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mathmari

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Hey!

I am looking the follwong exercise:

Using the method of Quine-McCluskey, determine the prime implicants for the following switching function and find a disjunctive minimal form. If available, also specify all other disjoint minimal forms.

The switching function is:

\begin{align*}f(x_1, x_2, x_3, x_4, x_5)&=\bar{x}_1\bar{x}_2\bar{x}_3\bar{x}_4\bar{x}_5\lor \bar{x}_1x_2\bar{x}_3\bar{x}_4\bar{x}_5\lor \bar{x}_1\bar{x}_2\bar{x}_3\bar{x}_4x_5\lor \bar{x}_1x_2\bar{x}_3x_4\bar{x}_5 \\ & \lor \bar{x}_1x_2\bar{x}_3\bar{x}_4x_5 \lor x_1x_2\bar{x}_3x_4\bar{x}_5\lor \bar{x}_1x_2x_3x_4\bar{x}_5 \lor \bar{x}_1x_2x_3x_4x_5 \\ & \lor x_1x_2\bar{x}_3x_4x_5\end{align*} I have done the following:

It holds that $\bar{x}=x^0$ and $x=x_1$. So we get the following:

\begin{align*}f(x_1, x_2, x_3, x_4, x_5)&=x_1^0x_2^0x_3^0x_4^0x_5^0\lor x_1^0x_2^1x_3^0x_4^0x_5^0\lor x_1^0x_2^0x_3^0x_4^0x_5^1\lor x_1^0x_2^1x_3^0x_4^1x_5^0 \\ & \lor x_1^0x_2^1x_3^0x_4^0x_5^1 \lor x_1^1x_2^1x_3^0x_4^1x_5^0\lor x_1^0x_2^1x_3^1x_4^1x_5^0\lor x_1^0x_2^1x_3^1x_4^1x_5^1 \\ & \lor x_1^1x_2^1x_3^0x_4^1x_5^1\end{align*}

We create the following table using the weights and we try to merge the midterms.

View attachment 9290

Therefore we get six primterms:

\begin{align*}&p_1=m_2+m_4=x_1^0x_2^1x_3^0x_5^0=\bar{x}_1x_2\bar{x}_3\bar{x}_5 \\ &p_2=m_4+m_6=x_2\bar{x}_3x_4\bar{x}_5 \\ &p_3=m_4+m_7=\bar{x}_1x_2x_4\bar{x}_5 \\ &p_4=m_6+m_9=x_1x_2\bar{x}_3x_4 \\ &p_5=m_7+m_8=\bar{x}_1x_2x_3x_4 \\ &p_6=m_1+m_2+m_3+m_5=\bar{x}_1\bar{x}_3\bar{x}_4\end{align*} The primterm table is then the following:

View attachment 9291

We can delete the columns $m_2, m_3, m_5$ because of $m_1$. We can delete also the columns $m_4,m_9$ because of $m_6$. We can delete also the columns $m_7$ because of $m_8$.

Then the table looks as follows:

View attachment 9292

We can delete the rows $p_1$ and $p_3$ since these are empty. We can delete also the row $p_2$ because of $p_4$ and so we get:

View attachment 9293

Do we get from that that the primterms are $p_4, p_5, p_6$? Are these the essential prime implicants?

How do we get the disjunctive minimal form from that? Do the selected primary terms simply have to be linked by disjunction? (Wondering)

I am looking the follwong exercise:

Using the method of Quine-McCluskey, determine the prime implicants for the following switching function and find a disjunctive minimal form. If available, also specify all other disjoint minimal forms.

The switching function is:

\begin{align*}f(x_1, x_2, x_3, x_4, x_5)&=\bar{x}_1\bar{x}_2\bar{x}_3\bar{x}_4\bar{x}_5\lor \bar{x}_1x_2\bar{x}_3\bar{x}_4\bar{x}_5\lor \bar{x}_1\bar{x}_2\bar{x}_3\bar{x}_4x_5\lor \bar{x}_1x_2\bar{x}_3x_4\bar{x}_5 \\ & \lor \bar{x}_1x_2\bar{x}_3\bar{x}_4x_5 \lor x_1x_2\bar{x}_3x_4\bar{x}_5\lor \bar{x}_1x_2x_3x_4\bar{x}_5 \lor \bar{x}_1x_2x_3x_4x_5 \\ & \lor x_1x_2\bar{x}_3x_4x_5\end{align*} I have done the following:

It holds that $\bar{x}=x^0$ and $x=x_1$. So we get the following:

\begin{align*}f(x_1, x_2, x_3, x_4, x_5)&=x_1^0x_2^0x_3^0x_4^0x_5^0\lor x_1^0x_2^1x_3^0x_4^0x_5^0\lor x_1^0x_2^0x_3^0x_4^0x_5^1\lor x_1^0x_2^1x_3^0x_4^1x_5^0 \\ & \lor x_1^0x_2^1x_3^0x_4^0x_5^1 \lor x_1^1x_2^1x_3^0x_4^1x_5^0\lor x_1^0x_2^1x_3^1x_4^1x_5^0\lor x_1^0x_2^1x_3^1x_4^1x_5^1 \\ & \lor x_1^1x_2^1x_3^0x_4^1x_5^1\end{align*}

We create the following table using the weights and we try to merge the midterms.

View attachment 9290

Therefore we get six primterms:

\begin{align*}&p_1=m_2+m_4=x_1^0x_2^1x_3^0x_5^0=\bar{x}_1x_2\bar{x}_3\bar{x}_5 \\ &p_2=m_4+m_6=x_2\bar{x}_3x_4\bar{x}_5 \\ &p_3=m_4+m_7=\bar{x}_1x_2x_4\bar{x}_5 \\ &p_4=m_6+m_9=x_1x_2\bar{x}_3x_4 \\ &p_5=m_7+m_8=\bar{x}_1x_2x_3x_4 \\ &p_6=m_1+m_2+m_3+m_5=\bar{x}_1\bar{x}_3\bar{x}_4\end{align*} The primterm table is then the following:

View attachment 9291

We can delete the columns $m_2, m_3, m_5$ because of $m_1$. We can delete also the columns $m_4,m_9$ because of $m_6$. We can delete also the columns $m_7$ because of $m_8$.

Then the table looks as follows:

View attachment 9292

We can delete the rows $p_1$ and $p_3$ since these are empty. We can delete also the row $p_2$ because of $p_4$ and so we get:

View attachment 9293

Do we get from that that the primterms are $p_4, p_5, p_6$? Are these the essential prime implicants?

How do we get the disjunctive minimal form from that? Do the selected primary terms simply have to be linked by disjunction? (Wondering)

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