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- Thread starter rattus
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One more thing: You have one thing right. Impedance has no meaning in instantaneous equations.
drbond24
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Ok, I actually have a serious question to ask, although I commend your inattention to my attempts at heckling you.
I still say you're college professor material. Asking loaded questions, ignoring the peanut gallery, and never admitting that you're wrong. That sums up every professor I've ever had! :grin:Doesn't every ratio of two variables potentially vary from 0 to infinity? How does that make this particular ratio useless?
For instance, consider the MPG on my truck. Usually it is around 11 (I have a big truck), but if I ran the engine sitting still it would be 0 and if I had someone push me with my engine off it would be infinity. That does not make the 11 that it usually is any less valid.
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Ok, I actually have a serious question to ask, although I commend your inattention to my attempts at heckling you.
Doesn't every ratio of two variables potentially vary from 0 to infinity? How does that make this particular ratio useless?
For instance, consider the MPG on my truck. Usually it is around 11 (I have a big truck), but if I ran the engine sitting still it would be 0 and if I had someone push me with my engine off it would be infinity. That does not make the 11 that it usually is any less valid.
One does not expect the mileage rating of a vehicle to be constant.
One does expect an impedance to be constant, not a function of time as Smart is proposing with his z(t). Furthermore, impedance is undefined for instantaneous equations.
Now,
V @ 0/I @ phi is a valid expression, but that was not the question!
drbond24
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I'm addressing your post #29. Forget everything else that has been said. You say the expression you posted is utter nonsense based on the fact that it varies from 0 to infinity. I'm asking how that is a reason it is nonsense using the example of the MPG on my truck. It varies from 0 to infinity and is NOT utter nonsense. I want to know how, in this particular case, the fact that the ratio varies from 0 to infinity is a reason to call it nonsense.
Your mpg formula is valid and useful and the results can and do vary.
But it is nonsense to call v(t)/i(t) an impedance.
It is more nonsense to say this varying ratio is useful in circuit analysis.
I would say useless rather than nonsense.
Anyway, in my first post to this topic I stated that I errantly called it impedance in that other thread. Instead I refer to it as the resistance equivalent. Would you not agree that the units of v(t)/i(t) is ohms? Perhaps rather than z(t) we should denote it Ω(t)...
drbond24
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Your post was an answer to the OP's question, which means you were playing the professor's game.
I (and I believe steve is with me) am deriding the thread as a whole, not answering the question.
That is the difference. Of course, if you'd care to join our club, you'd get a membership card and a t-shirt. The entrance fee is one post making fun of the overall thread. As a whole, this is like playing a game with my daughter that she made up on the spot. As soon as I figure something out and start to do well, she changes the rules to define my current state as losing and hers as winning.
Now, for something that the professor might actually respond to.
rattus said:
Perhaps I just haven't understood what exactly v(t) and i(t) are. They aren't phasors, they aren't complex, etc. What are they? That may help me to understand under what circumstances a voltage divided by a current doesn't yield an impedance.
Hameedulla-Ekhlas
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Now please read this
Suppose we launch a step voltage wave of 1 volt into the line. For example, we
connect to the front end of the line, between the signal path and the return path,
a 1 volt battery. The instant we make the connection, this voltage wave signal
begins to move down the line at the speed of light in the dielectric medium,
typically about 6 inches/nsec. (Why it moves so fast and is not the speed of the
electrons, which is closer to 1 cm/sec, might be the topic of a future article). The
?signal?, of course, is really the voltage difference between the signal line and the
return path, measured between any point along the signal line and the adjacent
point on the return path.
In a Zen sort of way, ?be the signal?, and travel down this transmission line at 6
inches/nsec. What do you see? In the first 10 psec, you have ?walked? about
0.06 inches down the line. Let?s freeze time and look at the line. Behind you, you
have left a wake of a constant voltage of 1 volt between the signal line and an
adjacent point on the return path. This means there is some excess ?+? charge
on the trace behind you and some excess ??? charge on the return path behind
you. It?s this excess charge difference that sustains the 1 volt signal between
these two conductors that make up a capacitor.
Ahead of you, the line doesn?t know you are coming and the voltage between the
signal line and the return path is still zero. In another 10 psec, with your next
step, you will transform the voltage in the 0.06 inch section under your foot from
0 volts to 1 volt. To do this, you must add some ?+? charge to the signal line and
add some ??? charge to the return path line. For each 0.06 inch step you take,
you are adding more ?+? charge to the signal line and more ?-? charge to the
return path. Every 10 psec, another section of the transmission line gets charged
up and the signal propagates down the line. But where does this charge come
from?
Answer: it comes from the source- the battery we attached to the front of the
transmission line to launch the initial signal. As we, the signal, march down the
transmission line, we are charging up successive sections of the line as we go,
leaving behind a 1 volt different between the signal and return path lines. Every
10psec, we take a step and draw another small charge, dQ, from the battery.
This constant amount of charge, dQ, flowing out of the battery, in constant time
interval, dt, is a constant current. There is + current flowing into the signal line
and at the same time, - current flowing into the return path.
The ?-? current flowing into the return path is really the same as ?+? current
flowing out of the return path. And, right at the wave front of the signal, the (AC)
current flows through the capacitance between the top line to the bottom line,
closing the loop.
Impedance of the Line
From the battery?s perspective, it sees that as soon as you connected its leads to
the front of the transmission line, there is a constant draw of current, for its
constant applied voltage. It you were to ask it, what sort of circuit element
behaves this way- has a constant current draw for constant applied voltage, of
course, it would say a resistor.
To the battery, while the signal is moving down the transmission line, (emphasis
added) charging up successive 0.06 inch sections every 10 psec, drawing a
constant current from the source, the transmission line looks like a resistor and
has a constant value of resistance, which we call the ?surge? impedance of the
line.
Likewise, for you as the signal, walking down the line, with each step you take,
you are constantly probing the electrical environment of the line and asking ?what
is the impedance of my next step?? As you put your foot down on the line, you
are electrically asking, what is the current required to charge up this footstep to
1v, in the 10 psec I have until my next step? This is the instantaneous impedance
you are probing for.
As we saw with the battery, if you move down the line at a steady pace, and the
line has the same cross section, then taking each step requires the same amount
of charge in 10 psec, for the same signal voltage. With every step you take, you
will draw the same current from your foot (and the battery), at the same voltage,
and you too will see the same instantaneous impedance as you walk down the
line.
If the line has the same signal velocity down its length and it has the same
capacitance per length down its length then with every step you, the signal, take
you will see the same instantaneous impedance as you move down the line.
Because this impedance is constant along the line, we give it the special name
that says it is characteristic for this particular transmission line. We call it the
?characteristic impedance? of the line. Characteristic impedance is the
instantaneous impedance a signal sees as it moves down the line. If, as the
signal moves down the line, it sees the same characteristic impedance with every
step, we can label the line as a controlled impedance line.
What?s so Important about the Characteristic Impedance of a
Transmission Line?
For optimal signal quality, the goal in interconnect design is to keep the
impedance the signal sees as constant as possible. This means primarily, keep
the characteristic impedance of the line constant. Hence the growing importance
in manufacturing controlled impedance boards. All the other tricks like minimize
stub lengths, terminate the ends, daisy chain rather than branch, are all designed
to keep the instantaneous impedance the signal sees constant.
Calculating Characteristic Impedance
From this simple model we can derive a value for the characteristic impedance,
which is the instantaneous impedance the signal sees as it walks down the line.
The impedance seen in each step, Z, is just the basic definition of impedance:
Z = V/I
The voltage is the voltage of the signal launched into the line, V, and the current,
I, is the charge that flows out of your foot, dQ, in the time for each step, dt:
I = dQ/dt
The charge that flows out of your foot (coming ultimately from the battery), is the
charge needed to charge up the capacitance, dC of one footprint, to the voltage
of the signal, V:
dQ = dC V
We can relate the capacitance of one footprint to the capacitance per length, CL,
of the transmission line and the speed of the signal, v, down the line. We need to
remember that the length of a footprint is our speed, v, times the time to take
each step, dt.
dC = CL v dt
Combining all the pieces, we can write the instantaneous impedance as:
Z = V/I = V/(dQ/dt) = V/(dC V/dt) = V/(CL v dt V/dt) = 1/(CL v)
We see that the instantaneous impedance is related to the capacitance per
length of the transmission line and the speed of the signal. This is also the
definition of the characteristic impedance of the line. To distinguish the term
characteristic impedance from the actual impedance, Z, we add a small zero to it.
We have just derived the characteristic impedance of a transmission line as:
Z0 = 1/(CL v)
Now we the unit of Z0 or characteristic impedance is ohm and you can give any number to CL ( capacitance per length and you will get still real number with ohm unit and there is no imaginary part.
That is why it is called v(t)/i(t) = R
Now who can answer my question?
what about the inductance of the transmission line?
Doesn?t it affect the characteristic impedance of the line??
Suppose we launch a step voltage wave of 1 volt into the line. For example, we
connect to the front end of the line, between the signal path and the return path,
a 1 volt battery. The instant we make the connection, this voltage wave signal
begins to move down the line at the speed of light in the dielectric medium,
typically about 6 inches/nsec. (Why it moves so fast and is not the speed of the
electrons, which is closer to 1 cm/sec, might be the topic of a future article). The
?signal?, of course, is really the voltage difference between the signal line and the
return path, measured between any point along the signal line and the adjacent
point on the return path.
In a Zen sort of way, ?be the signal?, and travel down this transmission line at 6
inches/nsec. What do you see? In the first 10 psec, you have ?walked? about
0.06 inches down the line. Let?s freeze time and look at the line. Behind you, you
have left a wake of a constant voltage of 1 volt between the signal line and an
adjacent point on the return path. This means there is some excess ?+? charge
on the trace behind you and some excess ??? charge on the return path behind
you. It?s this excess charge difference that sustains the 1 volt signal between
these two conductors that make up a capacitor.
Ahead of you, the line doesn?t know you are coming and the voltage between the
signal line and the return path is still zero. In another 10 psec, with your next
step, you will transform the voltage in the 0.06 inch section under your foot from
0 volts to 1 volt. To do this, you must add some ?+? charge to the signal line and
add some ??? charge to the return path line. For each 0.06 inch step you take,
you are adding more ?+? charge to the signal line and more ?-? charge to the
return path. Every 10 psec, another section of the transmission line gets charged
up and the signal propagates down the line. But where does this charge come
from?
Answer: it comes from the source- the battery we attached to the front of the
transmission line to launch the initial signal. As we, the signal, march down the
transmission line, we are charging up successive sections of the line as we go,
leaving behind a 1 volt different between the signal and return path lines. Every
10psec, we take a step and draw another small charge, dQ, from the battery.
This constant amount of charge, dQ, flowing out of the battery, in constant time
interval, dt, is a constant current. There is + current flowing into the signal line
and at the same time, - current flowing into the return path.
The ?-? current flowing into the return path is really the same as ?+? current
flowing out of the return path. And, right at the wave front of the signal, the (AC)
current flows through the capacitance between the top line to the bottom line,
closing the loop.
Impedance of the Line
From the battery?s perspective, it sees that as soon as you connected its leads to
the front of the transmission line, there is a constant draw of current, for its
constant applied voltage. It you were to ask it, what sort of circuit element
behaves this way- has a constant current draw for constant applied voltage, of
course, it would say a resistor.
To the battery, while the signal is moving down the transmission line, (emphasis
added) charging up successive 0.06 inch sections every 10 psec, drawing a
constant current from the source, the transmission line looks like a resistor and
has a constant value of resistance, which we call the ?surge? impedance of the
line.
Likewise, for you as the signal, walking down the line, with each step you take,
you are constantly probing the electrical environment of the line and asking ?what
is the impedance of my next step?? As you put your foot down on the line, you
are electrically asking, what is the current required to charge up this footstep to
1v, in the 10 psec I have until my next step? This is the instantaneous impedance
you are probing for.
As we saw with the battery, if you move down the line at a steady pace, and the
line has the same cross section, then taking each step requires the same amount
of charge in 10 psec, for the same signal voltage. With every step you take, you
will draw the same current from your foot (and the battery), at the same voltage,
and you too will see the same instantaneous impedance as you walk down the
line.
If the line has the same signal velocity down its length and it has the same
capacitance per length down its length then with every step you, the signal, take
you will see the same instantaneous impedance as you move down the line.
Because this impedance is constant along the line, we give it the special name
that says it is characteristic for this particular transmission line. We call it the
?characteristic impedance? of the line. Characteristic impedance is the
instantaneous impedance a signal sees as it moves down the line. If, as the
signal moves down the line, it sees the same characteristic impedance with every
step, we can label the line as a controlled impedance line.
What?s so Important about the Characteristic Impedance of a
Transmission Line?
For optimal signal quality, the goal in interconnect design is to keep the
impedance the signal sees as constant as possible. This means primarily, keep
the characteristic impedance of the line constant. Hence the growing importance
in manufacturing controlled impedance boards. All the other tricks like minimize
stub lengths, terminate the ends, daisy chain rather than branch, are all designed
to keep the instantaneous impedance the signal sees constant.
Calculating Characteristic Impedance
From this simple model we can derive a value for the characteristic impedance,
which is the instantaneous impedance the signal sees as it walks down the line.
The impedance seen in each step, Z, is just the basic definition of impedance:
Z = V/I
The voltage is the voltage of the signal launched into the line, V, and the current,
I, is the charge that flows out of your foot, dQ, in the time for each step, dt:
I = dQ/dt
The charge that flows out of your foot (coming ultimately from the battery), is the
charge needed to charge up the capacitance, dC of one footprint, to the voltage
of the signal, V:
dQ = dC V
We can relate the capacitance of one footprint to the capacitance per length, CL,
of the transmission line and the speed of the signal, v, down the line. We need to
remember that the length of a footprint is our speed, v, times the time to take
each step, dt.
dC = CL v dt
Combining all the pieces, we can write the instantaneous impedance as:
Z = V/I = V/(dQ/dt) = V/(dC V/dt) = V/(CL v dt V/dt) = 1/(CL v)
We see that the instantaneous impedance is related to the capacitance per
length of the transmission line and the speed of the signal. This is also the
definition of the characteristic impedance of the line. To distinguish the term
characteristic impedance from the actual impedance, Z, we add a small zero to it.
We have just derived the characteristic impedance of a transmission line as:
Z0 = 1/(CL v)
Now we the unit of Z0 or characteristic impedance is ohm and you can give any number to CL ( capacitance per length and you will get still real number with ohm unit and there is no imaginary part.
That is why it is called v(t)/i(t) = R
Now who can answer my question?
what about the inductance of the transmission line?
Doesn?t it affect the characteristic impedance of the line??
Cold Fusion
Senior Member
- Location
- way north
Ah.... Duh. That was apparent a few threads ago. The frisbe finally got big enough as soon as the posts included noetic definitions of impedance, instantaneous, instantaneous average
confused, absloute averageconfused:), instantaneous values have a phase angle (why? It is because we defined it that way )The part that leaves me completely clueless is why I tried to insert a helpfull comments in this post (post 7, 12). Even the passing thought that I could assist with understanding in this post is utterly gormless.
Now that I have derided the post, you, me ... extend to all other posters of this thread and similar previous threads (I try to be equal opportunity derisive) - I expect two t-shirts
cf
Cold Fusion
Senior Member
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Note to all posters - this is only for Dave Bond's eyes. I am desparately not trying to convince any of the regular posters of anything.
Take a look at post 12. See if that helps any.
v(t) = Voltage as a function of time. If the function is relatively steady state sinusiodal, It relates to phasors through Euler's equation.
That is a pretty simplistic explanation, so our theoretical mathematicians should have lots of fun with that.
cf
I would say useless rather than nonsense.
Anyway, in my first post to this topic I stated that I errantly called it impedance in that other thread. Instead I refer to it as the resistance equivalent. Would you not agree that the units of v(t)/i(t) is ohms? Perhaps rather than z(t) we should denote it Ω(t)...
How about "useless nonsense"?
Yes, the units come out to be ohms, but this limits us to the specific case where the load is resistive. The notion of a "resistive equivalent" is nonsense because the value is not constant and this notion is not supported by the physics. And, even if you say it in Greek, it is still nonsense.
Cold Fusion
Senior Member
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Again for Dave's eyes only.
Another way to look at this is: (and yes, I know you are aware of all of this)
v = L(di/dt)
i = C(dv/dt)
This is true regardless if the driving voltage (or current) is sinusiodal or not.
So, to get v(t)/i(t) one needs to solve the differential equations. Dividing the instantaneous voltage by the instantaneous current isn't defined (well, except for smart's definition)
For relatively steady state sinusoidal, this is usually through euler's equation and phasors.
Or for the last iteration of unit impulses and exponetial increasing currents, one might look at Laplace transforms. But this would be more in rattus' world than mine. Not much here to do with power engineering or protective relays. For any exponential current (or voltage) in my world, just wait a few milliseconds. Either the protective relay takes the equipment off line, or it blows up. Either way, what ever did it gets fixed so that doesn't happen any more.
cf
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drbond24
Senior Member
- Location
- Barboursville, West Virginia
That's what I thought, but due to the confusion I had started to think I was assuming it to be more simple than it was.
I can live with voltage as a function of time.Cold Fusion said:
Ok, here is my hang-up. I've divided v(t) by i(t) before using the differential equations. I got graded on it and got an A, as a matter of fact.
Therefore, I don't see where it is nonsense.Is it just nonsense because one of the prerequisites was 'no differential equations?' Otherwise I'm still confused.
Of course, but:
Of course, but:
Of course if v(t) and i(t)are sinusoidal, they can be converted to phasors, but the question did not ask about the ratio of phasors! If that were the case, I would have written it as such!
The expanded question is,
"Does the equation,
v(t)/i(t) = z(t),
as written, with no transformation, no advanced math, have any meaning?"
The answer is "NO" because,
1. Impedance is undefined for instantaneous equations.
2. Impedance is assumed to be constant in linear circuit analyses.
Now, if the question were,
"Does the equation,
(V @ phi)/(I @ theta) = Z(phi - theta)
have any meaning," the answer would be YES, but one more time,
That is not the question!
Of course, but:
Of course if v(t) and i(t)are sinusoidal, they can be converted to phasors, but the question did not ask about the ratio of phasors! If that were the case, I would have written it as such!
The expanded question is,
"Does the equation,
v(t)/i(t) = z(t),
as written, with no transformation, no advanced math, have any meaning?"
The answer is "NO" because,
1. Impedance is undefined for instantaneous equations.
2. Impedance is assumed to be constant in linear circuit analyses.
Now, if the question were,
"Does the equation,
(V @ phi)/(I @ theta) = Z(phi - theta)
have any meaning," the answer would be YES, but one more time,
That is not the question!
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Cold Fusion
Senior Member
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Somethig is radically wrong here:-?. My posts are susposed to be invisible to all but Dave.
rattus -
Until I get this fixed, stop reading my posts
cf
rattus -
Until I get this fixed, stop reading my posts
cf
You're entitled to your opinion.
How so, kimosabe? Reactance is in units of ohms. Are you saying reactance cannot be determined from instantaneous values. That would contradict being able to separate real and reactive components of instantaneous power.
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Who says resistance has to be constant. Take a thermistor, for example. It's resistance is not constant. Practically any real world resistor does not have constant resistance. There's the physics to contradict your claim. Besides, I'm not saying it is a resistor, but rather the measure of ... well I can't say impedance... perhaps inverse conductance, but even that would not be accurate.
Granted there really isn't a defined term for this that I know of, that's why I dubbed it thus only for the sake of conversation, not that I plan on writing up a technical paper and getting ostricized forever. You've dubbed it nonsense, so what makes your terminology any more valid than mine if it's not defined.
Cold Fusion
Senior Member
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Encrypted lifted for this post only...Ok, here is my hang-up. I've divided v(t) by i(t) before using the differential equations. I got graded on it and got an A, as a matter of fact.
Is it just nonsense because one of the prerequisites was 'no differential equations?' Otherwise I'm still confused.
I don't know the context of the class. if you were looking at other than sinusiodal waveforms and generally resistive loads, then .... maybe - I don't know.
As I said in post twelve, divide this:
(sin(wt))/(sin(wt + a)) = ???
I can't get anywhere with it.
Dealing with inductors, capacitors, is definitely differential calculus.
cf
Easy way:
Easy way:
Look, all you have to do is evaluate sin(wt) and sin(wt + phi) over a cylce, divide the results, and plot the ratios over a cycle. No DE, no phasors, just a little trig and arithmetic.
Phi can be anything but zero because that would indicate a resistive load, and we want the general case.
You don't even have to do that because it is clear the the ratio reaches infinity when the current is zero.
cf, if you want privacy, send a PM.
Easy way:
Look, all you have to do is evaluate sin(wt) and sin(wt + phi) over a cylce, divide the results, and plot the ratios over a cycle. No DE, no phasors, just a little trig and arithmetic.
Phi can be anything but zero because that would indicate a resistive load, and we want the general case.
You don't even have to do that because it is clear the the ratio reaches infinity when the current is zero.
cf, if you want privacy, send a PM.
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