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Unformatted text preview: ELEC 326: Homework 2 Kartik Mohanram Due in DH 3029 by 5pm on Friday, September 25th, 2009 Graded problems You may work together with others from the class to solve the problems on the homework, though each of
you has to submit solutions that are “individually” written up. Solutions that are “near” copies of each
other will not be awarded any credit. All solutions from past offerings are off limits under the honor code. 1. (From a past exam) This question is split into three parts: (a) What is the Boolean function that this circuit implements? What are its disadvantages? One quick way to derive the Boolean function is to apply all combinations of inputs (8 in this
case) to derive the truth table. A faster way to do this is to observe that the intermediate nodes
realize their own Boolean expressions, as annotated in the ﬁgure. In the same way that you use
logical expressions to determine what a netlist of gates implements, you can combine the logical
expressions to derive the Boolean function at the output. This is illustrated below. (b) Draw the optimum CMOS network for the Boolean function. Hint: My current solution uses 10
transistors with only one level of logic, so these are good numbers to shoot for. 4%.: aE+aa+ 7:35 2. (From a past exam) Use exactly ONE INVERTER and any number of AND and OR gates to imple—
ment the XOR function of three inputs a, b, and c. Note that you are given the inputs a, b, and c in
uncomplemented form only, i.e., (—1., b, and E, are not available. XOR(a, b, c) = abc+ ab5+ ab5+ abc. The ﬁrst term abc is easily generated using an AND gate, so what
remains is the generation of the remaining three terms that are symmetric. Consider the expression: g(a,b,c) = (ab+bc+ca)' = db+bé+dé Implementing g(a, b, c) consumes ONE INVERTER and we can generate the three terms required for
the XOR using three AND gates as: abé == g(a, b, c) a
Elba = g(a, b7 c)  b
Elba = g(a, b, c)  c 3. Convert the following Boolean functions into CMOS networks. You are given the inputs (1, b, and c
in uncomplemented form only, Le, a, b, and F: are not available and have to be generated if necessary.
Try to minimize the number of transistors and levels of logic that your solution requires. (b) f = a5 + 56: Again, recognize that f equals E(& + 5) and that this is (ab + c). So, the imple
mentation is straight—forward with just 6 transistors. (c) f = a5 + Etc: This function is inesting, since the absence of complements forc es you to
generate them locally using inverters. 35.: (ﬁlthy (mam4%) a 4. Implement the Boolean function shown in the Kmap below using as few xor gates as possible. 5. Problem 2.18 from the textbook See solution set posted on class website 6. (From a past exam) Give an example of a Boolean function which has more prime implicants than minterms. (Draw a picture, showing the primes and minterms, rather than giving a Boolean formula
for the function.) 7. Prove or show a counterexample to each of the following statements: (a) If the pairwise product of all prime implicants of f is 0, then f has a unique minimum expression.
Proof by induction: The case with 1 prime implicant is trivial. With two prime implicants whose
product (i.e., Boolean and) is 0, it is clear that each prime implicant covers at least one 1 and that
this ,1 is a distinguished 1—cell. (In fact, all ls covered by each prime implicant are distinguished
lcells.) This in turn means that each of the two prime implicants is an essential prime implicant,
and both are part of the SOP representation for f. This makes the SOP for f unique. For the general case, argue that each prime implicant covers is that are not covered by any of the
other prime implicants (if the Is were covered by other prime implicants, the pairwise product equals 0 property will be violated). Each prime implicant is essential, and hence f has a unique
minimum expression. (b) If a function f has a unique minimum sum—of—products expression, then f has a unique minimum
product~of—sums expression. This is false. Look at the Boolean function below. 8, (From a past exam) The onehot code on four inputs a, b, c, and d has exactly four valid input combi—
nations 1000, 0100, 0010, and 0001. (a) Derive the Kmap for an error detection circuit for the onehot code on four inputs, where only 03) singlebit, unidirectional 0 —> 1 errors can occur. In other words, your implementation must
detect (i.e., output 1) on all erroneous input combinations on a, b, c, and d (i) when a single bit
is in error and (ii) when these singlebit errors are unidirectional, i.e., they cause a 0 to become a
1. For example, 1100 is an erroneous input satisfying the singlebit, unidirectional O —> 1 error
condition. Warning: There are don’t cares in the K—map and they must be clearly identiﬁed in
the K—map for full credit. Derive optimal sumofproducts and product—ofsums expressions for your implementation. optima; $010 = (abmwawmbw wt) 0am P05 = (a+b+c)(a+b+a)(a+c+4)(b+cr—a)
1(a+c+4) (b+c+4) (c) You are given the inputs a, b, c, and d in uncomplemented form only, i.e., EL, l), J, and J are not
available and have to be generated if necessary. Draw the Optimal transistor—level schematic for
the CMOS implementation. Please refresh your memory of what a CMOS circuit is before you
answer this question. Hint: My current solution uses 18 transistors with two levels of logic. F: 01th +m+ bc+ m+m
: mm) + mm) + MLHJC
= (a+d)(b+c)+m+ Ibo
Attomatdgo W W» W W {23W 9% P05
am (Wain mam). mm 0 Practice problems (not graded)
9. Draw the optimal (minimum number of transistors) CMOS network for the Boolean function F: (ab+bc+cd+da) 10. Work out example (not problem) 2.10 from the textbook. 11. Work out example (not problem) 2.11 from the textbook. 12. There are 1024 (hmm...) bottles of wine, of which 1 bottle is adulterated. It takes a day for the
symptoms of consuming the adulterated wine to appear, and the only way to identify the adulterated
bottle is to have someone drink from it. If you have exactly one day to identify the adulterated bottle, what is the minimum number of volunteers that you will need to accomplish the task? (Hint: Try to
use binary numbers as identities.) 10 volunteers given by [log2(1024)] are required. The trick is to make a single volunteer consume
from more than a single bottle of wine in the following manner. Number the bottles 0 through 1023 using the 210 combinations of bits. Let the 10 volunteers represent
the 10 bits of a binaryencoded number from 0 to 1023. Each volunteer consumes wine only from
those bottles that have a bit value of 1 that corresponds to their position. Thus, each volunteer con
sumes wine from exactly half the bottles in the lot (think about why this is so). After a day, the bottle
that is the equivalent of the binary number constructed from the the volunteers who show symptoms
is the adulterated one. For example, if bottle 7 is adulterated, it is consumed by volunteers 1, 2, and 3 an noone else. Note that bottle 0 is untouched by all volunteers and so, if all volunteers exhibit no
symptoms, it is clearly the one that is adulterated. Now, if I had two days to discover the adulterated bottle, what is the minimum number of volunteers
that I need? 3 days? 4 days? 13. 14. When comparing the performance of gates implemented in different technologies or circuit styles, it is
important to not confuse the picture by including parameters such as local factors (i.e., capacitance),
fanin, and fanout. A uniform way of measuring gate delay so that technologies can be judged on an
equal footing is desirable. The defacto standard circuit for delay measurements is the ring oscillator,
which consists of an odd number of inverters connected in a circular chain. Due to the odd number
of inversiOns, the circuit does not have a stable operating point and oscillates. Determine the period,
i.e., time between two successive rising or falling transitions at the output of an inverter in the ring
oscillator as a function of the number of inverters n and the gate delay of an inverter A. What is the
frequency of the clock that can be tapped at the output (last inverter) of the ring oscillator? The period of oscillation is determined by the propagation time of a signal transition through the
complete chain, i.e., T=2xA><n (l) The factor 2 results from the observation that a full cycle (one period) requires both a lowtohigh and a
hightolow transition around the ring oscillator. The ring oscillator is a highly idealized circuit where
each gate has a fanin and fanout of exactly 1, and parasitics are minimal. Real circuits have larger
fanin and fanout values, and are also subject to larger parasitics. As a result, the achievable clock
frequency is 10—100 times slower than the frequency obtained from ring oscillator measurements. (Challenge problem from a past exam) Recent studies have indicated that a good diet should contain
adequate amount s of proteins (P), vitamins (V), fats (F), and cookies (C). An astronaut has to choose from a menu of ﬁve different preparations with the following FDAapproved nutritional information
labels. (a) Preparation 1 contains V and P,
(b) Preparation 2 contains V and F,
(c) Preparation 3 contains P and F,
(d) Preparation 4 contains V, and
(e) Preparation 5 contains C. Can the astronaut have a balanced diet with only two preparations? You can, of course, eyeball the
solution but there is no fun in that. What I would like to see is a Boolean formulation for this problem,
and a rational explanation for arriving at the solution. Hint: Think K—maps and covering minterms. This problem is a covering problem where preparations 1 through 5 are equivalent to prime implicants
and P, V, F, and C are equivalent to minterms. There is no Kmap, though you may envision a covering
table of the form shown here. Note that in this scenario, preparation 5 is equivalent to an essential prime implicant. To cover the other dietary requirements, you can go with preparations 1&2 or 1&3
or 2&3 or 3&4. ...
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 Boolean Algebra, Boolean function, Logical connective, Elba

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