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+clc;

+//page 697

+//problem 13.12

+

+//The qeneratr matrix requires impulse response of the coder.

+//This is the ourput generated when the initially reset coder is fed with a single 1.

+//The no of bits in the output code = 2(1+3) = 8

+

+//Taking in, s1, s2 , s3, v1 & v2 as row matrix where each column represents its corresponding input or output, in means input

+in = [0 1 0 0 0]

+

+s1 = zeros(1,5)

+s2 = zeros(1,5)

+s3 = zeros(1,5)

+v1 = zeros(1,5)

+v2 = zeros(1,5)

+

+

+for i = 2:5

+    s3(i) = s2(i-1)

+    s2(i) = s1(i-1)

+    s1(i) = in(i-1)

+    v1(i-1) = bitxor(s1(i),bitxor(s2(i),s3(i)))

+    v2(i-1) = bitxor(s1(i),s3(i))

+end

+

+//Output matrix is out

+out = zeros(1,8)

+for i = [1 3 5 7]

+    out(i) = v1((i+3)/2)

+    out(i+1) = v2((i+3)/2)

+end

+

+disp('impulse response is')

+disp(out)

+

+//Then generator matrix is G

+G = [1 1 1 0 1 1 0 0 0 0 0 0;0 0 1 1 1 0 1 1 0 0 0 0;0 0 0 0 1 1 1 0 1 1 0 0]

+

+//Note that, in G, impulse responses appear in staggered apper in a staggered manner in each row while the rest of the elements are 0.

+

+//Now, output is b_o = b_i*G where input b_i =[1 0 1] 

+b_i = [1 0 1]

+

+b_o = b_i*G

+

+//Here multiplication means Exor operation so whereever two occurs it should be changed to 1

+

+for i = 1:12

+        if  b_o(i) > 1 then

+            b_o(i) = 0;

+        end

+end

+

+disp('output is ')

+disp(b_o)

+disp('The output obtained is exactly same as example 13.1')

+