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Subject: Design of Sequential Circuits - Example 1.3
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<HTML><HEAD><TITLE>Design of Sequential Circuits - Example 1.3</TITLE>
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<P><BIG><BIG><STRONG>Design of Sequential =
Circuits</STRONG></BIG></BIG></P>
<P>This example is taken from M. M. Mano, <EM>Digital Design</EM>, =
Prentice=20
Hall, 1984, p.235.</P>
<P><BIG><FONT color=3D#0000ff><STRONG>Example 1.3&nbsp;&nbsp;=20
</STRONG></FONT></BIG><FONT color=3D#000000 face=3DArial>We wish to =
design a=20
synchronous sequential circuit whose state diagram is shown in Figure =
13. The=20
type of flip-flop to be use is J-K.</FONT></P>
<TABLE border=3D0 width=3D"56%" height=3D208>
  <TBODY>
  <TR>
    <TD height=3D204 width=3D"54%">
      <P align=3Dright><IMG alt=3D"fig13.gif (4506 bytes)" align=3Dleft=20
      =
src=3D"http://www.eelab.usyd.edu.au/digital_tutorial/part3/fig-13.gif"=20
      width=3D267 height=3D190></P></TD>
    <TD height=3D204 width=3D"46%"><BIG>Figure 13. State=20
  diagram</BIG></TD></TR></TBODY></TABLE>
<P><FONT face=3DArial>From the state diagram, we can generate the state =
table=20
shown in Table 9. Note that there is no output section for this circuit. =
Two=20
flip-flops are needed to represent the four states and are designated =
Q0Q1. The=20
input variable is labelled x.</FONT></P>
<TABLE border=3D1 width=3D"47%">
  <TBODY>
  <TR>
    <TD bgColor=3D#ff8040 width=3D"50%" align=3Dmiddle>Present State
      <TABLE border=3D0 width=3D"100%">
        <TBODY>
        <TR>
          <TD width=3D"100%">
            <P align=3Dcenter>Q0 Q1</P></TD></TR></TBODY></TABLE></TD>
    <TD bgColor=3D#ff8040 width=3D"50%" align=3Dmiddle>Next State
      <TABLE border=3D0 width=3D"100%">
        <TBODY>
        <TR>
          <TD width=3D"50%" align=3Dmiddle>x =3D 0</TD>
          <TD width=3D"50%" align=3Dmiddle>x =3D =
1</TD></TR></TBODY></TABLE></TD></TR>
  <TR>
    <TD width=3D"50%">
      <TABLE border=3D0 width=3D"100%">
        <TBODY>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0 0</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0 1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1 0</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1 =
1</TD></TR></TBODY></TABLE></TD>
    <TD width=3D"50%">
      <TABLE border=3D0 width=3D"100%">
        <TBODY>
        <TR>
          <TD width=3D"50%" align=3Dmiddle>0 0</TD>
          <TD width=3D"50%" align=3Dmiddle>0 1</TD></TR>
        <TR>
          <TD width=3D"50%" align=3Dmiddle>1 0</TD>
          <TD width=3D"50%" align=3Dmiddle>0 1</TD></TR>
        <TR>
          <TD width=3D"50%" align=3Dmiddle>1 0</TD>
          <TD width=3D"50%" align=3Dmiddle>1 1</TD></TR>
        <TR>
          <TD width=3D"50%" align=3Dmiddle>1 1</TD>
          <TD width=3D"50%" align=3Dmiddle>0=20
0</TD></TR></TBODY></TABLE></TD></TR></TBODY></TABLE>
<P><BIG>Table 9. State table.</BIG></P>
<P><FONT face=3DArial>We shall now derive the excitation table and the=20
combinational structure. The table is now arranged in a different form =
shown in=20
Table 11, where the present state and input variables are arranged in =
the form=20
of a truth table. Remember, the excitable for the JK flip-flop was =
derive in <A=20
href=3D"http://www.eelab.usyd.edu.au/digital_tutorial/part3/fl-types.htm"=
>Table=20
1</A></FONT>.</P>
<P>Table 10. Excitation table for JK flip-flop</P>
<TABLE border=3D1 width=3D"47%" height=3D87>
  <TBODY>
  <TR>
    <TD bgColor=3D#ff8040 height=3D52 width=3D"50%" =
align=3Dmiddle>Output Transitions
      <TABLE border=3D0 width=3D"100%">
        <TBODY>
        <TR>
          <TD width=3D"100%">
            <P align=3Dcenter>Q&nbsp;<FONT=20
            =
face=3DWingdings>=E0</FONT>Q<SMALL>(next)</SMALL></P></TD></TR></TBODY></=
TABLE></TD>
    <TD bgColor=3D#ff8040 height=3D52 width=3D"50%" =
align=3Dmiddle>Flip-flop inputs
      <TABLE border=3D0 width=3D"100%">
        <TBODY>
        <TR>
          <TD width=3D"100%">
            <P align=3Dcenter>J =
K</P></TD></TR></TBODY></TABLE></TD></TR>
  <TR>
    <TD height=3D27 width=3D"50%">
      <TABLE border=3D0 width=3D"100%">
        <TBODY>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0&nbsp;&nbsp; <FONT=20
            face=3DWingdings>=E0</FONT> &nbsp; 0</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0 &nbsp; <FONT =
face=3DWingdings>=E0</FONT>=20
            &nbsp; 1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1 &nbsp; <FONT =
face=3DWingdings>=E0</FONT>=20
            &nbsp;&nbsp;0</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1&nbsp;&nbsp; <FONT=20
            face=3DWingdings>=E0</FONT>&nbsp;&nbsp; =
1</TD></TR></TBODY></TABLE></TD>
    <TD height=3D27 width=3D"50%">
      <TABLE border=3D0 width=3D"100%">
        <TBODY>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0&nbsp; X</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1&nbsp; X</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>X&nbsp; 1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>X&nbsp;=20
  0</TD></TR></TBODY></TABLE></TD></TR></TBODY></TABLE>
<P>&nbsp;</P>
<P>Table 11. Excitation table of the circuit</P>
<TABLE border=3D1 width=3D"68%">
  <TBODY>
  <TR>
    <TD bgColor=3D#ff8040 width=3D"28%" align=3Dmiddle>Present State
      <TABLE border=3D0 width=3D"100%">
        <TBODY>
        <TR>
          <TD width=3D"100%">
            <P align=3Dcenter>Q0 Q1</P></TD></TR></TBODY></TABLE></TD>
    <TD bgColor=3D#ff8040 width=3D"29%" align=3Dmiddle>Next State
      <TABLE border=3D0 width=3D"100%">
        <TBODY>
        <TR>
          <TD width=3D"100%">
            <P align=3Dcenter>Q0 Q1</P></TD></TR></TBODY></TABLE></TD>
    <TD bgColor=3D#ff8040 width=3D"13%" align=3Dmiddle>Input
      <TABLE border=3D0 width=3D"100%">
        <TBODY>
        <TR>
          <TD width=3D"100%">
            <P align=3Dcenter>x</P></TD></TR></TBODY></TABLE></TD>
    <TD bgColor=3D#ff8040 width=3D"32%" align=3Dmiddle>Flip-flop Inputs
      <TABLE border=3D0 width=3D"100%">
        <TBODY>
        <TR>
          <TD width=3D"50%" align=3Dmiddle>J0K0</TD>
          <TD width=3D"50%" =
align=3Dmiddle>J1K1</TD></TR></TBODY></TABLE></TD></TR>
  <TR>
    <TD width=3D"28%" align=3Dmiddle>
      <TABLE border=3D0 width=3D"100%">
        <TBODY>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0 0</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0 0</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0 1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0 1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1 0</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1 0</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1 1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1 =
1</TD></TR></TBODY></TABLE></TD>
    <TD width=3D"29%" align=3Dmiddle>
      <TABLE border=3D0 width=3D"100%">
        <TBODY>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0 0</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0 1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1 0</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0 1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1 0</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1 1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1 1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0 =
0</TD></TR></TBODY></TABLE></TD>
    <TD width=3D"13%" align=3Dmiddle>
      <TABLE border=3D0 width=3D"100%">
        <TBODY>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>1</TD></TR>
        <TR>
          <TD width=3D"100%" align=3Dmiddle>0</TD></TR>
        <TR>
          <TD width=3D"100%" =
align=3Dmiddle>1</TD></TR></TBODY></TABLE></TD>
    <TD width=3D"32%" align=3Dmiddle>
      <TABLE border=3D0 width=3D"100%">
        <TBODY>
        <TR>
          <TD width=3D"50%" align=3Dmiddle>0 X</TD>
          <TD width=3D"50%" align=3Dmiddle>0 X</TD></TR>
        <TR>
          <TD width=3D"50%" align=3Dmiddle>0 X</TD>
          <TD width=3D"50%" align=3Dmiddle>1 X</TD></TR>
        <TR>
          <TD width=3D"50%" align=3Dmiddle>1 X</TD>
          <TD width=3D"50%" align=3Dmiddle>X 1</TD></TR>
        <TR>
          <TD width=3D"50%" align=3Dmiddle>0 X</TD>
          <TD width=3D"50%" align=3Dmiddle>X 0</TD></TR>
        <TR>
          <TD width=3D"50%" align=3Dmiddle>X 0</TD>
          <TD width=3D"50%" align=3Dmiddle>0 X</TD></TR>
        <TR>
          <TD width=3D"50%" align=3Dmiddle>X 0</TD>
          <TD width=3D"50%" align=3Dmiddle>1 X</TD></TR>
        <TR>
          <TD width=3D"50%" align=3Dmiddle>X 0</TD>
          <TD width=3D"50%" align=3Dmiddle>X 0</TD></TR>
        <TR>
          <TD width=3D"50%" align=3Dmiddle>X 1</TD>
          <TD width=3D"50%" align=3Dmiddle>X=20
1</TD></TR></TBODY></TABLE></TD></TR></TBODY></TABLE>
<P><FONT face=3DArial>In the first row of Table 11, we have a transition =
for=20
flip-flop Q0 from 0 in the present state to 0 in the next state. In =
Table 10 we=20
find that a transition of states from 0 to 0 requires that input J =3D 0 =
and input=20
K =3D X. So 0 and X are copied in the first row under J0 and K0 =
respectively.=20
Since the first row also shows a transition for the flip-flop Q1 from 0 =
in the=20
present state to 0 in the next state, 0 and X are copied in the first =
row under=20
J1 and K1. This process is continued for each row of the table and for =
each=20
flip-flop, with the input conditions as specified in Table =
10.</FONT></P>
<P><FONT face=3DArial>The simplified Boolean functions for the =
combinational=20
circuit can now be derived. The input variables are Q0, Q1, and x; the =
output=20
are the variables J0, K0, J1 and K1. The information from the truth =
table is=20
plotted on the Karnaugh maps shown in Figure 14.</FONT></P>
<P><IMG alt=3D"fig14.gif (10507 bytes)"=20
src=3D"http://www.eelab.usyd.edu.au/digital_tutorial/part3/fig-14.gif" =
width=3D562=20
height=3D196> </P>
<P><BIG>Figure 14. Karnaugh Maps</BIG></P>
<P><FONT face=3DArial>The flip-flop input functions are =
derived:</FONT></P>
<DIV align=3Dleft>
<TABLE border=3D0 width=3D511 height=3D1>
  <TBODY>
  <TR>
    <TD height=3D1 width=3D511><STRONG><BIG>J0 =3D=20
      Q1*x'&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; K0 =3D=20
  Q1*x</BIG></STRONG></TD></TR>
  <TR>
    <TD height=3D1 width=3D511><STRONG><BIG>J1 =3D x=20
      &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
      &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;K1 =3D Q0'*x' + Q0*x =3D =

      Q0</BIG><FONT=20
face=3DWingdings>=A4</FONT><BIG>x</BIG></STRONG></TD></TR></TBODY></TABLE=
></DIV>
<P>Note: the symbol <FONT face=3DWingdings>=A4</FONT> is =
exclusive-NOR.</P>
<P><FONT face=3DArial>The logic diagram is drawn in Figure =
15.</FONT></P>
<TABLE border=3D0 width=3D"62%">
  <TBODY>
  <TR>
    <TD width=3D"48%"><IMG alt=3D"fig-15.gif (3885 bytes)"=20
      =
src=3D"http://www.eelab.usyd.edu.au/digital_tutorial/part3/fig-15.gif"=20
      width=3D304 height=3D214></TD>
    <TD width=3D"52%">&nbsp;
      <P><BIG>Figure 15. Logic diagram of the sequential=20
  circuit.</BIG></P></TD></TR></TBODY></TABLE>
<P>&nbsp;</P>
<P><BIG><A=20
href=3D"http://www.eelab.usyd.edu.au/digital_tutorial/part3/example1-4.ht=
m"><STRONG>Example=20
1.4</STRONG></A></BIG></P>
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