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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')
+