Digitale Schaltungstechnik/ Zähler/ Synchron/ Umwandlung/ D-JK/ Bsp. 2

Titelseite
Synchronzähler D-Flipflop Vorwärtszähler Umschaltbar beliebige Zählfolge JK-Flipflop Vorwärtszähler beliebige Zählfolge umschaltbare Zählfolge T Flipflop Umwandlung D-JK Beispiel 1 Beispiel 2 Blockschaltbild Umschaltbar Kaskadieren Umkodierung Aufgaben Exkurs: Anwendungen

Beispiel: beliebige Zählfolge

Aufgabe

2 12 8 3 6 7 0

Wahrheitstabelle

	Eingang				Ausgang
	$Q_{n}$				$Q_{n+1}$
dez	$Q_{3}$	$Q_{2}$	$Q_{1}$	$Q_{0}$	$Q_{3}$	$Q_{2}$	$Q_{1}$	$Q_{0}$
0	0	0	0	0	0	0	1	0
1	0	0	0	1	x	x	x	x
2	0	0	1	0	1	1	0	0
3	0	0	1	1	0	1	1	0
4	0	1	0	0	x	x	x	x
5	0	1	0	1	x	x	x	x
6	0	1	1	0	0	1	1	1
7	0	1	1	1	0	0	0	0
8	1	0	0	0	0	0	1	1
9	1	0	0	1	x	x	x	x
10	1	0	1	0	x	x	x	x
11	1	0	1	1	x	x	x	x
12	1	1	0	0	1	0	0	0
13	1	1	0	1	x	x	x	x
14	1	1	1	0	x	x	x	x
15	1	1	1	1	x	x	x	x

KV-Diagramme

$Q_{3n+1}$	$Q_{3}Q_{2}$	$Q_{3}{\overline {Q_{2}}}$	${\overline {Q_{3}}}\ {\overline {Q_{2}}}$	${\overline {Q_{3}}}\ Q_{2}$
$Q_{1}\ Q_{0}$	15   X	11  X	3  0	7  0
$Q_{1}\ {\overline {Q_{0}}}$	14  X	10  X	2   1	6  0
${\overline {Q_{1}}}\ {\overline {Q_{0}}}$	12  1	8  0	0   0	4  X
${\overline {Q_{1}}}\ Q_{0}$	13  X	9  X	1  X	5   X

 $Q_{3n+1}=Q_{3}Q_{2}\lor {\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}}$ 

$Q_{2n+1}$	$Q_{3}Q_{2}$	$Q_{3}{\overline {Q_{2}}}$	${\overline {Q_{3}}}\ {\overline {Q_{2}}}$	${\overline {Q_{3}}}\ Q_{2}$
$Q_{1}\ Q_{0}$	15   X	11  X	3  1	7  0
$Q_{1}\ {\overline {Q_{0}}}$	14  X	10  X	2   1	6  1
${\overline {Q_{1}}}\ {\overline {Q_{0}}}$	12  0	8  0	0   0	4  X
${\overline {Q_{1}}}\ Q_{0}$	13  X	9  X	1  X	5   X

 $Q_{2n+1}=Q_{1}{\overline {Q_{0}}}\lor {\overline {Q_{2}}}Q_{1}$ 

$Q_{1n+1}$	$Q_{3}Q_{2}$	$Q_{3}{\overline {Q_{2}}}$	${\overline {Q_{3}}}\ {\overline {Q_{2}}}$	${\overline {Q_{3}}}\ Q_{2}$
$Q_{1}\ Q_{0}$	15   X	11  X	3  1	7  0
$Q_{1}\ {\overline {Q_{0}}}$	14  X	10  X	2   0	6  1
${\overline {Q_{1}}}\ {\overline {Q_{0}}}$	12  0	8  1	0   1	4  X
${\overline {Q_{1}}}\ Q_{0}$	13  X	9  X	1  X	5   X

 $Q_{1n+1}={\overline {Q_{1}}}{\overline {Q_{2}}}\lor {\overline {Q_{2}}}Q_{0}\lor {\overline {Q_{3}}}Q_{2}{\overline {Q_{0}}}$ 

$Q_{0n+1}$	$Q_{3}Q_{2}$	$Q_{3}{\overline {Q_{2}}}$	${\overline {Q_{3}}}\ {\overline {Q_{2}}}$	${\overline {Q_{3}}}\ Q_{2}$
$Q_{1}\ Q_{0}$	15   X	11  X	3  0	7  0
$Q_{1}\ {\overline {Q_{0}}}$	14  X	10  X	2   0	6  1
${\overline {Q_{1}}}\ {\overline {Q_{0}}}$	12  0	8  1	0   0	4  X
${\overline {Q_{1}}}\ Q_{0}$	13  X	9  X	1  X	5   X

 $Q_{0n+1}=Q_{3}{\overline {Q_{2}}}\lor {\overline {Q_{3}}}Q_{2}{\overline {Q_{0}}}$ 

Gleichung 3

$Q_{3n+1}=Q_{3}Q_{2}\lor {\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}}$ 

$Q_{3n+1}=J{\overline {Q_{3n}}}\lor {\overline {K}}Q_{3n}$

Der Ausdruck ${\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}}$ lässt sich nicht direkt zuordnen. Um es dennoch zuzuordnen verwenden wir einen Trick:

$Q_{3}\lor {\overline {Q_{3}}}=1$

$1\land {\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}}$

$1\land ({\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}})$

$(Q_{3}\lor {\overline {Q_{3}}})({\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}})$

$Q_{3}{\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}}\lor {\overline {Q_{3}}}\ {\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}}$

$Q_{3n+1}=Q_{3}Q_{2}\lor Q_{3}{\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}}\lor {\overline {Q_{3}}}\ {\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}}$	$Q_{3}$ ausklammern
$Q_{3n+1}=Q_{3}(Q_{2}\lor {\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}})\lor {\overline {Q_{3}}}\ {\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}}$	${\overline {Q_{3}}}$ ausklammern
$Q_{3n+1}=Q_{3}(Q_{2}\lor {\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}})\lor {\overline {Q_{3}}}({\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}})$	die Form der Gleichung stimmt nun überein:
$Q_{3n+1}=Q_{3}{\color {Red}(Q_{2}\lor {\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}})}\lor {\overline {Q_{3}}}{\color {blue}({\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}})}$ $Q_{3n+1}={\overline {Q_{3n}}}{\color {Red}J_{3}}\lor Q_{3n}{\color {Blue}{\overline {K_{3}}}}$	auslesen von $J_{3}$ und ${\overline {K_{3}}}$
$J_{3}=Q_{2}\lor {\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}}$ ${\overline {K_{3}}}={\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}}$	vereinfachen mittels Dermorgan
$J_{3}=Q_{2}\lor {\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}}$ $K_{3}=Q_{2}\lor {\overline {Q_{1}}}\lor Q_{0}$	Lösung

$J_{3}=Q_{2}\lor {\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}}$
$K_{3}=Q_{2}\lor {\overline {Q_{1}}}\lor Q_{0}$

Gleichung 2

$Q_{2n+1}=Q_{1}{\overline {Q_{0}}}\lor {\overline {Q_{2}}}Q_{1}$ 

$Q_{2n+1}=J{\overline {Q_{2n}}}\lor {\overline {K}}Q_{2n}$

Beim Teilausdruck

$Q_{1}{\overline {Q_{0}}}$

müssen wir Analog dem Oberen Beispiel vorgehen:

${\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}}\lor Q_{2}Q_{1}{\overline {Q_{0}}}$

$Q_{2n+1}={\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}}\lor Q_{2}Q_{1}{\overline {Q_{0}}}\lor {\overline {Q_{2}}}Q_{1}$

$Q_{2n+1}={\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}}\lor Q_{2}Q_{1}{\overline {Q_{0}}}\lor {\overline {Q_{2}}}Q_{1}$	$Q_{2}$ ausklammern
$Q_{2n+1}={\overline {Q_{2}}}Q_{1}{\overline {Q_{0}}}\lor Q_{2}(Q_{1}{\overline {Q_{0}}})\lor {\overline {Q_{2}}}Q_{1}$	${\overline {Q_{2}}}$ ausklammern
$Q_{2n+1}={\overline {Q_{2}}}(Q_{1}{\overline {Q_{0}}}\lor Q_{1})\lor Q_{2}(Q_{1}{\overline {Q_{0}}})$	die Form der Gleichung stimmt nun überein:
$Q_{2n+1}={\overline {Q_{2}}}{\color {Red}(Q_{1}{\overline {Q_{0}}}\lor Q_{1})}\lor Q_{2}{\color {Blue}(Q_{1}{\overline {Q_{0}}})}$ $Q_{2n+1}={\overline {Q_{2n}}}{\color {Red}J_{2}}\lor Q_{2n}{\color {Blue}{\overline {K_{2}}}}$	auslesen von $J_{2}$ und ${\overline {K_{2}}}$
$J_{2}=Q_{1}{\overline {Q_{0}}}\lor Q_{1}$ ${\overline {K_{2}}}=Q_{1}{\overline {Q_{0}}}$	vereinfachen
$J_{2}=Q_{1}$ ${\overline {K_{2}}}=Q_{1}{\overline {Q_{0}}}$	Vereinfachung mittels Demorgan
$J_{2}=Q_{1}$ $K_{2}={\overline {Q_{1}}}\lor Q_{0}$	Lösung

Gleichung 1

$Q_{1n+1}={\overline {Q_{1}}}\ {\overline {Q_{2}}}\lor {\overline {Q_{2}}}Q_{0}\lor {\overline {Q_{3}}}Q_{2}{\overline {Q_{0}}}$ 

Analog dem vorherigen Beispiel;

$Q_{1n+1}={\overline {Q_{1}}}\ {\overline {Q_{2}}}\lor Q_{1}{\overline {Q_{2}}}Q_{0}\lor {\overline {Q_{1}}}\ {\overline {Q_{2}}}Q_{0}\lor Q_{1}{\overline {Q_{3}}}Q_{2}{\overline {Q_{0}}}\lor {\overline {Q_{1}}}\ {\overline {Q_{3}}}Q_{2}{\overline {Q_{0}}}$ 

$Q_{1n+1}=J{\overline {Q_{1n}}}\lor {\overline {K}}Q_{1n}$

$Q_{1n+1}={\overline {Q_{1}}}\ {\overline {Q_{2}}}\lor Q_{1}{\overline {Q_{2}}}Q_{0}\lor {\overline {Q_{1}}}\ {\overline {Q_{2}}}Q_{0}\lor Q_{1}{\overline {Q_{3}}}Q_{2}{\overline {Q_{0}}}\lor {\overline {Q_{1}}}\ {\overline {Q_{3}}}Q_{2}{\overline {Q_{0}}}$	$Q_{1}$ ausklammern
$Q_{1n+1}=Q_{1}({\overline {Q_{2}}}Q_{0}\lor {\overline {Q_{3}}}Q_{2}{\overline {Q_{0}}})\lor {\overline {Q_{1}}}\ {\overline {Q_{2}}}\lor {\overline {Q_{1}}}\ {\overline {Q_{2}}}Q_{0}\lor {\overline {Q_{1}}}\ {\overline {Q_{3}}}Q_{2}{\overline {Q_{0}}}$	${\overline {Q_{1}}}$ ausklammern
$Q_{1n+1}=Q_{1}({\overline {Q_{2}}}Q_{0}\lor {\overline {Q_{3}}}Q_{2}{\overline {Q_{0}}})\lor {\overline {Q_{1}}}({\overline {Q_{2}}}\lor {\overline {Q_{2}}}Q_{0}\lor {\overline {Q_{3}}}Q_{2}{\overline {Q_{0}}})$	die Form der Gleichung stimmt nun überein:
$Q_{1n+1}=Q_{1}({\overline {Q_{2}}}Q_{0}\lor {\overline {Q_{3}}}Q_{2}{\overline {Q_{0}}})\lor {\overline {Q_{1}}}({\overline {Q_{2}}}\lor {\overline {Q_{2}}}Q_{0}\lor {\overline {Q_{3}}}Q_{2}{\overline {Q_{0}}})$ $Q_{1n+1}=J{\overline {Q_{1n}}}\lor {\overline {K}}Q_{1n}$	auslesen von $J_{1}$ und ${\overline {K_{1}}}$
$J_{1}={\overline {Q_{2}}}\lor {\overline {Q_{2}}}Q_{0}\lor {\overline {Q_{3}}}Q_{2}{\overline {Q_{0}}}$ ${\overline {K_{1}}}={\overline {Q_{2}}}Q_{0}\lor {\overline {Q_{3}}}Q_{2}{\overline {Q_{0}}}$	vereinfachen
$J_{1}={\overline {Q_{2}}}\lor {\overline {Q_{3}}}Q_{2}{\overline {Q_{0}}}$ ${\overline {K_{1}}}={\overline {Q_{2}}}Q_{0}\lor {\overline {Q_{3}}}Q_{2}{\overline {Q_{0}}}$	Resultat

$J_{1}=$
$K_{1}=$

Gleichung 0

$Q_{0n+1}=Q_{3}{\overline {Q_{2}}}\lor {\overline {Q_{3}}}Q_{2}{\overline {Q_{0}}}$ 

$Q_{0n+1}=Q_{3}{\overline {Q_{2}}}Q_{0}\lor Q_{3}{\overline {Q_{2}}}\ {\overline {Q_{0}}}\lor {\overline {Q_{3}}}Q_{2}{\overline {Q_{0}}}$ 

$Q_{0n+1}=J{\overline {Q_{0n}}}\lor {\overline {K}}Q_{0n}$

$Q_{0n+1}=Q_{3}{\overline {Q_{2}}}Q_{0}\lor Q_{3}{\overline {Q_{2}}}\ {\overline {Q_{0}}}\lor {\overline {Q_{3}}}Q_{2}{\overline {Q_{0}}}$	$Q_{0}$ ausklammern
$Q_{0n+1}=Q_{0}(Q_{3}{\overline {Q_{2}}})\lor Q_{3}{\overline {Q_{2}}}\ {\overline {Q_{0}}}\lor {\overline {Q_{3}}}Q_{2}{\overline {Q_{0}}}$	${\overline {Q_{0}}}$ ausklammern
$Q_{0n+1}=Q_{0}(Q_{3}{\overline {Q_{2}}})\lor {\overline {Q_{0}}}(Q_{3}{\overline {Q_{2}}}\ \lor {\overline {Q_{3}}}Q_{2})$	die Form der Gleichung stimmt nun überein:
$Q_{0n+1}=Q_{0}(Q_{3}{\overline {Q_{2}}})\lor {\overline {Q_{0}}}(Q_{3}{\overline {Q_{2}}}\ \lor {\overline {Q_{3}}}Q_{2})$ $Q_{0n+1}=J{\overline {Q_{0n}}}\lor {\overline {K}}Q_{0n}$	auslesen von $J_{0}$ und ${\overline {K_{0}}}$
$J_{0}=Q_{3}{\overline {Q_{2}}}\ \lor {\overline {Q_{3}}}Q_{2}$ ${\overline {K_{0}}}=Q_{3}{\overline {Q_{2}}}$	vereinfachen mittels Dermorgan
$J_{0}=Q_{3}{\overline {Q_{2}}}\ \lor {\overline {Q_{3}}}Q_{2}$ $K_{0}={\overline {Q_{3}}}\lor Q_{2}$	vereinfachung

$J_{0}=Q_{3}{\overline {Q_{2}}}\ \lor {\overline {Q_{3}}}Q_{2}$
$K_{0}={\overline {Q_{3}}}\lor Q_{2}$