MIT EECS: 6.003 Signals and Systems lecture notes (Spring...
Transcript of MIT EECS: 6.003 Signals and Systems lecture notes (Spring...
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6.003: Signals and Systems
Discrete-Time Systems
February 4, 2010
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Discrete-Time Systems
We start with discrete-time (DT) systems because they
• are conceptually simpler than continuous-time systems
• illustrate same important modes of thinking as continuous-time
• are increasingly important (digital electronics and computation)
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Multiple Representations of Discrete-Time Systems
Systems can be represented in different ways to more easily address
different types of issues.
Verbal description: ‘To reduce the number of bits needed to store
a sequence of large numbers that are nearly equal, record the first
number, and then record successive differences.’
Difference equation:
y[n] = x[n]− x[n− 1]
Block diagram:
−1 Delay
+x[n] y[n]
We will exploit particular strengths of each of these representations.
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Difference Equations
Difference equations are mathematically precise and compact.
Example:
y[n] = x[n]− x[n− 1]
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Difference Equations
Difference equations are mathematically precise and compact.
Example:
y[n] = x[n]− x[n− 1]
Let x[n] equal the “unit sample” signal δ[n],
δ[n] ={ 1, if n = 0;
0, otherwise.
−1 0 1 2 3 4n
x[n] = δ[n]
We will use the unit sample as a “primitive” (building-block signal)
to construct more complex signals.
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Difference Equations
Difference equations are mathematically precise and compact.
Example:
y[n] = x[n]− x[n− 1]
Let x[n] equal the “unit sample” signal δ[n],
δ[n] ={ 1, if n = 0;
0, otherwise.
−1 0 1 2 3 4n
x[n] = δ[n]
We will use the unit sample as a “primitive” (building-block signal)
to construct more complex signals.
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Step-By-Step Solutions
Difference equations are convenient for step-by-step analysis.
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
Find y[n] given x[n] = δ[n]: y[n] = x[n]− x[n− 1]
y[−1] = x[−1]− x[−2] = 0− 0 = 0y[0] = x[0]− x[−1] = 1− 0 = 1y[1] = x[1]− x[0] = 0− 1 = −1y[2] = x[2]− x[1] = 0− 0 = 0y[3] = x[3]− x[2] = 0− 0 = 0. . .
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Step-By-Step Solutions
Difference equations are convenient for step-by-step analysis.
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
Find y[n] given x[n] = δ[n]: y[n] = x[n]− x[n− 1]
y[−1] = x[−1]− x[−2] = 0− 0 = 0y[0] = x[0]− x[−1] = 1− 0 = 1y[1] = x[1]− x[0] = 0− 1 = −1y[2] = x[2]− x[1] = 0− 0 = 0y[3] = x[3]− x[2] = 0− 0 = 0. . .
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Step-By-Step Solutions
Difference equations are convenient for step-by-step analysis.
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
Find y[n] given x[n] = δ[n]: y[n] = x[n]− x[n− 1]
y[−1] = x[−1]− x[−2] = 0− 0 = 0y[0] = x[0]− x[−1] = 1− 0 = 1y[1] = x[1]− x[0] = 0− 1 = −1y[2] = x[2]− x[1] = 0− 0 = 0y[3] = x[3]− x[2] = 0− 0 = 0. . .
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Step-By-Step Solutions
Difference equations are convenient for step-by-step analysis.
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
Find y[n] given x[n] = δ[n]: y[n] = x[n]− x[n− 1]
y[−1] = x[−1]− x[−2] = 0− 0 = 0y[0] = x[0]− x[−1] = 1− 0 = 1y[1] = x[1]− x[0] = 0− 1 = −1y[2] = x[2]− x[1] = 0− 0 = 0y[3] = x[3]− x[2] = 0− 0 = 0. . .
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Step-By-Step Solutions
Difference equations are convenient for step-by-step analysis.
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
Find y[n] given x[n] = δ[n]: y[n] = x[n]− x[n− 1]
y[−1] = x[−1]− x[−2] = 0− 0 = 0y[0] = x[0]− x[−1] = 1− 0 = 1y[1] = x[1]− x[0] = 0− 1 = −1y[2] = x[2]− x[1] = 0− 0 = 0y[3] = x[3]− x[2] = 0− 0 = 0. . .
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Step-By-Step Solutions
Difference equations are convenient for step-by-step analysis.
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
Find y[n] given x[n] = δ[n]: y[n] = x[n]− x[n− 1]
y[−1] = x[−1]− x[−2] = 0− 0 = 0y[0] = x[0]− x[−1] = 1− 0 = 1y[1] = x[1]− x[0] = 0− 1 = −1y[2] = x[2]− x[1] = 0− 0 = 0y[3] = x[3]− x[2] = 0− 0 = 0. . .
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Step-By-Step Solutions
Difference equations are convenient for step-by-step analysis.
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
Find y[n] given x[n] = δ[n]: y[n] = x[n]− x[n− 1]
y[−1] = x[−1]− x[−2] = 0− 0 = 0y[0] = x[0]− x[−1] = 1− 0 = 1y[1] = x[1]− x[0] = 0− 1 = −1y[2] = x[2]− x[1] = 0− 0 = 0y[3] = x[3]− x[2] = 0− 0 = 0. . .
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Step-By-Step Solutions
Block diagrams are also useful for step-by-step analysis.
Represent y[n] = x[n]− x[n− 1] with a block diagram:
−1 Delay
+x[n] y[n]
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
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Step-By-Step Solutions
Block diagrams are also useful for step-by-step analysis.
Represent y[n] = x[n]− x[n− 1] with a block diagram: start “at rest”
−1 Delay
+x[n] y[n]
0
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
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Step-By-Step Solutions
Block diagrams are also useful for step-by-step analysis.
Represent y[n] = x[n]− x[n− 1] with a block diagram: start “at rest”
−1 Delay
+1 1
−10
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
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Step-By-Step Solutions
Block diagrams are also useful for step-by-step analysis.
Represent y[n] = x[n]− x[n− 1] with a block diagram: start “at rest”
−1 Delay
+1→ 0
−10→ −1
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
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Step-By-Step Solutions
Block diagrams are also useful for step-by-step analysis.
Represent y[n] = x[n]− x[n− 1] with a block diagram: start “at rest”
−1 Delay
+1→ 0 −1
00→ −1
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
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Step-By-Step Solutions
Block diagrams are also useful for step-by-step analysis.
Represent y[n] = x[n]− x[n− 1] with a block diagram: start “at rest”
−1 Delay
+0 −1
0−1
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
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Step-By-Step Solutions
Block diagrams are also useful for step-by-step analysis.
Represent y[n] = x[n]− x[n− 1] with a block diagram: start “at rest”
−1 Delay
+0 −1
0−1→ 0
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
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Step-By-Step Solutions
Block diagrams are also useful for step-by-step analysis.
Represent y[n] = x[n]− x[n− 1] with a block diagram: start “at rest”
−1 Delay
+0 0
0−1→ 0
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
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Step-By-Step Solutions
Block diagrams are also useful for step-by-step analysis.
Represent y[n] = x[n]− x[n− 1] with a block diagram: start “at rest”
−1 Delay
+0 0
00
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
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Step-By-Step Solutions
Block diagrams are also useful for step-by-step analysis.
Represent y[n] = x[n]− x[n− 1] with a block diagram: start “at rest”
−1 Delay
+0 0
00
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
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Step-By-Step Solutions
Block diagrams are also useful for step-by-step analysis.
Represent y[n] = x[n]− x[n− 1] with a block diagram: start “at rest”
−1 Delay
+0 0
00
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
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Check Yourself
DT systems can be described by difference equations and/or
block diagrams.
Difference equation:
y[n] = x[n]− x[n− 1]
Block diagram:
−1 Delay
+x[n] y[n]
In what ways are these representations different?
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Check Yourself
In what ways are difference equations different from block diagrams?
Difference equation:
y[n] = x[n]− x[n− 1]
Difference equations are “declarative.”
They tell you rules that the system obeys.
Block diagram:
−1 Delay
+x[n] y[n]
Block diagrams are “imperative.”
They tell you what to do.
Block diagrams contain more information than the corresponding
difference equation (e.g., what is the input? what is the output?)
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From Samples to Signals
Lumping all of the (possibly infinite) samples into a single object —
the signal — simplifies its manipulation.
This lumping is an abstraction that is analogous to
• representing coordinates in three-space as points
• representing lists of numbers as vectors in linear algebra
• creating an object in Python
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From Samples to Signals
Operators manipulate signals rather than individual samples.
−1 Delay
+x[n] y[n]
Nodes represent whole signals (e.g., X and Y ).
The boxes operate on those signals:
• Delay = shift whole signal to right 1 time step
• Add = sum two signals
• −1: multiply by −1
Signals are the primitives.
Operators are the means of combination.
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Operator Notation
Symbols can now compactly represent diagrams.
Let R represent the right-shift operator:
Y = R{X} ≡ RX
where X represents the whole input signal (x[n] for all n) and Y
represents the whole output signal (y[n] for all n)
Representing the difference machine
−1 Delay
+x[n] y[n]
with R leads to the equivalent representation
Y = X −RX = (1−R)X
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Operator Notation: Check Yourself
Let Y = RX. Which of the following is/are true:
1. y[n] = x[n] for all n
2. y[n+ 1] = x[n] for all n
3. y[n] = x[n+ 1] for all n
4. y[n− 1] = x[n] for all n
5. none of the above
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Check Yourself
Consider a simple signal:
−1 0 1 2 3 4n
X
Then
−1 0 1 2 3 4n
Y = RX
Clearly y[1] = x[0]. Equivalently, if n = 0, then y[n+ 1] = x[n].
The same sort of argument works for all other n.
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Operator Notation: Check Yourself
Let Y = RX. Which of the following is/are true:
1. y[n] = x[n] for all n
2. y[n+ 1] = x[n] for all n
3. y[n] = x[n+ 1] for all n
4. y[n− 1] = x[n] for all n
5. none of the above
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Operator Representation of a Cascaded System
System operations have simple operator representations.
Cascade systems → multiply operator expressions.
−1 Delay
+
−1 Delay
+XY1
Y2
Using operator notation:
Y1 = (1−R)X
Y2 = (1−R)Y1
Substituting for Y1:
Y2 = (1−R)(1−R)X
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Operator Algebra
Operator expressions can be manipulated as polynomials.
−1 Delay
+
−1 Delay
+XY1
Y2
Using difference equations:
y2[n] = y1[n]− y1[n− 1]
= (x[n]− x[n− 1])− (x[n− 1]− x[n− 2])
= x[n]− 2x[n− 1] + x[n− 2]
Using operator notation:
Y2 = (1−R)Y1 = (1−R)(1−R)X
= (1−R)2X
= (1− 2R+R2)X
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Operator Approach
Applies your existing expertise with polynomials to understand block
diagrams, and thereby understand systems.
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Operator Algebra
Operator notation facilitates seeing relations among systems.
“Equivalent” block diagrams (assuming both initially at rest):
−1 Delay
+
−1 Delay
+XY1
Y2
Delay
Delay
−2
+X Y
Equivalent operator expressions:
(1−R)(1−R) = 1− 2R+R2
The operator equivalence is much easier to see.
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Check Yourself
Operator expressions for these “equivalent” systems
(if started “at rest”) obey what mathematical property?
Delay−1
+ DelayX Y
Delay
Delay Delay−1
+X Y
1. commutate 2. associative
3. distributive 4. transitive
5. none of the above
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Check Yourself
Delay−1
+ DelayX Y
Y = R(1−R)X
Delay
Delay Delay−1
+X Y
Y = (R−R2)X
Multiplication by R distributes over addition.
![Page 39: MIT EECS: 6.003 Signals and Systems lecture notes (Spring ...web.mit.edu/6.003/S10/www/handouts/lec02.pdf6.003: Signals and Systems Discrete-Time Systems February 4, 2010 Discrete-Time](https://reader036.fdocuments.in/reader036/viewer/2022081411/60b21bb6a5a3b65b0d13af9a/html5/thumbnails/39.jpg)
Check Yourself
Operator expressions for these “equivalent” systems
(if started “at rest”) obey what mathematical property? 3
Delay−1
+ DelayX Y
Delay
Delay Delay−1
+X Y
1. commutate 2. associative
3. distributive 4. transitive
5. none of the above
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Check Yourself
How many of the following systems are equivalent to
Y = (4R2 + 4R+ 1)X ?
Delay 2 + Delay 2 +X Y
Delay + Delay 4 +X Y
Delay 4 +
Delay
+X Y
![Page 41: MIT EECS: 6.003 Signals and Systems lecture notes (Spring ...web.mit.edu/6.003/S10/www/handouts/lec02.pdf6.003: Signals and Systems Discrete-Time Systems February 4, 2010 Discrete-Time](https://reader036.fdocuments.in/reader036/viewer/2022081411/60b21bb6a5a3b65b0d13af9a/html5/thumbnails/41.jpg)
Check Yourself
Delay 2 + Delay 2 +X Y
Y = (2R+ 1)(2R+ 1)X—————————————————————————–
Delay + Delay 4 +X Y
Y = (4R2 + 4R+ 1)X—————————————————————————–
Delay 4 +
Delay
+X Y
Y = (4R2 + 4R+ 1)X—————————————————————————–
All implement Y = (4R2 + 4R+ 1)X
![Page 42: MIT EECS: 6.003 Signals and Systems lecture notes (Spring ...web.mit.edu/6.003/S10/www/handouts/lec02.pdf6.003: Signals and Systems Discrete-Time Systems February 4, 2010 Discrete-Time](https://reader036.fdocuments.in/reader036/viewer/2022081411/60b21bb6a5a3b65b0d13af9a/html5/thumbnails/42.jpg)
Check Yourself
How many of the following systems are equivalent to
Y = (4R2 + 4R+ 1)X ? 3
Delay 2 + Delay 2 +X Y
Delay + Delay 4 +X Y
Delay 4 +
Delay
+X Y
![Page 43: MIT EECS: 6.003 Signals and Systems lecture notes (Spring ...web.mit.edu/6.003/S10/www/handouts/lec02.pdf6.003: Signals and Systems Discrete-Time Systems February 4, 2010 Discrete-Time](https://reader036.fdocuments.in/reader036/viewer/2022081411/60b21bb6a5a3b65b0d13af9a/html5/thumbnails/43.jpg)
Operator Algebra: Explicit and Implicit Rules
Recipes versus constraints.
Recipe: subtract a right-shifted version of the input signal from a
copy of the input signal.
−1 Delay
+X Y
Y = (1−R)X
Constraint: the difference between Y and RY is X.
Delay
+X Y
Y = RY +X
(1−R)Y = X
But how does one solve such a constraint?
![Page 44: MIT EECS: 6.003 Signals and Systems lecture notes (Spring ...web.mit.edu/6.003/S10/www/handouts/lec02.pdf6.003: Signals and Systems Discrete-Time Systems February 4, 2010 Discrete-Time](https://reader036.fdocuments.in/reader036/viewer/2022081411/60b21bb6a5a3b65b0d13af9a/html5/thumbnails/44.jpg)
Example: Accumulator
Try step-by-step analysis: it always works. Start “at rest.”
+
Delay
x[n] y[n]
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
Find y[n] given x[n] = δ[n]: y[n] = x[n] + y[n− 1]
y[0] = x[0] + y[−1] = 1 + 0 = 1y[1] = x[1] + y[0] = 0 + 1 = 1y[2] = x[2] + y[1] = 0 + 1 = 1. . .
![Page 45: MIT EECS: 6.003 Signals and Systems lecture notes (Spring ...web.mit.edu/6.003/S10/www/handouts/lec02.pdf6.003: Signals and Systems Discrete-Time Systems February 4, 2010 Discrete-Time](https://reader036.fdocuments.in/reader036/viewer/2022081411/60b21bb6a5a3b65b0d13af9a/html5/thumbnails/45.jpg)
Example: Accumulator
Try step-by-step analysis: it always works. Start “at rest.”
+
Delay
x[n] y[n]
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
Find y[n] given x[n] = δ[n]: y[n] = x[n] + y[n− 1]
y[0] = x[0] + y[−1] = 1 + 0 = 1y[1] = x[1] + y[0] = 0 + 1 = 1y[2] = x[2] + y[1] = 0 + 1 = 1. . .
![Page 46: MIT EECS: 6.003 Signals and Systems lecture notes (Spring ...web.mit.edu/6.003/S10/www/handouts/lec02.pdf6.003: Signals and Systems Discrete-Time Systems February 4, 2010 Discrete-Time](https://reader036.fdocuments.in/reader036/viewer/2022081411/60b21bb6a5a3b65b0d13af9a/html5/thumbnails/46.jpg)
Example: Accumulator
Try step-by-step analysis: it always works. Start “at rest.”
+
Delay
x[n] y[n]
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
Find y[n] given x[n] = δ[n]: y[n] = x[n] + y[n− 1]
y[0] = x[0] + y[−1] = 1 + 0 = 1y[1] = x[1] + y[0] = 0 + 1 = 1y[2] = x[2] + y[1] = 0 + 1 = 1. . .
![Page 47: MIT EECS: 6.003 Signals and Systems lecture notes (Spring ...web.mit.edu/6.003/S10/www/handouts/lec02.pdf6.003: Signals and Systems Discrete-Time Systems February 4, 2010 Discrete-Time](https://reader036.fdocuments.in/reader036/viewer/2022081411/60b21bb6a5a3b65b0d13af9a/html5/thumbnails/47.jpg)
Example: Accumulator
Try step-by-step analysis: it always works. Start “at rest.”
+
Delay
x[n] y[n]
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
Find y[n] given x[n] = δ[n]: y[n] = x[n] + y[n− 1]
y[0] = x[0] + y[−1] = 1 + 0 = 1y[1] = x[1] + y[0] = 0 + 1 = 1y[2] = x[2] + y[1] = 0 + 1 = 1. . .
![Page 48: MIT EECS: 6.003 Signals and Systems lecture notes (Spring ...web.mit.edu/6.003/S10/www/handouts/lec02.pdf6.003: Signals and Systems Discrete-Time Systems February 4, 2010 Discrete-Time](https://reader036.fdocuments.in/reader036/viewer/2022081411/60b21bb6a5a3b65b0d13af9a/html5/thumbnails/48.jpg)
Example: Accumulator
Try step-by-step analysis: it always works. Start “at rest.”
+
Delay
x[n] y[n]
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
Find y[n] given x[n] = δ[n]: y[n] = x[n] + y[n− 1]
y[0] = x[0] + y[−1] = 1 + 0 = 1y[1] = x[1] + y[0] = 0 + 1 = 1y[2] = x[2] + y[1] = 0 + 1 = 1. . .
![Page 49: MIT EECS: 6.003 Signals and Systems lecture notes (Spring ...web.mit.edu/6.003/S10/www/handouts/lec02.pdf6.003: Signals and Systems Discrete-Time Systems February 4, 2010 Discrete-Time](https://reader036.fdocuments.in/reader036/viewer/2022081411/60b21bb6a5a3b65b0d13af9a/html5/thumbnails/49.jpg)
Example: Accumulator
Try step-by-step analysis: it always works. Start “at rest.”
+
Delay
x[n] y[n]
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
Find y[n] given x[n] = δ[n]: y[n] = x[n] + y[n− 1]
y[0] = x[0] + y[−1] = 1 + 0 = 1y[1] = x[1] + y[0] = 0 + 1 = 1y[2] = x[2] + y[1] = 0 + 1 = 1. . .
Persistent response to a transient input!
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Example: Accumulator
The response of the accumulator system could also be generated by
a system with infinitely many paths from input to output, each with
one unit of delay more than the previous.
Delay
Delay Delay
Delay Delay Delay
+
... ...
X Y
Y = (1 +R+R2 +R3 + · · ·)X
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Example: Accumulator
These systems are equivalent in the sense that if each is initially at
rest, they will produce identical outputs from the same input.
(1−R)Y1 = X1 ⇔ ? Y2 = (1 +R+R2 +R3 + · · ·)X2
Proof: Assume X2 = X1:
Y2 = (1 +R+R2 +R3 + · · ·)X2
= (1 +R+R2 +R3 + · · ·)X1
= (1 +R+R2 +R3 + · · ·) (1−R)Y1
= ((1 +R+R2 +R3 + · · ·)− (R+R2 +R3 + · · ·))Y1
= Y1
It follows that Y2 = Y1.
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Example: Accumulator
These systems are equivalent in the sense that if each is initially at
rest, they will produce identical outputs from the same input.
(1−R)Y1 = X1 ⇔ ? Y2 = (1 +R+R2 +R3 + · · ·)X2
Proof: Assume X2 = X1:
Y2 = (1 +R+R2 +R3 + · · ·)X2
= (1 +R+R2 +R3 + · · ·)X1
= (1 +R+R2 +R3 + · · ·) (1−R)Y1
= ((1 +R+R2 +R3 + · · ·)− (R+R2 +R3 + · · ·))Y1
= Y1
It follows that Y2 = Y1.
It also follows that (1−R) and (1 +R+R2 +R3 + · · ·) are reciprocals.
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Example: Accumulator
The reciprocal of 1−R can also be evaluated using synthetic division.
1 +R +R2 +R3 + · · ·1−R 1
1 −RRR −R2
R2
R2 −R3
R3
R3 −R4· · ·
Therefore
11−R
= 1 +R+R2 +R3 +R4 + · · ·
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Feedback
Systems with signals that depend on previous values of the same
signal are said to have feedback.
Example: The accumulator system has feedback.
Delay
+X Y
By contrast, the difference machine does not have feedback.
−1 Delay
+x[n] y[n]
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Cyclic Signal Paths, Feedback, and Modes
Block diagrams help visualize feedback.
Feedback occurs when there is a cyclic signal flow path.
R
R−2
+X Y
Delay
+X Y
acyclic cyclic
Acyclic: all paths through system go from input to output with no
cycles.
Cyclic: at least one cycle.
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Feedback, Cyclic Signal Paths, and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths.
Delay
+
p0
X Y
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output.
![Page 57: MIT EECS: 6.003 Signals and Systems lecture notes (Spring ...web.mit.edu/6.003/S10/www/handouts/lec02.pdf6.003: Signals and Systems Discrete-Time Systems February 4, 2010 Discrete-Time](https://reader036.fdocuments.in/reader036/viewer/2022081411/60b21bb6a5a3b65b0d13af9a/html5/thumbnails/57.jpg)
Feedback, Cyclic Signal Paths, and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths.
Delay
+
p0
X Y
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output.
![Page 58: MIT EECS: 6.003 Signals and Systems lecture notes (Spring ...web.mit.edu/6.003/S10/www/handouts/lec02.pdf6.003: Signals and Systems Discrete-Time Systems February 4, 2010 Discrete-Time](https://reader036.fdocuments.in/reader036/viewer/2022081411/60b21bb6a5a3b65b0d13af9a/html5/thumbnails/58.jpg)
Feedback, Cyclic Signal Paths, and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths.
Delay
+
p0
X Y
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output.
![Page 59: MIT EECS: 6.003 Signals and Systems lecture notes (Spring ...web.mit.edu/6.003/S10/www/handouts/lec02.pdf6.003: Signals and Systems Discrete-Time Systems February 4, 2010 Discrete-Time](https://reader036.fdocuments.in/reader036/viewer/2022081411/60b21bb6a5a3b65b0d13af9a/html5/thumbnails/59.jpg)
Feedback, Cyclic Signal Paths, and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths.
Delay
+
p0
X Y
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output.
![Page 60: MIT EECS: 6.003 Signals and Systems lecture notes (Spring ...web.mit.edu/6.003/S10/www/handouts/lec02.pdf6.003: Signals and Systems Discrete-Time Systems February 4, 2010 Discrete-Time](https://reader036.fdocuments.in/reader036/viewer/2022081411/60b21bb6a5a3b65b0d13af9a/html5/thumbnails/60.jpg)
Feedback, Cyclic Signal Paths, and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths.
Delay
+
p0
X Y
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output.
![Page 61: MIT EECS: 6.003 Signals and Systems lecture notes (Spring ...web.mit.edu/6.003/S10/www/handouts/lec02.pdf6.003: Signals and Systems Discrete-Time Systems February 4, 2010 Discrete-Time](https://reader036.fdocuments.in/reader036/viewer/2022081411/60b21bb6a5a3b65b0d13af9a/html5/thumbnails/61.jpg)
Feedback, Cyclic Signal Paths, and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths.
Delay
+
p0
X Y
−1 0 1 2 3 4n
x[n] = δ[n]
−1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output.
The response will persist even though the input is transient.
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Check Yourself
How many of the following systems have cyclic signal
paths?
R
R+ +X Y + +
R R
X Y
R
+ +X Y + +
R R
X Y
![Page 63: MIT EECS: 6.003 Signals and Systems lecture notes (Spring ...web.mit.edu/6.003/S10/www/handouts/lec02.pdf6.003: Signals and Systems Discrete-Time Systems February 4, 2010 Discrete-Time](https://reader036.fdocuments.in/reader036/viewer/2022081411/60b21bb6a5a3b65b0d13af9a/html5/thumbnails/63.jpg)
Check Yourself
How many of the following systems have cyclic signal
paths? 3
R
R+ +X Y + +
R R
X Y
R
+ +X Y + +
R R
X Y
![Page 64: MIT EECS: 6.003 Signals and Systems lecture notes (Spring ...web.mit.edu/6.003/S10/www/handouts/lec02.pdf6.003: Signals and Systems Discrete-Time Systems February 4, 2010 Discrete-Time](https://reader036.fdocuments.in/reader036/viewer/2022081411/60b21bb6a5a3b65b0d13af9a/html5/thumbnails/64.jpg)
Finite and Infinite Impulse Responses
The impulse response of an acyclic system has finite duration, while
that of a cyclic system can have infinite duration.
−1 Delay
+X Y
Delay
+X Y
−1 0 1 2 3 4n
−1 0 1 2 3 4n
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Analysis of Cyclic Systems: Geometric Growth
If traversing the cycle decreases or increases the magnitude of the
signal, then the fundamental mode will decay or grow, respectively.
If the response decays toward zero, then we say that it converges.
Otherwise, we it diverges.
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Check Yourself
How many of these systems have divergent unit-sample
responses?
Delay0.5
+X Y
Delay1.2
+X Y
Delay0.5
+ 1.2DelayX Y
![Page 67: MIT EECS: 6.003 Signals and Systems lecture notes (Spring ...web.mit.edu/6.003/S10/www/handouts/lec02.pdf6.003: Signals and Systems Discrete-Time Systems February 4, 2010 Discrete-Time](https://reader036.fdocuments.in/reader036/viewer/2022081411/60b21bb6a5a3b65b0d13af9a/html5/thumbnails/67.jpg)
Check Yourself
Delay0.5
+X Y
−1 0 1 2 3 4n
y[n]
Delay1.2
+X Y
−1 0 1 2 3 4n
y[n]
Delay0.5
+ 1.2DelayX Y
−1 0 1 2 3 4n
y[n]
![Page 68: MIT EECS: 6.003 Signals and Systems lecture notes (Spring ...web.mit.edu/6.003/S10/www/handouts/lec02.pdf6.003: Signals and Systems Discrete-Time Systems February 4, 2010 Discrete-Time](https://reader036.fdocuments.in/reader036/viewer/2022081411/60b21bb6a5a3b65b0d13af9a/html5/thumbnails/68.jpg)
Check Yourself
Delay0.5
+X Y
−1 0 1 2 3 4n
y[n]
X
Delay1.2
+X Y
−1 0 1 2 3 4n
y[n]
√
Delay0.5
+ 1.2DelayX Y
−1 0 1 2 3 4n
y[n]
X
![Page 69: MIT EECS: 6.003 Signals and Systems lecture notes (Spring ...web.mit.edu/6.003/S10/www/handouts/lec02.pdf6.003: Signals and Systems Discrete-Time Systems February 4, 2010 Discrete-Time](https://reader036.fdocuments.in/reader036/viewer/2022081411/60b21bb6a5a3b65b0d13af9a/html5/thumbnails/69.jpg)
Check Yourself
How many of these systems have divergent unit-sample
responses? 1
Delay0.5
+X YX
Delay1.2
+X Y √
Delay0.5
+ 1.2DelayX Y
X
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Cyclic Systems: Geometric Growth
If traversing the cycle decreases or increases the magnitude of the
signal, then the fundamental mode will decay or grow, respectively.
Delay0.5
+X Y
Delay1.2
+X Y
−1 0 1 2 3 4n
y[n]
−1 0 1 2 3 4n
y[n]
These are geometric sequences: y[n] = (0.5)n and (1.2)n for n ≥ 0.
These geometric sequences are called fundamental modes.
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Multiple Representations of Discrete-Time Systems
Now you know four representations of discrete-time systems.
Verbal descriptions: preserve the rationale.
“To reduce the number of bits needed to store a sequence of
large numbers that are nearly equal, record the first number,
and then record successive differences.”
Difference equations: mathematically compact.
y[n] = x[n]− x[n− 1]
Block diagrams: illustrate signal flow paths.
−1 Delay
+x[n] y[n]
Operator representations: analyze systems as polynomials.
Y = (1−R)X