University of California, Berkeley
EE 42/100
Spring 2012
Prof. A. Niknejad
Problem Set 8
Solutions
Please note that these are merely suggested solutions. Many of these problems can be
approached in different ways.
1. (a) In decimal, we have B5216 = 11 × 162 + 5 × 161 + 2 × 160 = 2898. In binary, this
becomes B5216 = 101101010010b . Clearly, 12 bits are required.
(b) In hexadecimal form, 1010110b = 5616 . In decimal, 5616 = 5 × 161 + 6 × 160 = 86.
(c) Using two’s complement, 34 = 0100010b and −57 = 1000111b . Adding them
together gives us 34 − 57 = −23, which in binary is 1101001b .
2. (a) Using a truth table, only the rows with two or three 1’s should result in F = 1.
A B
0 0
0 0
0 1
0 1
1 0
1 0
1 1
1 1
C
0
1
0
1
0
1
0
1
F
0
0
0
1
0
1
1
1
(b) F = ABC + ABC + ABC + ABC = AB + BC + AC
(c) Below is the logic circuit for the simpler expression (if you did it for an
equivalent but more complicated expression, that’s ok too).
A
B
F
C
3. To prove the identity using Boolean algebra, we can call the expression F and show
that the results are the same as the A input column.
A
0
0
0
0
1
1
1
1
B
0
0
1
1
0
0
1
1
C
0
1
0
1
0
1
0
1
F
0
0
0
0
1
1
1
1
Using Boolean algebra, we first combine the first term with the last term, as well as
the middle two terms:
ABC +ABC +AB C +ABC = AC(B +B)+AC(B +B) = AC +AC = A(C +C) = A
4. (a) To find a minimal expression, we can simplify using Boolean algebra:
F = ABC D + ABCD + ABCD + ABCD = ABD(C + C) + BCD(A + A) =
ABD + BCD
C
D
B
F
A
(b) One way to find a POS expression is to invert F and write the SOP expression:
F = A D + B + CD. Now invert the expression once more and apply
DeMorgan’s laws: F = (A + D)B(C + D).
5. Note that F is high if the first switch is open AND at least one of the two right
switches is open: F = A(B + C).
B
C
A
F
6. (a) To multiply by 2n in binary, simply shift the number to the left by n bits and add
in trailing 0s (just like how we do for multiplying by powers of 10 in decimal). To
divide, add a decimal point to the end of the number and shift the number to the
right by n bits. (b) For a circuit with n input variables, there are 2n different
n
combinations of inputs. But if we also count outputs, then we have 2(2 )
combinations. To see this, imagine that each of the 2n rows of inputs has a
corresponding output. All together then, the output is a binary string of length 2n .
2
How many possible realization of this bit string do we have? A bit string of length M
n
can take on 2M values, and now M = 2n , which gives us the desired result of 2(2 ) .
This grows very quickly. While for 2 inputs, it’s simply 16 functions, but for 10
inputs it’s a whopping number:
1797693134862315907729305190789024733617976978942306572734300811577326758055
0096313270847732240753602112011387987139335765878976881441662249284743063947
4124377767893424865485276302219601246094119453082952085005768838150682342462
8814739131105408272371633505106845862982399472459384797163048353563296242241
37216
3