Gentle Introduction to Digital Design using Logisim
Dive into the basics of digital design with Logisim, covering topics such as the basic gates (AND, OR, NOT), solving problems using logic gates, accessing, downloading, and using Logisim, and exploring more complex scenarios. Learn how to apply these concepts practically and enhance your understanding of digital design principles.
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Presentation Transcript
A Very Gentle Introduction to Digital Design and Logisim
What well see: 1. The Basic Gates: AND, OR, NOT 2. A Simple Problem: (p q) r 3. Using Logisim a. Accessing and Downloading b. Using the Menu 4. A More Difficult Problem: (r (p q)) (r p)) 5. (If Time) A 7-Segment Digital Display
How to remember those: AN O R N or N T T
A Simple Problem: Our Security Light A security light in our house is to be lit when it is 1. between 8 PM and 11 PM or 2. there is a serious storm Let p q r represent The time is after 8 PM represent The time is before 11 PM represent There is a serious storm then (p q) r represents the times when the security is to be lit.
A Simple Problem: (pq)r p q (p q) r r
Now Lets 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 tutorial under Help.)
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
A 7-segment display 1 3 6 7
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
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 q r
0 will be 0 0 0: 4 will be 1 0 0: 5 will be 1 0 1: 6 will be 1 1 0: think of this as p q r think of this as p q r think of this as p q r think of this as p q r p q r So the final expression is: ( p q r) (p q r) (p q r ) (p q r )
(pq r)(p q r) (p q r ) (p q r ) p q r
