Exercise DET-03.04 – Repeated calls during Do Not Disturb

Prof. Dr. Thorsten Jungmann (last updated 2026-10-19)

Relates to DET-03 Representation and minimization of logic functions

Task

Your smartphone can be set to Do Not Disturb, for example at night. It rings audibly only if silent mode is off. Do Not Disturb silences incoming calls, with two exceptions: calls from your favourites, and a repeated call (German: wiederholter Anruf), when the same person calls again shortly after a first call. Silent mode has priority over both exceptions. The inputs and the output are:

• $A$: silent mode is on
• $B$: the caller is one of your favourites
• $C$: Do Not Disturb is on
• $D$: the call is a repeated call
• $Y$: the ringtone is audible

a) Set up the truth table of $Y$.

b) Give the minterm list and the maxterm list of $Y$.

c) Minimize $Y$ with a Karnaugh map, once with loops of ones and once with loops of zeros. Which of the two results needs fewer gates?

d) Draw the circuit of the result with fewer gates, according to the standard of the International Electrotechnical Commission (IEC), IEC 60617.

e) Without repeated calls, the circuit $Y_{\mathrm{old}}=\overline{A}\cdot \left(B+\overline{C}\right)$ fulfilled the description. Show with a counterexample that it does not fulfil the new description.

Solution

a) a 1 in rows 0, 1, 3, 4, 5, 6 and 7, a 0 in row 2 and in rows 8 to 15
b) $Y(A,B,C,D)=\sum m(0,1,3,4,5,6,7)=\prod M(2,8,9,10,11,12,13,14,15)$
c) loops of ones: $Y=\overline{A}\cdot B+\overline{A}\cdot \overline{C}+\overline{A}\cdot D$; loops of zeros: $Y=\overline{A}\cdot \left(B+\overline{C}+D\right)$, with fewer gates: two instead of four
d) inverters for $A$ and $C$, an OR gate with the inputs $B$, $\overline{C}$ and $D$, and an AND gate with the inputs $\overline{A}$ and the output of the OR gate
e) row 3, with $A=0$, $B=0$, $C=1$ and $D=1$: a repeated call from someone who is not a favourite; $Y=1$, but $Y_{\mathrm{old}}=0$

▶️ Full worked solution