Solution: The design in question is a combinatorial design with 32-bit input (given) and a 4-bit output as shown in figure 1 below. How the output is 4-bit is a bit tricky. For this, we have to understand the problem. We have to count the number of non-overlapping "101" sequences in 32-bit input. Thus, "10101" counts as only a single occurrence, and "101101" counts as two occurrences. So, the maximum number of such patterns will occur when "101101" is repeated, which comes out to be 10 in a 32-bit number. 10 can be represented by a 4-bit number.
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| Figure 1: Design representation |
One of the solutions, of course is to make a truth-table and then find a solution using logic equation solving. But the number of combinations possible here is huge and practically impossible to find a solution. So, we need to follow a modular approach here.
We can divide the problem into two parts, detecting the required pattern and then counting how many patterns actually were detected. We are introducing an intermediate 32-bit output, each bit (Nth bit) detecting if the pattern was found with Nth bit of input as the middle symbol of pattern. To detect non-overlapping "101" pattern, we need to look into 2 bits on each side, thereby making a combinational logic comprising of 7 bits. There will be special cases for terminal bits (here bit-0, bit-1, bit-30 and bit-31) where we know that there are less than 2 bits on one side. So, we need to have special logic for these bits.
The overall combinational logic will look like as shown in figure below. Int-N (Nth bit of intermediate output) is a resultant of (Bit-N-3 to Bit-N+3). The number of 1's in the intermediate output will tell how many patterns were detected, which, as discussed earlier, will be maximum 10.
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| Figure 2: Block diagram representation of design |
Let us, now, proceed for a generic logic for Nth intermediate output. As discussed, Nth output (On) depends upon 7 bits, 3 bits on the up and 3 on down. The Nth bit of output will show "1"only for following cases: X1101XX & 00101XX. As 10101XX will be detected as "1" for N+2 bit of output. If we denote the variables involved as G,F,E,D,C,A, then the output expression becomes
On = FED'C + G'F'ED'C
On = ED'C (F+G')
For bit 31, the upper two bits do not exist. 101XX is the expression for getting output as 1. Thus, the equation, on a similar note, can be expressed as:
O31 = ED'C
For bit 30, the expression comes out to be X101XX. So, O30 also has same logic as O31.
For O1 and O0, we get the same expression as we get for On. The logic diagram for obtaining intermediate outputs is shown in figure 3 below:
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| Figure 3: Circuit with logic for intermediate output |
Now, we have got the intermediate outputs showing which bits have the specified pattern detected. The number of "1"in the final output is our final answer. So, we need a special circuit to count the number of 1's combinationally in a bus as shown in figure 3 above. "Combinationally count number of 1's in a bus" explains how we can do this.
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