Help with this truth table of logic circuit

Question

6. A logic circuit has four switches – A,B,C, and D—as inputs. The output of the circuit is supposed to be high if A and B or C and D are depressed. One of the switches, A, B, C, or D, is not functioning properly. The truth table shown below lists the actual output of the circuit. Define the problem and develop a solution plan to determine which switch has failed. Has it failed open or closed?

The above part is what came out of the book

here is the truth table for the actual results

0=off (open) 1=on (closed)

Switch A B C D output

0 0 0 0 0

0 0 0 1 1

0 0 1 0 0

0 0 1 1 1

0 1 0 0 0

0 1 0 1 1

0 1 1 0 0

0 1 1 1 1

1 0 0 0 0

1 0 0 1 1

1 0 1 0 0

1 0 1 1 1

1 1 0 0 1

1 1 0 1 1

1 1 1 0 1

1 1 1 1 1

So you have to figure out which switch is faulty and is it failing while it's open or closed.

The question confuses me because of the statement

"The output of the circuit is supposed to be high if A and B or C and D are depressed"

I'm not sure what that means. When they say depressed does that mean the switches are on or off? I'm guessing the equation look like this

(AB CD)not=output

So when I did the k-map for the truth table given I got AB D=output

So my guess is C is the faulty switch or is D. I don't know. Is this even how you figure it out?

If your up for the challenge I would really appreciate it. Sorry about the hard to read truth table but for some reason it would not let me put spaces between each number. If you can help me with that problem too it would be splendid.

Answer 1

Depressed means it is a momentary contact switch, and it is actuated, i.e. on.

A*B + C*D = output

Check each line of the output table to verify operation. For example, look at the second line on the truth table. You can see that only D is on, and the output is high. Thus, C must also be high.

Looking through the table, all results are consistent with C being high, and all other switches working properly.

Thus, C is failed as a short, i.e. always closed.

PS, your K-map solution is correct, circuit actually performing AB+D instead of AB+CD. Comparing, C must be a 1 to change from the design to the actual function. This verifies the results of the procedure above.