impedance of RLC circuit from phasor

[hanhan]

I want to ask about some basic concepts.
For example, I have a RLC series circuit like this:

And here is the phasor diagram to calculate impedance of the circuit:

Why we know that the hypotenuse of the right triangle Z with two sides R and XL - XC is the total impedance of the circuit?

 

[MrAl]

Hi,


There are probably a lot of ways to look at this, here is one way...

Since it is a series circuit, the impedance is the sum of all the complex impedances:
Z=zR+zC+zL

and the impedances for the individual elements are:
zR=R
zC=1/jwC
zL=jwL

so summing these we get:
Z=R+1/jwC+jwL

and 1/jwC=-j/wC so we have:
Z=R+jwL-j/wC

and we can recognize that the reactance for L and C are:
xL=wL
xC=1/wC

so we can write the impedance as:
Z=R+j*xL-j*xC

and factoring slightly we get:
Z=R+j*(xL-xC)

and this is in complex form with the real part being R and the imaginary part being xL-xC:
Real=R
Imag=xL-xC

and we know that the magnitude of this pair is:
A=sqrt(Real^2+Imag^2)

which can also be looked at as the hypotenuse.

So we get:
|Z|=sqrt(Read^2+Imag^2)=sqrt(R^2+(xL-xC)^2)

You might want to note that this is the magnitude of the impedance Z not the complex impedance which is just Z=R+j*(xL-xC).

So the magnitude of Z is:
|Z|=sqrt(R^2+(xL-xC)^2)

and the complex impedance is:
Z=R+j*(xL-xC)

The phase is:
TH=atan(Imag/Real)=atan((xL-xC)/R)

 

[Ratchit]

anhnha,

Why we know that the hypotenuse of the right triangle Z with two sides R and XL - XC is the total impedance of the circuit?

Because the voltages existing across XL and XC are orthogonal (at right angles) phasewise with respect to the voltage across R. Therefore, the component voltages are plotted as a right triangle. The rest is simple geometry.

Ratch

 

[hanhan]

Thanks for help.
I meant that why we know for sure that |Z|=sqrt(R^2+(xL-xC)^2) is the impedance of the circuit not something like this |Z|= R^2+(xL-xC)^2 or something else. On what basis can we say that this value sqrt(R^2+(xL-xC)^2) is the impedance of the circuit?
Maybe, I miss some basic concepts here, please be kind.

 

[Ratchit]

anhnha,

I meant that why we know for sure that |Z|=sqrt(R^2+(xL-xC)^2) is the impedance of the circuit not something like this |Z|= R^2+(xL-xC)^2 or something else. On what basis can we say that this value sqrt(R^2+(xL-xC)^2) is the impedance of the circuit?
Maybe, I miss some basic concepts here, please be kind.

Which one of us are you asking? MrAl, Ratch, or both of us?

If you understand what impedance is, you would not have to ask such a question. What is impedance? In a sinusoidal series circuit, it is the opposition to current existence through a device or circuit when a sinusoidal voltage is applied. For series circuits, the current is in phase with the voltage across any resistance. The current is 90° out of phase with respect to the voltage in any L or C component in the circuit. In a series circuit, the current is the same throughout, and is used as a reference to specify the phase of the voltage. You build the voltage triangle by plotting the in phase voltages and the orthogonal voltages. That gives you a right angle triangle with the resistor voltages on the x-axis and the reactance voltages on the y-axis. Then you can easily make a impedance triangle by dividing the voltages by the current of the circuit. The impedance will be the hypotenuse of the impedance triangle by simple geometry, just as you stated above. It cannot be anything else. In conclusion, you get the impedance of a series circuit by summing resistance directly, and adding reactance orthogonally. The result will be a complex number called the impedance.

Ratch

 

[MrAl]

Thanks for help.
I meant that why we know for sure that |Z|=sqrt(R^2+(xL-xC)^2) is the impedance of the circuit not something like this |Z|= R^2+(xL-xC)^2 or something else. On what basis can we say that this value sqrt(R^2+(xL-xC)^2) is the impedance of the circuit?
Maybe, I miss some basic concepts here, please be kind.

Hi,

We are not really sure what you are asking but that's ok, if you keep at it we will eventually know what you want to know

The short answer is because the magnitude of the impedance is the norm of the complex impedance. And the complex impedance is like a vector with two components, one real and one imaginary. The norm is the square root of the sum of the squares of the real and imag parts of the complex impedance, so it is the same as the hypotenuse. This idea comes from vector analysis where we could think of the real part being in another dimension as the imaginary part so they are 90 degrees out of phase, but when we find the norm we are finding the magnitude of the vector and applied to impedances that means it is the magnitude of the impedance.

We sometimes call the magnitude of the impedance just simply the impedance, and the true impedance the complex impedance, but really the complex impedance tells us more because we can also get the phase shift from that whereas we can not get the phase shift from knowing just the magnitude.

What you are looking for is the magnitude of the impedance which is written |Z|, and that is just the norm of the vector made from:
v=[realpart,imagpart]

which is also represented as:
v=realpart+imagpart*i

or that lower case 'i' is made as 'j' in electronic work so it is not confused with the current 'i'.

When we find the magnitude of that vector we take the square root of the sum of the squares of the two components:
|v|=sqrt(realpart^2+imagpart^2)

and that's norm of the vector and that is the way we do it with all vectors. The complex impedance is like a vector so we do it with that too.

Note that it is not anything else like just realpart^2+imagpart^2 because that is not the norm. We have to find the norm not just the sum of squares.

If this is still not clear then try to rephrase your question or add to it, or even supply another example.

 

[hanhan]

Hi, sorry for the late reply! To be honest, I really can't explain myself clearly. In general, I want to completely understand how complex number get into electrical engineering in terms of mathematics. I want to know the the formula zL = jωL and zC = -j1/ωC are formed. At present, I think I am familiar with the use of complex number in solving circuits. But I don't really how it work.

 

[MrAl]

Hi,

DC circuits can have variables like voltage represented by a single number like 10, 12, 15, etc, for 10v, 12v, 15v, etc. But AC circuits have voltages which must be represented by BOTH a voltage and a phase. These quantities are in general unrelated to each other so they have to be specified by two different numbers like 10 volts 90 degrees, or 12 volts with a phase of pi/2 radians. Note we are stuck having to use TWO numbers rather than just one to specify the voltage now.
With DC we have just say 10 volts, but with AC we have 10 volts at 90 degrees.
So with DC we have just "10", but with AC we have "10" and also "90" there.

Now complex numbers have two parts also, the real part and the imaginary part. And it is interesting that we can show a phase shift by noting the relationship between the real part and imaginary part:
Phase=atan(imag/real)

and amplitude by the square root of the sum of squares:
Ampl=sqrt(real^2+imag^2)

So we can represent an AC voltage which has amplitude A and phase shift TH as:
Vac=(A,TH) {sometimes written as "A angle TH")

or we can just state the real and imaginary parts:
Vac=(R,I)

or we can just use the syntax for a complex number:
Vac=R+I*j

So you see now that the AC voltage with amplitude and phase shift has an equal representation as a complex number. And this is true of other AC quantities also such as current and impedance.

Complex numbers are introduced so that we can use complex math to calculate circuit quantities in AC circuits with the same techniques we use to calculate quantities in DC circuits, so it makes sense to use them when we do AC circuits.

 

[Ratchit]

anhnha,

At present, I think I am familiar with the use of complex number in solving circuits. But I don't really how it work.

That is typical of "plug 'n chug" learning.

Complex numbers could also be called duplex numbers. They have a real part and a orthogonal (90° direction from the real part). You would do well to read this thread so as to fix in your mind what a "complex" number is. https://www.electro-tech-online.com/threads/why-dont-imaginary-numbers-make-so-much-sense.121331/

Complex numbers are useful in sinusoidal voltage and current calculations because in a circuit, the inductive current lags the voltage by 90°, and the capacitive current leads the voltage by 90°. That makes sinusoidal voltage/current easy to plot by magnitude and phase (phasors).

I want to know the the formula zL = jωL and zC = -j1/ωC are formed.

Any good physics book will tell you how Xc and Xl is derived.

Ratch

 

[hanhan]

Thanks for replies!
MrAl,

DC circuits can have variables like voltage represented by a single number like 10, 12, 15, etc, for 10v, 12v, 15v, etc. But AC circuits have voltages which must be represented by BOTH a voltage and a phase. These quantities are in general unrelated to each other so they have to be specified by two different numbers like 10 volts 90 degrees, or 12 volts with a phase of pi/2 radians. Note we are stuck having to use TWO numbers rather than just one to specify the voltage now.
With DC we have just say 10 volts, but with AC we have 10 volts at 90 degrees.
So with DC we have just "10", but with AC we have "10" and also "90" there.

Here is my understanding about complex number: Complex numbers are really just pairs or numbers with a system of math attached to them that defines things like addition and multiplication in a manner that makes them useful to model two dimensional phenomenon.
More: https://mathforum.org/library/drmath/view/53809.html
Does the bold mean that we can represent most of two unrelated things by a complex number? For example, in your example it is amplitude and phase. Maybe, we can represent man and woman, length and width of an object,...
Is that right?
MrAl and Ratch,
Please help me explanning the part from wikipedia: https://en.wikipedia.org/wiki/Electrical_impedance

What is the definition of ZL here?
To me it seems that ZL = ωL*e^(jωt)*e^j π/2 / e^(jωt) = ωL*e^(jπ/2) = jωL
But it doesn't sound right to me. Why we need a complex impedance? We are dealing with all real quantities but now the complex impedance is introduced. I can't understand the connection between them.
Why we don't use the definition?
The formula is intuitive and seems logical to me but it is not used.
ZL = vL(t)/iL(t) = ωL*sin(ωt + π/2) / sin(ωt) = Im{ ωL*e^(jωt)*e^j π/2} / Im{e^(jωt)}
However, this cannot be simplified more.

 

[hanhan]

Hi,
I think my problem is the part "validity of complex representation" here: https://bmia.bmt.tue.nl/people/BRomeny/Courses/8C120/pdf/ElectricalImpedance.pdf

For example, let consider RLC circuit as a system that has a transfer function y = f(x) where x is input and y is the output of the system.
Now we consider the case when the input is a sinusoid x = cos (ωt + θ).
According to Euler's formula:
x = cos (ωt + θ) = 1/2 ( e^(jωt + θ) + e^-(jωt + θ))
By the principle of superposition, we may analyse the behaviour of the sinusoid on the left hand side by analysing the behaviour of the two complex terms on the right-hand side. Given the symmetry, we only need to perform the analysis for one right-hand term; the results will be identical for the other.
This means that we only need to find the output of one input x1 = e^(jωt + θ):
y1 = f(x1) = a + jb
Given the symmetry, we only need to perform the analysis
for one right-hand term; the results will be identical for the other.
This means that with x2 = e^-(jωt + θ) we have y2 = f(x2) = a- jb

Then y = y1 + y2 = (a +jb) + (a - jb) = 2a.
My question is:
With the system above y = f(x)
we have y1 = f(x1) = a + jb where x1 = e^(jωt + θ)
Then with x2 = e^-(jωt + θ) why we know that the output y will have the following form?
y2 = f(x2) = a - jb

 

[Jony130]

Try to watch this video lectures, start from lecture 12.
https://ocw.mit.edu/courses/electri...s-and-electronics-spring-2007/video-lectures/

 

[hanhan]

Hello Jony, long time!
Don't know why I can't load the link. But I watched all the videos of the course and don't see where he mentions about this. He used Laplace transform. My confusion is about the validity of complex representation.

 

[MrAl]

Hi again,


The short answer is that one term is the conjugate of the other, so when you add the two you get the real part times 2 and dividing by 2 you get the real part alone. So this is equivalent to just taking the real part of one of the terms, period. So where "Real(x)" represents taking the real part of x and f* is the conjugate of f we have:
f(t)+f*(t)=Real(f(t))=Real(f*(t))

So we could look at it like this:
((a+bj)+(a+bj)*)/2=((a+bj)+(a-bj))/2=a=Real(a+bj)=Real(a-bj)

So nature threw us a real nicety for a change

 

[hanhan]

Hi,
Let me explain my understanding about the validity of complex representation more clearly.
The bold are the words quoted from the text "Validity of complex representation" in the link above.
Considering RLC circuit as a system that has a transfer function y = f(x) where x is input and y is the output of the system.
if x = cos(ωt + θ):
y = f(cos(ωt + θ)) = f(1/2*e^j(ωt + θ) + 1/2* e^-j(ωt + θ))
By the principle of superposition, we may analyse the behaviour of the sinusoid on the left hand side by analysing the behaviour of the two complex terms on the right-hand side.
This means that y = f(cos(ωt + θ)) = f(1/2*e^j(ωt + θ) + 1/2* e^-j(ωt + θ)) = f(1/2*e^j(ωt + θ)) + f(1/2* e^-j(ωt + θ))
My first question: Why this operation give the correct answer? f(x + y) = f(x) + f(y)???
Given the symmetry, we only need to perform the analysis for one right-hand term; the results will be identical for the other.
This means that we only need to compute f(1/2*e^j(ωt + θ)); the result for f(1/2*e^-j(ωt + θ)) will be inferred from f(1/2*e^j(ωt + θ)).
My second question: What is the relation between f(1/2*e^j(ωt + θ)) and f(1/2*e^-j(ωt + θ))?

 

[MrAl]

Hello again,


Your second question reveals why this doesnt make sense to you i think. Lets see if this helps...

For your first question, you wrote:
Why this works: f(x+y)=f(x)+f(y)

but that's not really the question you should be asking if you knew the answer to your second question, i think. You would have instead asked this:

Why this works: f(0.5*(x+x'))=f(0.5*x)+f(0.5*x') where x' is the complex conjugate of x.

So maybe now you have already guessed what is coming next

e^-j(wt+θ) is the complex conjugate of e^j(wt+θ) , and that should answer your second question.

So as i said in my previous post, the imaginary parts cancel out in the final operation so we only get the real part remaining, and this is equivalent to using just one of the two and then later taking the real part of the total result. And just to make this clear, when we take the real part of the total result that means we just cancel the imaginary part almost like pretending it isnt there anymore. An example is for X=a+bj:
Real(a+bj)=a

so the real part is just 'a'. This is equivalent to adding the guts to the complex conjugate of the guts and then dividing by 2.

So i think your main problem is that you are viewing the two terms as exponential as in e^iX so it is harder to see the relationships but instead you should be looking at the Euler equivalents:

e^ix=cos(x)+i*sin(x)
e^-ix=cos(x)-i*sin(x)

and from this we can easily see that they are the complex conjugates of each other. Replace 'x' with wt+θ and you get the identities you want.

We then also see that we have these identities:
f(0.5*(x+x'))=Real(f(x))
f(0.5*(x+x'))=Real(f(x'))

and to be complete:
Real(f(x))=Real(f(x'))

 

[hanhan]

Thanks MrAl,
I see clearly that e^-j(wt+θ) is the complex conjugate of e^j(wt+θ). But I may be mistaken to make generally that f(x + y) = f(x) + f(y).
I still don't see why f(0.5*(x+x*))=0.5*f(x)+0.5*f(x*) will be a real number.
It would be a real number if f(0.5*(x+x*))= 0.5(f(x) + f*(x)). How can you know that f(x) and f(x*) will have the imaginary parts cancel out?

 

[MrAl]

Hello again,


When working in the time domain if we dont see the imaginary parts cancel out then we made a mistake somewhere because in this context imaginary really does mean that it does not exist.

Consider the function 'F' as in y=F(t),

If we get as result: y=5.4 that's not a problem because 5.4 is a constant so this tells us that the voltage or current is a constant 5.4 volts or amps.
If we get as result: y=2+3*t that should not be a problem because we just have time in that equation and there is nothing wrong with that, and that just means that the current or voltage is changing with time, so no big deal.
But what if we as result something like this: y=3+7*i, what does this mean? This means we got a two dimensional current or voltage when we only wanted a single dimensional current or voltage. How do we even measure this on a current or voltage meter?
Or what if we got the result: y=3+4*t+5*t*i, what does this mean? We have time running in two different directions?

So for any real problem in the time domain the imaginary part cancels out or else we made a mistake. In the frequency domain we are allowed to have imaginary parts because that is a different representation of nature. But terms like e^iwt are time domain representations.

If you do a few experiments with the Euler equivalents you'll see how this works much better i think.

 

[hanhan]

Hi,

So for any real problem in the time domain the imaginary part cancels out or else we made a mistake.

I want to prove that.
In the my case I want to prove that the result of f(x) + f(x*) is a real number.
( I want to understand the explanation of "Validity of complex representation" above.)
I think there is a theorem or something that says: f*(x) = f(x*). Of course, there are some constrains that I don't know.
Then f(x) + f(x*) = f(x) + f*(x) = 2Re{f(x)}

 

[MrAl]

Hi,

Well if it does not cancel out then how do we interpret the result?

Also, using an apostrophe to indicate the conjugate, if we have
f(w)=1+i*w

then:
f(w')=1+i*w'=1+i*w

but the conjugate of the whole function is 1-i*w. Is that what you meant?