Trapezoidal Rule: Mapping CT poles to DT poles
Trapezoidal Map:
s
→
z=
1+ sT
2
1− sT
2
0
1
− T1
1
3
− T2
0
−∞
−1
jω
2+jωT
2−jωT
s
0
z
z → 2+sT
2−sT
−1
1
The entire left-half plane maps inside the unit circle.
The jω axis maps onto the unit circle
35
Mapping s to z: Leaky-Tank System
Forward Euler Method
s
z
1
T
− T2
− T1
z → 1 + sT
−1
1
Backward Euler Method
s
0
1
z → 1−sT
−1
z
1
Trapezoidal Rule
s
0
z → 2+sT
2−sT
36
−1
z
1
Mapping s to z: Mass and Spring System
Forward Euler Method
s
z
1
T
− T2
− T1
z → 1 + sT
−1
1
Backward Euler Method
s
0
1
z → 1−sT
−1
z
1
Trapezoidal Rule
s
0
z → 2+sT
2−sT
37
−1
z
1
Mapping s to z: Mass and Spring System
Forward Euler Method
s
z
1
T
− T2
− T1
z → 1 + sT
−1
1
Backward Euler Method
s
0
1
z → 1−sT
−1
z
1
Trapezoidal Rule
s
0
z → 2+sT
2−sT
38
−1
z
1
Concept Map
Relations between CT and DT representations.
Delay → R
Block Diagram
+
X
+
Delay
DT
+
−
1
τ
R
T
D
Y
A
=
X
A+τ
Y
x(t)
h(t) = τ1 e−t/τ u(t)
CT
Differential Equation
τ ṙ1 (t) = r0 (t) − r1 (t)
System Function
Y (z)
z2
= 2
X(z)
z −z−1
H(z) =
AX
Impulse Response
ẋ(t)
h[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .
Difference Equation
System Functional
R
Delay
y[n] = x[n] + y[n−1] + y[n−2]
Block Diagram
X
Y
1
= H(R) =
X
1 − R − R2
Unit-Sample Response
index shift
CX T R
System Functional
Y
System Function
Y (s)
1
H(s) =
=
X(s)
1 + τs
39
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6.003 Signals and Systems
Fall 2011
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