1

ENGG 1203 Tutorial

Sequential Logic (II) and Electrical Circuit (I)

22 Feb

Learning Objectives

Design a finite state machine

Analysis circuits through circuit laws (Ohm’s Law, KCL

and KVL)

News

HW1 (Feb 22, 2013, 11:55pm)

Ack.: ISU CprE 281x, HKU ELEC1008, MIT 6.111,

MIT 6.01

2

Quick Checking

NOT

always

true

Always

True

If , then

If , then

32

4 5

RRR R

=

60i =

2 3 4 5i i i i+ = +

2 6 3i i i+ =

( )4

1 0

2 4

Re V

R R=

+

60i =

( ) ( )32

2 4 3 5

RRR R R R

=+ +

A FSM design for a Vending machine

(Revisited) 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

3

Solution

We are designing a state machine which 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

4

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

5

Solution

6

Solution

7

S3511/110 S35

10/011 S00

01/001 S00

11/110

11/000

01/000

10/000

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)

Input

Output

Next state

00/000 S35

Solution

8

S3511/110 S35

10/011 S00

01/001 S00

11/110

11/000

01/000

10/000

e.g. 10/011: If we insert a dime (10), then the machine

should return one nickel and one product (011)

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)

00/000 S35

A Parking Ticket FSM

At Back Bay garage, Don and Larry are thinking of using

an automated parking ticket machine to control the

number of guest cars that a member can bring. The card

reader tells the controller whether the car is a member or

a guest car. Only one guest car is allowed per member

at a discount rate only when s/he follows out the member

at the exit (within the allotted time). The second guest

must pay the regular parking fees. You have been hired

to implement the control system for the machine which is

located at the exit.

Using your expertise on FSMs, design a FSM for the

control system.

9

Solution

Specifications

Signals from the card reader: MEMBER and GUEST

Signals from the toll booth: TOKEN (meaning one toke

received), EXP (time for discounted guest payment

has expired).

Signal to the gate: OPEN.

Fee: Members are free, Guest with a Member is 1

Token, Regular Guest is 2 Tokens.

10

11

Solution

The truth table that corresponds to the FSM

The state labels can be mapped to a three bit state

variable. All entries not entered below are illegal.

12

Solution

13 14

Rules Governing Currents and

Voltages Rule 1: Currents flow in loops

The same amount of current flows into the bulb (top path) and out of the bulb (bottom path)

Rule 2: Like the flow of water, the flow of electrical

current (charged particles) is incompressible

Kirchoff’s Current Law (KCL): the sum of the currents into a node is zero

Rule 3: Voltages accumulate in loops

Kirchoff’s Voltage Law (KVL): the sum of the voltages around a closed loop is zero

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Parallel/Series Combinations of

Resistance To simplify the circuit for analysis

1 2

1 2

s

s

v R i R i

v R i

R R R

= +

=

∴ = +

1 2

1 2

1 2

1 2

1

1 1

//

p

R RR

R RR R

R R

= =++

=

Series

Parallel

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Voltage/Current Divider

1 2

1

1 1

1 2

2

2 2

1 2

VI

R R

RV R I V

R R

RV R I V

R R

=+

= =+

= =+

( )1 2

1 2 2

1

1 1 1 2

1

2

1 2

//

//

V R R I

R R RVI I I

R R R R

RI I

R R

=

= = =+

=+

Voltage Divider

Current Divider

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Question: Potential Difference

Assume all resistors have the same resistance, R. Determine the voltage vAB.

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Solution

Determine VAB

We assign VG=0

2

1 2

4

3 4

5 2.5

3 1.5

A

B

RV V

R R

RV V

R R

= ⋅ =+

= − ⋅ = −+

( )2.5 1.5 4AB A B

V V V V= − = − − =

For the circuit in the figure, determine i1 to i5.

19

Question: Current Calculation using

Parallel/Series Combinations

20

Solution

21 // 2

3Ω Ω = Ω

2 44 //

3 7Ω Ω = Ω

4 253 //

7 7Ω Ω = Ω

40V

3Ω

4Ω 1Ω 2Ω

40V

3Ω

4/7Ω

(i)

(iii)

(ii)

(iv)

We apply:

V = IR

Series / Parallel Combinations

Current Divider

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Solution

1 1

2540 11.2

7

V IR

i i A

=

= ⇒ =( )

1 2 3

2 1

3

213 11.2 1.6

2 743

11.2 1.6 9.6

i i i

i i A

i A

= +

= = =

+

= − =

( )( )

( )( )

3 4 5

4

5

2 9.6 6.43

1 9.6 3.23

i i i

i A

i A

= +

= =

= =

(v)

(vi)

(vii)

40V

3Ω

4Ω 1Ω 2Ω

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Analyzing Circuits

Assign node voltage variables to every node except ground (whose voltage is arbitrarily taken as zero)

Assign component current variables to every component in the circuit

Write one constructive relation for each component in terms of the component current variable and the component voltage

Express KCL at each node except ground in terms of the component currents

Solve the resulting equations

Power = IV = I2R = V2/R

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R1 = 80Ω, R2 = 10Ω, R3 = 20Ω,R4 = 90Ω, R5 = 100Ω

Battery: V1 = 12V, V2 = 24V, V3 = 36V

Resistor: I1, I2, …, I5 = ? P1, P2, …, P5 = ?

Question: Circuit Analysis

Step 1, Step 2

24

Solution a

VN = 0

I1: M R5 V1 R1 B

I2: M V3 R3 R2 B

I4: M V2 R4 B

Step 1, Step 2

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Solution a

VM – VB = R5I1 + V1 + R1I1 I1 = (VM – VB – V1)/(R5 + R1) = (24 – VB)/180

Step 3

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Solution a

VN – VB = R2I2 + R3I2 I2 = (VN – VB)/(R2 + R3) = – VB/30

Step 3

27

Solution a

VM – VB = V2 + R4I4 I4 = (VM – VB – V2)/R4 = (12 – VB)/90

We get three relationships now (I1, I2, I4)

Step 3

28

Solution a

KCL of Node B: I1 + I4 + I2 = 0

(24 – VB)/180 + (12 – VB)/90 – VB/30 = 0

VB = 16/3 V Step 4, Step 5

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Solution a

I1 = (24 – VB)/180 = 14/135 A = 0.104A

I4 = (12 – VB)/90 = 2/27 A = 0.074A

I2 = – VB/30 = – 8/45 A = – 0.178A

Step 5

30

Solution a

P = I2R =

P1 = (0.104)2 80 = 0.86528W

P4 = (0.074)2 90 = 0.49284W = VR42 / R

(6.66V, 90Ω)

31

Quick Checking

NOT

always

true

Always

True

If , then

If , then

32

4 5

RRR R

=

60i =

2 3 4 5i i i i+ = +

2 6 3i i i+ =

( )4

1 0

2 4

Re V

R R=

+

60i =

( ) ( )32

2 4 3 5

RRR R R R

=+ +

32

Quick Checking

NOT

always

true

Always

True

If , then

If , then

32

4 5

RRR R

=

60i =

2 3 4 5i i i i+ = +

2 6 3i i i+ =

( )4

1 0

2 4

Re V

R R=

+

60i =

( ) ( )32

2 4 3 5

RRR R R R

=+ +

√

√

√

√

√

(Appendix) Boolean Algebra

33

(Appendix) Boolean Algebra

34

35

(Appendix) Question: Voltage

Calculation Find V2 using single loop analysis

Without simplifying the circuit

Simplifying the circuit

1 2 3 1 2 32 , 2 , 2 , 1 , 2 , 4

s s sV v V v V v R R R= = = = Ω = Ω = Ω

R1

Vs1

Vs3

Vs2

R3

-R2+

36

Solution

Choose loop current

Apply KVL

Replace V2 by R2I

Find V2

R1

Vs1

Vs3

Vs2

R3

-R2+

2 1 2 3 3 10

2

7

s s sV R I R I R I V V

I A

+ + + + + − =

⇒ = −

2 2

4

7V R I v= − =

37

Solution

Simplify the circuit with one voltage source

and one resistor

Req. = R1 + R2 + R3 = 7 ohm

Veq. = Vs1 + Vs2 + Vs3 = -2 + 2 + 2 = 2 V

I = Veq. / Req. = 2/7 A

V2 = 4/7 vVeq.

Req.