Solutions to the Exam in Digitalteknik, EIT020, 16 december 2011, kl On input x 0 x 1 x 2 = 010, the multiplexer

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1 Solutions to the Exam in Digitalteknik, EIT020, 6 december 20, kl 8-3 Problem (a) DNF = x x 2x 3 x x 2 x 3 x x 2 x 3 x x 2 x 3. (b) MDF = x x 2 x 3. (c) MDF = (x 2 x 3 )(x x 3 ). (d) We can use the result rom (c) since we included no don t care-terms. RSE = (x 2 x 3 )(x x 3 ) = (x 2 x 3 x 2 x 3 )((x ) x 3 (x )x 3 ) = x 2 x 3 x x 2 x 2 x 3 x x 2 x 3 (e) NAND ROM: x V x x DD 0 2 On input x 0 x x 2 = 00, the multiplexer outputs a logical zero on the ourth line and logical ones on all others. This means that the transistor on that line will be closed, giving ininite resistance between and ground (logical zero). Thereore, will take the value o V DD i.e. a logical one. DEC8 Problem 2 (a) Starting in the lower let corner, the sequence becomes s = 000. (b) Deriving the gcd as we get gcd( D 2 D 4 D 5 D 7 D 8, D 9 ) = D s(d) = D2 D 4 D 5 D 7 D 8 D 9 ( D D 4 D 7 )( D) = ( D D 2 D 3 D 4 D 5 D 6 D 7 D 8 )( D) D D 4 D 7 = D D 2 D 3 D 4 D 5 D 6 D 7 D 8 (c) and C(D) = D D 2 D 3 D 4 D 5 D 6 D 7 D 8. The LFSR is shown in Figure 2.. Use the BM algorithm

2 Solutions to the Exam in Digitalteknik, EIT020, 6 december 20, kl i s i δ T (D) C(D) L C p (D) l D 0 D 2 2 D D 2 D 2 D 3 4 D D 2 D 3 D D STOP Hence C(D) = D D 2 and L = 4. The LFSR is shown in Figure Problem 3 Figure 2.: LFSRs or Problem 2 a and b. I we ollow the idea rom Lab 3, we modiy a modulo 6 counter to become a module 9 counter. This can be achieved by a parallel load with 7 = 0 and connecting TC with PE to distinguish between loading and increasing. I the output o this modiied counter is either or 4, UPP will be become, otherwise 0. Similar is the modiied counter is 5, RESET will be become, otherwise 0. We get the ollowing table or the mapping rom state to the unctions UPP and RESET, and the corresponding Karnaugh map (or UPP) and unction expressions (clearly, RESET is given directly by TC output rom the counter). State UPP RESET UPP UPP = Q 2 Q Q 0 Q 2Q Q 0 RESET = TC

3 Solutions to the Exam in Digitalteknik, EIT020, 6 december 20, kl SR 6 VCC 0 CP P0 P P2 P3 CEP GND D6- SO6- SO6-7 E TC Q0 Q Q2 Q3 CET PE RESET UPP Problem 4 In this problem we have three inputs and three outputs. Inputs: From the card reader or entrance From the card reader or exit G rom an optical circuit at the bar Outputs: L in To the light or the entrance L ut To the light or the exit Bom To control the bar. Assume that and will not be at the same time. The problem can be solved directly drawing a graph, but it will require ive states and three inputs and three outputs. This results in six unctions with six variables, which is not very tempting. (Actually, at the exam there were also a solution with only our states that should work. However, this graph is not straight orward to construct). So, instead the problem can be split in parts. There are two natural ways to do this. Below are the corresponding block diagram and graphs shown. Alternative First take care o the bar and then the lights. The bar does not depend on the direction o the car, hence we can use a new variable, R =. The block diagram then becomes L ut L in R G M B Bom The graphs or the dierent parts becomes

4 Solutions to the Exam in Digitalteknik, EIT020, 6 december 20, kl M B R,G/Bom,Bom/L ut 0 0/ 00/,Bom/L in s 0 s 0-/0 0/ 0 0/ / t 0 t 0 s 2 0/ Alternative 2 In the second approach the two directions are separated and then the signal Bom is generated rom that. The main idea is that while a light is on the bar should be lited. The block diagram is L ut G Bom L in The graphs or the dierent parts becomes R,G/L ut in 0-/0,G/L Mut in 0/ 00/ 0/ 0/ s 0 s s 2 0 Then, or each o the versions, the graphs should be realized by choosing an appropriate state assignment, drawing Karnaugh maps, getting unction expressions and drawing the circuits. But this is more or less standard procedures and is not shown here. Problem 5 Let M be any inite state machine with n states. Let I = {0, } (a) Assume that there exists a inite state machine M with n states such that the output never becomes periodic. The state machine is deterministic, i.e. it behaves the same way every time it visits a state. Thereore, ater n s, we must have visited n distinct states. Now consider the nth input -symbol. By the pigeonhole principle, we must visit one among the already visited states. Contradiction. (b) We can output at most n s beore entering the periodic state. Example with n = 7, which produces the sequence s = [] / start s0 s s2 s3 s4 s5 s6 which produces the sequence []

5 / Solutions start s0 to the Exam in Digitalteknik, EIT020, 6 december 20, kl s s2 s3 s4 s5 / s6 start s0 s s2 s3 s4 s5 s6 (c) The maximum length o the period is the number o states n. Example which produces withthe n sequence = 7, which [] produces the sequence s = [000000] which produces the sequence [] start s0 s start s0 s / s2 / s2 s6 s3 s6 s5 s4 s3 (d) s5 s4 The only way we can insert the all-zero state (without parallel load) is by making an which produces the sequence [000000] non-linear transition between states and So i we are in state 00000, which produces the sequence [000000] we want to go to state and then to We get the ollowing igure, where (x, x 2, x 3, x 4, x 5 ) = x x 2x 3x 4x 5 where (x x 2 x 3 x 4 x 5 ) = x x 2x 3x 4x 5. where (x, x 2, x 3, x 4, x 5 ) = x x 2x 3x 4x 5