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Unformatted text preview: 2L . EE 306 FALL loo? PROISLEM SET 1 Sownom of XOR gates can be. used ta) test A 50163
ace, an odd number of is. to problem Star m “HHS PVOMUY‘ Set /Hso refer NOT(A AND (B 0K0) ‘rL 3 . KNOWS) M33 C) OR A AL‘IEKNATIVE SOLU‘HON “NOT A) AND KNor 6) ANDC) OK (A ﬁrm? (NOT 8) RN]; 0101;»
OK (9 PxNID (NOT 3) RN29 Q) 0R (A AND B ANy (NO—r C3) 0K (A AND 3 AND a)
fc'qued by Jrauw'na Out
ﬂahéistor 9‘14? This a tr’uth “fable, CCYCul'i", {for Ha 6 Y ‘—' (A AND C) OK (NOTA
AND a) 0" CD) A 3‘(hput OR gate has {'he SQMQ Huck table . H, , AW V A:
w m’fﬁgiiﬁzmam / ﬁg???“ Threext'nPM AND gate
r<o\
5 rd
(«/4 Three, 'mPU‘t
0R Safel gate. Each row can be represented with one 3 input AND gate. The logical equation for
this gate level logic circuit is: A’B’C’+A’BC+AB’C+ABC’. 9. A) The output of the circuit when select line S is 0 is equal to the input A, that is for A=l,
output is l and for A=0 output is 0. B) If the S switches from O to l, the output depends on the last propagated value of A, i.e.
if the previous output was 0 the output is going to stay 0, and if the previous output was
1, the new output is going to be 1. c) This is a storage element, since it stores the value that A had in the moment when S
changed from 0 to l. 10. For implementation of a 4to—1 mux we need to understand what is the 4tol mux doing.
The idea is that with 2 selection bits we can distinguish 4 different inputs in the
multiplexor and by assigning the values to the selection bits we are building a
programmable switch that connects output to one of the 4 inputs. The short table for
representing this is: Sl S
O
l
O
l z
0 A
0 B
L c
1 1) If we look at the table more closely, we can see that selection bit 81 makes the choice
will one of the signals in the group (A,B) be on the output or a signal from group (C,D).
In each of this groups signal SO makes the precise decision which one will be used. This
is presented on the ﬁgure So, the output values are: SlSOABCDOUT SlSOABCDOUT
0000000 100000T0
0 0 0001i0 10 00010
0 0 00100 10 00101
0 0 001k10 10 0L0111
0001ﬁ000 1001000
0 0 01010 10 01010
0 0 0110+0 10 01101
0001110 1001111
0010001 1010000
00i1.0011 1010010
0010101 1010101
0010111 1010111
0011001 1011000
0011011l1011010
0011101 1011101 H _.‘~ _.‘_. _ ,_.‘_. ﬂ‘o l 1.
a) The value of X controls will we do A+B or A+C. If the input of X is 0 then we will
add A+B, if it is equal tol we will add A+C.
b) What we have is circuit that does only addition, if we want it to do subtraction we need
to convert our number B into a 2’5 complement negative number. The way we do it is by
taking the inputs for B, invert it, and put it back into C. What we are still missing is the 1
that we need to add to ﬂipped B to make it a 2’3 complement, however we can use X as
an input into the Cin bit of the circuit, so we actually added 1 if we are doing subtraction
but we add 0 if we are doing addition. The ﬁgure of the circuit is: 3‘ O 903 t >_ $1 ..
as ca. E x
 12. a) The propagation of mux is three gate delays since the longest path is inverting S bit,
doing 8’ AND A, and in the end we have an OR gate. b) The propagation delay for one full adder is 3 gate delays, so for 4 bit adder it is 12. c) we can reduce it to 3 gate delays by grouping the terms ((A and B) and (C and D) ) and
E. ._n
b) l :43;
F1
N W l
n O
O O
O
HO
HO 0'0
0‘0
9 h—A O
O
H O
elelolololol El c‘c H‘t—‘c clelololc‘ ﬁlo OlU
'—‘ O elelolo
Ololcle 0} o ﬁe} o ,_. O l O‘O t—t‘H .—
O >—‘ O‘H
_. _‘¢‘_ O
_. l l H a clH‘H l O‘OOOOH
HOG o»4 O l ,__t
HHH
Lid
J H0
°l~~ l r—‘h—ir—4
Ht—‘OOO ._. ._..
,__. l ._‘ l >4 H‘H r—i #H‘H H HHH‘M‘Q elole Olelolelo ._.t
,_;
_; b) All the combinations that give the output 1 have an odd number of Is and all the
combinations that give output 0 have an even number of Is. ...
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 Spring '07
 Ambler

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