2011 midterm

2011 midterm - 1; Test #2 Section B ECE 1521 Winter 2011...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1; Test #2 Section B ECE 1521 Winter 2011 (6p) 1. Implement the following function f(a, b, 0, cl) using NOR gates only. Primary inputs are available in positive and negative polarity. f(a, b; c, d) = 2m(1,3,5,7,10,11,15) Test #2 February 17, 201 l l 2. a) (7p) Complete the timing diagrams for the circuit below. Numbers inside the gates represent their delays. omzoioziuéuéutogodotbugge (b) (5p) Can this circuit have a static hazard on f when one input is switching? If yes, show the input values and indicate the switching input. If it can’t, explain, why. 1% Mpomc 79:0 '1 such/wag, eve/vol: em 0. ed‘H’UC (Ci; 9) =(oio) 90 390(1)!) Or (cue): 0,0 90 eo(o,o) ‘3.»an Pmtcooo W flvxmwg, a.ng £73159 gem flu, W99, 9.9“va Dace-534‘ CIA—c, W SfiCo-moi doe/ym'fij Atkietw 3&9qu M)£{\La We 7Cng COM/34A fl“- M Ab ~\ #4») Qrm¢_ Test #2 February 17, 2011 2 3. The function F(a,b,c,d,e) = Zm(0,2,4,5,l0,12,13,24,25,26,28,29) has 8 prime impli- cants: Pl=bc’de’; P2=bcd’, P3Za’cd’; P4=a’b’d’e’; P5=a’b’c’e’, P6=a’c’de’, P7:abd’, P8=abc’e’. a. (4p) List minterms covered by each prime implicant: a 3.4010: 010104- HOIO : nod); P2, = - “0* = 0H00+ one; 4—):100 + «Hot ; t2+t$+2€+39 P3, 2 0‘ ‘0‘ 500‘“) +0010! +onoo+bltol 341+§Hu13 (3“: 00-00 3 @0000 1.0910030 +9 Pg: coo—o : 00°00 +00mto :04-1 P6: 0*010 =- Com {0 +0!O[O:;+lo “~0— = \\ 000 +uoo¢ +Iuoo+l1101339+zs+58+25 P1}: : “0—0 =1‘0‘O‘04wllOLOfi5-29‘plg P8 b. (4p) Build the prime implicat chart fr F(a,b,c,d,e): V ,1 P u e ' il i w 2 [31" i; [873 i 0:2 S llIKllllllllllzlll ?» cm is» ' lung-mammal: 7 _ -_n_-_-==5:§s=:g=s:==::5=:=- p w. Illlllllllllllllllli P5 C 0 1“ HIIIIIIIIIIIIIIEIIII ? a, .o) =lllllllllllll5llll -llllIllllllrri§l:l-II— P " f ‘ “ “-llll-llI-nnuzal§liilnl P2 (emu lllllllllllllllfilllll 0. (6p) Find a minimum cost sum-of—product implementation of F(a,b,e,d,e). (Hint: The cost of a prime implicant = #literals +l if #literals >l; if #literals =1 then Cost(Pl) =1 .) Pg :10me (’5 .9 0°41 Co. )2“. dmrpul PW aovwe‘. : P5 +\°~+ +i°5 P‘ Glome Pg ) 5’8 CO» 15L dawn-o! Mxflpvdr 2 P3 +— 19? {,fy +f’ O'r 2w Pwou’) New .- to: CPA+PS)CPS+6)CP,+& Keefe): wg-leplfi‘Pé 4—ngng3 Beg ‘ hex COYQC X’l +95 4- Psr+él d. (213) If any among the prime implicants of F are essential, list them F3 1 Test #2 March 1, 2011 3 4. Implement the function f(a,b,c,d) : Z m(0,9,10,11,12,13,14,15) (a) (6p) Using only an 8:1 multiplexer and inverters (0) (5p) What is the smallest number of 2:1 multiplexes in a multiplexer-inverter circuit sufficient to implement f? Show this realization. 4 C (2 n. (T‘Eflfllkjfli I Test #2 February 17, 2011 4 ...
View Full Document

Page1 / 4

2011 midterm - 1; Test #2 Section B ECE 1521 Winter 2011...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online