(-4Xl2 + Xc*Xl2) + j*(6*Xl2)/ (6) + j*(4+Xl2-Xc) * (6)-j*(4+Xl2-Xc)/ (6)-j*(4+Xl2-Xc)

( -24Xl2 + 6*Xc*Xl2) + 36*j*Xl2) + (-4Xl2* (-j*(4+Xl2-Xc)) + Xc*Xl2(-j*(4+Xl2-Xc) + j*(6*Xl2)*(-j*(4+Xl2-Xc) numerator
( -24Xl2 + 6*Xc*Xl2) + 36*j*Xl2) + 16Xl2j + 4Xl2²j - 4*Xl2*Xc - Xc*Xl2*j*4 - Xc*Xl2² - Xc²Xl2 + 24Xl2 + Xl2²*6 - 6Xl2*Xc

+ 36*j*Xl2 + 16Xl2j + 4Xl2²j - 4*Xl2*Xc - Xc*Xl2*j*4 - Xc*Xl2² - Xc²Xl2 + Xl2²*6
- 4*Xl2*Xc - Xc*Xl2² - Xc²Xl2 +j(36Xl2+ 16Xl2 - Xc*Xl2*j*4 + Xl2²*6)
- 4*Xl2*Xc - Xc*Xl2² - Xc²Xl2 +j(52Xl2 - Xc*Xl2*j*4 + Xl2²*6)
..........................
.....................

 

(-4Xl2 + Xc*Xl2) + j*(6*Xl2)/ (6) + j*(4+Xl2-Xc) * (6)-j*(4+Xl2-Xc)/ (6)-j*(4+Xl2-Xc)

( -24Xl2 + 6*Xc*Xl2) + 36*j*Xl2) + (-4Xl2* (-j*(4+Xl2-Xc)) + Xc*Xl2(-j*(4+Xl2-Xc) + j*(6*Xl2)*(-j*(4+Xl2-Xc) numerator
( -24Xl2 + 6*Xc*Xl2) + 36*j*Xl2) + 16Xl2j + 4Xl2²j - 4*Xl2*Xc- Xc*Xl2*j*4 - Xc*Xl2² - Xc²Xl2 + 24Xl2 + Xl2²*6 - 6Xl2*Xc

+ 36*j*Xl2 + 16Xl2j + 4Xl2²j - 4*Xl2*Xc - Xc*Xl2*j*4 - Xc*Xl2² - Xc²Xl2 + Xl2²*6
- 4*Xl2*Xc - Xc*Xl2² - Xc²Xl2 +j(36Xl2+ 16Xl2 - Xc*Xl2*j*4 + Xl2²*6)
- 4*Xl2*Xc - Xc*Xl2² - Xc²Xl2 +j(52Xl2 - Xc*Xl2*j*4 + Xl2²*6)
..........................
.....................

I see some errors shown in red:

( -24Xl2 + 6*Xc*Xl2) + 36*j*Xl2) + (-4Xl2* (-j*(4+Xl2-Xc)) + Xc*Xl2(-j*(4+Xl2-Xc) + j*(6*Xl2)*(-j*(4+Xl2-Xc) numerator
( -24Xl2 + 6*Xc*Xl2) + 36*j*Xl2) + 16Xl2j + 4Xl2²j - 4*Xl2*Xc*j - Xc*Xl2*j*4 - Xc*Xl2² *j - Xc²Xl2*j + 24Xl2 + Xl2²*6 - 6Xl2*Xc

Also, don't overlook the fact that the Xl2²*6 term is NOT multiplied by j.

 

I see some errors shown in red:

( -24Xl2 + 6*Xc*Xl2) + 36*j*Xl2) + (-4Xl2* (-j*(4+Xl2-Xc)) + Xc*Xl2(-j*(4+Xl2-Xc) + j*(6*Xl2)*(-j*(4+Xl2-Xc) numerator
( -24Xl2 + 6*Xc*Xl2) + 36*j*Xl2) + 16Xl2j + 4Xl2²j - 4*Xl2*Xc*j - Xc*Xl2*j*4 - Xc*Xl2² *j - Xc²Xl2*j + 24Xl2 + Xl2²*6 - 6Xl2*Xc

Also, don't overlook the fact that the Xl2²*6 term is NOT multiplied by j.

It is very difficult to solve any technique.

 

It is very difficult to solve any technique.

You are very close to a correct result for the numerator. If you make the corrections I showed in red, and collect terms, you should end up with this:

6 Xl2^2 + j (52 Xl2 - 8 Xc Xl2 + Xc^2 Xl2 + 4 Xl2^2 - Xc Xl2^2)

Next, work out the multiplication for the denominator.

 

( -24Xl2 + 6*Xc*Xl2) + 36*j*Xl2) + (-4Xl2* (-j*(4+Xl2-Xc)) + Xc*Xl2(-j*(4+Xl2-Xc) + j*(6*Xl2)*(-j*(4+Xl2-Xc) numerator
( -24Xl2+ 6*Xc*Xl2) + 36*j*Xl2) + 16Xl2j + 4Xl2²j +4*Xl2*Xc*j - Xc*Xl2*j*4 - Xc*Xl2² *j+ Xc²Xl2*j + 24Xl2 + Xl2²*6 - 6Xl2*Xc




6 Xl2^2 + j (52 Xl2 - 8 Xc Xl2 + Xc^2 Xl2 + 4 Xl2^2 - Xc Xl2^2)

 

@The Electrician

My apologies!

It seems RRitesh has been carrying on a private conversation paralleling your posts here -- I missed the fact that two of his posts were to your thread

Misplaced posts deleted!

Most sincerely
HP

 

( -24Xl2 + 6*Xc*Xl2) + 36*j*Xl2) + (-4Xl2* (-j*(4+Xl2-Xc)) + Xc*Xl2(-j*(4+Xl2-Xc) + j*(6*Xl2)*(-j*(4+Xl2-Xc) numerator
( -24Xl2+ 6*Xc*Xl2) + 36*j*Xl2) + 16Xl2j + 4Xl2²j +4*Xl2*Xc*j - Xc*Xl2*j*4 - Xc*Xl2² *j+ Xc²Xl2*j + 24Xl2 + Xl2²*6 - 6Xl2*Xc




6 Xl2^2 + j (52 Xl2 - 8 Xc Xl2 + Xc^2 Xl2 + 4 Xl2^2 - Xc Xl2^2)

Now can you expand the denominator, which is the product:

[(6) + j*(4+Xl2-Xc)] *[ (6)-j*(4+Xl2-Xc)]

 

I have no problem with others pointing out genuine errors such as math errors, formatting errors, typos, etc., and I'm sure the mods wouldn't find any such to be OT. In fact, I would appreciate it.

 

@The Electrician how you do careful calculation?


[(6) + j*(4+Xl2-Xc)] *[ (6)-j*(4+Xl2-Xc)]
36+24j+ 6jXl2 - 6jXc + j*(4+Xl2-Xc)*-j*(4+Xl2-Xc)
36+24j+ 6jXl2 - 6jXc + 4j + jXl2 - Xcj - 4j - Xl2j +Xcj
36+24j

 

I have no problem with others pointing out genuine errors such as math errors, formatting errors, typos, etc., and I'm sure the mods wouldn't find any such to be OT. In fact, I would appreciate it.

Sincere thanks you for courtesy!

Again, I am impressed with RK's progress!

Best regards
HP

 

@The Electrician how you do careful calculation?


[(6) + j*(4+Xl2-Xc)] *[ (6)-j*(4+Xl2-Xc)]
36+24j+ 6jXl2 - 6jXc + j*(4+Xl2-Xc)*-j*(4+Xl2-Xc)
36+24j+ 6jXl2 - 6jXc + 4j + jXl2 - Xcj - 4j - Xl2j +Xcj
36+24j

This is a special property of complex arithmetic.

You probably remember one of the special products you learned in your first algebra course:

(a + b)*(a - b) = a^2 - b^2

With complex algebra, we get:

(a + jb)*(a - jb) = a^2 + b^2

Notice that (a - jb) is the conjugate of (a + jb), and in the expansion we're working on now
a = 6
b = (4+Xl2-Xc)

Proceed like this:

[(6) + j*(4+Xl2-Xc)] *[ (6)-j*(4+Xl2-Xc)]
36+24j+ 6jXl2 - 6jXc - 24j - 6jXl2 + 6jXc + j*(4+Xl2-Xc)*-j*(4+Xl2-Xc)

The blue and red terms cancel, and the last two parentheticals are multiplied with j and -j involved, so we get:

16 - 8 Xc + Xc^2 + 8 Xl2 - 2 Xc Xl2 + Xl2^2 for them.

The whole thing ends up being:

[(6) + j*(4+Xl2-Xc)] *[ (6)-j*(4+Xl2-Xc)] = (6)^2 + (4+Xl2-Xc)^2

Do you understand so far?

 

16 - 8 Xc + Xc^2 + 8 Xl2 - 2 Xc Xl2 + Xl2^2 for them.

Where 36 has gone having some problem.

j*(4+Xl2-Xc)*-j*(4+Xl2-Xc)
(4+Xl2-Xc)*(4+Xl2-Xc) as j²=-1
16+4Xl2-4Xc+4Xl2 + Xl2²- Xl2*Xlc-4Xc-Xl2*Xc+Xc²
16 + 8Xl2 - 8Xc + Xl2² - 2*Xl2*Xlc + Xc²+Xl²

 

Where 36 has gone having some problem.

j*(4+Xl2-Xc)*-j*(4+Xl2-Xc)
(4+Xl2-Xc)*(4+Xl2-Xc) as j²=-1
16+4Xl2-4Xc+4Xl2 + Xl2²- Xl2*Xlc-4Xc-Xl2*Xc+Xc²
16 + 8Xl2 - 8Xc + Xl2² - 2*Xl2*Xlc + Xc²+Xl²

The 36 didn't go anywhere.

This whole problem can be expanded like this:

[(6) + j*(4+Xl2-Xc)] *[ (6)-j*(4+Xl2-Xc)]

36+24j+ 6jXl2 - 6jXc - 24j - 6jXl2 + 6jXc + j*(4+Xl2-Xc)*-j*(4+Xl2-Xc)

The green part is the product of 6*6. The blue and red parts are the cross products and they cancel.

The purple part is another product and it gives us 16 - 8 Xc + Xc^2 + 8 Xl2 - 2 Xc Xl2 + Xl2^2

The full result is:

36 + 16 - 8 Xc + Xc^2 + 8 Xl2 - 2 Xc Xl2 + Xl2^2

Which is the same as (6)^2 + (4+Xl2-Xc)^2

Does that help?

 

yes ok.

 

Now I'm going to summarize what you've done. We chose a topology in post #54. We realized that the output of the circuit as shown in post #60 must be 50 ohms in order to match the 50 ohm load resistance.

There are two components C1 and L2 whose value (actually their reactance) must be determined to give us a match.

We have obtained an expression for the output impedance of the circuit in symbolic form; C1 and L2 are unknown at this point.

Let's examine our results to this point. in post #60 we have an expression for the output impedance and we rationalized the denominator:



The reason we've done all this is so we can solve for the values of C1 and L2. How can we proceed now? We know that we want the output of the circuit to look like a pure 50 ohm resistance. That means the imaginary part must be zero. This is very important to understand.

We have two components C1 and L2 that we need to determine, and we need two equations to do this. The fact that the output impedance has a real part and an imaginary part gives us the two equations.

We get one equation by setting the expression for the imaginary part (the red part above) equal to zero, For our second equation, we set the real part to 50. Let's start by setting the imaginary part equal to zero and solving for one of the unknown components:



Can you solve for Xl2 from the last equation above? Show your solution.

Do you understand why we are doing this?

 

Do you understand why we are doing this?

to get more power at output no mismatch

 

(4 - Xc ) Xl2 + (Xc² - 8Xc +52)=0
Xl2*4 - Xc*Xl2 + Xc² - 8Xc +52=0

 

to get more power at output no mismatch

That's our overall goal, but what I meant is do you understand why we are setting the imaginary part to zero? It's because the 50 ohm output impedance of the circuit must match the 50 ohm load, but the 50 ohm load is pure real, without any reactance. This fact, that the 50 ohm load has no reactive part (imaginary part) gives us one of the equations we need.

Now, can you show the solution for Xl2 derived from the last equation?

 

That's our overall goal, but what I meant is do you understand why we are setting the imaginary part to zero? It's because the 50 ohm output impedance of the circuit must match the 50 ohm load, but the 50 ohm load is pure real, without any reactance. This fact, that the 50 ohm load has no reactive part (imaginary part) gives us one of the equations we need.

Now, can you show the solution for Xl2 derived from the last equation?

We need a result that looks like this: Xl2 = (something)/(something)

 

(4 - Xc ) Xl2 + (Xc² - 8Xc +52)=0

Xl2=- (Xc² - 8Xc +52) / (4 - Xc )