2 bit counter

A = lsb ; B = msb

up count:
B A
0 0
0 1
1 0
1 1
0 0

down count:
B A
1 1
1 0
0 1
0 0
1 1

where the functs:
/A := A ;always toggle

/B := /B * /A * UP ; hold on up
+ /B * A * /UP ; hold on down
+ B * A * UP ; toggle up
+ B * /A * /UP ; toggle down



i can get how the A comes, but i still did not get how the /B funct becomes like that. actually i had got the answer by guessing. but, i felt i did something that unsure, not an "official" way to solve the equations. could anyone help me out to explain, where the B equ comes from?

thanks a lot
Posted on 2003-11-08 05:00:05 by dion
Both A and B are driven by feedbacks.
Since the equations are registered, a change only happens on a clock edge.

From the equations alone, you don't know the initial conditions of either a or b. That depends on the device.
Posted on 2003-11-09 04:20:21 by eet_1024
i mean how can i get the answer for /B equation.
Posted on 2003-11-14 18:56:02 by dion