15 files changed, 401 insertions, 0 deletions
diff --git a/45/CH10/EX10.1/example_10_1.sce b/45/CH10/EX10.1/example_10_1.sce
new file mode 100755
index 000000000..dcd7c8130
--- /dev/null
+++ b/45/CH10/EX10.1/example_10_1.sce
@@ -0,0 +1,10 @@
+//example 10.1
+clc;
+clear;
+//c= input('Enter the period of the waveform at C in micro seconds : ');
+c=24;// given period of waveform 
+clk= c/8;
+clkf = 1/(clk*10^-3);
+printf('The clock period is %f micro seconds \n',clk);//displaying the results 
+printf('The clock frequenc must be %f  KHz ', clkf);
+
diff --git a/45/CH10/EX10.10/example_10_10.sce b/45/CH10/EX10.10/example_10_10.sce
new file mode 100755
index 000000000..1dde98ea0
--- /dev/null
+++ b/45/CH10/EX10.10/example_10_10.sce
@@ -0,0 +1,4 @@
+//example 10.10
+clc;
+clear;
+printf('The correct expresion is ""8"" = Qd Qc'' Qb Qa''');
+\ No newline at end of file
diff --git a/45/CH10/EX10.12/example_10_12.sce b/45/CH10/EX10.12/example_10_12.sce
new file mode 100755
index 000000000..258fea5ba
--- /dev/null
+++ b/45/CH10/EX10.12/example_10_12.sce
@@ -0,0 +1,42 @@
+//example 10.12
+clc
+clear
+//pro= ('Enter the value to whic counter should progress:');
+pro =11; // given input 
+q=1;
+aa=pro;
+for i=1:4             //converting the given number in to binary 
+    x=modulo(aa,2);
+    b(q)=x;
+    aa=aa/2;
+    aa=floor(aa);
+    q=q+1; 
+end
+           
+bi=' ';   // then printing the NAND gate inputs 
+for i=1:4
+    if i==1 & b(i)==1 then
+        bi=strcat([bi 'Qa']);
+    elseif i==1 & b(i)== 0 ;
+        bi=strcat([bi 'Qa''']);   
+    end
+    if i==2 & b(i)==1 then
+        bi=strcat([bi 'Qb']);
+    elseif i==2 & b(i)== 0 ;
+        bi=strcat([bi 'Qb''']);   
+    end
+    if i==3 & b(i)==1 then
+        bi=strcat([bi 'Qc']);
+    elseif i==3 & b(i)== 0 ;
+        bi=strcat([bi 'Qc''']);   
+    end
+    if i==4 & b(i)==1 then
+        bi=strcat([bi 'Qd']);
+    elseif i==4 & b(i)== 0 ;
+        bi=strcat([bi 'Qd''']);   
+    end
+
+end
+disp('The NAND gate inputs must be :'); 
+disp(bi)
+
diff --git a/45/CH10/EX10.13/example_10_13.sce b/45/CH10/EX10.13/example_10_13.sce
new file mode 100755
index 000000000..6e5020d7d
--- /dev/null
+++ b/45/CH10/EX10.13/example_10_13.sce
@@ -0,0 +1,28 @@
+//example 10.13
+clc;
+clear
+//pre=input("Enter the number where the counter is preset");
+pre = 1001; // given preset value  
+q=1;
+b=0;
+f=0;                
+a=pre;
+while(a>0) //converting to decimal 
+    r=modulo(a,10);
+    b(1,q)=r;
+    a=a/10;
+    a=floor(a);
+    q=q+1;
+end
+for m=1:q-1
+     c=m-1
+    f = f + b(1,m)*(2^c);
+end
+disp("The counter will count down to 15 , Then preset back to %d, The resulting state diagram is shown below");
+for k=1:3
+for i=9:-1:0  // this will print the states 
+printf('%d  ',i);
+end;
+printf('15  ');
+
+end;
+\ No newline at end of file
diff --git a/45/CH10/EX10.14/example_10_14.sce b/45/CH10/EX10.14/example_10_14.sce
new file mode 100755
index 000000000..4a5ecfc4e
--- /dev/null
+++ b/45/CH10/EX10.14/example_10_14.sce
@@ -0,0 +1,63 @@
+//example 10.14
+//this program uses the following functions 
+//kmap3.sci
+//noof.sci and noof0.sci
+//the above programs should be executed before executing these programs  
+clc;
+n= [ 0 0 0;
+     0 0 1;
+     0 1 0;
+     0 1 1;
+     1 0 0;
+     1 0 1;
+     1 1 0;
+     1 1 1];
+     for i= 1 : 5
+         n1(i,:)= n(i+1,:) 
+     end
+     for i=6:8
+     n1(i,:)=[0 0 0]
+  end;
+  p=1;
+  for i= 1:3  //making the state table 
+      for j = 1:8
+          if n(j,i)== 0 
+              jf(j,p)= n1(j,i);
+              jf(j,p+1)= 2;
+          elseif n(j,i) == 1
+              jf(j,p)=2;
+              jf(j,p+1)=bitcmp(n1(j,i),1);
+      end
+  end
+  p=p+2
+end;
+disp('State tabel for mod 6 counter:'); //displaying the state table 
+di= [n n1 jf];
+disp('   Cn    Bn    An   Cn1   Bn1    An1   Jc    Kc    Jb    Kb    Ja    Ka');
+disp(di);
+disp('Here ''2'' represents a don''t care condition');
+disp('These below Karnaugh maps give the design equations');
+
+jc=[0 0 1 0;2 2 2 2] //Krnaugh Maps for the design equations 
+JC=kmap3(jc);   //calling the 3-variable kmap 
+printf('\n\nJC = %s \n',JC); //displaying the result
+
+kc=[2 2 2 2;0 1 1 1]
+KC=kmap3(kc);//calling the 3-variable kmap 
+printf('\n\nKC = %s \n',KC); //displaying the result
+
+jb=[0 1 2 2;0 0 2 2]
+JB=kmap3(jb);//calling the 3-variable kmap 
+printf('\n\nJB = %s \n',JB); //displaying the result
+
+kb=[2 2 1 0;2 2 1 1]
+KB=kmap3(kb);//calling the 3-variable kmap 
+printf('\n\nKB = %s \n',KB); //displaying the result
+
+ja=[1 2 2 1;1 2 2 0]
+JA=kmap3(ja);//calling the 3-variable kmap 
+printf('\n\nJA = %s \n',JA); //displaying the result
+
+ka=[2 1 1 2;2 1 1 2]
+KA=kmap3(ka);//calling the 3-variable kmap 
+printf('\n\nKA = %s \n',KA); //displaying the result
diff --git a/45/CH10/EX10.15/example_10_15.sce b/45/CH10/EX10.15/example_10_15.sce
new file mode 100755
index 000000000..3c69197bb
--- /dev/null
+++ b/45/CH10/EX10.15/example_10_15.sce
@@ -0,0 +1,18 @@
+//example 10.15 
+//this program use kmap2.sci 
+//kmap2.sci should be executed before executing this program
+clc;
+a= [0 0 1 1]
+b= [0 1 0 1]
+y= [1 1 0 1]
+k= [1 1 ; 0 1]
+bi = kmap2(k); // calling 2-variable kmap
+disp("   A     B     Y");
+for i=1:3
+    Y(i,1) = a(i);
+    Y(i,2) = b(i);
+    Y(i,3) = y(i);
+end
+disp(Y);
+disp('The minimised expression from karnaugh map is');// displaying the result
+disp(bi);
+\ No newline at end of file
diff --git a/45/CH10/EX10.2/example_10_2.sce b/45/CH10/EX10.2/example_10_2.sce
new file mode 100755
index 000000000..6994f5600
--- /dev/null
+++ b/45/CH10/EX10.2/example_10_2.sce
@@ -0,0 +1,17 @@
+//example 10.2
+clc;
+clc
+c=128; // given counters 
+d=32;
+e=64;
+fc=log2(c);//making necessar calculations 
+fd=log2(d);
+printf('A mod-128 conter should have %d flipflops\n', fc);
+printf(' A mod-32 conter should have %d flipflops\n', fd); //displaying the results 
+fe=log2(e);
+n=2^fe - 1 ;
+printf(' The largest decimal no that can be stored in a mod-64 counter is %d',n);
+
+
+
+
diff --git a/45/CH10/EX10.3/10_3.jpg b/45/CH10/EX10.3/10_3.jpg
new file mode 100755
index 000000000..efe9c9562
--- /dev/null
+++ b/45/CH10/EX10.3/10_3.jpg
diff --git a/45/CH10/EX10.3/example_10_3.sce b/45/CH10/EX10.3/example_10_3.sce
new file mode 100755
index 000000000..c27633205
--- /dev/null
+++ b/45/CH10/EX10.3/example_10_3.sce
@@ -0,0 +1,96 @@
+clc;
+clear;
+close;
+t=320; 
+p=1;
+while p<t*1 //making arrays for ploting 
+    if p==1 | modulo(p,t)==0 then
+        for k=1:t/2
+            qd(p+k)=0;
+        end
+         p=p+t/2;
+        else 
+            qd(p)=1;
+            p=p+1;
+        end
+end
+t=160;
+p=1;
+while p<t*2
+    if p==1 | modulo(p,t)==0 then
+        for k=1:t/2
+            qc(p+k)=0;
+        end
+         p=p+t/2;
+        else 
+          qc(p)=1;
+            p=p+1;
+        end
+end
+t=80;
+p=1;
+while p<t*4
+    if p==1 | modulo(p,t)==0 then
+        for k=1:t/2
+            qb(p+k)=0;
+        end
+         p=p+t/2;
+        else 
+          qb(p)=1;
+            p=p+1;
+        end
+end
+t=40;
+p=1;
+while p<t*8
+    if p==1 | modulo(p,t)==0 then
+        for k=1:t/2
+            qa(p+k)=0;
+        end
+         p=p+t/2;
+        else 
+          qa(p)=1;
+            p=p+1;
+        end
+end
+t=20;
+p=1;
+while p<t*16
+    if p==1 | modulo(p,t)==0 then
+        for k=1:t/2
+            clk(p+k)=0;
+        end
+         p=p+t/2;
+        else 
+          clk(p)=1;
+            p=p+1;
+        end
+end
+for i=320:350
+    clk(i)=0;
+    qa(i)=0;
+    qb(i)=0;
+    qc(i)=0;
+    qd(i)=0;
+end;
+y=[3 3]; //ploting the graphs  
+subplot(5,1,1)
+title('Clock')
+plot(clk)
+plot(y)
+subplot(5,1,2)
+title('Qa')
+plot(qa)
+plot(y)
+subplot(5,1,3)
+title('Qb')
+plot(qb)
+plot(y)
+subplot(5,1,4)
+title('Qc')
+plot(qc)
+plot(y)
+subplot(5,1,5)
+title('Qd')
+plot(qd)
+plot(y)
+\ No newline at end of file
diff --git a/45/CH10/EX10.5/example_10_5.sce b/45/CH10/EX10.5/example_10_5.sce
new file mode 100755
index 000000000..43a79cc5d
--- /dev/null
+++ b/45/CH10/EX10.5/example_10_5.sce
@@ -0,0 +1,4 @@
+//example 10.5
+clc;
+clear;
+printf('The correct expression is : (count-up clock)''(Qa)(Qb)(Qc)');
+\ No newline at end of file
diff --git a/45/CH10/EX10.6/example_10_6.sce b/45/CH10/EX10.6/example_10_6.sce
new file mode 100755
index 000000000..eaf9d8e75
--- /dev/null
+++ b/45/CH10/EX10.6/example_10_6.sce
@@ -0,0 +1,5 @@
+//example 10.6 
+
+clc;
+clear;
+printf('The correct logic expression is : (down-up)''(Qa)(Qb)(Qc)(enable)''');
+\ No newline at end of file
diff --git a/45/CH10/EX10.7/example_10_7.sce b/45/CH10/EX10.7/example_10_7.sce
new file mode 100755
index 000000000..45edcce0f
--- /dev/null
+++ b/45/CH10/EX10.7/example_10_7.sce
@@ -0,0 +1,16 @@
+//example 10.7
+clc;
+clear;
+mod = input("Enter the n value in your desired mod-n counter:");//taking the input
+m=mod;
+while 1
+ n= log2(mod); //checking whether the given number is a power of 2 
+ k=modulo(n,1);
+ if k==0  then
+     printf('The number of flip flops used in mod-%d counter are:',m); // if yes the print th outpu.
+     printf('%d',n);
+     return;
+ end
+ mod =mod+1;
+ end
+
diff --git a/45/CH10/EX10.8/example_10_8.sce b/45/CH10/EX10.8/example_10_8.sce
new file mode 100755
index 000000000..6da38fcd5
--- /dev/null
+++ b/45/CH10/EX10.8/example_10_8.sce
@@ -0,0 +1,12 @@
+//example 10.8 
+clc;
+clear;
+//ff = input('Enter the no of flip-flops :');
+ff=4; //given input 
+k=2^ff;
+if(k==2) then  //output display
+    printf('With given flipflop we can only count 2, we can have a modulus 2 counter'); 
+    else
+printf('With given number of flip-flops the counter will have a natural count of %d\n',k);
+printf('We can thus construct any counter that has a modulus between %d and 2',k )
+end;
+\ No newline at end of file
diff --git a/45/CH10/EX10.9/10_9.jpg b/45/CH10/EX10.9/10_9.jpg
new file mode 100755
index 000000000..6f3c02999
--- /dev/null
+++ b/45/CH10/EX10.9/10_9.jpg
diff --git a/45/CH10/EX10.9/example_10_9.sce b/45/CH10/EX10.9/example_10_9.sce
new file mode 100755
index 000000000..d59245d48
--- /dev/null
+++ b/45/CH10/EX10.9/example_10_9.sce
@@ -0,0 +1,86 @@
+//example 10.9
+
+clc;
+clear;
+close;
+c = [0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0]; //taking the values for a mod -6 counter 
+q = [0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0];
+a = [0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0];
+b = [0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 0];
+y1=q;
+y2=a;
+y3=b;
+y11p=1;
+y22p=1;
+y33p=1;
+y44p=1;
+cp=1;
+yf1p=1;
+for i=1:25    // making arrays to draw the output 
+    if y1(i)==1 then
+        for o=1:100
+        y11(y11p)=1;
+        y11p=y11p+1;
+        end
+    else
+        for o=1:100
+        y11(y11p)=0;
+        y11p=y11p+1;
+        end
+
+end
+if y2(i)==1 then
+        for o=1:100
+        y21(y22p)=1;
+        y22p=y22p+1;
+        end
+    else
+        for o=1:100
+        y21(y22p)=0;
+        y22p=y22p+1;
+        end
+
+end
+if y3(i)==1 then
+        for o=1:100
+        y31(y33p)=1;
+        y33p=y33p+1;
+        end
+    else
+        for o=1:100
+        y31(y33p)=0;
+        y33p=y33p+1;
+        end
+
+end
+if c(i)==1 then
+        for o=1:100
+        c1(cp)=1;
+        cp=cp+1;
+        end
+    else
+        for o=1:100
+        c1(cp)=0;
+        cp=cp+1;
+        end
+end
+
+end
+z=[2 2];   
+subplot(4,1,1); //ploting the out put 
+title('Timing Diagram');
+plot(c1);
+plot(z);
+ylabel('C');
+subplot(4,1,2);
+plot(y11);
+ylabel('Q');
+plot(z);
+subplot(4,1,3);
+plot(y21);
+ylabel('A');
+plot(z);
+subplot(4,1,4);
+plot(z);
+ylabel('B');
+plot(y31);