Question
Download Solution PDFLet k = 2n. A circuit is built by giving the output of an n-bit binary counter as input to an n-to-2n bit decoder. This circuit is equivalent to a
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFLet’s take small example: 2-bit binary counter and 2 × 4 decoder
2 bit binary counter output: 00, 01, 10, and 11. These outputs are inputs for decoder.
|
Decoder input |
Decoder output |
|
00 |
1000 (Line 1 will be activated and all other will be deactivated) |
|
01 |
0100 (Line 2 will be activated and all other will be deactivated) |
|
10 |
0010 (Line 3 will be activated and all other will be deactivated) |
|
11 |
0001 (Line 4 will be activated and all other will be deactivated) |
Decoder output is similar to 2n ring counter here n is 2 so 4 ring counter outputs:
|
State |
Q0 |
Q1 |
Q2 |
Q3 |
|
0 |
1 |
0 |
0 |
0 |
|
1 |
0 |
1 |
0 |
0 |
|
2 |
0 |
0 |
1 |
0 |
|
3 |
0 |
0 |
0 |
1 |
|
0 |
1 |
0 |
0 |
0 |
|
1 |
0 |
1 |
0 |
0 |
|
2 |
0 |
0 |
1 |
0 |
|
3 |
0 |
0 |
0 |
1 |
Johnson counter counts 2k states but ring counter counts k states.
Hence k bit ring counter is the correct answer.
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