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A Very Gentle Introduction to Digital Design and Logisim

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.

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A Very Gentle Introduction to Digital Design and Logisim

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  1. A Very Gentle Introduction toDigital Design and Logisim

  2. 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

  3. The Basic Gates

  4. How to remember those: AN O R N or N T T

  5. A Simple Problem: Our Security Light A security light in our house is to be lit when it is • between 8 PM and 11 PM or 2. 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 (pq)r represents the times when the security is to be lit.

  6. A Simple Problem: (pq)r p (pq)r q r

  7. Now Let’s Do It with Logisim • Download the code: go to http://www.cburch.com/logisim/ then click Download Logisim thenclickLogisim'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 tutorial under Help.)

  8. (pq)r

  9. A More Difficult Problem: (r(pq))(rp) (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

  10. (r  (p  q))  (r  p))

  11. A 7-segment display 6 3 1 7

  12. Which numbers from 0 to 7 need the f segment lit? So we need to turn on the f segment for the numbers 0, 4, 5, and 6 0, 1, 2, 3, 4, 5, 6, 7

  13. 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 p r q

  14. 0 will be 0 0 0: think of this as pq  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: (pq  r)(p q  r) (p  q  r ) (p q  r ) p r q

  15. (pq  r)(p  q  r) (p  q  r ) (p q  r ) p q r

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