Example K-map simplification
•
•
•
Let’s consider simplifying f(x,y,z) = xy + y’z + xz.
First, you should convert the expression into a sum of minterms form, if
it’s not already.
– The easiest way to do this is to make a truth table for the function,
and then read off the minterms.
– You can either write out the literals or use the minterm shorthand.
Here is the truth table and sum of minterms for our example:
x
y
z
f(x,y,z)
0
0
0
0
0
0
1
1
0
1
0
1
0
1
0
0
1
1
1
1
0
0
1
1
0
1
0
1
0
1
1
1
June 17, 2002
f(x,y,z) = x’y’z + xy’z + xyz’ + xyz
= m1 + m5 + m6 + m7
Basic circuit analysis and design
1
Unsimplifying expressions
•
You can also convert the expression to a sum of minterms with Boolean
algebra.
– Apply the distributive law in reverse to add in missing variables.
– Very few people actually do this, but it’s occasionally useful.
xy + y’z + xz = (xy  1) + (y’z  1) + (xz  1)
= (xy  (z’ + z)) + (y’z  (x’ + x)) + (xz  (y’ + y))
= (xyz’ + xyz) + (x’y’z + xy’z) + (xy’z + xyz)
= xyz’ + xyz + x’y’z + xy’z
•
In both cases, we’re actually “unsimplifying” our example expression.
– The resulting expression is larger than the original one!
– But having all the individual minterms makes it easy to combine
them together with the K-map.
June 17, 2002
Basic circuit analysis and design
2
Example: Simplify m0+m2+m5+m8+m10+m13
•
The expression is already a sum of minterms, so here’s the K-map:
Y
W
1
0
0
1
0
1
1
0
0
0
0
0
Y
1
0
0
1
X
W
m0
m4
m12
m8
m1
m5
m13
m9
m2
m6
m14
m10
X
Z
Z
•
m3
m7
m15
m11
We can make the following groups, resulting in the MSP x’z’ + xy’z.
Y
W
1
0
0
1
0
1
1
0
0
0
0
0
Y
1
0
0
1
X
W
w’x’y’z’
w’xy’z’
wxy’z’
wx’y’z’
w’x’y’z
w’xy’z
wxy’z
wx’y’z
Z
June 17, 2002
w’x’yz
w’xyz
wxyz
wx’yz
w’x’yz’
w’xyz’
X
wxyz’
wx’yz’
Z
Basic circuit analysis and design
3
K-maps can be tricky!
•
There may not necessarily be a unique MSP. The K-map below yields two
valid and equivalent MSPs, because there are two possible ways to
include minterm m7.
Y
0
0
X
1
1
0
1
1
1
Z
Y
X
•
0
0
1
1
0
1
Y
1
1
X
0
0
1
1
0
1
Z
Z
y’z + yz’ + xy
y’z + yz’ + xz
1
1
Remember that overlapping groups is possible, as shown above.
June 17, 2002
Basic circuit analysis and design
4
Prime implicants
•
•
•
•
The challenge in using K-maps is selecting the right groups. If you don’t
minimize the number of groups and maximize the size of each group:
– Your resulting expression will still be equivalent to the original one.
– But it won’t be a minimal sum of products.
What’s a good approach to finding an actual MSP?
First find all of the largest possible groupings of 1s.
– These are called the prime implicants.
– The final MSP will contain a subset of these prime implicants.
Here is an example Karnaugh map with prime implicants marked:
Y
W
1
1
0
0
1
1
1
0
0
0
1
1
0
0
0
1
X
Z
June 17, 2002
Basic circuit analysis and design
5
Essential prime implicants
Y
W
1
1
0
0
1
1
1
0
0
0
1
1
0
0
0
1
X
Z
•
•
•
If any group contains a minterm that is not also covered by another
overlapping group, then that is an essential prime implicant.
Essential prime implicants must appear in the MSP, since they contain
minterms that no other terms include.
Our example has just two essential prime implicants:
– The red group (w’y’) is essential, because of m0, m1 and m4.
– The green group (wx’y) is essential, because of m10.
June 17, 2002
Basic circuit analysis and design
6
Covering the other minterms
Y
W
1
1
0
0
1
1
1
0
0
0
1
1
0
0
0
1
X
Z
•
•
•
Finally pick as few other prime implicants as necessary to ensure that
all the minterms are covered.
After choosing the red and green rectangles in our example, there are
just two minterms left to be covered, m13 and m15.
– These are both included in the blue prime implicant, wxz.
– The resulting MSP is w’y’ + wxz + wx’y.
The black and yellow groups are not needed, since all the minterms are
covered by the other three groups.
June 17, 2002
Basic circuit analysis and design
7
Practice K-map 2
•
Simplify for the following K-map:
Y
W
0
1
1
0
0
0
1
0
1
1
1
1
0
1
X
1
0
Z
June 17, 2002
Basic circuit analysis and design
8
Solutions for practice K-map 2
•
Simplify for the following K-map:
Y
W
0
1
1
0
0
0
1
0
1
1
1
1
0
1
X
1
0
Z
All prime implicants are circled.
Essential prime implicants are xz’, wx and yz.
The MSP is xz’ + wx + yz.
(Including the group xy would be redundant.)
June 17, 2002
Basic circuit analysis and design
9
I don’t care!
•
•
•
You don’t always need all 2n input combinations in an n-variable function.
– If you can guarantee that certain input combinations never occur.
– If some outputs aren’t used in the rest of the circuit.
We mark don’t-care outputs in truth tables and K-maps with Xs.
x
y
z
f(x,y,z)
0
0
0
0
0
0
1
1
0
1
0
1
0
1
X
0
1
1
1
1
0
0
1
1
0
1
0
1
0
1
X
1
Within a K-map, each X can be considered as either 0 or 1. You should
pick the interpretation that allows for the most simplification.
June 17, 2002
Basic circuit analysis and design
10
Example: Seven Segment Display
Input: digit encoded as 4 bits: ABCD
a
f
e
g
d
Table for e
b
Assumption: Input
represents a legal
digit (0-9)
c
CD
AB
00
01
11
10
00
1
0
0
1
01
0
0
0
1
11
X
X
X
X
10
1
0
X
X
CD’ + B’D’
June 17, 2002
Basic circuit analysis and design
A
B
C
D
e
0
0
0
0
0
1
1
0
0
0
1
0
2
0
0
1
0
1
3
0
0
1
1
0
4
0
1
0
0
0
5
0
1
0
1
0
6
0
1
1
0
1
7
0
1
1
1
0
8
1
0
0
0
1
9
1
0
0
1
0
X
X
X
X
X
X
X
X
X
X
X
X
11
Example: Seven Segment Display
a
f
Table for a
b
g
e
d
c
CD
AB
00
00
1
01
01
11
10
1
1
1
1
1
11
X
X
X
X
10
1
1
X
X
A + C + BD + B’D’
June 17, 2002
Basic circuit analysis and design
A
B
C
D
a
0
0
0
0
0
1
1
0
0
0
1
0
2
0
0
1
0
1
3
0
0
1
1
1
4
0
1
0
0
0
5
0
1
0
1
1
6
0
1
1
0
1
7
0
1
1
1
1
8
1
0
0
0
1
9
1
0
0
1
1
X
X
X
X
X
X
X
X
X
X
X
X
12
Practice K-map 3
•
Find a MSP for
f(w,x,y,z) = m(0,2,4,5,8,14,15), d(w,x,y,z) = m(7,10,13)
This notation means that input combinations wxyz = 0111, 1010 and 1101
(corresponding to minterms m7, m10 and m13) are unused.
Y
W
1
1
0
1
0
1
x
0
0
x
1
0
1
0
X
1
x
Z
June 17, 2002
Basic circuit analysis and design
13
Solutions for practice K-map 3
•
Find a MSP for:
f(w,x,y,z) = m(0,2,4,5,8,14,15), d(w,x,y,z) = m(7,10,13)
W
1
1
0
1
0
1
x
0
Z
Y
0
x
1
0
1
0 X
1
x
All prime implicants are circled. We can treat X’s as 1s if we want, so
the red group includes two X’s, and the light blue group includes one X.
The only essential prime implicant is x’z’. The red group is not essential
because the minterms in it also appear in other groups.
The MSP is x’z’ + wxy + w’xy’. It turns out the red group is redundant;
we can cover all of the minterms in the map without it.
June 17, 2002
Basic circuit analysis and design
14
Summary
•
•
K-maps are an alternative to algebra for simplifying expressions.
– The result is a minimal sum of products, which leads to a minimal
two-level circuit.
– It’s easy to handle don’t-care conditions.
– K-maps are really only good for manual simplification of small
expressions... but that’s good enough for CS231!
Things to keep in mind:
– Remember the correct order of minterms on the K-map.
– When grouping, you can wrap around all sides of the K-map, and your
groups can overlap.
– Make as few rectangles as possible, but make each of them as large
as possible. This leads to fewer, but simpler, product terms.
– There may be more than one valid solution.
June 17, 2002
Basic circuit analysis and design
15
Basic circuit analysis and design
•
•
W have learned all the prerequisite material:
– Truth tables and Boolean expressions describe functions.
– Expressions can be converted into hardware circuits.
– Boolean algebra and K-maps help simplify expressions and circuits.
Now, let us put all of these foundations to good use, to analyze and
design some larger circuits.
June 17, 2002
©2000-2002 Howard Huang
16
Circuit analysis
•
•
Circuit analysis involves figuring out what some circuit does.
– Every circuit computes some function, which can be described with
Boolean expressions or truth tables.
– So, the goal is to find an expression or truth table for the circuit.
The first thing to do is figure out what the inputs and outputs of the
overall circuit are.
– This step is often overlooked!
– The example circuit here has three inputs x, y, z and one output f.
June 17, 2002
Basic circuit analysis and design
17
Write algebraic expressions...
•
•
Next, write expressions for the outputs of each individual gate, based
on that gate’s inputs.
– Start from the inputs and work towards the outputs.
– It might help to do some algebraic simplification along the way.
Here is the example again.
– We did a little simplification for the top AND gate.
– You can see the circuit computes f(x,y,z) = xz + y’z + x’yz’
June 17, 2002
Basic circuit analysis and design
18
...or make a truth table
•
•
It’s also possible to find a truth table directly from the circuit.
Once you know the number of inputs and outputs, list all the possible
input combinations in your truth table.
– A circuit with n inputs should have a truth table with 2n rows.
– Our example has three inputs, so the truth table will have 23 = 8
rows. All the possible input combinations are shown.
June 17, 2002
Basic circuit analysis and design
x
y
z
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
f
19
Simulating the circuit
•
•
Then you can simulate the circuit, either by hand or with a program like
LogicWorks, to find the output for each possible combination of inputs.
For example, when xyz = 101, the gate outputs would be as shown below.
– Use truth tables for AND, OR and NOT to find the gate outputs.
– For the final output, we find that f(1,0,1) = 1.
1
0
1
1
1
1
1
0
0
June 17, 2002
0
Basic circuit analysis and design
x
y
z
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
f
1
20
Finishing the truth table
•
•
Doing the same thing for all the other input combinations yields the
complete truth table.
This is simple, but tedious.
June 17, 2002
Basic circuit analysis and design
x
y
z
f
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
0
1
1
0
0
1
0
1
21
Expressions and truth tables
•
•
Remember that if you already have a Boolean expression, you can use
that to easily make a truth table.
For example, since we already found that the circuit computes the
function f(x,y,z) = xz + y’z + x’yz’, we can use that to fill in a table:
– We show intermediate columns for the terms xz, y’z and x’yz’.
– Then, f is obtained by just OR’ing the intermediate columns.
June 17, 2002
x
y
z
xz
y’z
x’yz’
f
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
0
0
0
0
0
1
0
1
0
1
0
0
0
1
0
0
0
0
1
0
0
0
0
0
0
1
1
0
0
1
0
1
Basic circuit analysis and design
22
Truth tables and expressions
•
•
•
The opposite is also true: it’s easy to come up with an expression if you
already have a truth table.
We saw that you can quickly convert a truth table into a sum of
minterms expression. The minterms correspond to the truth table rows
where the output is 1.
x
y
z
f
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
0
1
1
0
0
1
0
1
f(x,y,z) = x’y’z + x’yz’ + xy’z + xyz
= m1 + m 2 + m5 + m7
You can then simplify this sum of minterms if desired—using a K-map,
for example.
June 17, 2002
Basic circuit analysis and design
23
Circuit analysis summary
•
•
After finding the circuit inputs and outputs, you can come up with
either an expression or a truth table to describe what the circuit does.
You can easily convert between expressions and truth tables.
Find the circuit’s
inputs and outputs
Find a Boolean
expression
for the circuit
June 17, 2002
Find a truth table
for the circuit
Basic circuit analysis and design
24
Basic circuit design
•
•
The goal of circuit design is to build hardware that computes some
given function.
The basic idea is to write the function as a Boolean expression, and
then convert that to a circuit.
Step 1:
Figure out how many inputs and outputs you have.
Step 2:
Make sure you have a description of the function, either as a
truth table or a Boolean expression.
Step 3:
Convert this into a simplified Boolean expression. (For this course,
we’ll expect you to find MSPs, unless otherwise stated.)
Step 4:
Build the circuit based on your simplified expression.
June 17, 2002
Basic circuit analysis and design
25
Design example: Comparing 2-bit numbers
•
•
Let’s design a circuit that compares two 2-bit numbers, A and B. The
circuit should have three outputs:
– G (“Greater”) should be 1 only when A > B.
– E (“Equal”) should be 1 only when A = B.
– L (“Lesser”) should be 1 only when A < B.
Make sure you understand the problem.
– Inputs A and B will be 00, 01, 10, or 11 (0, 1, 2 or 3 in decimal).
– For any inputs A and B, exactly one of the three outputs will be 1.
June 17, 2002
Basic circuit analysis and design
26
Step 1: How many inputs and outputs?
•
•
Two 2-bit numbers means a total of four inputs.
– We should name each of them.
– Let’s say the first number consists of digits A1 and A0 from left to
right, and the second number is B1 and B0.
The problem specifies three outputs: G, E and L.
•
•
Here is a block diagram that shows the inputs and outputs explicitly.
Now we just have to design the circuitry that goes into the box.
June 17, 2002
Basic circuit analysis and design
27
Step 2: Functional specification
•
•
•
•
For this problem, it’s probably easiest
to start with a truth table. This way,
we can explicitly show the relationship
(>, =, <) between inputs.
A four-input function has a sixteenrow truth table.
It’s usually clearest to put the truth
table rows in binary numeric order; in
this case, from 0000 to 1111 for A1,
A0, B1 and B0.
Example: 01 < 10, so the sixth row of
the truth table (corresponding to
inputs A=01 and B=10) shows that
output L=1, while G and E are both 0.
June 17, 2002
A1
A0
B1
B0
G
E
L
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
0
0
0
0
1
1
1
1
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
0
0
0
1
0
0
0
1
1
0
0
1
1
1
0
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
0
1
1
1
0
0
1
1
0
0
0
1
0
0
0
0
Basic circuit analysis and design
28
Step 3: Simplified Boolean expressions
•
Let’s use K-maps. There are three functions (each with the same inputs
A1 A0 B1 B0), so we need three K-maps.
B1
A1
0
1
1
1
0
0
1
1
0
0
0
0
B1
0
0
A0
1
0
B0
G(A1,A0,B1,B0) =
A1 A0 B0’ +
A0 B1’ B0’ +
A1 B1’
June 17, 2002
A1
1
0
0
0
0
1
0
0
0
0
1
0
B1
0
0
A0
0
1
B0
E(A1,A0,B1,B0) =
A1’ A0’ B1’ B0’ +
A1’ A0 B1’ B0 +
A1 A0 B1 B0 +
A1 A0’ B1 B0’
Basic circuit analysis and design
A1
0
0
0
0
1
0
0
0
1
1
0
1
1
1
A0
0
0
B0
L(A1,A0,B1,B0) =
A1’ A0’ B0 +
A0’ B1 B0 +
A1’ B1
29
Step 4: Drawing the circuits
G = A1 A0 B0’ + A0 B1’ B0’ + A1 B1’
E = A1’ A0’ B1’ B0’ + A1’ A0 B1’ B0 + A1 A0 B1 B0 + A1 A0’ B1 B0’
L = A1’ A0’ B0 + A0’ B1 B0 + A1’ B1
LogicWorks has gates
with NOTs attached
(small bubbles) for
clearer diagrams.
June 17, 2002
Basic circuit analysis and design
30
Testing this in LogicWorks
•
•
Where do the inputs come from? Binary switches, in LogicWorks
How do you view outputs? Use binary probes.
probe
switches
June 17, 2002
Basic circuit analysis and design
31
Example wrap-up
•
•
•
Data representations.
– We used three outputs, one for each possible scenario of the
numbers being greater, equal or less than each other.
– This is sometimes called a “one out of three” code.
K-map advantages and limitations.
– Our circuits are two-level implementations, which are relatively easy
to draw and follow.
– But, E(A1,A0,B1,B0) couldn’t be simplified at all via K-maps. Can you
do better using Boolean algebra?
Extensibility.
– We used a brute-force approach, listing all possible inputs and
outputs. This makes it difficult to extend our circuit to compare
three-bit numbers, for instance.
– We’ll have a better solution after we talk about computer
arithmetic.
June 17, 2002
Basic circuit analysis and design
32
Summary
•
•
•
•
Functions can be represented with expressions,
truth tables or circuits. These are all equivalent,
and we can arbitrarily transform between them.
Circuit analysis involves finding an expression or
truth table from a given logic diagram.
Designing a circuit requires you to first find a
(simplified) Boolean expression for the function
you want to compute. You can then convert the
expression into a circuit.
Next time we’ll talk about some building blocks
for making larger combinational circuits, and the
role of abstraction in designing large systems.
June 17, 2002
Basic circuit analysis and design
33