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wbishop256

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It seems I put my question in the wrong section, so I'm reposting it here:

Hello!

Here's a question from the Pratice Exam I'd like some help on:

PE1.pngPE2.png

The correct answer is C. I understand that only CS4 and A15-13 matter for the answer... I just don't understand how. Could someone please walk me through this problem? Explain it to me like I'm idiot?

Thanks.
 
List out the address range spaces 15 to 0, then start putting a 1 or 0 as needed for each space. For CS4 to go low, A15 = 1, A14 = 0, A13 = 0, so those are defined as such. This leaves A12 to A0 as don't care bits. When coupled with 1 0 0 for bits 15-12, if A12 through A0 are all 0 to find the bottom end of the range, you get 1000/0000/0000/0000, which is address 8000H. When coupled with 1 0 0 for bits 15-12, if A12 through A0 are all 1 to find the upper end of the range, you get 1001/1111/1111/1111, which is address 9FFFH.
 
OOOOH! It's in groups of four! I figured it had to be 8 from 1000, but I didn't know where the fourth bit came from!

Where does the last letter in the chain, "H" come from?
 
The H is a way of indicating each group of four digits is in hexadecimal.
 
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