What does my teacher want me to do?

[david90]

Design a sequential circuit to decode a given serial sequence of bits. This is basically a state machine problem. When the given sequence occurs a LED is turned on. 11011001.

So there is an input to this problem?

 

[Janitor]

Don't know if I will be of any help on this. Are you allowed to assume that your circuit is always in some "ready" state (or whatever you call it) when it is fed an 8-bit byte? Or could those particular eight bits be buried anywhere within a continuing stream of bits, such that your circuit has to be ready to detect that pattern even when it is embedded in a bunch of other bits before and after it?

 

[chroot]

I'm assuming he's covered the design of finite state machines with you already? This is not a trivial sort of thing to solve, but it's straightforward once you know how. I believe he's telling you to design a finite state machine (either Mealy or Moore, as you choose, apparently) with one input and one output. When the input has followed 11011001 for 8 clock cycles, the state machine should be in a state where the output is asserted.

- Warren

 

[david90]

Finite machine works with a clk input right?

 

[chroot]

Yes, FSMs are sequential logic, and must have a clock.

- Warren

 

[david90]

I've worked with state machine before that counts in a certain sequence but that doesn't have an input. You just turn on the clock and the circuit counts. The only input was the clk.

In this assigment in addition to the clk, I need to have another input to enter squence of numbers but how do I create this input??

 

[david90]

nevermind about my last msg ^. I think I got it. So I guess it requires 5 FFs right? Correct me if I'm wrong please.

My teacher said that in the real world, they don't use K-maps to figure out the boolean equations; so what alternate method do they use? show me please

 

[Guybrush Threepwood]

after the sequence is detected the LED remains on?

 

[david90]

The quarter is over. Final grades are in. Thanks for replying though. :)