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Part II Exercise Key
| 1. |
Various Answers |
| 2 a). |
10101100
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| 2.b) |
01100110
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| 2.c) |
10010011
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| 2.d) |
00110001
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| 3.a) |
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| 3.b) |
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| 3.c) |
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| 3.d) |
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| 4. |
The 8 bit sequense counts from “0” to “255.” this is a total of 256
digits. The pattern for binary counting should be very clear.
Each extra bit “switch” doubles the count. The following clearly
shows the layout of the 9 bit binary.
| _1_ _1_ _1_ _1_ _1_ _1_
_1_ _1_ _1_ 111111111 |
| 256+128+ 64 +32 +16 + 8
+ 4 + 2 + 1 = 511 |
The count from 0 to 511 is a total
of 512 digits. |
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