ENGG 1015 Tutorial - University of Hong Kong

ENGG 1015 Tutorial - University of Hong Kong

ENGG 1203 Tutorial Combinational Logic (II) and Sequential Logic (I) 8 Feb Learning Objectives News Apply Karnaugh map for logic simplification Design a finite state machine HW1 (Feb 22, 2013, 11:55pm) Ack.: HKU ELEC1008, ISU CprE 281x, PSU CMPEN270, Wikipedia 1 Simplification using K-map

Simplify the Boolean expression of the circuit Change each NAND gate in the circuit to a NOR gate, and simplify the Boolean expression of the circuit M N Q x 2 Solution (a) M 0 0 0 0 1 1 1 1 N 0 0 1 1 0

0 1 1 Q 0 1 0 1 0 1 0 1 x 0 0 0 1 0 1 0 1 From truth table to K-map NQ M

00 01 11 10 0 0 0 1 0 1 0 1 1 0 x MQ NQ M

N Q A B x C 3 Solution (b) M 0 0 0 0 1 1 1 1 N 0 0 1 1 0 0

1 1 Q 0 1 0 1 0 1 0 1 x 0 1 0 1 0 1 1 1 NQ M 00 01

11 10 0 0 1 1 0 1 0 1 1 1 x MN Q M N Q A

B x C 4 Finite State Machine (FSM) State transition diagram Truth table K-Map Circuit State Present state: before the register Next state: after the register State transition: during clock 2n states: n FFs 5 A simple Finite State Machine (FSM)

Turnstile Control access Depositing a token in a slot on the turnstile unlocks the arms, allowing a single customer to push through. After the customer passes through, the arms are locked again until another coin is inserted. Current State Locked Unlocked Input coin push coin push Next State Unlocked Locked Unlocked Locked Output

Release turnstile so customer can push through None None When customer has pushed through, lock turnstile 6 A simple FSM Current State Locked Unlocked Input Next State Output coin Release turnstile so customer can push through Unlocked push Locked None coin None

Unlocked push Locked When customer has pushed through, lock turnstile Specification FSM Arm: 0 State Arm: 1 Transition Transition condition 7 Steps in designing a state machine Draw a state transition diagram An initial state

Other states to keep track of various activities Transitions Generate a state transition table and a output table Write state transition table and output table in binary State assignment, i.e., the code used for each state Derive canonical sum-of-product expressions Draw the circuit 8 From state transition diagram to truth table Four states Two-bit state q: Present state q*: Next state z: Output

Condition/Output 9 From truth table to K-map A B DA DB DA DB 10 From K-map to circuit State register Logic for output Logic for state transition 11 A simple FSM design Design a state machine that will repeatedly display in binary values 1 (001), 3 (011), 5 (101), and 7 (111)

How many states we need? S0, S1, S2, S3 Simplified state transition diagram? 12 Solution Current Output Current state Output X Y L2 L1 L0 S0 (00)

0 0 0 0 1 1 0 1 1 0 1 0 1 1 1

1 1 1 (001) Output table S1 (01) 3 (011) 0 5 (101) L2 = XY'+XY = X S2 (10) 1 7 (111) 1 L1 = X'Y+XY = Y S3 (11) L0 = X'Y'+X'Y+XY'+XY = 1 State transition table Current state Next state X = X'Y+XY' S0 (00) S1 (01) Y = X'Y+XY' = Y' Current Next

X Y X Y 0 0 0 1 S1 (01) S2 (10) 0 1 1 0 S2 (10)

S3 (11) 1 0 1 1 S3 (11) S0 (00) 1 1 0 0 13 A complicated FSM design Vending Machine

Collect money, deliver product and change Vending machine may get three inputs Inputs are nickel (5c), dime (10c), and quarter (25c) Only one coin input at a time Product cost is 40c Does not accept more than 50c Returns 5c or 10c back Exact change appreciated 14 Solution We are designing a Mealy state machine (i.e., output depends on both current state and inputs). Suppose we ask the machine to directly return the coin if

it cannot accept an input coin. Input specification: I1 I2 Represent the coin inserted 00 - no coin (0 cent), 01 nickel (5 cents), 10 dime (10 cents), 11 quarter (25 cents) Output specification: C1C2P C1C2 represent the coin returned 00, 01, 10, 11 P indicates whether to deliver product 0, 1 15 Solution States: S1S2S3 Represent the money inside the machine now 3 bits are enough to encode the states

S00 (0 cents) 000 S05 (5 cents) 001 S10 010 S15 011 S20 100 S25 101 S30 110 S35 111 16 Solution 17 Solution 11/110 11/000 01/000 10/000

Next state Input Output S35 11/110 S35 10/011 S00 01/001 S00 00/000 S35 S35: Currently the machine has 35 cents e.g. 11/110 : If we insert a quarter (11), then the machine should return one quarter and zero product (110) 35c (35 cents inside the machine now) + 25c (insert 25 cents) = 35c (35 cents inside the machine in the next state) + 25c (return 25 cents) + 0c (return no product) 18 Solution S35

11/110 S35 10/011 S00 01/001 S00 00/000 S35 e.g. 10/011: If we insert a dime (10), then the machine should return one nickel and one product (011) 11/110 11/000 01/000 10/000 35c (35 cents inside the machine now) + 10c (insert 10 cents) = 0c (zero cent inside the machine in the next state) + 5c (return 5 cents) + 40c (return one product) e.g. 01/001: If we insert a nickel (01), then the machine should return zero coin and one product (001) 35c (35 cents inside the machine now) + 5c (insert 5 cents) = 0c (zero cent inside the machine in the next state) + 0c (return zero cent) + 40c (return one product) 19

(Appendix) Simplification using K-map Simplify the following Boolean expressions using Karnaugh map. i) A B A B ii) B BC ABC AB 20 Solution i) A/B 0 1 0 0

0 1 1 1 ii) A/BC A B A B A 00 01 11 10 0 0 0 1 1

1 1 1 1 1 B BC ABC AB B A 21 (Appendix) Counter Figure a) shows a complete four-bit parallel adder with registers and b) shows the signals used to add binary numbers from memory and store their sum in the accumulator. Suppose the numbers being added are 1001 and 0101. Also assume that Co=0. Describe what happen at t1, t2, t3, t4 and t5. 22

Solution 23 At time t1, is active low FF at the bottom will be cleared 24 At time t2, load is active high Set A numbers will be loaded into the upper register At time t3, transfer is active high Adder process between A3A2A1A0 = 0000 and B3B2B1B0 = 1001 The sum S3S2S1S0 = 1001 are transferred to register A on PGT due to this transfer pulse at t 3 25

At time t4, the load is active high, the set B numbers will be loaded into register B on PGT of LOAD pulse B3B2B1B0 = 0101 At time t5, A3A2A1A0 = 1001 and B3B2B1B0 = 0101, the adder produces S3S2S1S0 = 1110. This sum is transferred into register A when TRANSFER pulse occur at t5. 26 (Appendix) State changing in FSM Design a 2-bit counter with input x that can be A down counter when x = 0 (1110010011) A Johnson counter when x = 1 ( 0001111000) 27 Solution 28

(Appendix) A typical FSM FSM Truth table Circuit State register Logic for state transition Logic for output 29

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