Neural Networks And Fuzzy System

Neural Networks And Fuzzy System Referencing Styles : Harvard | Pages : 32  ONE [ The Generalised XOR Problem ] [ 50 marks ] A…

Neural Networks And Fuzzy System

Referencing Styles :
Harvard | Pages :
32

 ONE [ The Generalised XOR Problem ] [ 50 marks ]

A neural network model with two input neurons, three hidden neurons in a single

hidden layer, and an output neuron is used to learn the decision surface of the wellknown

generalised XOR problem:

sgn( ) 1 2 d  x x

The input range is [-1,1]. The desired output value is either –1 (corresponds to logical

zero) or 1 (corresponds to logical one).

This generalised XOR problem has been utilised recently to compare the

effectiveness of various neural network learning algorithms. A simplified version is

used in this question. Eight input samples 1 2 7 8 x ,x ,,x ,x and their corresponding

target vectors 1 2 7 8 d ,d ,,d ,d in the training set are:





 



  0.5242

0.7826

x1 , 



 





  0.5377

0.9003

x2 , 



 







  0.9630

0.0871

x3 , 



 



  0.7873

0.1795

x4 ,





 



  0.5839

0.2309

x5 , 



 





  0.0280

0.2137

x6 , 



 







  0.1886

0.6475

x7 , 



 



  0.2076

0.6026

x8

 1 d1  ,  1 d2   , d3   1 ,  1 d4
  , 1 d5  ,  1 d6   , d7   1 , d8
   1

Assume that the network as shown in Figure 1 has 3 hidden neurons, 1 output neuron,

and all continuous perceptrons use the bipolar activation function f e

e

2

1

1 ( ) 



  





 .

Note that due to the necessary augmentation of inputs and of the hidden layer by one

fixed input, the trained network should have 3 input nodes, 4 hidden neurons, and 1

output neuron. Assign -1 to all augmented inputs.

1.1 Assume that the learning constant is   0.2, and the initial random output

layer weight matrix W( ) 1 and hidden layer weight matrix W ( ) 1 are

W (1)    0.3443 0.6762  0.9607 0.3626























 

 



0.0056 0.3908 0.3644

0.6636 0.1422 0.6131

0.2410 0.4189 0.6207

Using the error back propagation training, calculate the next weight updates

W W ( ), ( ) 2 2 .

 [ 25 marks ]

1.2 The above training set was trained with the same set of initial random output

layer weight matrix W( ) 1 and hidden layer weight matrix W ( ) 1 as above, and

a learning constant of   0.2. The training set was recycled when necessary.

Determine the final weight matrices W W (4001) f  and W W (4001) f  after

500 cycles. Plot the cycle error curve for this training exercise.

 [ 20 marks ]

Consider two particular input vectors of the test set





 





  0.9803

0.6263

xt1 and 



 



  0.0500

0.0700

xt 2

Classify these two test input vectors using the final weight matrices and

discuss the results.

 [ 5 marks ] 

QUESTION TWO [Truck-Backer Upper Control] [ 50 marks ]

Backing up a truck to a loading dock is a nonlinear control problem. The truck and

loading zone are shown in Figure 2.1. The truck position is exactly determined by the

three state variables , , x y where  is the angle of the truck with the horizontal.

Control to the truck is the angle  .

Only backing up is considered. The truck moves backward by a fixed unit distance

every stage. For simplicity, assume that there is enough clearance between the truck

and the loading dock such that y does not have to be considered as an input. The task

here is to design a control system, whose inputs are   90 270 0 20   ,, , x and

whose output is     40 40 , such that the final stages will be  ,     10,90 f f x  .

The dynamics of the truck backer-upper procedure can be approximated by:

     

    





 



   

    

    



b

k k k

y k y k k k k k

x k x k k k k k

2sin[ ( )] ( 1) ( ) sin

( 1) ( ) sin ( ) ( ) sin ( ) cos ( )

( 1) ( ) cos ( ) ( ) sin ( ) sin ( )

1   

   

   

where b is the length of the truck. Assume that b  4.

Fuzzy logic is required for this truck backer-upper control. In this simple fuzzy logic

controller, a set of linguistic variables is chosen to represent 5 degrees of truck angle

   error        90 70 90 110 270 ,,, , , 5 degrees of truck position   x error

  0 7 10 13 20 mm m m m ,, , , , and 5 degrees of control angle
        40 10 0 10 40 , ,, , 

as shown in Figure 2.2. The generic rule set in the form of “Fuzzy Associative

Memories” is shown in Figure 2.3.

The initial states of this truck are assumed to be ) ( (1), x(1), y(1)) (35 ,15m,15m    .

2.1 If the Centre of Area (COA) defuzzification strategy is used with the fire

strength i of the i-th rule calculated from

 i XX   i i  min( ( ), ( )) x x 1 2 1 2

determine the defuzzified control angle  (1) and the next state

[(2), x(2), y(2)].

 [ 20 marks ]

If the Mean of Maximum (MOM) defuzzification strategy is used with the fire

strength i of the i-th rule calculated from

( ). ( ) 1 1 2 2 x x X i X i

i   

determine the defuzzified control angle  (1) and the next state

[(2), x(2), y(2)]. Then continue and calculate  (2) and [x(3),(3), y(3)].

Write a computer program to calculate the system state vector

[x(k 1),(k 1), y(k 1)]

and the defuzzified control angle  (k) for 100

consecutive sampling points. Plot the corresponding vertical truck position

y(k) against the horizontal truck position x(k) for these 100 sampling points.

Plot the defuzzified control angle  (k) for these 100 sampling points.

 [ 20 marks ]

Find the dominant rule which contributes the highest fire strength to the

control action for the defuzzified control angle (1) . If softer control action

(for slower response) is required, modify this dominant rule and recalculate

the new defuzzified control angle ) (1 *  and the next state vector

[(2), x(2), y(2)]. Using the modified FAM table, plot the corresponding

vertical truck position y(k) against the horizontal truck position x(k) for

these 100 sampling points. Plot the new defuzzified control angle ) (

*  k for

these 100 sampling points.

 [ 10 marks ] 
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