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If there are $m$ input lines and $n$ output lines for a decoder that is used to uniquely address a byte addressable $1$ KB RAM, then the minimum value of $m+n$ is ________ .
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23 votes
23 votes
Given that we need to address every byte of a $1\; KB$ RAM.
Therefore $1 K$ addresses are needed.

By using decoder, output lines should be $n=1\; K=1024.$
This means we should have number of input lines, $m = \log_2 1024 =10.$

Thus, $m+n = 10+1024 = 1034.$
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20 votes
20 votes

So when decoder has n input then it has $2^{n}$ ouput.

So, for n=10 it generate $2^{10} = 1024$ outputs.

So, n = 10 and m= outputs = 1024 

Answer n + m = 10 + 1024 = 1034

3 votes
3 votes

For  m input lines, the decoder can have maximum 2^m output lines

Output lines notated as n, so n = 2^m. 

And each output line uniquely identifying a cell of memory, so n = #memory cells.

memory capacity is given as 1KB, since it is Byte addressable memory, we can bifurcate the capacity as, 1KB = 1K × 1B.

So, we get #cells = 1K = n = 2^m

=> m = log(1024) = 10 (on base 2)

So, we got m & n  as,

m=10 & n=1024 

m+n=1034.

1 votes
1 votes

IN THIS QUESTION, for n=10 it generate 2^10=1024outputs.

So, n = 10 and m= 1024 

HERE M= output

so we give 10 input and we get 1024 s output.

Answer—( n + m ) =( 10 + 1024 )

= 1034

Answer:

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