Fundamental Hardware Elements of Computers: Building circuits

A common question in the exam is to be given some boolean algebra and be asked to express it as logic gates. Let's take a look at an addition and subtraction example that you should be familiar with:

$9-(7+1)$

First we are going to deal with the inner-most brackets

$(7+1)=8$

Finally we combine this answer with the $9-$

$=9-(8)=1$

It will work exactly in the same way for boolean algebra, but instead of using numbers to store our results, we'll use logic gates:

Example: Building circuits $C.(A+B)$ As with any equation, we are going to deal with the inner-most brackets first$(A+B)$, then combine this answer with the $C.$

Exercise: Building circuits Draw the circuit diagrams for the following. (Remember: do we deal with AND or OR first?): $A.B+C$ Answer: $({\overline {A}}+{\overline {B}}).C$ Answer: ${\overline {(A+B)}}.C$ Answer: $(A.{\overline {B}})\oplus (B.C)$ Answer: ${\overline {(A.B).(C+{\overline {D}})}}$ Answer:

A common question in the exam is to give you a description of a system. You'll then be asked to create a boolean statement from this description, and finally build a logic gate circuit to show this system:

Example: Building circuits Using boolean algebra describe the following scenario: “ A car alarm is set off if a window is broken or if it senses something moving inside car, and the car is not being towed, or the engine is not on. ” Where: A = being towed, B = window broken, C = engine on, D = senses movement Before you rush into answering a question like this, let's try and break it down into its components. The questioner will often be trying to trick you. The two occasions that the alarm will sound are: $B+D$ but there is a caveat, the alarm will sound if either of these are true AND two things are also true, namely the engine is NOT on, and the car is NOT being towed: ${\overline {A}}+{\overline {C}}$ Combining both we get (remember the brackets!): $(B+D).({\overline {A}}+{\overline {C}})$ The next step is to create a diagram out of this:

Exercise: Building circuits A security system allows people of two different clearance levels access to a building. Either they have low privileges and they have a card and they are not carrying a mobile. Alternatively they have a key and are allowed to carry a mobile. The inputs available are: A = carrying a card B = carrying a mobile phone C = carrying a key Write down the boolean equation to express this: Answer: $(A.{\overline {B}})+(C)$ If you wrote: $(A.{\overline {B}})+(C.B)$ You'd be wrong! The reason being the text says: Alternatively they have a key and are allowed to carry a mobile. This doesn't mean $C.B$, it means they can carry a mobile, or they can choose not to: $C.(B+{\overline {B}})$, which simplifies to: $C.(B+{\overline {B}})=C.(1)=C$. Draw the logic gate diagram to solve this: Answer: