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Drawing Truth Tables etc. Some clarification please?

shamieh

Active member
Sep 13, 2013
539
1a)Draw a truth table corresponding to f(X,Y,Z) = \(\displaystyle \sum\)m(0,2,5,7)

Apparently since I know that \(\displaystyle \sum\)m(0,2,5,7), I also intuitively know that\(\displaystyle \pi\)m(1,3,4,6) NOTE: I HAVE NO IDEA WHAT THIS MEANS


so I know that I have 3 inputs. So I know I have 2^3 rows, starting at 0. So this part is easy, I know I have

  1. x y z| f
  2. 0 0 0|1 <--- How do they get a 1 here? Isn't 0 AND 0 AND 0 = 0??
  3. 0 0 1|0 <- How do they get a 0 here?Isn't 0 and 0 = 1. Then 1 AND 0 =0 ??]
  4. 0 1 0|
  5. 0 1 1|
  6. 1 0 0|
  7. 1 0 1|
  8. 1 1 0|
  9. 1 1 1|

BUT, I'm not sure how they are getting the F column!
 
Last edited:

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
$f(x,y,z)=\sum m(0,2,5,7)$ is the definition of the function $f$. It means that those rows of the truth table get 1's, and the other rows get 0's. You could also write the function $f$ as follows:
$$f(x,y,z)= \bar{x} \bar{y} \bar{z}+ \bar{x}y \bar{z}+x \bar{y}z+xyz.$$
 

shamieh

Active member
Sep 13, 2013
539
$f(x,y,z)=\sum m(0,2,5,7)$ is the definition of the function $f$. It means that those rows of the truth table get 1's, and the other rows get 0's. You could also write the function $f$ as follows:
$$f(x,y,z)= \bar{x} \bar{y} \bar{z}+ \bar{x}y \bar{z}+x \bar{y}z+xyz.$$
That makes complete sense! so $f$ would be 1 0 1 0 0 1 0 1.

So now when it asks me - for the next question to -- Write out the canonical sum of products (SOP) expression for $$f(X,Y,Z,)$$" I would write:

$$ \bar{x} \bar{y} \bar{z} + \bar{x}y \bar{z} + x \bar{y}z + xyz$$

correct? So, my next question Ach.. Is when it says Minimize the expression of what I just got above, how do I go about minimizing that expression? What is the simplest way?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
That makes complete sense! so $f$ would be 1 0 1 0 0 1 0 1.

So now when it asks me - for the next question to -- Write out the canonical sum of products (SOP) expression for $$f(X,Y,Z,)$$" I would write:

$$ \bar{x} \bar{y} \bar{z} + \bar{x}y \bar{z} + x \bar{y}z + xyz$$

correct? So, my next question Ach.. Is when it says Minimize the expression of what I just got above, how do I go about minimizing that expression? What is the simplest way?
Well, my favorite way is Karnaugh Maps, but I don't know if you've learned that, yet. I'd probably go this route:
\begin{align*}
f&=\bar{x} \bar{y} \bar{z} + \bar{x}y \bar{z} + x \bar{y}z + xyz \\
&= \bar{x} \bar{z}(y+ \bar{y})+xz(y+ \bar{y}).
\end{align*}
Can you continue?
 

shamieh

Active member
Sep 13, 2013
539
Well, my favorite way is Karnaugh Maps, but I don't know if you've learned that, yet. I'd probably go this route:
\begin{align*}
f&=\bar{x} \bar{y} \bar{z} + \bar{x}y \bar{z} + x \bar{y}z + xyz \\
&= \bar{x} \bar{z}(y+ \bar{y})+xz(y+ \bar{y}).
\end{align*}
Can you continue?
We haven't learned Karnaugh Maps yet. Ahh I see! Factor by grouping and then y + y! = 1 right? so you're just left with \(\displaystyle \bar{x} \bar{z} + xz\) correct?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
We haven't learned Karnaugh Maps yet. Ahh I see! Factor by grouping and then y + y! = 1 right? so you're just left with \(\displaystyle \bar{x} \bar{z} + xz\) correct?
Right. I don't think there's anything else you can do with that.
 

shamieh

Active member
Sep 13, 2013
539
Okay awesome, but now it says: 1b)Write out the canonical sum of products (SOP) expression for $f$(X,Y,Z) of 1a. (This was the truth table we just drew and solved above). So for this one my teacher had something like this in the $f$ column...so what would be going on here?

  1. x y z | f
  2. 0 0 0| 1 <--- how are these numbers being implemented? this must be different from
  3. 0 0 1| 1 what we just did right?
  4. 0 1 0| 0
  5. 0 1 1| 1
  6. 1 0 0| 0
  7. 1 0 1| 1
  8. 1 1 0| 0
  9. 1 1 1| 1
 
Last edited:

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Okay awesome, but now it says: Write out the canonical sum of products (SOP) expression for $f$(X,Y,Z) of 1a. (This was the truth table we just drew and solved above). So for this one my teacher had something like this in the $f$ column...so what would be going on here?

  1. x y z | f
  2. 0 0 0| 1 <--- how are these numbers being implemented? this must be different from
  3. 0 0 1| 1 what we just did right?
  4. 0 1 0| 0
  5. 0 1 1| 1
  6. 1 0 0| 0
  7. 1 0 1| 1
  8. 1 1 0| 0
  9. 1 1 1| 1
If you're being asked to write an SOP for a given function, all you have to do is read out the row numbers. For each row with a $1$ in the $f$ column, add a term that represents the values of $x$, $y$, and $z$ for that row. If the $x$-value is a $0$, then put in $\bar{x}$. If it is a $1$, then put in an $x$. So, your next problem here will have to have
$$f=\underbrace{\bar{x} \bar{y} \bar{z}}_{\text{Row 2}}+ \underbrace{\bar{x} \bar{y} z}_{\text{Row 3}}+\dots.$$
Can you finish?
 

shamieh

Active member
Sep 13, 2013
539
So is this correct? Can someone check my work? That's essentially what I am asking.

1a) Draw the truth table corresponding to $f$((X,Y,Z,) = \(\displaystyle \sum\)m(0,2,5,7)
ANSWER:
x y z| f
0 0 0|1
0 0 1|0
0 1 0|1
0 1 1|0
1 0 0|0
1 0 1|1
1 1 0|0
1 1 1|1


1b)Write the canonical sum of products (SOP) expression for $f$(X,Y,Z,) of 1a.
ANSWER:
x!y!z! + x!yz! + xy!z + xyz

1c) Minimize the expression of 1b.
ANSWER:
x!z!(y! +y) + xz(y! + y) = x!z! + xz.
 
Last edited:

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,502
Yes, this is correct.
 

shamieh

Active member
Sep 13, 2013
539