Re: CPF University: EE Course (17)
Hello again and welcome!
This time the answers (and some hints) will be provided with the
self test questions so you can check your work immediately instead
of having to wait for the next part.
Answers for last time
1. In fig 9c, if R1=10 and R26=10 and I1=1 amp, what is I2?
ans: I2=I1*R1/(R1+R26)=0.5 amps
2. In fig 9c, if R1=10 and R26=30 and I1=1 amp, what is I3?
ans: I3=I1*R26/(R1+R26)=0.75 amps
3. In fig 9d, if R1=100 and R2=100 and V=10 volts, what is V2?
ans: V2=V*R2/(R1+R2)=5 volts
4. In fig 9d, if R1=100 and R2=300 and V=4 volts, what is V2 and what is V1?
answers:
V2=V*R2/(R1+R2)=3 volts
V1=V*R1/(R1+R2)=1 volt
For today:
Serial: EE-Ch0002-p103-{A939C0B7-558E-4687-9271-AB92E4935FD0}
Today we'll have to talk a little about simultaneous equations. We are
used to solving one equation and extending that to doing more then one
equation at the same time wont be too much harder. We'll find that
using a calculator will be needed most of the time for equations of
maybe 3 or greater, but it ends up being so simple because all
we end up doing is copying the coefficients of the equations to
the calculator in a specific format and pressing a few buttons to
get the answer even to fairly complex systems of equations. If for
some reason you find this section difficult, hold on 'till the end
where the calculator again makes everything fairly simple. If you are
willing to pay attention you'll find you dont have to do much real
work to get to the answers.
Before we start, i'd just like to mention that sometimes we use a shortcut
notation for multiplication. A term such as 2*a can be written as simply
2a
where the 'a' is written right next to the '2'.
This means the same as "2*a".
Also, since there are two major calculator types being used for the
calculations there will be two entries for finding answers: one for
the TI89 and one for the HP49G. You only have to refer to the line
that contains the calculator you are using. For example:
TI89: (2+3)^(-1) <enter>
HP49G: INV(2+3) <enter>
means if you are using the TI89 you enter (2+3) and then take it up to
the (-1) power, but if you are using an HP49G then you use the function
called "INV" built in, which is available on the 'CAT' menu. Both these
entries come out to the same answer, but the HP is a little different then
the TI so their entries have to be done a little differently.
Simultaneous Equations
We've already looked at equations of the form:
5=2a+b (as noted above, this is the same as 5=2*a+b)
where we can solve for either a or b, but there also exists
a form of equation that actually contains more then one equation
to be solved as a 'system' of equations. That is, all the
equations are taken to belong to the same problem and are
written just for the purpose of solving one single problem.
That problem involves more then one variable and we'll need
to find the values for all of the variables.
Take the following system of equations for example:
5=2a+b
8=2a+4b
This is a system in two unknowns, a and b, and the equations
are usually just written one after the other just as shown, one
equation on one line and the next equation of the set on the next line.
The difference here is that these two equations form a system for
solving for BOTH variables a and b, while previously we had only solved
for one variable, either a or b, using only one equation.
We first look at adding and subtracting whole equations...
Lets say we want to add the first equation to the second one above:
[8=2a+4b]+[5=2a+b]=?
In order to get the answer to this, we need to realize that when
adding equations we have to keep the individual terms separate,
and all we have to do is add the 'like' terms to each other to
form the result. The result will also be an equation.
'Like' terms are those terms which appear in both equations
that have the same variable:
2a is on one side and the other so they are 'like'
4b is 'like' b because they both have 'b'
8 is 'like' 5 because they both have no variable
Thus, when we do the adding we only add the like terms:
2a+2a=4a
4b+b=5b
8+5=13
and these new terms form our resulting equation:
13=4a+5b
Note all we did here is add the individual terms separately,
then form a new equation. Nothing strange there.
The whole thing written out looks like this:
[8=2a+4b]+[5=2a+b]=[13=4a+5b]
Now the whole idea is to find out what a AND b both are, and
adding the two didnt seem to help, because we ended up with
an equation which STILL has both variables a and b in it:
13=4a+5b
Let's try subtracting the two:
[8=2a+4b]-[5=2a+b]=?
[8=2a+4b]-[5=2a+b]=[3=3b]
Now we've gotten somewhere! We got rid of the terms with 'a' in them
so now we can find out what b is equal to.
Taking our resulting equation and dividing both sides by 3 we get
1=b
Now all we need to do is substitute 1 in for b in one of the equations
to solve for a, so say we chose 5=2a+b ...
knowing b=1 now we can substitute and solve for a:
5=2a+b
5=2a+1
4=2a
so
a=2
We therefore have solved for both a and b using the two simultaneous equations
and we're done.
Now lets say we have a system of three equations in three unknowns:
8=2a+b+c
17=2a+4b+3c
12=4a+b+c
We begin by noting that the first and third equations both have (b+c) in them,
so subtracting we'll end up with an equation in 'a' alone so we can solve for a:
[12=4a+b+c]-[8=2a+b+c]=[4=2a], a=2
We next take two of the equations (one of which wasnt used yet)
and replace their 'a' with what we found for a, 2, and proceed to
solve for one of the other variables b or c. We have to work
with two equations at the same time as we did above for the system
of two equations:
17=2a+4b+3c
8=2a+b+c
17=4+4b+3c
8=4+b+c
13=4b+3c
4=b+c
Now we can get rid of the c variables in both equations
if we multiply the second one by 3 and then subtract...
We end up with these two equations:
13=4b+3c
12=3b+3c
Subtracting the second from the first, we get rid of the '3c' term in
both equations so we can get the value for b:
1=b
Now all that's left to do is substitute the values we found for a and b
into one of the original equations and solve for c. Thus, we solved
for a, b, and c given three equations in the three unknowns.
It's good to know that we wont always have to do this by hand, as
there are other methods that use matrix algebra to solve for the
unknowns, and some calculators can handle this.
For example, the system of two we did above can be written more compactly...
5=2a+b
8=2a+4b
can also be written like this:
[[2,1][2,4]] for the right hand side and
[[5][8]] for the left hand side.
Note a matrix is used for the side that contains the variables and
a vector for the other side.
Now to solve for the variables a and b we simply enter this data
into the calculator calling [[2,1][2,4]] a matrix and calling
[[5][8]] a vector (or column matrix).
We then have two choices...
We can either use the 'simult' function on our calculator, or
we can do the following:
Take the inverse of the matrix and multiply this by the vector, which
gives us the values of a and b.
To take the inverse of the matrix we might just enter it like this:
TI89: [[2,1][2,4]]^(-1) <press enter>
HP49G: INV([[2,1][2,4]]) <press enter>
which will give us another matrix, which we then multiply by
the vector to get the values of both a and b:
TI89: [[2,1][2,4]]^(-1) * [[5][8]] <press enter>
HP49G: INV([[2,1][2,4]]) * [[5][8]] <press enter>
we get: [[2][1]]
(note on the HP49G calculator [[2][1]] may appear vertically)
As you can see, the calculator makes life much easier
Let's look now at our three equation system we did above...
8=2a+b+c
17=2a+4b+3c
12=4a+b+c
This can also be entered into a calculator and solved by forming the matrix:
[[2,1,1][2,4,3][4,1,1]]
and the vector:
[[8][17][12]]
Note all we do is enter the coefficients on the right hand side of the
equations into a matrix and enter the ones from the left hand side
of the equations into a vector to get the proper forms. That's almost
too easy
Then, to get the answer again we use the 'simult' function or take the
inverse of the matrix and multiply the result by the vector:
TI89: [[2,1,1],[2,4,3],[4,1,1]]^(-1) * [[8][17][12]]
HP49G: INV([[2,1,1],[2,4,3],[4,1,1]]) * [[8][17][12]]
After pressing enter we get the values of a, b, and c in a vector
which looks like this:
[[2][1][3]]
which are the values for a, b, and c respectively.
Note that the form used for the calculators are very specific,
in that you must order the coefficients correctly and use
the brackets '[' and ']' exactly as shown. Each equation
forms a new 'row' in the matrix which is entered in
brackets like this: [1,2,3]
while a new 'column' is formed in a new set of brackets.
For example, the first equation right side makes a row:
[2,1,1] (note the 1's come from having an implied '1' as coefficient)
the second row from the second equation:
[2,4,3]
and the third from the third equation:
[4,1,1]
Put them all on the same line and you get:
[2,1,1][2,4,3][4,1,1]
then add more brackets around the whole thing:
[[2,1,1][2,4,3][4,1,1]]
and you now have the proper form for the matrix.
The 'vector' is formed in the same way, only using the
left side of the equations for the numbers:
[8]
[17]
[12]
put all together:
[8][17][12]
then add brackets around:
[[8][17][12]]
and now you have the proper form for the vector.
All you do now is take the matrix up to the power -1 and
multiply the result times the vector to get the answer.
The above works the same on the TI89 and HP49G, but
with the HP49G you must use the built in function "INV"
as we have been doing above.
The general form for the TI89 is:
Matrix^(-1) * Vector
while for the HP49G it is:
INV(Matrix)*Vector
On the TI89 you can look up the function used to
solve simultaneous systems and use that if you wish,
while the HP49G uses a editing screen. You'll have to
look these up the the manuals if you wish to use them
instead of taking the inverse and multiplying as we did above.
Note that the TI85 might accept another form for the vector.
As a final note, once you get the values of all your
variables it's a good idea to plug them into one or more
of the equations to check to see that they are correct.
Self Test Questions
(hints and answers below...look at hints first if you have to)
1. Solve the following system of equations for a,b, and c:
a+2b+3c=7
6a+5b+4c=14
7a+2b+2c=21
2. Solve the following system of equations for a,b, and c:
4a+2b+3c=203
6a+5b+4c=406
7a+2b+2c=609
Hints for questions only if you need them:
1. TI89: [[1,2,3][6,5,4][7,2,2]]^(-1) * [[7][14][21]]
HP49G: INV([[1,2,3][6,5,4][7,2,2]])) * [[7][14][21]]
2. TI89: [[4,2,3][6,5,4][7,2,2]]^(-1) * [[203][406][609]]
HP49G: INV([[4,2,3][6,5,4][7,2,2]]) * [[203][406][609]]
Answers (use to check your work):
1. The system
a+2b+3c=7
6a+5b+4c=14
7a+2b+2c=21
can be put into matrix form for the calculator as:
TI89: [[1,2,3][6,5,4][7,2,2]]^(-1) * [[7][14][21]]
HP49G: INV([[1,2,3][6,5,4][7,2,2]]) * [[7][14][21]]
which after hitting enter comes out with:
[[3][-4][4]]
so
a=3
b=-4
c=4
2.
4a+2b+3c=203
6a+5b+4c=406
7a+2b+2c=609
TI89: [[4,2,3][6,5,4][7,2,2]]^(-1) * [[203][406][609]]
HP49G: INV([[4,2,3][6,5,4][7,2,2]]) * [[203][406][609]]
which comes out to:
[[105][28][-91]]
so
a=105
b=28
c=-91
See you next time!
Take care,
MrAl