This is a normal design of 4-bit counter using D flip-flops
The outputs Q0 to Q3 are connected to diodes. I want to modify this counter in such a way that it follows the following counting sequence and that it could be set back to 9 asynchronously (maybe through the Reset and Preset inputs of the flip-flops?) anytime.
Any help would be much appreciated, thank you.
Simple table provides what's needed:
$$\begin{array}{c|c} \text{Beginning State} & \text{Ending State}\\\\ {\begin{array}{cccc} Q_D & Q_C & Q_B & Q_A\\\\ 0&0&0&0\\ 0&0&1&1\\ 0&1&1&0\\ 1&0&0&1\\ 1&1&0&0\\ 1&1&1&1\\ 0&0&1&0\\ 0&1&0&1\\ 1&0&0&0\\ 1&0&1&1\\\\ 0&0&0&1\\ 0&1&0&0\\ 0&1&1&1\\ 1&0&1&0\\ 1&1&0&1\\ 1&1&1&0\\ \end{array}} & {\begin{array}{cccc} D_D & D_C & D_B & D_A\\\\ 0&0&1&1\\ 0&1&1&0\\ 1&0&0&1\\ 1&1&0&0\\ 1&1&1&1\\ 0&0&1&0\\ 0&1&0&1\\ 1&0&0&0\\ 1&0&1&1\\ 0&0&0&0\\\\ x&x&x&x\\ x&x&x&x\\ x&x&x&x\\ x&x&x&x\\ x&x&x&x\\ x&x&x&x\\ \end{array}} \end{array}$$
Now the four K-map tables.
$$\begin{array}{rl} \begin{smallmatrix}\begin{array}{r|cccc} D_D&\overline{Q_B}\:\overline{Q_A}&\overline{Q_B}\: Q_A&Q_B \:Q_A&Q_B \:\overline{Q_A}\\ \hline \overline{Q_D}\:\overline{Q_C}&0&x&0&0\\ \overline{Q_D}\:Q_C&x&1&x&1\\ Q_D\: Q_C&1&x&0&x\\ Q_D\:\overline{Q_C}&1&1&0&x \end{array}\end{smallmatrix} & \begin{smallmatrix}\begin{array}{r|cccc} D_C&\overline{Q_B}\:\overline{Q_A}&\overline{Q_B}\: Q_A&Q_B \:Q_A&Q_B \:\overline{Q_A}\\ \hline \overline{Q_D}\:\overline{Q_C}&0&x&1&1\\ \overline{Q_D}\:Q_C&x&0&x&0\\ Q_D\: Q_C&1&x&0&x\\ Q_D\:\overline{Q_C}&0&1&0&x \end{array}\end{smallmatrix}\\\\ \begin{smallmatrix}\begin{array}{r|cccc} D_B&\overline{Q_B}\:\overline{Q_A}&\overline{Q_B}\: Q_A&Q_B \:Q_A&Q_B \:\overline{Q_A}\\ \hline \overline{Q_D}\:\overline{Q_C}&1&x&1&0\\ \overline{Q_D}\:Q_C&x&0&x&0\\ Q_D\: Q_C&1&x&1&x\\ Q_D\:\overline{Q_C}&1&0&0&x \end{array}\end{smallmatrix} & \begin{smallmatrix}\begin{array}{r|cccc} D_A&\overline{Q_B}\:\overline{Q_A}&\overline{Q_B}\: Q_A&Q_B \:Q_A&Q_B \:\overline{Q_A}\\ \hline \overline{Q_D}\:\overline{Q_C}&1&x&0&1\\ \overline{Q_D}\:Q_C&x&0&x&1\\ Q_D\: Q_C&1&x&0&x\\ Q_D\:\overline{Q_C}&1&0&0&x \end{array}\end{smallmatrix} \end{array}$$
Let's start with \$D_A\$ and just follow along to see how I changed the \$x\$ values. Here we get: \$D_A=\overline{Q_A}\$:
$$\begin{array}{rl} \begin{smallmatrix}\begin{array}{r|cccc} D_A&\overline{Q_B}\:\overline{Q_A}&\overline{Q_B}\: Q_A&Q_B \:Q_A&Q_B \:\overline{Q_A}\\ \hline \overline{Q_D}\:\overline{Q_C}&1&0&0&1\\ \overline{Q_D}\:Q_C&1&0&0&1\\ Q_D\: Q_C&1&0&0&1\\ Q_D\:\overline{Q_C}&1&0&0&1 \end{array}\end{smallmatrix} \end{array}$$
Next is \$D_B\$. Again, spot my changes to \$x\$. See that: \$D_B=\overline{Q_A}\:\overline{Q_B}+Q_C\: Q_D+Q_A\: Q_B\:\overline{Q_D}\$:
$$\begin{array}{rl} \begin{smallmatrix}\begin{array}{r|cccc} D_B&\overline{Q_B}\:\overline{Q_A}&\overline{Q_B}\: Q_A&Q_B \:Q_A&Q_B \:\overline{Q_A}\\ \hline \overline{Q_D}\:\overline{Q_C}&1&0&1&0\\ \overline{Q_D}\:Q_C&1&0&1&0\\ Q_D\: Q_C&1&1&1&1\\ Q_D\:\overline{Q_C}&1&0&0&0 \end{array}\end{smallmatrix} \end{array}$$
Now for \$D_C\$. Spot changes and see: \$D_C=\overline{Q_A}\:\overline{Q_B}\:Q_C+Q_A\:\overline{Q_B}\:\overline{Q_C}+Q_B\:\overline{Q_C}\:\overline{Q_D}\$:
$$\begin{array}{rl} \begin{smallmatrix}\begin{array}{r|cccc} D_C&\overline{Q_B}\:\overline{Q_A}&\overline{Q_B}\: Q_A&Q_B \:Q_A&Q_B \:\overline{Q_A}\\ \hline \overline{Q_D}\:\overline{Q_C}&0&1&1&1\\ \overline{Q_D}\:Q_C&1&0&0&0\\ Q_D\: Q_C&1&0&0&0\\ Q_D\:\overline{Q_C}&0&1&0&0 \end{array}\end{smallmatrix} \end{array}$$
And \$D_D\$: \$D_D=\overline{Q_B}\:Q_D+ Q_C\:\overline{Q_D}\$:
$$\begin{array}{rl} \begin{smallmatrix}\begin{array}{r|cccc} D_D&\overline{Q_B}\:\overline{Q_A}&\overline{Q_B}\: Q_A&Q_B \:Q_A&Q_B \:\overline{Q_A}\\ \hline \overline{Q_D}\:\overline{Q_C}&0&0&0&0\\ \overline{Q_D}\:Q_C&1&1&1&1\\ Q_D\: Q_C&1&1&0&0\\ Q_D\:\overline{Q_C}&1&1&0&0 \end{array}\end{smallmatrix} \end{array}$$
So the equation summary from the above work is:
$$\begin{align*} D_A&=\overline{Q_A}\\ D_B&=\overline{Q_A}\:\overline{Q_B}+Q_C\: Q_D+Q_A\: Q_B\:\overline{Q_D}\\ D_C&=\overline{Q_A}\:\overline{Q_B}\:Q_C+Q_A\:\overline{Q_B}\:\overline{Q_C}+Q_B\:\overline{Q_C}\:\overline{Q_D}\\ D_D&=\overline{Q_B}\:Q_D+ Q_C\:\overline{Q_D} \end{align*}$$
Just note that the above isn't the only possible arrangements. By choosing differently for the \$x\$ values, you might come up with different (but equivalent) equations. If you put in a little time formulating them in several ways, you might find a better arrangement for the final circuit (fewer gates.) But I've only so much time to apply and I'm stopping it here.
Clearly, \$D_A\$ is free and \$D_D\$ is just a mux (if allowed.) The other two will involve a little more logic. But if you wire it up, it should work.
Here's how I implemented it in Neemann's Digital:
Worked exactly as expected.
The one thing that is missing is the reset to a specific point. I'll leave that to you.
The async reset to 9 was simple to add. So here it is despite "leaving it to you":
