Proving Maximum power transfer theorem

[hanhan]

Hi, I am reading about "Maximum power transfer theorem" in the page: https://en.wikipedia.org/wiki/Maximum_power_transfer_theorem
Here is the part that I don't understand. Please help me. Thanks.

 

[MrAl]

Hello there,


I am happy to see you are still looking at these kinds of problems.

I will give a description in words of one way you can prove this and you can try it and see if you can get the results you are looking for. See far below for problems that came up however.

First, we'll declare the equation they give as f(x,y) and assume it is correct as is:

f(XS,XL)=(1/2)*|Vs|^2*RL/[(RS+RL)^2+(XS+XL)^2]

and since the only variables we need to look at are XS and XL everything else can be lumped into constants, so we end up with:
f(x,y)=A/[B^2+(x+y)^2]

where A and B are the constants, and we replaced the two variables by x and y to make it easier to read.

Next, we take first partials of f(x,y) and because there are two vars we get two first partials we'll call Px and Py. So we have:

Px= ∂f/∂x
Py= ∂f/∂y


Next, we set those two equal to zero:
Px=0
Py=0

then simplify if possible. We then solve this set of simultaneous equations for x and y.
If everything worked out ok, we will get as results 'x' equal to some constant K and 'y' equal to the negative of that constant:
x=K
y=-K

But we are not done yet because that is just a critical point (K,-K) and not necessarily a min or max.

Next we would take second partials of f(x,y) and we would get what we'll call PPx and PPy and PPxy:
PPx=∂(∂f/∂x)/∂x
PPy=∂(∂f/∂y)/∂y
PPxy=∂(∂f/∂y)/∂x

We then define:
D=PPx(x0,y0)*PPy(x0,y0)-(PPxy(x0,y0))^2

and substitute the values of K for x0 and -K for y0 (we found above).

Next we determine the values of that D and use the following criteria:
a) If D>0 and PPx(x0,y0)>0 then f has a relative min at (x0,y0).
b) If D>0 and PPx(x0,y0)<0 then f has a relative max at (x0,y0).
c) If D<0 then f has a saddle point at (x0,y0).
d) If D=0 then no conclusion can be drawn.

What you should find in this case is that (b) is satisfied when x0=K and y0=-K which means we have a maximum at that point. And that maximum will be A/B^2. That means one reactance has to be the negative of the other.

A problem came up however in that we seem to get D=0 for this problem. That's interesting because x=K and y=-K seem to work just fine.

What we could do here is graph the function f(x,y) in 3d and look for a max visually and see if it matches our critical point. We do know that the only critical point is at (K,-K) so we could test that numerically too.

Another idea is to make y=N and look for a max with x:
f(x)=A/(B^2+(N+x)^2)
f'(x)=-(2*A*(N+x))/(B^2+(N+x)^2)^2

set that equal to zero:
-(2*A*(N+x))/(B^2+(N+x)^2)^2=0

solve for x:
x=-N

So again we get one value is the negative of the other.

 

[Ratchit]

anhnha,

You are making the problem more complicated than it is. The bottom of your attachment summarizes the conditions for max power transfer if the source impedance is fixed and the load impedance is variable. There are two conditions that have to be satisified as listed at the end of the attachment. The first is that Xs = -Xl . You can easily see that if the reactances are equal and negatives of each other, then they will cancel out each other giving Xs+Xl=0, and the demonimator of the third equation of the attachment will be smaller. A smaller demoninator will give a higher Pl. The other conditon is that Rl=Rs. You have already seen proofs that this condition will result in the highest power Pl to the load.

Ratch

 

[hanhan]

Thank you, MrAl, for detailed explanation.
I have just followed your explanation and indeed, D = 0 and it means that no conclusion can be drawn.

A problem came up however in that we seem to get D=0 for this problem. That's interesting because x=K and y=-K seem to work just fine.

What we could do here is graph the function f(x,y) in 3d and look for a max visually and see if it matches our critical point.

Do you mean that we has to graph the function and find its maximum visually? If so, I think we can't get the exact value and if the function is very complicated it is very hard to graph too.

We do know that the only critical point is at (K,-K) so we could test that numerically too.

Does this mean that the critical point can be maximum or minimum and therefore we can test it by substituting them into the function and check it?

Another idea is to make y=N and look for a max with x:
f(x)=A/(B^2+(N+x)^2)
f'(x)=-(2*A*(N+x))/(B^2+(N+x)^2)^2

set that equal to zero:
-(2*A*(N+x))/(B^2+(N+x)^2)^2=0

solve for x:
x=-N

So again we get one value is the negative of the other.

This is what I am confused. Here we have a two-variable function and the maximum value is found by making a variable constant and the function is simplified to one variable. Thus, we can find the value of the variable for the function get maximum.
The next step is similar by exchange the role of two variable.
However, how can you know that this method will work?

PS.

First, we'll declare the equation they give as f(x,y) and assume it is correct as is:

f(XS,XL)=(1/2)*|Vs|^2*RL/[(RS+RL)^2+(XS+XL)^2]

and since the only variables we need to look at are XS and XL everything else can be lumped into constants, so we end up with:
f(x,y)=A/[B^2+(x+y)^2]

where A and B are the constants, and we replaced the two variables by x and y to make it easier to read.

I think the variables here is RL and XL not Xs and XL. Could you explain why?

 

[hanhan]

anhnha,

You are making the problem more complicated than it is. The bottom of your attachment summarizes the conditions for max power transfer if the source impedance is fixed and the load impedance is variable. There are two conditions that have to be satisified as listed at the end of the attachment. The first is that Xs = -Xl . You can easily see that if the reactances are equal and negatives of each other, then they will cancel out each other giving Xs+Xl=0, and the demonimator of the third equation of the attachment will be smaller. A smaller demoninator will give a higher Pl. The other conditon is that Rl=Rs. You have already seen proofs that this condition will result in the highest power Pl to the load.

Ratch

Thank you, Ratch.
Yes, you are right. I am wondering why I didn't see it.

 

[MrAl]

Thank you, MrAl, for detailed explanation.
I have just followed your explanation and indeed, D = 0 and it means that no conclusion can be drawn.

Do you mean that we has to graph the function and find its maximum visually? If so, I think we can't get the exact value and if the function is very complicated it is very hard to graph too.

Does this mean that the critical point can be maximum or minimum and therefore we can test it by substituting them into the function and check it?

This is what I am confused. Here we have a two-variable function and the maximum value is found by making a variable constant and the function is simplified to one variable. Thus, we can find the value of the variable for the function get maximum.
The next step is similar by exchange the role of two variable.
However, how can you know that this method will work?

PS.

I think the variables here is RL and XL not Xs and XL. Could you explain why?

Hello again,


Yes we could test the function in the regions close to the critical point, but i am never really happy until i see the whole graph. For this you would either need graphic software or you could use a program you write yourself or use a scientific programmable calculator. It's not very hard really because the function isnt that big, character count wise.
All you have to do is set up a double loop so that you loop through the variables with some relatively small step, and print out the results in a 2d space, which would look like a matrix.
What we would see is something like this (small data set here just for illustration):

3 2 2 1 1
2 3 2 2 1
1 2 3 2 2
1 1 2 3 2
1 1 1 2 3

and that is where x is swept across the bottom and y across the left hand side (the x axis and y axis) and the y axis runs from say 0 to minus something.
Notice that you can follow the 3's in the graph with your eyes and they suggest a peak to the graph that is the max value. And notice that whereever the 3's appear y=-x.
That is a fictitious graph but that is close to what we would see, and no matter how small we make the increment we'd always see that peak of 3. So the function looks like a "tent" where the peak is the top ridge of the tent.
That would mean that A/B^2=3.
I could throw together a program to do this but it will have to wait till tomorrow.

Variables...

Well the variables in question are XS and XL, and has nothing to do with RL and RS. We wanted to find the max with respect to the two variables XS and XL not the two R's.

 

[hanhan]

Thanks MrAl.

Yes we could test the function in the regions close to the critical point, but i am never really happy until i see the whole graph.

I also like to see 3D graph too.

What we would see is something like this (small data set here just for illustration):

3 2 2 1 1
2 3 2 2 1
1 2 3 2 2
1 1 2 3 2
1 1 1 2 3

and that is where x is swept across the bottom and y across the left hand side (the x axis and y axis) and the y axis runs from say 0 to minus something.
Notice that you can follow the 3's in the graph with your eyes and they suggest a peak to the graph that is the max value. And notice that whereever the 3's appear y=-x.
That is a fictitious graph but that is close to what we would see, and no matter how small we make the increment we'd always see that peak of 3. So the function looks like a "tent" where the peak is the top ridge of the tent.
That would mean that A/B^2=3.

Interesting. Now I didn't understand the matrix until reading your explanation.
I like to see the result both by calculation and seeing the graph.

Variables...

Well the variables in question are XS and XL, and has nothing to do with RL and RS. We wanted to find the max with respect to the two variables XS and XL not the two R's.

Sorry but I still don't get it. We have source impedance Zs is fixed meaning that Rs and Xs are fixed. We need to transform load impedance ZL to Zs* (conjugate of Zs) and therefore we have two variable RL and XL.

 

[MrAl]

Hello again,

I'll repeat the equation we are using here just for reference:
z=A/(B^2+(x+y)^2)

and we are looking for z being a max with x and y. Note it does not matter at this point what A and B are. We just want to see if z has a max for some values of x and y or not.

First we step from -500 to +500 and we see a clear max value of 3 running diagonally, then we step from -5 to 5 and we see what looks like a value of 3 as the max, then step from -0.5 to +0.5 and see the same max, then step from -50 to +50 (wider view) and see the same max.

So whenever x has the opposite sign of y we see a max, and because of the constants that max here is 3.00 which comes from A/B^2=3.

We also see an anomaly where x=0 and y=0 and we still have a max, but that still satisfies x+y=0 so it just means that if both reactances were equal to zero we'd still see a max. Not a problem really except in the real world we might never see that.

Running x and y from -500 to +500 and stepping by 100:

         y
         |
+500.00  |   +3.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00  
+400.00  |   +0.00    +3.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00  
+300.00  |   +0.00    +0.00    +3.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00  
+200.00  |   +0.00    +0.00    +0.00    +3.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00  
+100.00  |   +0.00    +0.00    +0.00    +0.00    +3.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00  
   0.00  |   +0.00    +0.00    +0.00    +0.00    +0.00    +3.00    +0.00    +0.00    +0.00    +0.00    +0.00  
-100.00  |   +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +3.00    +0.00    +0.00    +0.00    +0.00  
-200.00  |   +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +3.00    +0.00    +0.00    +0.00  
-300.00  |   +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +3.00    +0.00    +0.00  
-400.00  |   +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +3.00    +0.00  
-500.00  |   +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +0.00    +3.00  
          ------------------------------------------------------------------------------------------------------ x
           -500.00  -400.00  -300.00  -200.00  -100.00     0.00  +100.00  +200.00  +300.00  +400.00  +500.00     






Running x and y from -5 to 5 and stepping by 1:

       y
       |
+5.00  | +3.00  +1.50  +0.60  +0.30  +0.18  +0.12  +0.08  +0.06  +0.05  +0.04  +0.03  
+4.00  | +1.50  +3.00  +1.50  +0.60  +0.30  +0.18  +0.12  +0.08  +0.06  +0.05  +0.04  
+3.00  | +0.60  +1.50  +3.00  +1.50  +0.60  +0.30  +0.18  +0.12  +0.08  +0.06  +0.05  
+2.00  | +0.30  +0.60  +1.50  +3.00  +1.50  +0.60  +0.30  +0.18  +0.12  +0.08  +0.06  
+1.00  | +0.18  +0.30  +0.60  +1.50  +3.00  +1.50  +0.60  +0.30  +0.18  +0.12  +0.08  
 0.00  | +0.12  +0.18  +0.30  +0.60  +1.50  +3.00  +1.50  +0.60  +0.30  +0.18  +0.12  
-1.00  | +0.08  +0.12  +0.18  +0.30  +0.60  +1.50  +3.00  +1.50  +0.60  +0.30  +0.18  
-2.00  | +0.06  +0.08  +0.12  +0.18  +0.30  +0.60  +1.50  +3.00  +1.50  +0.60  +0.30  
-3.00  | +0.05  +0.06  +0.08  +0.12  +0.18  +0.30  +0.60  +1.50  +3.00  +1.50  +0.60  
-4.00  | +0.04  +0.05  +0.06  +0.08  +0.12  +0.18  +0.30  +0.60  +1.50  +3.00  +1.50  
-5.00  | +0.03  +0.04  +0.05  +0.06  +0.08  +0.12  +0.18  +0.30  +0.60  +1.50  +3.00  
        -------------------------------------------------------------------------------- x
         -5.00  -4.00  -3.00  -2.00  -1.00   0.00  +1.00  +2.00  +3.00  +4.00  +5.00     




Running x and y from -0.5 to +0.5 stepping by 0.1:

       y
       |
+0.50  | +3.00  +2.97  +2.88  +2.75  +2.59  +2.40  +2.21  +2.01  +1.83  +1.66  +1.50  
+0.40  | +2.97  +3.00  +2.97  +2.88  +2.75  +2.59  +2.40  +2.21  +2.01  +1.83  +1.66  
+0.30  | +2.88  +2.97  +3.00  +2.97  +2.88  +2.75  +2.59  +2.40  +2.21  +2.01  +1.83  
+0.20  | +2.75  +2.88  +2.97  +3.00  +2.97  +2.88  +2.75  +2.59  +2.40  +2.21  +2.01  
+0.10  | +2.59  +2.75  +2.88  +2.97  +3.00  +2.97  +2.88  +2.75  +2.59  +2.40  +2.21  
 0.00  | +2.40  +2.59  +2.75  +2.88  +2.97  +3.00  +2.97  +2.88  +2.75  +2.59  +2.40  
-0.10  | +2.21  +2.40  +2.59  +2.75  +2.88  +2.97  +3.00  +2.97  +2.88  +2.75  +2.59  
-0.20  | +2.01  +2.21  +2.40  +2.59  +2.75  +2.88  +2.97  +3.00  +2.97  +2.88  +2.75  
-0.30  | +1.83  +2.01  +2.21  +2.40  +2.59  +2.75  +2.88  +2.97  +3.00  +2.97  +2.88  
-0.40  | +1.66  +1.83  +2.01  +2.21  +2.40  +2.59  +2.75  +2.88  +2.97  +3.00  +2.97  
-0.50  | +1.50  +1.66  +1.83  +2.01  +2.21  +2.40  +2.59  +2.75  +2.88  +2.97  +3.00  
        -------------------------------------------------------------------------------- x
         -0.50  -0.40  -0.30  -0.20  -0.10   0.00  +0.10  +0.20  +0.30  +0.40  +0.50     



Running x and y from -50 to +50 and stepping by 10:

        y
        |
+50.00  |  +3.00   +0.03   +0.01   +0.00   +0.00   +0.00   +0.00   +0.00   +0.00   +0.00   +0.00  
+40.00  |  +0.03   +3.00   +0.03   +0.01   +0.00   +0.00   +0.00   +0.00   +0.00   +0.00   +0.00  
+30.00  |  +0.01   +0.03   +3.00   +0.03   +0.01   +0.00   +0.00   +0.00   +0.00   +0.00   +0.00  
+20.00  |  +0.00   +0.01   +0.03   +3.00   +0.03   +0.01   +0.00   +0.00   +0.00   +0.00   +0.00  
+10.00  |  +0.00   +0.00   +0.01   +0.03   +3.00   +0.03   +0.01   +0.00   +0.00   +0.00   +0.00  
  0.00  |  +0.00   +0.00   +0.00   +0.01   +0.03   +3.00   +0.03   +0.01   +0.00   +0.00   +0.00  
-10.00  |  +0.00   +0.00   +0.00   +0.00   +0.01   +0.03   +3.00   +0.03   +0.01   +0.00   +0.00  
-20.00  |  +0.00   +0.00   +0.00   +0.00   +0.00   +0.01   +0.03   +3.00   +0.03   +0.01   +0.00  
-30.00  |  +0.00   +0.00   +0.00   +0.00   +0.00   +0.00   +0.01   +0.03   +3.00   +0.03   +0.01  
-40.00  |  +0.00   +0.00   +0.00   +0.00   +0.00   +0.00   +0.00   +0.01   +0.03   +3.00   +0.03  
-50.00  |  +0.00   +0.00   +0.00   +0.00   +0.00   +0.00   +0.00   +0.00   +0.01   +0.03   +3.00  
         ------------------------------------------------------------------------------------------- x
          -50.00  -40.00  -30.00  -20.00  -10.00    0.00  +10.00  +20.00  +30.00  +40.00  +50.00

 

[hanhan]

Thanks a lot, MrAl, for nice illustrations. I like them a lot Just out of curiosity: Did you use a software to calculate them? Just for fun, I guess in your equation A = 3 and B =1.

 

[MrAl]

Hi again anhnha,

You're welcome.

Yes i used software that i wrote myself using a free programming language. It only takes a few minutes to set it up to print out all those values because it's just a set of nested "for next" loops. And yes i kept it simple with those values for A and B.

Here is the main set of loops that do all the work. Equ(k,i) is just the equation we've been using with A and B.
BTW you should be doing this too with some programming language of your choice.

for i=ymax to ymin by -incy do
  Printf(1,PFormat&"  | ",{i})
  for k=xmin to xmax by incx do
    z=Equ(k,i)
    Printf(1,PFormat&"  ",{z})
  end for
  Printf(1,"%s\n",{""})
end for

Hey you know what else, maybe if we rotate the equation we can determine more analytically. It might become a function of one variable but we'll have to see.

A LITTLE LATER:

Yes, finding an angle that eliminates one of the variables was easy due to the previous plots. The angle is of course pi/4, and rotating in the x,y plane the original equation :
z=A/(B^2+(x+y)^2)

eliminates y so we end up with a brand new equation in one variable, x:
z=A/(B^2+2*x^2)

and solving that for a critical point we get:
x=0

and plugging that back into the new equation we get:
max(z)=A/B^2

Q.E.D.

I've included a graph of this new function and it is clear that this is a new view of the function as if we were looking across the x,y plane down the middle of the 'tent' that is formed from the original equation, as if we were looking through the tent that is open on both ends.

Note that the graph uses A=3 and B=1 as before, but the math shown does not assume any particular values.