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//example 9.2
clc
clear
close
t1=100; // clock period
s=0100; //given serial input
for i = 4:-1:1
se(i) = modulo(s,10);
s=s/10;
s=round(s);
end
se(i+4)=0;
k=0;
for i = 2:6 //initially making a state table
clk(k+1) = k;
q(i)=se(i-1);
if i>1 then
r(i) = q(i-1);
else
r(i) =0;
end;
if i>2 then
s(i) = r(i-1);
else
s(i) =0;
end;
if i>3 then
t(i) =s(i-1);
else
t(i) =0;
end;
k=k+1;
end
for m=1:5 // drawing the graph
if(se(m)==1) then
v= ((m-1).*t1)
for u= 1: t1
se1(u+v)=1;
end
else
v= ((m-1)*t1)
for u= 1: t1
se1(u+v)=0;
end
end;
if(q(m)==1) then
v= ((m-1).*t1)
for u= 1: t1
q1(u+v)=1;
end
else
v= ((m-1)*t1)
for u= 1: t1
q1(u+v)=0;
end
end;
if(r(m)==1) then
v= ((m-1).*t1)
for u= 1: t1
r1(u+v)=1;
end
else
v= ((m-1)*t1)
for u= 1: t1
r1(u+v)=0;
end
end;
if(s(m)==1) then
v= ((m-1).*t1)
for u= 1: t1
s1(u+v)=1;
end
else
v= ((m-1)*t1)
for u= 1: t1
s1(u+v)=0;
end
end;
if(t(m)==1) then
v= ((m-1).*t1)
for u= 1: t1
t11(u+v)=1;
end
else
v= ((m-1)*t1)
for u= 1: t1
t11(u+v)=0;
end
end;
end;
p=1;
while p<t1*5
if p==1 | modulo(p,t1) == 1 then
for k=1:t1/2
cin(p+k)=0;
end
p=p+t1/2;
else
cin(p)=1;
p=p+1;
end
end
y=[3 3];
subplot(6,1,1) // making subplots to draw all graphs in a single window
title('Clock')
plot(cin)
plot(y)
subplot(6,1,2)
title('Serial input')
plot(se1)
plot(y)
subplot(6,1,3)
title('Q')
plot(q1)
plot(y)
subplot(6,1,4)
title('R')
plot(r1)
plot(y)
subplot(6,1,5)
title('S')
plot(s1)
plot(y)
subplot(6,1,6)
title('T')
plot(t11)
plot(y)
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