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Daniel Bernard ECE 1315 Digital System Design Lab #10: FSM Design and Interface with Stepper Motor 12/07/2010 Page 1 Lab #10: FSM Design and Interface with Stepper Motor The purpose of this investigation was to design and implement a Finite State Machine that controls a bi-directional stepper motor. There are three given input variables, 𝑥, 𝑦, and 𝑐𝑙𝑜𝑐𝑘. The controller should have four output variables, 𝑧3 , 𝑧2 , 𝑧1 , and 𝑧0 , one for each wire of the stepper motor. The inputs should give the function specified below based on their value. 𝑥 = {0, Motor stopped (controller disabled); 1, Motor active} 𝑦 = {0, Motor turns Clockwise; 1, Motor turns CC} 𝑐𝑙𝑜𝑐𝑘: (.5 second pulses) The operation can be shown by the pulses through the system summarized in Table 1. Table 1: Output patterns required for forward and reverse rotation 𝑦 = 0, Forward Operation: 𝑦 = 1, Reverse Operation 𝑧3 , 𝑧2 , 𝑧1 , 𝑧0 𝑧3 , 𝑧2 , 𝑧1 , 𝑧0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 0 0 1 1 0 0 0 Repeat Repeat The required states can be summarized in a simpler format, since there is only one high signal in each state, such that there are only two required input variables. In order to create the circuit in such a way that it would reverse if 𝑦 = 1, we need to have the inputs to the first FF based upon 𝑦 𝑥𝑜𝑟 [𝑖𝑛𝑝𝑢𝑡]. Table 2 shows the conversion from 𝑧3 , 𝑧2 , 𝑧1 , 𝑧0 to 𝑄1 , 𝑄0 . Table 2: Conversion Table Previous New 𝑧3 𝑧2 𝑧1 𝑧0 𝑄1 𝑄0 1 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 1 0 0 0 0 1 1 1 Diagram 1 shows the states given. Table 3 shows the State Transition Table, and Table 4 is the State Assign table, and shows the previous and next states for both JK flip-flops. Diagram 1: State Diagram y=0 y=0 A y=0 D C B y=1 y=0 y=1 y=1 y=1 Page 2 Table 3:State Transition Table Previous Next State State 𝑦=0 𝑦=1 A B D B C A C D B D A C 𝑧3 0 0 0 1 Output 𝑦 = 0 𝑧2 𝑧1 𝑧0 1 0 0 0 1 0 0 0 1 0 0 0 Table 4: State Assign Table 𝑧3 0 1 0 0 Output 𝑦 = 1 𝑧2 𝑧1 0 0 0 0 1 0 0 1 𝑧0 1 0 0 0 Previous State Next State Output ̅̅̅ ̅̅̅ 𝑄1 𝑄0 𝐽1 𝐽0 𝑄1 𝑄0 𝐾1 𝐾0 0 0 0 x 1 x 0 1 0 1 1 x x 0 1 0 1 0 x 1 1 x 1 1 1 1 x 0 x 0 0 0 From Table 4, we can create four Karnaugh maps to design the circuit implementation. Map 1: 𝐽1 Karnaugh Map 0 1 𝑄1 , 𝑄0 0 0 1 1 X X Map 2: ̅̅̅ 𝐾1 Karnaugh Map 0 1 𝑄1 , 𝑄0 0 X X 1 1 0 Map 3: 𝐽0 Karnaugh Map 0 1 𝑄1 , 𝑄0 0 1 X 1 1 X Map 4: ̅̅̅ 𝐾0 Karnaugh Map 0 1 𝑄1 , 𝑄0 0 X 0 1 X 0 From Map 1-4, the following functions can be extracted, and implemented in Diagram 2. 𝐽1 = 𝑄0 ̅̅̅ 𝐾1 = ̅̅̅ 𝑄0 𝐽0 = 1 ̅̅̅ 𝐾0 = 0 Page 3 Diagram 2: Circuit Implementation With the given circuit diagram, we were able to attach the outputs of 𝑧3 , 𝑧2 , 𝑧1 , 𝑧0 to the control of the stepper motor. Given such, we were able to make the motor switch direction based upon the input of 𝑦, and have it continuously go in the desired direction. Page 4