Exercise DET-03.04 – Repeated calls during Do Not Disturb (Solution)

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.

◀️ Back to the task

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Solution

Given

Given explicitly:

• the inputs $A$ (‘silent mode on’), $B$ (‘caller is a favourite’), $C$ (‘Do Not Disturb on’) and $D$ (‘repeated call’), and the output $Y$ (‘ringtone audible’)
• the rule: with silent mode on, the phone stays silent; Do Not Disturb silences calls, except calls from favourites and repeated calls
• e) the earlier circuit $Y_{\mathrm{old}}=\overline{A}\cdot \left(B+\overline{C}\right)$

Known:

• Steps 1 and 2: from the description to the truth table (DET-03)
• Minterm and maxterm lists (DET-03)
• The Karnaugh map with four inputs, loops of ones and loops of zeros (DET-03, DET-03, DET-03)
• Effort of a circuit (DET-03); from the minimized expression to the circuit (DET-03)
• Checking a circuit with a counterexample (DET-03)

Required

a) the truth table of $Y$
b) the minterm list and the maxterm list of $Y$
c) the minimized DNF and the minimized CNF of $Y$, and the one with fewer gates
d) the circuit of the result with fewer gates
e) a counterexample that shows that $Y_{\mathrm{old}}$ does not fulfil the new description

a) Truth table

With silent mode on, the phone stays silent whatever the other inputs say: a 0 in rows 8 to 15. With silent mode off, it rings unless Do Not Disturb blocks the call. Do Not Disturb blocks a call only if it comes neither from a favourite nor repeatedly: Do Not Disturb on, not a favourite, not repeated – that is row 2. In all other rows with silent mode off, the phone rings: without Do Not Disturb in rows 0, 1, 4 and 5, with Do Not Disturb for a repeated call in row 3 and for a favourite in rows 6 and 7 (DET-03, 1.2):

Row	$A$	$B$	$C$	$D$	$Y$
0	0	0	0	0	1
1	0	0	0	1	1
2	0	0	1	0	0
3	0	0	1	1	1
4	0	1	0	0	1
5	0	1	0	1	1
6	0	1	1	0	1
7	0	1	1	1	1
8	1	0	0	0	0
9	1	0	0	1	0
10	1	0	1	0	0
11	1	0	1	1	0
12	1	1	0	0	0
13	1	1	0	1	0
14	1	1	1	0	0
15	1	1	1	1	0

Check by counting: half of the 16 rows have silent mode on. Of the eight others, only row 2 has a blocked call, so the table has $8-1=7$ ones ✓

b) Minterm and maxterm lists

The minterm list names the seven rows with a 1, the maxterm list the nine rows with a 0 (DET-03, 2.5):

$$\begin{array}{rl}Y(A,B,C,D)& =\sum m(0,1,3,4,5,6,7)\\ & =\prod M(2,8,9,10,11,12,13,14,15)\end{array}$$

Check: together, the two lists contain each of the 16 row numbers exactly once ✓

c) Minimization with the Karnaugh map

With four inputs, the map has $A$ on the left of the rows and $D$ on the right of them, $B$ above and $C$ below the columns. Row by row, the cells contain the minterms 0, 2, 6 and 4; 1, 3, 7 and 5; 9, 11, 15 and 13; 8, 10, 14 and 12 (DET-03, 4.1). The ones lie in the two upper rows, the rows with $A=0$, except the cell of 2.

Loops of ones (figure): the seven ones form three loops of four (DET-03, 4.2):

• the cells of 4, 5, 6 and 7, a square in the two right columns of the upper rows: $\overline{A}\cdot B$
• the cells of 0, 1, 4 and 5 in the first and the last column of the upper rows, neighbours across the left and the right edge: $\overline{A}\cdot \overline{C}$
• the cells of 1, 3, 5 and 7, the whole second row: $\overline{A}\cdot D$

$$Y=\overline{A}\cdot B+\overline{A}\cdot \overline{C}+\overline{A}\cdot D$$

Loops of zeros (figure): the lower half of the map, the eight cells with $A=1$, forms a loop of eight; there, $A$ is 1, so the sum term is $\overline{A}$. The remaining zero, in the cell of 2 at the top of the second column, has a neighbour across the top and the bottom edge: the zero in the cell of 10 at the bottom of the same column. In both, $B$ is 0, $C$ is 1 and $D$ is 0, while $A$ changes, so the sum term is $B+\overline{C}+D$ (DET-03, 4.3):

$$Y=\overline{A}\cdot \left(B+\overline{C}+D\right)$$

Counted as in the lesson (DET-03, 3.1):

Minimized form	AND gates	OR gates	Gate inputs	Inverters
DNF (loops of ones)	3	1	9	2
CNF (loops of zeros)	1	1	5	2

The CNF needs two gates with five gate inputs instead of four gates with nine; both forms need inverters for $A$ and $C$. Read the CNF: the phone rings if silent mode is off and one of three things holds: the caller is a favourite, Do Not Disturb is off, or the call is repeated.

Check: factoring out $\overline{A}$ from the DNF gives the CNF:

$$\overline{A}\cdot B+\overline{A}\cdot \overline{C}+\overline{A}\cdot D=\overline{A}\cdot \left(B+\overline{C}+D\right)$$

✓

d) Circuit

The minimized CNF becomes a two-level circuit directly (DET-03, 5.1): an OR gate with three inputs forms the sum term $B+\overline{C}+D$, and an AND gate combines it with $\overline{A}$. The sum term $\overline{A}$ has a single literal and needs no OR gate; it comes directly from an inverter. A second inverter provides $\overline{C}$ (figure). As in the lesson, the intermediate node at the output of the OR gate is called $X$:

$$\begin{array}{rl}X& =B+\overline{C}+D\\ Y& =\overline{A}\cdot X=\overline{A}\cdot \left(B+\overline{C}+D\right)\end{array}$$

Check in row 2, the only row with silent mode off in which the phone stays silent. For $A=0$, $B=0$, $C=1$ and $D=0$:

$$\begin{array}{rl}X& =B+\overline{C}+D=0+\overline{1}+0=0\\ Y& =\overline{A}\cdot X=\overline{0}\cdot 0=0\end{array}$$

✓

e) Counterexample

$Y_{\mathrm{old}}$ does not contain $D$; it treats a repeated call like any other call. A difference can therefore only occur where a repeated call matters: silent mode off, Do Not Disturb on, the caller not a favourite and the call repeated. This is row 3 (DET-03, 5.3). For $A=0$, $B=0$, $C=1$ and $D=1$:

$$\begin{array}{rl}{Y}_{\mathrm{old}}& =\overline{A}\cdot \left(B+\overline{C}\right)=\overline{0}\cdot \left(0+\overline{1}\right)=1\cdot 0=0\\ Y& =\overline{A}\cdot \left(B+\overline{C}+D\right)=\overline{0}\cdot \left(0+\overline{1}+1\right)=1\cdot 1=1\end{array}$$

The description demands a 1: the same person calls again shortly after a first call, and Do Not Disturb lets the call through. The old circuit delivers a 0, and the phone stays silent. Row 3 is a counterexample, so $Y_{\mathrm{old}}$ does not fulfil the new description.

Check with the minterm lists: the old circuit is 1 where silent mode is off and either the caller is a favourite or Do Not Disturb is off, whatever $D$ is:

$$Y_{\mathrm{old}}(A,B,C,D)=\sum m(0,1,4,5,6,7)$$

The minterm list of $Y$ from b) contains row 3 as well; apart from it, the two lists agree ✓

The simulation shows both circuits, driven by the same four inputs: the new one with the output $Y$ above, the old one with the output $Y_{\mathrm{old}}$ below. It starts in row 3, the counterexample: $Y$ shows 1 and $Y_{\mathrm{old}}$ shows 0. Switch the inputs through all 16 rows: the two outputs differ only in row 3.

Open in the simulator

Circuit as text (File → Import From Text…)

<cir f="1" ts="0.000005" ic="1.500424758475255" cb="50" pb="50" vr="5" mts="5e-11">
  <L x="96 80 48 80" f="2"/>
  <L x="96 144 48 144" f="2"/>
  <L x="96 208 48 208" f="2" p="1"/>
  <L x="96 272 48 272" f="2" p="1"/>
  <w x="96 80 128 80" f="0"/>
  <I x="128 80 192 80" f="0" sl="0.5" hi="5"/>
  <w x="192 80 240 80" f="0"/>
  <w x="240 80 240 320" f="0"/>
  <w x="240 320 240 416" f="0"/>
  <w x="96 144 288 144" f="0"/>
  <w x="288 144 288 352" f="0"/>
  <w x="288 352 288 448" f="0"/>
  <w x="96 208 128 208" f="0"/>
  <I x="128 208 192 208" f="0" sl="0.5" hi="5"/>
  <w x="192 208 336 208" f="0"/>
  <w x="336 208 336 368" f="0"/>
  <w x="336 368 336 480" f="0"/>
  <w x="96 272 384 272" f="0"/>
  <w x="384 272 384 384" f="0"/>
  <Or x="480 368 624 368" f="0" in="3" o="5"/>
  <w x="288 352 480 352" f="0"/>
  <w x="336 368 480 368" f="0"/>
  <w x="384 384 480 384" f="0"/>
  <And x="688 336 832 336" f="0" o="5"/>
  <w x="240 320 688 320" f="0"/>
  <w x="624 368 656 368" f="0"/>
  <w x="656 368 656 352" f="0"/>
  <w x="656 352 688 352" f="0"/>
  <M x="832 336 880 336" f="2"/>
  <Or x="480 464 624 464" f="0"/>
  <w x="288 448 480 448" f="0"/>
  <w x="336 480 480 480" f="0"/>
  <And x="688 432 832 432" f="0"/>
  <w x="240 416 688 416" f="0"/>
  <w x="624 464 656 464" f="0"/>
  <w x="656 464 656 448" f="0"/>
  <w x="656 448 688 448" f="0"/>
  <M x="832 432 880 432" f="2"/>
  <x x="12 86 22 89" f="0" si="16" te="A"/>
  <x x="12 150 22 153" f="0" si="16" te="B"/>
  <x x="12 214 23 217" f="0" si="16" te="C"/>
  <x x="12 278 23 281" f="0" si="16" te="D"/>
  <x x="632 358 642 361" f="0" si="16" te="X"/>
  <x x="900 342 910 345" f="0" si="16" te="Y"/>
  <x x="900 438 936 441" f="0" si="16" te="Y old"/>
</cir>