Time for me to sign off for the nonce -- Please think about it! -- I'll post more clues, etc... tomorrow!

Chat with you tomorrow or next week (whatever works for you) -- Very best regards and kudos on your hard work!

Best regards
HP

 

ok, sure
thanks

 

ok, sure
thanks

While HP is away, you can do some learning on your own. Start here: https://en.wikipedia.org/wiki/Impedance_matching

Check out the references at the bottom of that page.

 

@RRITESH KAKKAR

Inasmuch as it seems the subject of port reactance compensation has become a ‘stumbling block’ to your study of the larger subject, I here offer a utilitarian ‘walk through’ of compensation procedure. --- Please understand that significantly ‘smoother’ means to said end are available and, in fact, in standard practice! – I feel this approach is of some educational benefit for its ‘transparency’ - note, for instance, that all formulae presented here are ‘variants’ of those applied to purely resistive networks…

Note that step 2 (below) is not elaborated inasmuch as it is what the exercise calls upon you to 'figure out'!!!

Finally -- Should you see something that makes no sense whatever – don’t despair! It may well be a 'typo'! – Although I’ve proofread this text, it was, nonetheless, composed in less than 30 minutes… IOW Please ask for clarification wherever!!!


So… To begin…
Inasmuch as ‘Smith Chart’ interpretation was the 'technique of choice' prior to the advent of readily available data processing devices, impedance specification preferred series equivalent form as a rule – while said ‘tradition’ endures to the present, parallel equivalent Z form specification is hardly ‘unheard of’…

The following offers an expedient (albeit much less than elegant) approach to port reactance compensation…

Given the complex impedance specification for each port and having selected the desired network configuration:

1) Where miscongress of network to port orientation is at issue, re-draft the affected port{s} as follows:

Where the reactive component of a parallel equivalent port impedance is to be subsumed by a series-connected network reactor, re-draft the port as series equivalent with component values as calculated below:
Rsport =(Rpport*Xpport^2)/(Rpport^2+Xpport^2)
Xsport = (Rpport^2*Xpport)/(Rpport^2+Xpport^2)

Where the reactive component of a series equivalent port impedance is to be subsumed by a parallel-connected network reactor, re-draft the port as parallel equivalent with component values as calculated below:
Rpport = (Rsport^2+Xsport^2)/Rsport
Xpport =(Rsport^2+Xsport^2)/Xsport

2) Determine the requisite network reactances sans compensation…
Apprehension of this step is the subject of the exercise!

3) Compensate network reactances as follows:

A) To compensate a series equivalent port reactance via adjustment of the series connected network reactor:
Xsadj=Xsnet-Xsport

B) To compensate a parallel equivalent port reactance via adjustment of the parallel connected network reactor:
Xpadj=(Xpnet*Xpport)/(Xpport-Xpnet)


Assignment of variables used in this discussion:
Rpport = The parallel equivalent port resistance.
Rsport = The series equivalent port resistance.

Xpport = The parallel equivalent port reactance.
Xsport = The series equivalent port reactance.

Xpnet = The unadjusted reactance of the parallel network reactor.
Xsnet = The unadjusted reactance of the series network reactor.

Xpadj =The adjusted reactance of the parallel connected network reactor.
Xsadj =The adjusted reactance of the series connected network reactor.

Best regards
HP

 

@RRITESH KAKKAR

So... Having had ample opportunity to explore the references suggested by @The Electrician in post #223 and to merely think about this ludicrously simple exercise -- Have you arrived at a solution? Have you so much as tried???

Seriously! -- Following the highly empirical (and, hence, intuitive) exposition of the compensation procedure (offered in post #224), the exercise has been reduced to nothing more than matching two non-reactive impedances with an L-Network!!! What, may I ask, is the problem!???

@RRITESH KAKKAR --- You have told us that you are a practicing EE -- Granting encumberment of a 'language barrier' taken with the basal nature of your inquires - it would seem the intent of your claim is that you are an aspiring engineer? -- Please be advised that conversance with impedance transformation is about as basic (and crucial) to said calling as it gets!

Respectfully
HP

 

HP It's time for you to ignore this jerk and stop making excuses for him! He says he is EE and no mistake cuz he says it in more ways! He says that but he doesn't know how to match impedance, he doesn't know how to apply Kirchhoff's law to ladder network, he doesn't understand linear algebra and even Gaussian elimination, fundamental calculus is over his head, he doesn't pay attention, he doesn't even try to think about problem and he is totally rude by ignoring posts and wasting time of anyone trying to teach him!

Either he is lying about being engineer or he is pretending to be ignorant of most basic requirements! Ether way he has proven that he is liar and I RESENT BEING LIED TO and so should you! HP I know you think all's well that ends well if he learns but don't you see? He doesn't want to learn he only wants to act like halfwit teen on FB and for my money he's doing good job of it!

HP I say it's good you don't to cave to peer pressure even from me! But I also say trying to get blood from turnip is just lame!

 

HP It's time for you to ignore this jerk and stop making excuses for him!

Indeed! I'm (finally) beginning to see that it's long past time!

Either he is lying about being engineer or he is pretending to be ignorant of most basic requirements! Ether way he has proven that he is liar and I RESENT BEING LIED TO and so should you

he is totally rude by ignoring posts and wasting time of anyone trying to teach him!

Time seems to have proven you (and many others) correct! -- FWIW: My, as it turns out, misplaced, solicitude owed to the observation that what passes for rudeness or deceit is often down to nothing more than misunderstanding or disparity of 'protocol' -- especially where language/cultural differences exist... It is, for instance, conceivable that the distinction between 'engineer' and 'technician' (as we understand the terms) may be 'blurred' in certain locales/cultures --- Even here - a 'railroad operator' is far nearer a technician than an 'engineer' in the broader sense of the term

HP I say it's good you don't to cave to peer pressure even from me!

My comments in post #180 were intended to 'assure' him that my burgeoning dubiety is based upon nothing beyond my observations of his behaviour --- To be clear - although I am my 'own person', as it were, I both accept and value input from others!

But I also say trying to get blood from turnip is just lame!

I tend to agree -- granting that one is certain a turnip is at issue!

All the best
HP

 

I am back i didn't get any material to study the impedance matching properly.

 

I am back i didn't get any material to study the impedance matching properly.

@RRITESH KAKKAR
We have nothing further to discuss prior to your satisfactory solution of the problem of post #113 -- I have supplied all the information required to solve the problem! -- Moreover, I suspect your renewed communicativeness owes more to flagging 'attention seeking prospects' than to an earnest desire to study --- Enough said! I will respond if and when, in my estimation, I see honest effort on your part!

Sincerely
HP

 

@RRITESH KAKKAR

You have, upon multiple occasions, indicated a desire for practically applicable problems -- So... I offer the following exercise in impedance transformation -- Please note that, owing to the stipulation of a two reactor network (and, hence, necessary dependence of Q) and to the fact that one of the termination impedances is pure real (i.e. pure resistive ) this is a trivially basic problem! --- Rest assured that you have demonstrated the requisite skills! -- That said you will need to THINK on how to apply said skills to this exercise!

So, to the exercize:
---------------------------------------------------------------------------------------------------------------------------------------------
Port 1 is is comprised of a 6Ω resistor in series with an inductor exhibiting 4Ω of reactance
Port 2 is merely a 50Ω resistor.

Match port 1 to port 2 using a network comprised of two lumped reactors... (said network to interviene terminals 'p1' and 'p2' as per the image (below)
What are the reactances of the network reactors and how are they connected?
---------------------------------------------------------------------------------------------------------------------------------------------
Special thanks to @Aleph(0) for her suggestion of an exercise of this type
View attachment 97483

Best regards
HP

@WBahn may help to understand.
the #113

 

@WBahn may help to understand.
the #113

If you receive help from others that's ok with me -- That said -- you already know how to solve the problem and could do so if you'd but think about it!!!

 

I have seen circuit many time but don't understand it how and what to do with it.
You said impedance matching but how to do?

 

@RRITESH KAKKAR

Below is a 'compilation' of hints, illustrations and examples offered on this thread -- these alone are more than sufficient to solve the exercise of post #113! --- Please review them carefully! - Then think about the problem! - I've shown you how to compensate port reactance in post #177 (quoted below) -- so you need concentrate only on transformation - if you take a moment to think you'll realize that you already know how to solve the problem! -- Please try!!!

Best regards
HP

Hint from post #117

Please note: the below quoted 'hint' verges on exposition!

Re: Two reactor networks -- As may should be obvious:
System Quality Factor (Q) = √(transformation_ratio-1) {transformation_ratio|transformation_ratio>1}
Please 'dwell' on it! --- The implications of this are 'huge'!!!

Discussion from post #145

@RRITESH KAKKAR

It occurs to me that the 'block' you seem to be experiencing with the current exercise may owe to misapprehension as regards impedance transformation via networks (arrangements) of discrete reactors...
Here I present an example of transformation of 10v to over 60V via a series resonant RLC circuit --- While not directly applicable to the exercise, this is intended to set you upon the right path!
Note that, with some qualification, impedance may be thought of as EMF to current ratio!

Please note that following is not to be confused with other impedance transformation schemes/phenomena (i.e. mutual inductance, 'buck/boost', etc...)

/////////////////////////////
Within the confines of this discussion, resonance is a condition wherein 'opposite' reactor types exhibit reactances of equal magnitude but opposite sign and, hence, 'cancel' -- Below is a representation of a series RLC circuit at resonance -- Please consider the following:

Because EMF lags current by 90° in a purely reactive, net capacitive circuit, capacitive reactance is said to be negative.
Because EMF leads current by 90° in a purely reactive, net inductive circuit, inductive reactance is said to be positive.

With reference to the below attached schematic:
Reactor 1 (The capacitor) exhibits a reactance of -2πΩ @ a frequency of 100MHz
Reactor 2 (The inductor) exhibits a reactance of 2πΩ @ a frequency of 100MHz
Hence the circuit is said to be resonant @ 100MHz

Owing to their series connection, the net reactance = the sum of the reactances = 0Ω. Thus the net impedance = the value of R = 1Ω
Thus it is that the current through the circuit is determined by the EMF of the power supply (1oV Peak) and the resistance of R (1Ω) → 10V (Peak)/1Ω= 10A (Peak)

Ohm's law tells us that the EMF across a pure reactance = I*X = 10A*2π ≈ 62.83V in the case of the inductor --- And 10A*2π ≈ -62.83V in the case of the capacitor (note that the signs indicate the respective phase shifts)
In theory any finite transformation ratio is possible via proper choice of the reactors and the resistor.

Illustration of a series RLC circuit at resonance -- Post continued below image...
View attachment 97569

As an aside please consider a parallel resonant circuit.

Via the technique of calculating the equivalent value of paralleled resistors or pure reactances:
1/(1/Xc+1/Xl)=Xnet but since Xc=-Xl at resonance: 1/(1/-x + 1/x) = 1/(0/x^2)=1/0 = ∞ (For these purposes)
Thus we see that a parallel LC circuit at resonance presents as an 'open circuit'

Example From post #177

@RRITESH KAKKAR
By way of illustration, I have created the following example of a typical problem in impedance transformation/matching - with correct solutions -- Please note that, although I have chosen a low-pass transformation network, such is not mandatory for the exercise of post #113

Inasmuch as 'prompting' discovery of your existing ability to solve this sort of problem is the 'point' of post #113, I do not offer explicit instructions/formulae -- I do, however, offer insight into port reactance compensation as well as exact solutions (such that you may 'behold' the subject from a conceptual/intuitive standpoint)

Here is the statement of the problem: Design an 'L-Network' to match port 1 to Port 2: --- (Scroll down to the next image to see the solution)
View attachment 97615


Here is a properly matched system: (Please see next image for some perspective on port reactance compensation)
View attachment 97616

Here is the system sans port reactances -- What do the differences in the network reactances (by comparison to the above image) tell you about compensation of port reactance??? (Post continued below image)
View attachment 97617

@RRITESH KAKKAR

Please explore the above examples! -- Further to that - a review of post #145 'through the lens' of your knowledge of resistance/reactance behaviour is highly advised!

Note that solution of the problem of post #113 is less involved than that of the above example in that only a single port is reactive and its impedance is in 'series form'...

 

OK, i will tell in step.
1st the frequency of both circuit are same?

2nd The Resistance will not matter as it is linear and there is no change in with frequency etc

3rd The PORT 1 has series capacitor which has negative voltage it will be cut down by series induct or , right?

4th the parallel inductor has +ve voltage which wil be cut by parallel capacitor , right?

and how the value is calculated of it reactance?

 

 

OK, i will tell in step.
1st the frequency of both circuit are same?
2nd The Resistance will not matter as it is linear and there is no change in with frequency etc
3rd The PORT 1 has series capacitor which has negative voltage it will be cut down by series induct or , right?
4th the parallel inductor has +ve voltage which wil be cut by parallel capacitor , right?
and how the value is calculated of it reactance?

1) Correct

2) Incorrect -- While it is true that non-reactive impedance (i.e. resistance) does not change with frequency, it is, nonetheless, a component of port impedance and, hence, an essential 'element' of the transformation ratio... --- Note: apprehension of frequency is not necessary to the solution of the exercise inasmuch as reactances are specified and sought...

3) Capacitive reactance is said to be negative (historically because EMF lags current in net capacitive circuits) -- While said assignment is essentially 'arbitrary' the sign analogy is valid inasmuch as (90°+(-90°)=180° -- Hence Xnet=Xl-Xc --- That said I recommend exclusive use of signed reactance (as opposed to 'Xc' and 'Xl') in computation...

4) Paralleled reactances are analogous to paralleled resistances -- as you'd know had you carefully reviewed post #177 -- also please 'drop' use of "+ve" it is both nonstandard and ambiguous...

Thank you for asking germane questions!

Best regards
HP

 

how these value are coming?

 

how these value are coming?

What part of:

Inasmuch as 'prompting' discovery of your existing ability to solve this sort of problem is the 'point' of post #113, I do not offer explicit instructions/formulae -- I do, however, offer insight into port reactance compensation as well as exact solutions (such that you may 'behold' the subject from a conceptual/intuitive standpoint)

Don't you understand!?

Jeeeze! I've explicitly shown you how to perform compensation - all you are being asked to 'do' is the transformation!!! -- What is the problem????

 

is this formula you were applying?

 

No i will do it.