# Discusisng imaginary part of impedance in L network impedance matching circuit....

#### Himanshoo

Joined Apr 3, 2015
260
Hi folks,
I went through this extremely helpful and informative thread.... post 3.... and knew about the physical significance of imaginary part of the impedance.
Here I am presenting a circuit which is an L network used for matching.On simplifying the circuit the equivalent circuit is shown below.
Now the text says that in the equivalent circuit the imaginary part of the impedance of inductor and capacitor cancels each and the circuit is matched.

Now I am not able to physically visualize both the imaginary impedance cancelling each other.
Can someone explain it in reference to post 3....

source of article...
thread url...Physical meaning behind an imaginary impedance

regards!

#### drc_567

Joined Dec 29, 2008
751
One way to visualize the complex impedance problem is to equate the real resistance as a horizontal vector. Next, at the extreme tip of this resistance vector, place the inductive reactance at a right angle, and in a positive sense. The capacitive reactance is co-linear with the inductive reactance, but oriented in the negative sense. The algebraic sum of the inductive and capacitive reactances is the value of interest. If the algebraic sum is zero, then the total impedance for the circuit will just be the value of the remaining resistive vector. However, if there is a portion of the orthogonal reactive vector remaining, being a quantity other than zero, then the circuit will have either an inductive component (+), or a capacitive component (-). The hypotenuse of the resistor-reactance triangle is the sum or total impedance, which is calculated geometrically, rather than using simple addition.
... This method of visualization works equally well with power circuits that utilize alternating current.

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• Himanshoo

#### crutschow

Joined Mar 14, 2008
23,725
An inductive impedance generates a leading voltage across it.
A capacitive impedance generates a lagging voltage across it.
When you connect them together in series they will subtract from each other (the pull from the lag is cancelled by the push from the lead).
Thus when the two are equal, the lag cancels the lead and we end up with zero impedance.

Below is the LTspice simulation of a simple series RLC circuit to show that.
The circuit is resonant at about 5kHz.
At that frequency you can see that the inductor voltage, V(1,2), is equal to, and 180 degrees out of phase with, the capacitor voltage, V(2,3).
This means the voltages cancel and the voltage across both of them, V(1,3) is zero.
Thus the voltage across the resistor, V(3) equals the source voltage, V(1) • Himanshoo

#### MrAl

Joined Jun 17, 2014
6,771
Hi folks,
I went through this extremely helpful and informative thread.... post 3.... and knew about the physical significance of imaginary part of the impedance.
Here I am presenting a circuit which is an L network used for matching.On simplifying the circuit the equivalent circuit is shown below.
Now the text says that in the equivalent circuit the imaginary part of the impedance of inductor and capacitor cancels each and the circuit is matched.

Now I am not able to physically visualize both the imaginary impedance cancelling each other.
Can someone explain it in reference to post 3....

source of article...
thread url...Physical meaning behind an imaginary impedance

regards!
Hi,

A simple way to look at it is to just look at the complex number representation of an impedance. It becomes immediately clear what happens.

A general complex number is represented by a real part and imaginary part:
z=a+b*j

In the above, 'z' is the complex impedance, 'a' is the real part, and 'b' is the imaginary part.

Now if we had two complex impedances z1 and z2 we would have:
z1=a1+b1*j
z2=a2+b2*j

and if these two were in series, then they would simply add so we would get:
z1+z2=a1+a2+(b1+b2)*j

It doesnt take much to notice that if b2=-b1 then the imaginary parts cancel and we are left with a completely real quantity:
z=a1+a2

However, if these two complex impedances are in parallel then we have a more complicated scenario:
z=z1*z2/(z1+z2)

where we have to solve:
b1*b2^2+b1^2*b2+a1^2*b2+a2^2*b1=0

in order to see cancelation of the imaginary parts.
Making b2=-b1 alone doesnt work anymore, but one solution is if a2=a1 then making b2=-b1 does work.
However, if a2=a1*2 for example then we have to solve that equation to find out what works. For example, if a1=1 and b1=4 we get two solutions:
b2=-(sqrt(33)+17)/8
b2=(sqrt(33)-17)/8

So in the series case it is quite easy to do, in the parallel a bit harder to do.

Also, the impedance then looks purely resistive when the two cancel.

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• Himanshoo

#### crutschow

Joined Mar 14, 2008
23,725
A simple way to look at it is to just look at the complex number representation of an impedance. It becomes immediately clear what happens.
..................
I, for one, never found anything about a complex mathematical representation of an impedance "simple" or that it makes clear what happens.
I can crank through the numbers and come up with the right answer, but that doesn't explain "how the circuit works".
It's just a mathematical representation of that which gives little physical insight as to what's actually happening.
And I think that's what the OP is looking for, which was the intent of my post #3.

• GopherT and Himanshoo

#### Himanshoo

Joined Apr 3, 2015
260
the pull from the lag is cancelled by the push from the lead
In order to rectify your statement I came to an article...which says....

*Assuming LC parallel circuit connected to source and load.
In the "C" part of the circuit the current leads the voltage and in the "L" part the current lags behind the voltage. If the values of inductance and capacitance are selected so that both offer the same impedance at the frequency of the alternating current supply, then the current through both "L" and "C" parts will be equal. But since one is a quarter of a cycle behind the voltage, and the other is a quarter of a cycle in front of the voltage, there is a difference of phase of a half cycle between the currents in the "L" and "C" parts. As one current is positive, the other current is negative (i.e., flowing in the opposite direction) and the same size as the positive current. So the two currents cancel each other out, and as a result no current flows out of the "L" and "C" combination, although there is a voltage connected across the pair of them.

So could you explain so called " push and pull of lag and lead" in reference to above text...

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#### MrAl

Joined Jun 17, 2014
6,771
I, for one, never found anything about a complex mathematical representation of an impedance "simple" or that it makes clear what happens.
I can crank through the numbers and come up with the right answer, but that doesn't explain "how the circuit works".
It's just a mathematical representation of that which gives little physical insight as to what's actually happening.
And I think that's what the OP is looking for, which was the intent of my post #3.
Hello,

Complex number representation of the quantities involved is an excellent way to show how they cancel if not what is happening physically. There may be different physical views and that could be one of them.
What you said about the phase lead and lag is really just the same. You have to calculate the phase that's the only difference, and then add the two just like i did with the complex numbers for the series case. The phase can be calculated from the complex representations. So there is not much difference.

To understand from a more physical viewpoint we usually have to look at the energy, because that's all we have that can be called 'physical'. The voltage itself is not enough nor is the current, and the phase alone isnt that much help either if you dont accept the complex representation.

The energy levels in the cap and inductor are somewhat unique when we have perfect cancelation. The sum of energies in L and C add to a constant over time, and the energy in the inductor completely transfers to the capacitor (and vice versa) during one cycle. So there is a complete exchange of energy between L and C that we dont see if there is not perfect cancelation, and the total energy is a constant that does not vary in the circuit. When we dont have perfect cancelation, either L or C has a bigger change of energy level over one cycle than the other, and the sum does not add to a constant but fluctuations just like the AC current or voltage itself, so the total energy varies.

So if you are looking for a better physical explanation then you should look at the energy in both L and C and see how they relate to each other when there is perfect cancelation versus when there is not perfect cancelation. It's resonance at it's best.

• Himanshoo

#### MrAl

Joined Jun 17, 2014
6,771
In order to rectify your statement I came to an article...which says....

*Assuming LC parallel circuit connected to source and load.
In the "C" part of the circuit the current leads the voltage and in the "L" part the current lags behind the voltage. If the values of inductance and capacitance are selected so that both offer the same impedance at the frequency of the alternating current supply, then the current through both "L" and "C" parts will be equal. But since one is a quarter of a cycle behind the voltage, and the other is a quarter of a cycle in front of the voltage, there is a difference of phase of a half cycle between the currents in the "L" and "C" parts. As one current is positive, the other current is negative (i.e., flowing in the opposite direction) and the same size as the positive current. So the two currents cancel each other out, and as a result no current flows out of the "L" and "C" combination, although there is a voltage connected across the pair of them.

So could you explain so called " push and pull of lag and lead" in reference to above text...
Hi,

So you want to talk about an L and C in direct parallel with no resistances (ie no 'real' parts) ?

• Himanshoo

#### Himanshoo

Joined Apr 3, 2015
260
So you want to talk about an L and C in direct parallel with no resistances (ie no 'real' parts) ?
No with real parts..

#### crutschow

Joined Mar 14, 2008
23,725
So could you explain so called " push and pull of lag and lead"
I was referring to my series circuit.
For a parallel LC circuit it's the currents that are leading or lagging the voltage.
Below is an LTspice simulation of the same circuit components with the L and C in parallel and at resonance.
You can see that the L and C currents are 180 degrees out of phase and thus cancel each other, with essentially no current going through the resistor.
The capacitor current is also leading the voltage across it, V(1,2), by 90° and the inductor current is lagging by 90° • Himanshoo

#### MrAl

Joined Jun 17, 2014
6,771
No with real parts..
Hi,

Ok then if you want a resistance in series with the inductor and a perhaps different resistor in series with the capacitor, then i gave the formula previously:
z=z1*z2/(z1+z2)

and when both real parts are equal then cancellation occurs when 2*pi*f=1/sqrt(L*C) again.
However, if both real parts are NOT the same then we have to solve that equation i gave in the previous post.
When cancellation does occur the inductor has more energy some of the time and the cap has more energy some of the time, and the sum does not equal a constant value but fluctuates.

In the series case when the inductor had max energy the cap had min energy and vice versa, but here the inductor has less than max energy when the cap has min energy and vice versa.

A reason for any of these cases is that the two elements can store energy and then later deliver it back to the circuit and they do this at regular intervals. The storing of energy and giving it back later at a different time is paramount to power factor correction. The whole thing about a circuit that is not power factor corrected is that it stores power at times we dont want it to do so, so by changing the timing of the storage and redelivery of energy we can control the power factor, and we usually do this by introducing new circuit elements such as extra capacitors or inductors into the circuit.

• Himanshoo

#### Himanshoo

Joined Apr 3, 2015
260

Basically I got the idea that..In series LC circuit voltage are out of phase i.e opposite in polarity and gets canceled and in parallel current cancels due to being opposite to each other in both L and C....
but I am looking for a diagram which illustrate voltage polarity for L and C in series circuit for each separate half cycles of input signal..

#### crutschow

Joined Mar 14, 2008
23,725
but I am looking for a diagram which illustrate voltage polarity for L and C in series circuit for each separate half cycles of input signal..
That's in my post #3.
V(1,2) shows the voltage across the inductor and V(2,3) is the voltage across the capacitor.
V(1) is the input signal.

• Himanshoo

#### Himanshoo

Joined Apr 3, 2015
260

#### MrAl

Joined Jun 17, 2014
6,771

Basically I got the idea that..In series LC circuit voltage are out of phase i.e opposite in polarity and gets canceled and in parallel current cancels due to being opposite to each other in both L and C....
but I am looking for a diagram which illustrate voltage polarity for L and C in series circuit for each separate half cycles of input signal..
Hi,

Also, do you use any circuit simulator? They can help a lot because you can draw a circuit in there and then run it, and look at any waveforms you wish to study. I highly recommend you doing this. A free simulator is the one commonly known as LT Spice.
You should also back up any conclusions you make with some math. The results from your math should match the results of the simulation in most cases.

• Himanshoo and Sinus23

#### Himanshoo

Joined Apr 3, 2015
260
You should also back up any conclusions you make with some math.
what I believe is that maths doesn't provide real understanding of the circuit..it make me hard to visualize the phenomenon using complex math...

Yes btw LTSpice is worth trying...

#### MrAl

Joined Jun 17, 2014
6,771
what I believe is that maths doesn't provide real understanding of the circuit..it make me hard to visualize the phenomenon using complex math...

Yes btw LTSpice is worth trying...
Hi again,

Well just to note, you should really strive to find the meld between math and the circuit. Math is more complete than intuition. Intuition provides the guide, math provides the exact answers.
Simulators use math that's how they get some good results.

• Himanshoo

#### crutschow

Joined Mar 14, 2008
23,725
Simulators use math that's how they get some good results.
True. Of course simulators generally use iterative math to converge to the (usually approximate) solution, not the closed form mathematics that humans usually use. • Himanshoo

#### MrAl

Joined Jun 17, 2014
6,771
True. Of course simulators generally use iterative math to converge to the (usually approximate) solution, not the closed form mathematics that humans usually use. Hi,

Well i am human and i use iterative methods Not to be pedantic here, but just a few little points...

The type of calculation depends on what method is being used. For example Taylor's is not iterative in nature while Gear's is. By not iterative i mean each solution point is computed in one step.
That's all math too though, adding, subtracting, multiplying, and dividing, exponentiation.

But more to the point, we normally talk about the simpler circuits used in learning courses where calculations done by hand are practical. That, and the estimation of more complex systems with simpler first or second order systems is also a good thing to know.
I stress math at least to some degree because it holds some secrets that simulators dont show and may never be able to show, although i think we would all be at a disadvantage if we didnt have them anymore. With that in mind, to paraphrase a great writer:
"Romeo, what makes thou hours so long?"
"Not having that, which having, makes them short."
Obviously Romeo cant find his simulator It's interesting that math can show a whole slew of solutions in one line which if interpreted correctly can tell a person a lot about a circuit, unlike a simulator which can show just one set of results which may or may not illustrate a key point about the circuit. For example, if we ran a simulation of a simple LC tank circuit and didnt know about resonance, we might actually mist one of the most important aspects of the circuit if we did not set the frequency range correctly. Knowing about resonance in advance allows us to be sure we cover that frequency at the very least, or at least know that we should look for it.

Granted many things must be learned and i have to press on for math, but also for the use of a decent simulator. I think they are both tools that are available to us in this day and age in many forms.

There is also much to be said about the 'after' effects of learning the math combined with the physics. To some folks they may not like to use the math that involves complex numbers, but to me that is the only way to do it right, and the physics unfolds from the math just as the math unfolds from the physics once we become familiar with the math forms required. For a quick example, feedback shows up in some equations so naturally it's hard to imagine a better way to see it illustrated. If someone doesnt yet understand the form yet, they may never have that opportunity to see that kind of beauty in math, so learning it is a good idea.

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• anhnha and Himanshoo