text stringlengths 16 3.88k | source stringlengths 60 201 |
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.
For the example shown, we will first check to see if n=0. Since
it does not, we take the list pointed to by joe, extract the rest
of that list by literally taking the pointer out of the cdr part of the first box, and call list-ref on that structure, with a
decrement of n. This recursive call will now have n=0, so... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/1239c38467fc8018b19c808d121bcf5d_lecture5webhand.pdf |
whatever we get by appending
the rest of the first list onto the second list. Using the same
closure properties, we see that this will create a list, and adjoin
will then put the first element on the front of this new list.
Notice how we are using the recursive properties of lists to build
this procedure, and in p... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/1239c38467fc8018b19c808d121bcf5d_lecture5webhand.pdf |
y-sum and how-many) are bound to the values of the expressions
following those names. Then within the confines of the let expression, those names are simply local names for
those values, and are substituted for just as we would in the standard substitution model.
6.001 Structure and Interpretation of Computer Progr... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/1239c38467fc8018b19c808d121bcf5d_lecture5webhand.pdf |
ly
separated from the use of that structure. This means that we rely
on the user showing discipline when applying procedures to data
structures, expecting them not to use procedures that directly
take advantage of the implementation of the structure.
Sometimes we are much better off imposing strong abstraction
ba... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/1239c38467fc8018b19c808d121bcf5d_lecture5webhand.pdf |
to drive home is the
separation of the details of implementation of an abstraction
from the use of that abstraction. We are going to place this solid
barrier between the implementation and use of an abstraction,
and we are going to see why having that barrier makes it much
easier for us to create useful systems.
... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/1239c38467fc8018b19c808d121bcf5d_lecture5webhand.pdf |
going to separate all of this from the actual implementation of rationals. We should ideally be
able to take any implementation of rationals that satisfies these contracts and build our own procedures to use those
objects, just by relying on the constructor and selectors, and the contract between them.
Slide 5.4.6
... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/1239c38467fc8018b19c808d121bcf5d_lecture5webhand.pdf |
by separating the numerator and
denominator by a "slash". The key question is can this procedure
tell which implementation of rationals we are using? The
answer, of course, is no. The procedure uses the selectors to get
out the pieces, and cannot tell which implementation is being
used so long as the right selecto... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/1239c38467fc8018b19c808d121bcf5d_lecture5webhand.pdf |
know by our previous work, that this should pull apart these two
rationals, manipulate the underlying integer pieces, then
reassemble a new rational that is the sum of these two.
Slide 5.4.13
This sounds pretty straightforward. What happens if we look at
the pieces of new? The numerator is 10, and the denominator ... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/1239c38467fc8018b19c808d121bcf5d_lecture5webhand.pdf |
. But
when we ask for the parts of the rational, we will first compute
the gcd of the numerator and denominator, then reduce the
selector part by that amount. This will fix the problem we saw in
the previous slide, always reducing a rational to its lowest
possible form.
But notice that since we built all of our o... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/1239c38467fc8018b19c808d121bcf5d_lecture5webhand.pdf |
have simply stripped off the abstraction barrier.
6.001 Structure and Interpretation of Computer Programs. Copyright © 2004 by Massachusetts Institute of Technology.
Slide 5.4.20
But now, suppose we decide to implement the idea of
incorporating the reduction by the gcd into the creation of
rationals. In this case... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/1239c38467fc8018b19c808d121bcf5d_lecture5webhand.pdf |
6.001 Structure and Interpretation of Computer Programs. Copyright © 2004 by Massachusetts Institute of Technology.
6.001 Notes: Section 4.1
Slide 4.1.1
In this lecture, we are going to take a careful look at the kinds
of procedures we can build. We will first go back to look very
carefully at the substitution mod... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/124c1c683236effa0fe4f7d4318ac9b7_lecture4webhand.pdf |
the procedure, and we
evaluate the operands to get the set of arguments. If we have a
primitive procedure, we are just going to “do the right thing”.
Otherwise we are going to replace this entire expression with
the body of the compound expression, with the arguments
substituted for their associated parameters.
S... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/124c1c683236effa0fe4f7d4318ac9b7_lecture4webhand.pdf |
..and here is a partial trace of fact using those rewrite rules.
We start with (fact 4). That reduces to evaluating an if
expression. Since the predicate is not true, this reduces to
evaluating the alternative statement, which is a combination of
a multiplication and another evaluation of fact.
Notice how the oran... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/124c1c683236effa0fe4f7d4318ac9b7_lecture4webhand.pdf |
amount of space needed by any rewrite stage is a linear multiple
of the size of the argument. It also has linear growth in time,
because as we saw it takes a number of basic steps that is also a
linear multiple of the size of the argument.
Slide 4.1.13
On the other hand, iterative-fact has no deferred
operations,... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/124c1c683236effa0fe4f7d4318ac9b7_lecture4webhand.pdf |
solve Fibonacci for any smaller
sized argument. Using that idea, we can then work out a
solution to the problem. With this in hand, it is clear that the
solution to the general problem is just to solve two smaller
sized problems, then just add the results together. Note that in
this case we are using wishful think... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/124c1c683236effa0fe4f7d4318ac9b7_lecture4webhand.pdf |
of size
3 and one of size 2, and each of these requires solving two
smaller problems, and so on.
This gives rise to a kind of tree of things we have to do, and
each recursive call gives rise to two subproblems of smaller
size. This leads to a different order of growth.
Slide 4.2.4
To measure the order of growth,... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/124c1c683236effa0fe4f7d4318ac9b7_lecture4webhand.pdf |
accomplish this?
Slide 4.3.2
So, recall the stages we used to solve problems like this: we
will use some wishful thinking to assume that solutions to
simpler versions of the problem exist; we will then decompose
the problem into a simpler version of the same problem, plus
some other simple operations, and we will... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/124c1c683236effa0fe4f7d4318ac9b7_lecture4webhand.pdf |
that is easy. Anything to the 0th power is just 1!
Slide 4.3.7
And thus we can add our base case to our procedure, creating
the same kind of procedure we saw earlier for factorial.
Note how the base case will stop unwinding the computation
into simpler cases.
6.001 Structure and Interpretation of Computer Progra... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/124c1c683236effa0fe4f7d4318ac9b7_lecture4webhand.pdf |
next step, we will multiply our current product by a
and keep track of that new result, plus the fact that we have one
less thing to do.
Slide 4.3.12
And this process we can continue until ...
Slide 4.3.13
... we reach a point where we have no further multiplies to do.
Slide 4.3.14
Now, what are the stages of t... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/124c1c683236effa0fe4f7d4318ac9b7_lecture4webhand.pdf |
.
6.001 Structure and Interpretation of Computer Programs. Copyright © 2004 by Massachusetts Institute of Technology.
6.001 Notes: Section 4.4
Slide 4.4.1
So now we have seen three very different kinds of procedures:
ones that give rise to constant behavior, ones that give rise to
linear growth, and ones that gi... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/124c1c683236effa0fe4f7d4318ac9b7_lecture4webhand.pdf |
multiplying a by the result of raising a to the
power b-1, that is to a simpler, or smaller, version of the same
problem.
Slide 4.4.5
Notice the effect of doing this. If b is odd, then in one step I
reduce the problem to the case where b is even, which means in
the next step, I reduce the problem size by half. Th... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/124c1c683236effa0fe4f7d4318ac9b7_lecture4webhand.pdf |
and thus
after 2k steps, the problem is reduced by a factor of 2^k, or cut
in half k times.
How do we find the number of times we need to do this? We
are done when the problem size is just 1, and that happens when
k = log n. So this procedure has a different behavior, it is
logarithmic. This is in fact a very eff... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/124c1c683236effa0fe4f7d4318ac9b7_lecture4webhand.pdf |
.2
So, let’s structure this problem a bit. Let’s order the rows, and
enumerate them, labeling the first row n=0, the second row n=1,
and so on (we will see that this choice of labeling leads to a
cleaner description of the problem). Using this labeling, we
also see that the n’th row has n+1 elements in it. Let’s u... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/124c1c683236effa0fe4f7d4318ac9b7_lecture4webhand.pdf |
space. We have already suggested that
exponential algorithms are costly. In the case of Fibonacci, we
didn’t look for any other way of structuring the problem, but let’s
try to do better for Pascal.
Slide 4.5.7
To do better, we have to go back to the original problem. A
little information from combinatorics tells... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/124c1c683236effa0fe4f7d4318ac9b7_lecture4webhand.pdf |
in the denominator. Notice how our procedural abstraction
nicely isolates the details of fact from its use in this case, and
the code here cleanly expresses the idea of computing Pascal
based on factorial.
Slide 4.5.10
So do we do any better with this version of Pascal?
Sure! We know that this version of fact is ... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/124c1c683236effa0fe4f7d4318ac9b7_lecture4webhand.pdf |
, this may be the best version, but theoretically it
has the same class of behavior as our previous version.
Slide 4.5.15
And that leads to our concluding point. First, we stress that the
same problem may have many different solutions, and that
these solutions may have very different computational
behaviors. Some... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/124c1c683236effa0fe4f7d4318ac9b7_lecture4webhand.pdf |
Chapter 5 Signals and Systems
6.01— Spring 2011— April 25, 2011
169
Chapter 5
Signals and Systems
Imagine that you are asked to design a system to steer a car straight down the middle of a lane.
It seems easy, right? You can figure out some way to sense the position of the car within its lane.
Then, if the car is r... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
system with one input and one output. Both the input and output
are signals. A signal is a mathematical function with an independent variable (most often it will
be time for the problems that we will study) and a dependent variable (that depends on the
independent variable). The system is described by the way that i... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
mode of our car steering system.
(t) as a function of time correspond to the oscillations of the car within its
Is po(t) the only important output signal from the car-steering system? The answer to this ques
tion depends on your goals. Analyzing a system with this single output is likely to give important
insights ... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
the center of the lane), and
our actual position in the lane po(t). Let e(t) = pi(t) − po(t). Thus we can think about the
steering controller as having an input e(t) and output φ(t).
In the composite system (in figure 5.3), the steering controller determines φ(t), which is the input
to the car. The car generates po(... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
sub
systems that can be combined to analyze the overall system.
5.1.2 Discrete-time signals and systems
This chapter focuses on signals whose independent variables are discrete (e.g., take on only inte
ger values). Some such signals are found in nature. For example, the primary structure of DNA
is described by a s... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
[n]po[n]+car−1steeringcontrollerφ(t)e(t)pi(t)po(t) Chapter 5 Signals and Systems
6.01— Spring 2011— April 25, 2011
173
5.1.3 Linear time-invariant systems
We already have a great way of specifying systems that operate on discrete-time signals: a state
machine transduces a discrete-time input signal into a discrete... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
interested in LTI systems because they can be analyzed mathematically, in a
way that lets us characterize some properties of their output signal for any possible input signal.
This is a much more powerful kind of insight than can be gained by trying a machine out with
several different inputs.
Another important pro... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
y[n] = c x[n]
·
·
.
That is, the resulting signal has a value at every index n that is c times the value of the original
signal at that location. Here are the signals 4∆ and −3.3∆.
The next operation is the delay operation. The result of delaying a signal X is a new signal RX
such that:
if Y = RX then y[n] = x[n ... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
then make a new signal Z = Y + 0.3RY, which would look like this:
Be sure you understand how the heights of the spikes are determined by the definition of Z.
Exercise 5.2.
Draw a picture of samples −1 through 4 of Y − RY.
It is important to remember that, because signals are infinite objects, these combination opera... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
and scaling satisfy the familiar algebraic properties of addition and multiplication: ad
dition is commutative and associative, scaling is commutative (in the sense that it doesn’t matter
whether we pre- or post-multiply) and scaling distributes over addition:
c (X1 + X2) = c X1 + c X2
·
·
·
,
which can be verifie... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
are appropriately scaled and shifted. We could similarly start with a family of discretely-
sampled sinusoids as our primitives, where
x[n] = cos(Ωn)
.
Here are plots of two primitives in this family:
cos(0.2n)
cos(1.0n)
The second plot may seem confusing, but it is just a sparsely sampled sinusoid. Note that sig... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
, delay, and addition on whole signals.
Consider a system that has an input signal X, and whose output signal is X − RX. We can describe
that system using the operator equation
Y = X − RX .
Using the algebraic properties of operators on signals described in section 5.2.3, we can rewrite
this as
Y = (1 − R)X ,
wh... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
x[1] − x[0]
= 0 − 1 = −1
y[2] = x[2] − x[1]
= 0 − 0 = 0
y[3] = x[3] − x[2]
= 0 − 0 = 0
· · ·
Block diagrams
Another way of describing a system is by drawing a block diagram, which is made up of com
ponents, connected by lines with arrows on them. The lines represent signals; all lines that are
connected to one an... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
our system
Y = X − RX
can be specified in Python as a state machine by:
class Diff(sm.SM):
def __init__(self, previousInput):
self.startState = previousInput
def getNextValues(self, state, inp):
return (inp, inp-state)
Here, the state is the value of the previous input. One important thing to notice is that, since
w... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
Z, and operator equation Z =
·
(Φ2 Φ1)X.
The product of polynomials is another polynomial, so Φ2 Φ1 is a polynomial in R. Furthermore,
because polynomial multiplication is commutative, cascade combination is commutative as well
(as long as the systems are at rest, which means that their initial states are 0).
·
S... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
2011— April 25, 2011
182
Combining cascade and parallel operations
Finally, the distributive law applies for cascade and parallel combination, for systems at rest, in
the same way that it applies for multiplication and addition of polynomials, so that if we have
three systems, with operator equations:
Y = Φ1X
U ... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
DelayDelayDelay−12−1++XYDelayDelayDelayDelay−12−1++XY Chapter 5 Signals and Systems
6.01— Spring 2011— April 25, 2011
184
Exercise 5.7.
Convince yourself that all of these systems are equivalent. One strategy is
to convert them all to operator equation representation.
5.4 Feedback Systems
So far, all of our examp... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
+ y[n − 1]
y[0] = x[0] + y[−1]
= 1 + 0 = 1
y[1] = x[1] + y[0]
= 0 + 1 = 1
y[2] = x[2] + y[1]
= 0 + 1 = 1
· · ·
Here are plots of the input signal X and the output signal Y:
This result may be somewhat surprising! In feedforward systems, we saw that the output was
always a finite sum of scaled and delayed versions of ... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
5 Signals and Systems
6.01— Spring 2011— April 25, 2011
186
These systems are equivalent in the sense that if each is initially at rest, they will produce identical
outputs from the same input. We can see this by taking the original definition and repeatedly
substituting in the definition of Y in for its occurrence ... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
R3 + · · ·
R + R2 + R3 + · · ·
OR =
So
O(1 − R) = 1
And so,
1
1 − R
=
1 + R + R2 + R3 +
· · ·
Exercise 5.8.
Check this derivation by showing that
(1 + R + R2 + R3 +
· · ·
)(1 − R) = 1
So, we can rewrite the operator equation for the accumulator as
Y =
1
1 − R
X .
We don’t have an intuitive way to int... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
sm.SM):
def __init__(self, dCoeffs, cCoeffs):
j = len(dCoeffs) - 1
k = len(cCoeffs)
self.cCoeffs = cCoeffs
self.dCoeffs = dCoeffs
self.startState = ([0.0]*j, [0.0]*k)
Chapter 5 Signals and Systems
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188
def getNextValues(self, state, input):
(inputs, outputs) = state
inputs = [input... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
, we have
Y = c0 RY + c1 R2Y + . . . + ck−1 RkY + d0 X + d1 RX + . . . + dj RjX
= (c0 R + c1 R2 + . . . + ck−1 Rk) Y + (d0 + d1 R + . . . + dj Rj) X .
We can rewrite this as
(1 − c0 R − c1 R2 − . . . − ck−1 Rk) Y = (d0 + d1 R + . . . + dj Rj) X ,
so
Y
X
=
d0 + d1R + d2R2 + d3R3 +
· · ·
1 − c0R − c1R2 − c2R3 − ... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
ients
of the denominator polynomial, ai, and the coefficients of the numerator polynomial, bi. It is
always possible to rewrite this in a form in which a0 = 1.
Feedforward systems have no dependence on previous values of Y, so they have D(R) = 1.
Feedback systems have persistent behavior, which is determined by D(R)... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
2011
190
where H = H1 + H2.
Cascade
The system function of the cascade of two systems is the product of their system functions. So,
given two systems with system functions H1 and H2, connected like this:
and letting
W = H1X
and
Y = H2W ,
we have
Y = H2W
= H2H1X
= HX ,
where H = H2H1. And note that, as was... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
1 + H1H2) = H1X
Y =
H1
1 + H1H2
X
Y = HX ,
where
H =
H1
1 + H1H2
.
Armed with this set of primitives and composition methods, we can specify a large class of ma
chines. Ultimately, we will want to construct systems with multiple inputs and outputs; such
systems are specified with a matrix of basic system functio... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
rst-order
systems, in which the denominator of the system function is a first-order polynomial (that is, it
only involves R, but not R2 or other higher powers of R.)
Let’s consider this very simple system
+Rp0XY Chapter 5 Signals and Systems
6.01— Spring 2011— April 25, 2011
192
for which we can write an operator ... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
ing the cycle decreases or increases the magnitude of the signal, then the sample values
will decay or grow, respectively, as time increases.
Delay+p0XYDelay+p0XYDelay+p0XY−101234ny[n]−101234ny[n]−101234ny[n] Chapter 5 Signals and Systems
6.01— Spring 2011— April 25, 2011
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For the first system, the unit sample r... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
April 25, 2011
194
We can describe it with the operator equation
Y = 1.6RY − 0.63R2Y + X ,
so the system function is
H =
1
1 − 1.6R + 0.63R2
.
Here is its response to a unit sample signal:
We can try to understand its behavior by decomposing it in different ways. First, let’s see if we
can see it as a cascade... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
1
(1 − 0.9R)(1 − 0.7R)
=
A
1 − 0.9R
= H1 + H2
+
.
B
1 − 0.7R
To find values for A and B, we start with
1
(1 − 0.9R)(1 − 0.7R)
=
A
1 − 0.9R
+
B
1 − 0.7R
,
multiply through by 1 − 1.6R + 0.63R2 to get
1 = A(1 − 0.7R) + B(1 − 0.9R)
,
and collect like terms:
1 = (A + B) − (0.7A + 0.9B)R
.
Equating the terms th... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
summing them to recover the response of H to the unit sample. In the next figure, the blue stem
plot is the overall signal, which is the sum of the green and red signals, which correspond to the
top and bottom parts of the diagram, respectively.
In this case, both of the poles (0.9 and 0.7) are less than 1, so the ma... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
1.1 and the blue dots are generated by the sum.
Chapter 5 Signals and Systems
6.01— Spring 2011— April 25, 2011
198
In these examples, as in the general case, the long-term unit-sample response of the entire
system is governed by the unit-sample response of the mode whose pole has the larger magni
tude. In the l... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
a wall might be described with difference equation:
do[n] = do[n − 1] + KT do[n − 2] − KT di[n − 1]
.
Difference equations with real-valued coefficients generate real-valued outputs from real-valued
inputs. But, like the difference equation
y[n] = 2x[n] + 2x[n − 1] − 2y[n − 1] − 4y[n − 1]
,
corresponding to syste... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
j sin(tan−1(b, a)))
=
=
(cid:112)
a2 + b2( √
a
a2 + b2
+ j √
b
a2 + b2
)
= a + bj
Why should we bother with this change of representation? There are some operations on complex
numbers that are much more straightforwardly done in the exponential representation. In partic
ular, let’s consider raising a complex ... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
· ·
.
For a complex pole p = rejΩ, pn = rnejΩn. So
1
1 − rejΩR
= 1 + rejΩR + r 2 ej2ΩR2 +
· · ·
What happens as n tends to infinity when p is complex? Think of pn as a point in the complex
plane with coordinates (rn, Ωn). The radius, rn, will grow or shrink depending on the mode’s
magnitude, r. And the angle, Ω... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
to 0; the second one diverges
(look carefully at the scales).
p = 0.85ejπ/8 ≈ 0.785 + 0.325j
p = 1.1ejπ/8 ≈ 1.016 + 0.421j
If complex modes occurred all by themselves, then our signals would have complex numbers in
them. But isolated complex pole can result only from a difference equation with complex-valued
coef... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
−jΩR)
Let’s slowly multiply out the denominator:
(cid:45)0.20.20.40.60.81.0(cid:45)0.20.20.40.60.81.0(cid:45)5000500010000(cid:45)5000500010000 Chapter 5 Signals and Systems
6.01— Spring 2011— April 25, 2011
202
(1 − rejΩR)(1 − re −jΩR) = 1 − rejΩR − re −jΩR + r 2 ejΩe −jΩR2
= 1 − r(ejΩ + e −jΩ)R + r 2 ejΩ−jΩR2
U... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
to do an additive decomposition. That means doing a partial fraction
expansion of
Y
X
=
1
(1 − rejΩR)(1 − re−jΩR)
.
It’s a little trickier than before, because we have complex numbers, but the method is the same.
The end result is that we can decompose this system additively to get
Y
X
=
1 (1 − j cot Ω)
2
1 ... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
out, and that real parts are equal. Thus, the real part of the output is going to be twice the real
part of these elements. The figure below shows the unit sample response of the entire system.
In the formula below,
y[n] = r n(cos nΩ + cot Ω sin nΩ)
,
we know that
(cid:112)
−
1 + cot2 Ω (cid:54) cos nΩ + cot Ω s... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
Both r and Ω tells us something very useful about the way in which the system behaves. In
the previous section, we derived an expression for the samples of the unit sample response for a
system with a pair of complex poles. It has the form
y[n] = r n(cos nΩ + α sin nΩ)
,
where α is a constant. We know that (cos nΩ... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
Poles and behavior: summary
In a second-order system, if we let p0 be the pole with the largest magnitude, then there is a time
step at which the behavior of the dominant pole begins to dominate; after that time step
•
If p0 is real and
• p0 < −1, the magnitude increases to infinity and the sign alternates.
• −1 <... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
b3R3 +
· · ·
1 + a1R + a2R2 + a3R3 + · · ·
Regrouping terms, we can write this as the operator equation:
Y = (b0 + b1R + b2R2 + b3R3 +
· · ·
) X − (a1R + a2R2 + a3R3 +
· · ·
) Y
and construct an equivalent block diagram:
Returning to the general polynomial ratio
Y
X
=
b0 + b1R + b2R2 + b3R3 + · · ·
1 + a1R ... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
modal decomposition leads us to an alternative block diagram:
RRRRRR−a1−a2b0b1b2++······XY Chapter 5 Signals and Systems
6.01— Spring 2011— April 25, 2011
207
We can fairly easily observe that the behavior is going to be the sum of the behaviors of the
individual modes, and that, as in the second-order case, the m... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
+ z 2 = 0 .
The roots are 3 and 4. If we go back to our original polynomial in R, we can see that:
12R2 − 7R + 1 = (1 − 4R)(1 − 3R)
.
so that our poles are 4 and 3. So, remember, the poles are not the roots of the polynomial in R, but
are the roots of the polynomial in the reciprocal of R.
+p0RC0+C1Rp1++D0D1R++··... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
at 0.5. In this case, the system has a repeated pole; it is still possible to perform an
additive decomposition, but it is somewhat trickier. Ultimately, however, it is still the magnitude
of the largest root that governs the long-term convergence properties of the system.
5.5.5 Superposition
The principle of super... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
, it is easy to see that once we understand the unit-sample response of a system,
we can see how it will respond to any finite input.
We might be interested in understanding how a system H responds to a step input signal. Let’s
just consider the basic step signal, U, defined as
if n (cid:62) 0
1
0 otherwise
u[n] =... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
’t study this in any further detail, but it’s useful to understand that the basis of our analy
sis of systems applies very broadly across LTI systems and inputs.
5.6 Designing systems
Will eventually include a discussion of root-locus plots.
For now, see section 5.8.3 for a discussion of picking k for the wall findi... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
after finitely many steps, begin to increase or decrease monotonically.
If the dominant pole is real and negative, then in response to a transient input, the signal will,
after finitely many steps, begin to alternate signs.
If the dominant pole is complex, then in response to a transient input, the signal will, after ... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
2y[n − 1]
dCoeffs: none, cCoeffs: 2
• Output at step n is the input at step n − 1 plus the output at step n − 2:
y[n] = x[n − 1] + y[n − 2]
dCoeffs: 0, 1, cCoeffs: 0, 1
5.8.2 Difference equations and block diagrams
Let H represent a system whose input is a signal X and whose output is a signal Y. The system H
is de... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
, combine them to get
Y = (1 − R)
X
1 − 2R
=
1 − R
1 − 2R
X
Showing that this system is the same as H.
This time, let’s name the signal that’s flowing between the two adders A. Now, we have equations
equivalent to H (yes/no)? NO
Y = A + 2RY
A = X + 2RA
Rewrite the first equation as:
Y =
A
1 − 2R
And the secon... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
” and that the input signal X is the unit sample
signal. Determine y[3].
y[3] =
4
++RRXY++RR−1XY Chapter 5 Signals and Systems
6.01— Spring 2011— April 25, 2011
214
n
−1
0
1
2
2
x[n]
z[n]
y[n]
0
1
0
0
0
0
0
1
2
4
0
1
1
2
4
Part c. Let po represent the dominant pole of H. Determine po.
Enter... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
sensor responds immediately, so that the robot’s com
manded velocity v[n] depends instantaneously on its actual distance do[n] from the wall. Of
di=desiredFrontdo=distanceFrontcontrollerplantsensor+XYerrorcommand− Chapter 5 Signals and Systems
6.01— Spring 2011— April 25, 2011
215
course this dependence can’t trul... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
−T kR
1 − (1 + T k)R
Di
We can solve for the poles analytically after substituting 1/z in for R:
z − (1 + T k) = 0 .
There is one pole at 1 + T k.
In order for the system to converge, we need
| 1 + T k | < 1
− 1 <
1 + T k < 1
− 2 <
−2
T
<
T k
< 0
k
< 0
Chapter 5 Signals and Systems
6.01— Spring 2011— April... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
ge as
Chapter 5 Signals and Systems
6.01— Spring 2011— April 25, 2011
217
does a gain less than −20. And for gain values between −10 and −20, it converges, but alternates
sign.
Here is what is called a root-locus plot. It shows how the poles of the system (in this case just one
pole) move in the complex plane a... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
and Systems
6.01— Spring 2011— April 25, 2011
218
Do(1 − R) = −T RV
Do(1 − R) = −T Rk(Di − RDo)
Do(1 − R − T kR2) = −T kRDi
−T kR
1 − R − T kR2
Do =
Di
We can solve for the poles analytically after substituting 1/z for R:
z 2 − z − T k = 0 .
We find that the roots of this polynomial are
1
2
±
1 √
2
1 + 4kT
.... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
with the same magnitude. On the right
is the maximum of the pole magnitudes. We can see that it is minimized at kT = −0.25.
Magnitudes of both poles
Magnitude of dominant pole
Here is a root-locus plot for this system. It shows how the poles of the system move in the complex
plane as we vary parameter k. In this fi... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
April 25, 2011
220
5.8.4 Cool difference equations
Newton’s law of cooling states that: the rate of change of the temperature of an object is propor
tional to the difference between its own temperature and the temperature of its surroundings.
We can model this process in discrete time, by assuming that the change i... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
the output of the gain of k2. Then we can write the following set of operator equations:
Z = k1(X − RY)
W = k2(Z − RY)
Y = RW
Eliminating Z and W, we have:
Y = Rk2(Z − RY)
= Rk2(k1(X − RY) − RY)
= k1k2RX − k1k2R2Y − k2R2Y
Reorganizing terms, we have
Y + k2(1 + k1)R2Y = k1k2RX
RRR++k1k2XY−− Chapter 5 Signals a... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
Let k1 = 1 and k2 = −2, determine the poles of H.
Enter poles or none if there are no poles:
2, -2
For these k values, the denominator polynomial is 1 − 4R2 . So, we need to find the roots of the
polynomial z2 − 4, which are ±2.
Part d. For each of the systems below indicate whether the system is equivalent to this... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
2 be W. Then we can write the following set of operator equations:
Equivalent to H (yes/no)? Yes
W = k2(k1X − (1 + k1)RY)
Y = RW
Eliminating W, we have:
Y = Rk2(k1X − (1 + k1)RY)
= k1k2RX − k1k2R2Y − k2R2Y
which is equal to the operator equation for the original system.
RR++k1k2k1XY−RRR++k1k2k1XY− Chapter 5 Sig... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
2]
The important thing to see here is that the values in the state are (y[n−2], y[n−1], x[n−2], x[n−1]),
so that the output is 2y[n − 2] + 3x[n − 2].
5.8.7 On the Verge
For each difference equation below, say whether, for a unit sample input signal:
•
•
the output of the system it describes will diverge or not, ... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
Find the roots of the polynomial in z = 1/R:
Z2 + Z + 10 = 0
−1 ±
Z =
√
1 − 100
2
Z = 0.5 ± 4.97j
The magnitude of the poles is 5, which is greater than 1, so it diverges. The poles are complex,
so the output will oscillate.
3.
Chapter 5 Signals and Systems
6.01— Spring 2011— April 25, 2011
227
y[n] = −0.6y[n − ... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
ality constant such that Ic is the rate of heat input.
The system is thus described by the following diagram:
1. a. Give the system function:
c
(1 − k1R)(1 + k2R)
++DelayDelay-ITk1k2c Chapter 5 Signals and Systems
6.01— Spring 2011— April 25, 2011
228
If we name the signal coming out of the first adder W, then we... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
plots of T [n] as a function of n for this system for c = 1 and different values of k1
and k2.
Chapter 5 Signals and Systems
6.01— Spring 2011— April 25, 2011
229
a
c
e
b
d
f
a.
Which of the plots above corresponds to k1 = 0.5 and k2 = 0 ?
Circle all correct answers:
a
b
c
d
e
f
none
The denominator as... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
Let k1 = 0.5, k2 = 3, and c = 1. Determine the poles of H, or none if there are no poles.
Looking at the factored form of the denominator, we can easily see that the poles are at k1
and −k2. If you didn’t see that factored form, then you could explicitly see that the poles are
roots of the equation z2 + (k2 − k1)z −... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
= ±π/2. We
can see that the magnitude is about 0.3 of the previous magnitude after 4 steps, which means
that the magnitude of the pole is about 0.75 (because 0.754 = 0.316). Pole 8 has angle −π/2
and magnitude 0.75.
1.51.00.50.51.01.51.51.00.50.51.01.512345678 Chapter 5 Signals and Systems
6.01— Spring 2011—... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
the larger system shown below.
10203040(cid:45)1.5(cid:45)1.0(cid:45)0.50.51.01.52.0102030400.51.01.52.0102030405000100001500010203040(cid:45)1.0(cid:45)0.50.51.01.52.010203040(cid:45)0.50.51.01.52.010203040(cid:45)0.50.51.01.52.010203040(cid:45)5050100150102030400.51.01.52.0 Chapter 5 Signals and Systems
6.01— Sprin... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
The pole on the negative real axis causes the unit sample
response to alternate. Thus there are no values of KO and KB for which there is monotonic
decay.
+H1RK0X0Y0−+RRKBX1Y1 MIT OpenCourseWare
http://ocw.mit.edu
6.01SC Introduction to Electrical Engineering and Computer Science
Spring 2011
For information abou... | https://ocw.mit.edu/courses/6-01sc-introduction-to-electrical-engineering-and-computer-science-i-spring-2011/1268f3289b19d628e9be3bd2ecfb4f44_MIT6_01SCS11_chap05.pdf |
Introduction to Robotics, H. Harry Asada
1
Chapter 9
Force and Compliance Controls
A class of simple tasks may need only trajectory control where the robot end-effecter is
moved merely along a prescribed time trajectory. However, a number of complex tasks, including
assembly of parts, manipulation of tools, and wa... | https://ocw.mit.edu/courses/2-12-introduction-to-robotics-fall-2005/127c560e6052cb02ed3f7adc8d3c1512_chapter9.pdf |
, two types of control loops
are combined in the hand control system, as illustrated in Figure 9.1.2.
z
O
y
Fz
x
Figure 9.1.1 Robot drawing a line with a pencil on a sheet of paper
Department of Mechanical Engineering
Massachusetts Institute of Technology
Introducti... | https://ocw.mit.edu/courses/2-12-introduction-to-robotics-fall-2005/127c560e6052cb02ed3f7adc8d3c1512_chapter9.pdf |
conflict with the
geometric constraints and physics. M. Mason addressed this issue in his seminal work on hybrid
position/force control. He called conditions dictated by physics Natural Constraints, and
conditions determined by task goals and objectives Artificial Constraints. Table 9.1.1 summarizes
these condition... | https://ocw.mit.edu/courses/2-12-introduction-to-robotics-fall-2005/127c560e6052cb02ed3f7adc8d3c1512_chapter9.pdf |
is friction-
less, no resistive force acts on the peg in the direction that is not constrained by geometry.
Therefore, the linear force in the z direction must be zero:
. The rotation about the z axis,
0=zf
too, is not constrained. Therefore, the torque about the z axis must be zero:
conditions are dictated by phys... | https://ocw.mit.edu/courses/2-12-introduction-to-robotics-fall-2005/127c560e6052cb02ed3f7adc8d3c1512_chapter9.pdf |
this next.
Let V6 be a six-dimensional vector space, and
Va ⊂
6V
be an admissible motion space,
that is, the entire collection of admissible motions conforming to the geometric constraints
involved in a given task. Let Vc be a constraint space that is the orthogonal complement to the
admissible motion space:
V
c... | https://ocw.mit.edu/courses/2-12-introduction-to-robotics-fall-2005/127c560e6052cb02ed3f7adc8d3c1512_chapter9.pdf |
a
is in a static equilibrium, the virtual work must vanish for all virtual displacements
p δ=
p
becomes a virtual displacement, and eq.(4) reduces to virtual work. Since the system
to be a virtual
. Then,
∆ cp
0=
p∆
apδ .
Department of Mechanical Engineering
Massachusetts Institute of Technology
... | https://ocw.mit.edu/courses/2-12-introduction-to-robotics-fall-2005/127c560e6052cb02ed3f7adc8d3c1512_chapter9.pdf |
will appreciate Mason’s Principle when considering the following exercise
problem, in which the partition between admissible motion space and constraint space cannot be
described by a simple division of C-frame axes. Rather the admissible motion space lies along an
axis where a translational axis and a rotational ax... | https://ocw.mit.edu/courses/2-12-introduction-to-robotics-fall-2005/127c560e6052cb02ed3f7adc8d3c1512_chapter9.pdf |
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