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CS2100 Lab #10: Using Logisim II Solution

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CS2100 Computer Organisation
Lab #10: Using Logisim II
[ This document is available on LumiNUS and module website http://www.comp.nus.edu.sg/~cs2100 ]



Name: __________________________________ Student No.: ____________________
Lab Group: _____
Objective:
In this experiment, you will use Logisim to analyse and design sequential circuits.

Part I
1. Run Logisim, open the file lab10part1.circ. The circuit is shown below.


2. The circuit consists of two JK flip-flop and an OR gate. Note the following:
 The outputs of the two JK flip-flops are labelled A and B, which form the state of the circuit.
 The Clock is connected to the clock inputs of the flip-flops.
 The logic constant 1 is connected to the Enable inputs of the flip-flops.
 The Clear switch is connected to the clear inputs of the flip-flops. Hence when Clear = 1, it clears the contents of both flip-flips to 0, bringing the circuit to the initial state of AB=00.
 The flip-flop inputs are as follows:
For flip-flop A: JA = A + B; KA = 0
For flip-flop B: JB = 1;
KB = A + B
Present state Flip-flop inputs Next state
A B JA KA JB KB A+ B+
0 0
0 1
1 0
1 1

4. Verify the correctness of your table above by testing the circuit in Logisim.
a) Click on “Clear” input to get 1. This clears both flip-flops to 0, bringing the circuit to the initial state of AB=00.
b) Click on “Clear” input to get 0 before you proceed. This puts the flip-flops in their normal operation mode.
c) Clicking the “Clock” input toggles its value. When the “Clock” value changes from 0 to 1 (i.e. a rising edge), the flip-flops react according to the commands at their J and K inputs.
d) Click the “Clock” input several times to simulate the square wave, and watch the outputs of the flip-flops change their values. Do the values follow your table above?
e) If at any point of time you want to reset the flip-flops to the initial state of 00, go to step (a) above.







Part II
6. You will design a sequential circuit using JK flip-flops. The flip-flop inputs are given below:
For flip-flop A: JA = 1; KA = A  B
For flip-flop B: JB = 0; KB = (A  B)'


Present state Flip-flop inputs Next state
A B JA KA JB KB A+ B+
0 0
0 1
1 0
1 1











10. As this is your final lab, your lab report will not be returned to you.





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