A Very Gentle Introduction to Digital Design and Logisim

15
A Very Gentle Introduction to Digital Design and Logisim

description

A Very Gentle Introduction to Digital Design and Logisim. What we’ll see:. The Basic Gates: AND, OR, NOT A Simple Problem: ( p q )r Using Logisim Accessing and Downloading Using the Menu A More Difficult Problem : (r  ( p  q ))  ( r  p )) (If Time) A 7-Segment Digital Display. - PowerPoint PPT Presentation

Transcript of A Very Gentle Introduction to Digital Design and Logisim

A Very Gentle Introduction toDigital Design and Logisim

What we’ll see:

1. The Basic Gates: AND, OR, NOT2. A Simple Problem: (pq)r3. Using Logisim

a. Accessing and Downloadingb. Using the Menu

4. A More Difficult Problem: (r(pq))(rp))5. (If Time) A 7-Segment Digital Display

The Basic Gates

How to remember those:

ANO R

N TN

T

or

A Simple Problem: Our Security Light

A security light in our house is to be lit when it is1. between 8 PM and 11 PM or2. there is a serious storm

Let p represent “The time is after 8 PM”q represent “The time is before 11 PM”r represent “There is a serious storm”

then (pq)r represents the times when the security is to be lit.

A Simple Problem: (pq)r

p

qr

(pq)r

Now Let’s Do It with Logisim

1. Download the code: go to http://www.cburch.com/logisim/ then click Download Logisim then click Logisim's SourceForge.net page then click Download

logism-win-2.7.1.exe

2. It should install itself.(3. … and there’s a very nice basic tutorialunder Help.)

(pq)r

A More Difficult Problem: (r(pq))(rp)

(This was actually one of your Quest problems.) We don’t have an IMPLIES gate so first let’s replace the implies using Conditional Disjunction.

(r (p q)) (r p)is the same as

(r (p q)) (r p)

Remember: antecedent consequent is the same as

antecedent consequent

(r (p q)) (r p))

A 7-segment display

1 63 7

Which numbers from 0 to 7 need the f segment lit?

0, 1, 2, 3, 4, 5, 6, 7

So we need to turn on

the f segment for the numbers 0, 4, 5, and 6

But the input will be arriving in three bit binary!

• 0 will be 0 0 0 • 4 will be 1 0 0• 5 will be 1 0 1 • 6 will be 1 1 0 Let’s call the leading bit p,

let’s call the middle bit q, and let’s call the trailing bit r

pq

r

• 0 will be 0 0 0: think of this as pq r

• 4 will be 1 0 0: think of this as p q r

• 5 will be 1 0 1: think of this as p q r

• 6 will be 1 1 0: think of this as p q r

So the final expression is: (pq r)(p q r) (p q r ) (p q r )

pq

r

(pq r)(p q r) (p q r ) (p q r )

p

r

q