{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 4 : Combinational And Sequential Logic Circuits"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example4_1,pg 488"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# find output and carry\n",
"# it is a half-adder circuit with the output 'a' and carry 'c' given by the boolean equations\n",
"\n",
"import math\n",
"#Variable declaration\n",
"b1 = 1 #input-1\n",
"b2 = 1 #input-2\n",
"\n",
"#Calculations\n",
"a=(b1 and (not b2))+((not b1) and b2) #sum\n",
"c=(b1 and b2) #carry\n",
"\n",
"#Result\n",
"print(\"sum:\")\n",
"print(\"a = %d\\n\"%a)\n",
"print(\"carry:\")\n",
"print(\"c = %.f\"%c) "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"sum:\n",
"a = 0\n",
"\n",
"carry:\n",
"c = 1\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example4_2,pg 489"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# find difference and borrow\n",
"#the circuit is that of a half subtractor\n",
"\n",
"import math\n",
"#Variable declaration\n",
"b1 = 1 #input-1, case-1\n",
"B1 = 0 #input-2 case-1\n",
"b2 = 1 #input-1, case 2\n",
"B2 = 1 #input-2, case 2\n",
"\n",
"\n",
"\n",
"#Calculations\n",
"#case-1\n",
"d1=(b1 and (not B1))+(B1 and (not b1)) #difference\n",
"r1=(b1 and (not B1)) #borrow\n",
"\n",
"#case-2\n",
"d2=(b2 and (not B2))+(B2 and (not b2))\n",
"r2=(b2 and (not B2))\n",
"\n",
"#Result\n",
"print(\"difference case-1:\")\n",
"print(\"d1 = %d\"%d1)\n",
"print(\"borrow case-1:\")\n",
"print(\"r1 = %.f\\n\"%r1)\n",
"print(\"difference case-2:\")\n",
"print(\"d2 = %.f\"%d2)\n",
"print(\"borrow case-2:\")\n",
"print(\"r2 = %.f\\n\"%r2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"difference case-1:\n",
"d1 = 1\n",
"borrow case-1:\n",
"r1 = 1\n",
"\n",
"difference case-2:\n",
"d2 = 0\n",
"borrow case-2:\n",
"r2 = 0\n",
"\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example4_3,pg 489"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# find final output\n",
"\n",
"import math \n",
"#Variable declaration\n",
"b=1 #input-1\n",
"B=0 #input-2\n",
"\n",
"#Calculations\n",
"y=(not((b or B)) or (b and B))\n",
"\n",
"#Result\n",
"print(\"final output:\")\n",
"print(\"y = %d\"%y)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"final output:\n",
"y = 0\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example4_4_a,pg 489"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# decoder output\n",
"\n",
"import math\n",
"#VAriable declaration\n",
"#initial conditions\n",
"b = 0 # input \n",
"Bi = 0 # initial value\n",
"Bf = 1 # final value\n",
"\n",
"#initial state of outputs\n",
"\n",
"y1i=not(b or Bi)\n",
"\n",
"y2i=not(b or (not Bi))\n",
"\n",
"y3i=not(Bi or (not(b)))\n",
"\n",
"y4i=not((not(Bi)) or (not(b)))\n",
"\n",
"#final state of outputs\n",
"\n",
"y1f=not(b or Bf)\n",
"\n",
"y2f=not(Bf or (not(b)))\n",
"\n",
"y3f=not(b or (not(Bf)))\n",
"\n",
"y4f=not((not(Bf)) or (not(b)))\n",
"\n",
"print(\"first: \")\n",
"print(\"y1 = %.f \"%y1i)\n",
"print(\"y2 = %.f \"%y2i)\n",
"print(\"y3 = %.f \"%y3i)\n",
"print(\"y4 = %.f\\n\"%y4i)\n",
"print(\"next: \")\n",
"print(\"y1 = %.f \"%y1f)\n",
"print(\"y2 = %.f \"%y2f)\n",
"print(\"y3 = %.f \"%y3f)\n",
"print(\"y4 = %.f\"%y4f)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"first: \n",
"y1 = 1 \n",
"y2 = 0 \n",
"y3 = 0 \n",
"y4 = 0\n",
"\n",
"next: \n",
"y1 = 0 \n",
"y2 = 0 \n",
"y3 = 1 \n",
"y4 = 0\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example4_4_b,pg 489"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# decoder output\n",
"\n",
"import math\n",
"#VAriable declaration\n",
"#initial conditions\n",
"b = 1 # input \n",
"Bi = 0 # initial value\n",
"Bf = 1 # final value\n",
"\n",
"#initial state of outputs\n",
"\n",
"y1i=not(b or Bi)\n",
"\n",
"y2i=not(Bi or (not(b)))\n",
"\n",
"y3i=not(b or (not Bi))\n",
"\n",
"y4i=not((not(Bi)) or (not(b)))\n",
"\n",
"#final state of outputs\n",
"\n",
"y1f=not(b or Bf)\n",
"\n",
"y2f=not(b or (not(Bf)))\n",
"\n",
"y3f=not(Bf or (not(b)))\n",
"\n",
"y4f=not((not(Bf)) or (not(b)))\n",
"\n",
"print(\"first: \")\n",
"print(\"y1=%.f \"%y1i)\n",
"print(\"y2=%.f \"%y2i)\n",
"print(\"y3=%.f \"%y3i)\n",
"print(\"y4=%.f\\n\"%y4i)\n",
"print(\"next: \")\n",
"print(\"y1=%.f \"%y1f)\n",
"print(\"y2=%.f \"%y2f)\n",
"print(\"y3=%.f \"%y3f)\n",
"print(\"y4=%.f\"%y4f)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"first: \n",
"y1=0 \n",
"y2=1 \n",
"y3=0 \n",
"y4=0\n",
"\n",
"next: \n",
"y1=0 \n",
"y2=0 \n",
"y3=0 \n",
"y4=1\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example4_5,pg 489"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# convert 8421 to 2421 code\n",
"\n",
"import math\n",
"#Variable declaration\n",
"#if A8,B8,C8,D8 is the binary in 8421 code, for 12 this would be 1100(DCBA)\n",
"#in 8421-code\n",
"A8 = 0\n",
"B8 = 0\n",
"C8 = 1\n",
"D8 = 1\n",
"\n",
"#Calculations\n",
"#in 2421-code\n",
"D2 = D8\n",
"C2 =(C8 or D8)\n",
"B2 =(B8 or D8)\n",
"A2=A8\n",
"\n",
"#Result\n",
"print(\"2421-code for 12(1100) is:\")\n",
"print(\"%d%d%d%d \"%(D2,C2,B2,A2))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"2421-code for 12(1100) is:\n",
"1110 \n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example4_6,pg 490"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Xcess 3 code\n",
"\n",
"import math\n",
"#Variable declaration\n",
"add = \"0011\" #binary-3 to be added\n",
"x = \"0010\" #binary-2\n",
"y = \"0100\" #binary-4\n",
"z = \"0111\" #binary-7\n",
"\n",
"#Calculations\n",
"x = int(x,2)\n",
"add = int(add,2)\n",
"XS31=x+add\n",
"XS31=bin(XS31)[2:]\n",
"y=int(y,2)\n",
"XS32=y+add\n",
"XS32=bin(XS32)[2:]\n",
"z=int(z,2)\n",
"XS33=z+add\n",
"XS33=bin(XS33)[2:]\n",
"\n",
"#Result\n",
"print(\"XS-3 for 2: %s\"%XS31)\n",
"print(\"XS-3 for 4: %s\"%XS32)\n",
"print(\"XS-3 for 7: %s\"%XS33)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"XS-3 for 2: 101\n",
"XS-3 for 4: 111\n",
"XS-3 for 7: 1010\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example4_7,pg 490"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# 8 to 1 MUX\n",
"\n",
"# one can see from relations of AND gate outputs in terms of address bits and input bit \n",
"# that there are possibilities depending on which input is low\n",
"# if I7=0, by making all address bits s1,s2,s3 as 1 one can have the conditions satisfied\n",
"\n",
"#Result\n",
"print(\"This requires all the outputs from AND gates should be low\")"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"This requires all the outputs from AND gates should be low\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example4_8,pg 490"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# truth table of RS flip flop\n",
"\n",
"import math\n",
"#Variable declaration\n",
"#enter binary 1-bit values only\n",
"\n",
"S = input(\"Enter value of S\\n\")\n",
"R = input(\"Enter value of R\\n\")\n",
"Qn = input(\"Enter previous value of Q\\n\")\n",
"En = input(\"Enter enable value\\n\")\n",
"print(\"RS flip-flop truth table:\")\n",
"if (En==0):\n",
" op=Qn\n",
" print(\"op = %d\"%op)\n",
"elif(( S==0) and (R==0)): \n",
" op=Qn\n",
" print(\"op = %d\"%op)\n",
"elif ((S==0) and (R==1)): \n",
" op=0\n",
" print(\"op = %d\"%op)\n",
"elif((S==1) and (R==0)): \n",
" op=1\n",
" print(\"op = %d\"%op)\n",
"elif ((S==1) and (R==1)): \n",
" print(\"output not determinable\")\n",
"\n",
"print(\"the relations are:\")\n",
"print(\"Qn=(R+Qn*)*\") #Q* = not(Q)\n",
"print(\"Qn*=(S+Qn)*\")"
],
"language": "python",
"metadata": {},
"outputs": [],
"prompt_number": "*"
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example4_9,pg 491"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# JK fiip flop\n",
"\n",
"# Q=(Q*+Q.K)* and Q*=(Q+Q*.J)*\n",
"# with J=K=0 Q=Q and Q*=Q*\n",
"# Q* is not(Q)\n",
"\n",
"#Result\n",
"print(\"operational equations:\\n\")\n",
"print(\"Q=(Q*+Q.K)* and Q*=(Q+Q*.J)*\\n\")\n",
"print(\"holds good where Q and Q* should be given appropriate values\")"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"operational equations:\n",
"\n",
"Q=(Q*+Q.K)* and Q*=(Q+Q*.J)*\n",
"\n",
"holds good where Q and Q* should be given appropriate values\n"
]
}
],
"prompt_number": 26
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example4_10,pg 491"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# 3 bit binary counter\n",
"\n",
"print(\"It remains positive from the falling edge of pulse-2, then falling edge of 4th, 6th and 8th pulses and so on....\")"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"It remains positive from the falling edge of pulse-2, then falling edge of 4th, 6th and 8th pulses and so on....\n"
]
}
],
"prompt_number": 27
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example4_11,pg 491"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# 6 modulo counter\n",
"\n",
"print(\"All modulo counters are basically scalars.A 6-modulo counter is a 6-scaler\")\n",
"print(\"so that after 6-input pulses the content of counter becomes 000(reset)\")"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"All modulo counters are basically scalars.A 6-modulo counter is a 6-scaler\n",
"so that after 6-input pulses the content of counter becomes 000(reset)\n"
]
}
],
"prompt_number": 30
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example4_12,pg 491"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# 3 bit 5 modulo counter \n",
"\n",
"print(\"Normal count would be 2^3=8, while 5-modulo counter would limit it to 5, so that illegitimate states are 8-5=3\")"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Normal count would be 2^3=8, while 5-modulo counter would limit it to 5, so that illegitimate states are 8-5=3\n"
]
}
],
"prompt_number": 32
}
],
"metadata": {}
}
]
}