3 bit counter like circuit using D flip flop

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cfreng2

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hi all! Can anyone help me analyze this problem? I'm confused with this.

Design a 3 bit counterlike circuit controlled by the input w. If w=1 then the counter adds 2 to its contents, wrapping around if the count reaches 8 or 9. Thus if the present state is 8 or 9, then the next state becomes 0 or 1, respectively. If w = 0 then the counter subtracts 1 from its contents, acting as a normal down-counter. Use D flip flop in your circuit.
 

Do it with VHDL and perform the synthesis f the code using a synthesis tool. You'll get what you want designed with dffs.
 

I'm not doing a simulation. I'm doing a manual design for this. I'm confused on the count from 0-8 or 1-9... Since there are only 8 states (2^3bits = 8). So we could only have a count from 0-7 or 1-8.What can I do to get 8?..I'm really confused about this...
 

Heynthe problem is simple... you have only 3 flip-flops...
But your adder can be 4-bit...i.e if you get an overflow in the adder then you reset the counter to 0 or 1 depending on whether the overflow was 8 or 9... Hope i'm clear...
 

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