Output) of the one machine into those of the other can convert the



One representation to the other.

Thus, in the example given, apply the one-one transformation P

The transformation P:

Uv

Y – x

Is a shorthand way of describing the one-one transformation that

Pairs off states in S and R thus:

In S,

           ,, ,,

           ,, ,,

           ,, ,,

I.e. ,, ,,

(Compare U of S.4/9.)

Result is

Against (– 3,2) in R

 (1,0) ,,(0,1) ,, ,,

 (4,5) ,,(– 5,4) ,, ,,

(– 3,0) ,,(0,– 3) ,, ,,

U,v) ,,(– v,u) ,, ,,

Apply P to all the description of S; the

                       δ ε g h j k

                 P: ↓ β α c a b d

To N’s table, applying it to the borders as well as to the body. The

Result is

P: ↓

c

a

b

d

D b a c

D a c c

This is essentially the same as M. Thus, c and β in the border give

β

α

                         y' = – y + x

                        – x' = – y – x

Which is algebraically identical with R. So R and S are isomorphic.

Ex. 1: What one-one transformation will show these absolute systems to be iso-

Morphic?

            a b c d ep q r s tY: ↓ c c d d bZ: ↓ r q q p r

Fig. 6/9/2

Hint: Try to identify some characteristic feature, such as a state of equilib-

Rium.)

Ex. 2: How many one-one transformations are there that will show these absolute

Systems to be isomorphic?

                      a b cB: ↓ p q rA: ↓ b c ar p q

*Ex. 3: Write the canonical equations of the two systems of Fig. 6/8/1 and show

That they are isomorphic. (Hint: How many variables are necessary if the sys-

Tem is to be a machine with input ?)

Ex. 4: Find a re-labelling of variables that will show the absolute systems A and

B to be isomorphic.

                 x' = – x2 + y u' = w2 + u

                            2A:  y' = – x – yB:  v' = – v2 + w

                          2 z' = y + z w' = – v2 – w

(Hint: On the right side of A one variable is mentioned only once; the same

Is true of B. Also, in A, only one of the variables depends on itself quadrat-

ically, i.e. if of the form a' = + a2 . . . ; the same is true of B.)

D in both. The isomorphism thus corresponds to the definition.

(The isomorphism can be seen more clearly if first the rows are

Interchanged, to

                 ↓ c a b d

D a c c

D b a c

And then the columns interchanged, to

α

β

a

b

c

d

A c d c

B a d c

But this re-arrangement is merely for visual convenience.)

When the states are defined by vectors the process is essentially

Unchanged. Suppose R and S are two absolute systems:

           R:  x' = x + yS:  u' = – u – v

            y' = x – y v' = – u + v

98

α

β

The previous section showed that two machines are isomor-

Phic if one can be made identical to the other by simple relabel-

Ling. The “re-labelling”, however, can have various degrees of

Complexity, as we will now see.

99

A N I N T R O D UC T I O N T O C Y B E R NE T I C S

TH E BL AC K B O X

The system that is specified only by states, as in the previous

Section, contains no direct reference either to parts or to variables.

In such a case, “re-labelling” can mean only “re-labelling the

States”. A system with parts or variables, however, can also be

Re-labelled at its variables— by no means the same thing. Relabel-

Ling the variables, in effect, re-labels the states but in a way sub-

Ject to considerable constraint (S.7/8), whereas the re-labelling of

States can be as arbitrary as we please. So a re-labelling of the

States is more general than a re-labelling of the variables.

Thus suppose a system has nine states; an arbitrary re-labelling

Of eight of the states does not restrict what label shall be given to

The ninth. Now suppose that the system has two variables, x and

Y, and that each can take three values: x1, x2, x3 and y1, y2, y3. Nine

States are possible, of which two are (x2,y3) and (x3,y1). Suppose

This system is re-labelled in its variables, thus

Ex. 1: (Ex. 6/9/4 continued.) Compare the diagram of immediate effects of A and

B.

Ex. 2: Mark the following properties of an absolute system as changed or

Unchanged by a re-labelling of its states: (i) The number of basins in its

Phase-space; (ii) whether it is reducible; (iii) its number of states of equilib-

Rium; (iv) whether feedback is present; (v) the number of cycles in its

Phase-space.

Ex. 3: (Continued.) How would they be affected by a re-labelling of variables?

X y

ξ η

The subject of isomorphism is extensive, and only an intro-

Duction to the subject can be given here. Before we leave it, how-


Дата добавления: 2019-11-16; просмотров: 213; Мы поможем в написании вашей работы!

Поделиться с друзьями:






Мы поможем в написании ваших работ!