6 ms·

Electrical impedance made easy - Part 1

58 points by xd 15y ago

kwantam 15y ago
Put simply, impedance is just complex resistance. We get it by taking the Laplace transform of the differential equation that defines the relationship between current and voltage in a capacitor or inductor. I = C dV/dt Laplace(dV/dt) = V*s I = C * V * s V / I = 1 / (C * s) V = L dI/dt Laplace(dI/dt) = I*s V / I = L * s Another way to look at this: since this is a linear system, we know its eigenvalues are complex exponentials, i.e., their real and imaginary parts are sinusoids of any frequency (by the Euler identity). So we can say V = Vm * exp(j*omega*t) -- using j = sqrt(-1) to avoid confusion with current dV/dt = Vm * j * omega * exp(j*omega*t) = j * omega * V V / I = 1 / (C * j * omega) | V / I | = 1 / (C * omega) arg(V/I) = -pi/2 -- voltage lags current by 90 degrees By a similar argument, | V / I | = L * omega arg(V/I) = pi/2 We can see that as expected, s=j*omega.
- haberman 15y ago
  All I can say is that if this is the "simple" explanation to you, you and I think in very different ways. :) To me, it's much easier to understand the visual where impedance is the hypotenuse of a triangle whose legs are resistance and reactance.
  - jules 15y ago
    That doesn't really explain impedance. First of all, you need to know what resistance and reactance are to get that explanation, and second it doesn't explain why it's useful or how to do calculations with impedance. On the other hand, kwantam's explanation is easy and clear, provided you know what a Laplace transform is (which you're going to need to understand reactance anyway). If you set up the system of differential equations for a linear circuit, how would you go about solving these equations? You take the Laplace transform, and you get the equivalent algebraic equations of the form V = ZI for each element. The numbers Z that appear are the impedances. From a mathematical viewpoint, impedance is a simpler and more natural concept than resistance and reactance, since it appears naturally when you solve linear circuits.
    - haberman 15y ago
      > That doesn't really explain impedance. First of all, you need to know what resistance and reactance are to get that explanation, and second it doesn't explain why it's useful or how to do calculations with impedance. It's amazing to me that we can see this in such opposite ways. To me kwantam's post doesn't explain impedance at all, it just formally defines it. It doesn't even begin to explain to me what it measures or how it applies to an actual circuit. What components of a circuit have an impedance? If I increased the impedance, what would happen? How would I even do that? kwantam's post doesn't begin to answer these questions. I don't get any intuition at all for what this concept actually means from a formal definition. That's why teaching methods that start with formalisms have always frustrated me. I bet you could dress up a really simple concept in a formalism that would be totally indecipherable. I actually think this would be fun to do sometime: describe some mundane concept in a really formal way and then ask a bunch of mathematical people "what am I describing?" And it turns out to be something boring like the rules for driving through an intersection. I guess my point is, formalisms seem obvious once you understand the concept intuitively, but are a terrible way to gain that intuition. The basic idea of resistance is simple (a resistor restricts the flow of electricity), reactance was new to me but I got a rough idea that it similarly resists the flow of electricity in an AC circuit and is a function of the frequency. So if I know that the impedance is a right-triangle combination of those two things, I've got a great start for getting some intuition: I know that anything that has resistance or reactance also has impedance, and something that has both has even more impedance, but in a DC circuit where the frequency is zero you'd get some edge condition where a non-zero reactance would, let's see, act as infinite resistance, I think? The formal definition doesn't plant any of these paths of understanding or inquiry in my brain.
      - mturmon 15y ago
        I understand the Laplace transform ideas above (used to TA the class where it's taught) and I agree with your that they are more definitional than explanatory. To provide actual explanatory power, you'd have to explain what the reactive portion ("imaginary part" in the phasor notation) of the impedance is actually doing in real terms. What it's doing is to cause the current to lead or lag the voltage in the AC signal. In a purely resistive circuit, current and voltage vary exactly the same way at every instant. You can actually go a long way with this just by understanding that, e.g., for an inductor, V = L dI/dt so that small current changes will imply very large voltage changes. If you're in to intuition, this means "voltage leads current" (i.e., the sinusoid of voltage is 90 degrees ahead of the sinusoid of current). The above equation is also why lights in your house dim when you turn on a motor (= inductor). There is a significant current change, and the voltage changes even more, so the lights dim. Or, you can be even more intuitive, and note that it's obvious that when you put a coil of wire (= motor = inductor) across the two sides of your house's electrical wiring, the coil of wire will pull the + and - conductors together in voltage at first. Until the magnetic field sets up to provide reactance, the coil of wire is basically a short circuit. And finally, a capacitor is just the dual of the inductor.
  - psycovic23 15y ago
    His method is a bit more complex, but you eventually need to describe how different frequencies change the characteristics of a system. Initially, it's nice to see impedance in terms of a + bi, but that's not easy to manipulate when you eventually want to know a circuit's behavior at various frequencies. This is because (a + bi) does not inherently describe the sinusoidal behavior of AC. By changing things into the frequency domain (where writing a phasor implies a driving rotational force at some variable frequency), we can now see how the impedance of an entire system varies as frequency changes. http://betterexplained.com/articles/intuitive-arithmetic-with-complex-numbers/ http://betterexplained.com/articles/intuitive-arithmetic-wit... http://en.wikipedia.org/wiki/Phasor http://en.wikipedia.org/wiki/Phasor It's really just a math abstraction. Maybe this analogy is a stretch, but you interface with stacks using push and pop, and there's an inherent size to the stack that you keep in the back of your mind. You don't care about the individual memory addresses and their values, you just want to manipulate the stack such that it doesn't overflow/underflow. Just like AC - you don't care where in the complex plane an impedance is at a given frequency (which would be in terms of a+bi), you'd rather just see how its behavior changes as frequencies vary. And why do you care about impedances for a whole range of frequencies? One example: http://www.ecoustics.com/electronics/products/articles/131062.html http://www.ecoustics.com/electronics/products/articles/13106...
- jwcacces 15y ago
  If you take the term "put simply" and then add the term "Laplace transform" you're doing something wrong.
haberman 15y ago
The only place where I've encountered the term "impedance" is in relation to the hi-z and lo-z instrument cables that you use to connect eg. electric guitars to amplifiers. You'd sometimes see these hi-z to lo-z converters. Is this the same thing? The video's explanation would lead me to believe that impedance only applies to AC circuits. But instrument cables don't run AC, do they?
- sparky 15y ago
  Yes, the 'z' in hi-z and lo-z cables is the standard variable name for impedance (like 'v' for voltage or 'i' for current). 'AC' in this context means 'not DC', which is to say 'varies over time'. Instrument cables certainly qualify under that definition, though it is definitely confusing to call it 'alternating' (which we normally think implies 'goes back and forth between 2 states'), where perhaps 'varying' or 'time-varying' would be more appropriate.
- tkok 15y ago
  Audio signals are alternating. The voltage corresponds to the sound pressure picked up by the microphone, which typically looks more or less like a bunch of sinusoids.
  - kwantam 15y ago
    It's not just typically a bunch of sinusoids: Fourier tells us that it is always the sum of a bunch of sinusoids. [1] http://en.wikipedia.org/wiki/Fourier_transform http://en.wikipedia.org/wiki/Fourier_transform
- kwantam 15y ago
  In the case you're talking about, impedance means "characteristic impedance", i.e., one parameter which (along with speed of light in the given medium and length) defines the behavior of a transmission line. [1] It is the same thing in the sense that mathematically both are a complex (i.e., Real plus Imaginary) relationship between voltage and current. However, a transmission line is a two-port network while a capacitor is a one-port network, so these two senses of impedance correspond to different physical phenomena. In a capacitor, the impedance defines the relationship between voltage and current at the port (i.e., between the two terminals of the cap; this is sometimes called the driving point impedance). For a transmission line, the characteristic impedance defines, among other things, the equivalent impedance seen at one end of the line when the line is either infinitely long or an integral number of wavelengths long and terminated at the other end by an impedance of the same value as the characteristic impedance. Instrument cables do indeed carry AC signals, inasmuch as the signal on the line corresponds to the frequency of the note(s) you're playing. AC is just "non-zero frequency"---any signal which is not a constant, unchanging (DC) value is considered AC. So, 50/60Hz line power is an AC signal, the 2.4GHz signals used by your 802.11 network card are AC, and middle C (261.6 Hz) is AC. [1] http://en.wikipedia.org/wiki/Characteristic_impedance http://en.wikipedia.org/wiki/Characteristic_impedance
xd 15y ago
Part 2: http://benkrasnow.blogspot.com/2011/06/tutorial-electrical-impedance-made-easy_19.html http://benkrasnow.blogspot.com/2011/06/tutorial-electrical-i...
mrothe 15y ago
I don't know about the rules in America, but this circuit design wouldn't be approved for use here in Europe. Circuits are not allowed to draw what is called 'reactive power', because it stresses the grid. Therefore you have to compensate it (in the simplest case using an inductor or a capacitor).

58 points by xd 15y ago

Put simply, impedance is just complex resistance. We get it by taking the Laplace transform of the differential equation that defines the relationship between current and voltage in a capacitor or inductor. I = C dV/dt Laplace(dV/dt) = V*s I = C * V * s V / I = 1 / (C * s) V = L dI/dt Laplace(dI/dt) = I*s V / I = L * s Another way to look at this: since this is a linear system, we know its eigenvalues are complex exponentials, i.e., their real and imaginary parts are sinusoids of any frequency (by the Euler identity). So we can say V = Vm * exp(j*omega*t) -- using j = sqrt(-1) to avoid confusion with current dV/dt = Vm * j * omega * exp(j*omega*t) = j * omega * V V / I = 1 / (C * j * omega) | V / I | = 1 / (C * omega) arg(V/I) = -pi/2 -- voltage lags current by 90 degrees By a similar argument, | V / I | = L * omega arg(V/I) = pi/2 We can see that as expected, s=j*omega.

haberman 15y ago
All I can say is that if this is the "simple" explanation to you, you and I think in very different ways. :) To me, it's much easier to understand the visual where impedance is the hypotenuse of a triangle whose legs are resistance and reactance.
- jules 15y ago
  That doesn't really explain impedance. First of all, you need to know what resistance and reactance are to get that explanation, and second it doesn't explain why it's useful or how to do calculations with impedance. On the other hand, kwantam's explanation is easy and clear, provided you know what a Laplace transform is (which you're going to need to understand reactance anyway). If you set up the system of differential equations for a linear circuit, how would you go about solving these equations? You take the Laplace transform, and you get the equivalent algebraic equations of the form V = ZI for each element. The numbers Z that appear are the impedances. From a mathematical viewpoint, impedance is a simpler and more natural concept than resistance and reactance, since it appears naturally when you solve linear circuits.
  - haberman 15y ago
    > That doesn't really explain impedance. First of all, you need to know what resistance and reactance are to get that explanation, and second it doesn't explain why it's useful or how to do calculations with impedance. It's amazing to me that we can see this in such opposite ways. To me kwantam's post doesn't explain impedance at all, it just formally defines it. It doesn't even begin to explain to me what it measures or how it applies to an actual circuit. What components of a circuit have an impedance? If I increased the impedance, what would happen? How would I even do that? kwantam's post doesn't begin to answer these questions. I don't get any intuition at all for what this concept actually means from a formal definition. That's why teaching methods that start with formalisms have always frustrated me. I bet you could dress up a really simple concept in a formalism that would be totally indecipherable. I actually think this would be fun to do sometime: describe some mundane concept in a really formal way and then ask a bunch of mathematical people "what am I describing?" And it turns out to be something boring like the rules for driving through an intersection. I guess my point is, formalisms seem obvious once you understand the concept intuitively, but are a terrible way to gain that intuition. The basic idea of resistance is simple (a resistor restricts the flow of electricity), reactance was new to me but I got a rough idea that it similarly resists the flow of electricity in an AC circuit and is a function of the frequency. So if I know that the impedance is a right-triangle combination of those two things, I've got a great start for getting some intuition: I know that anything that has resistance or reactance also has impedance, and something that has both has even more impedance, but in a DC circuit where the frequency is zero you'd get some edge condition where a non-zero reactance would, let's see, act as infinite resistance, I think? The formal definition doesn't plant any of these paths of understanding or inquiry in my brain.
    - mturmon 15y ago
      I understand the Laplace transform ideas above (used to TA the class where it's taught) and I agree with your that they are more definitional than explanatory. To provide actual explanatory power, you'd have to explain what the reactive portion ("imaginary part" in the phasor notation) of the impedance is actually doing in real terms. What it's doing is to cause the current to lead or lag the voltage in the AC signal. In a purely resistive circuit, current and voltage vary exactly the same way at every instant. You can actually go a long way with this just by understanding that, e.g., for an inductor, V = L dI/dt so that small current changes will imply very large voltage changes. If you're in to intuition, this means "voltage leads current" (i.e., the sinusoid of voltage is 90 degrees ahead of the sinusoid of current). The above equation is also why lights in your house dim when you turn on a motor (= inductor). There is a significant current change, and the voltage changes even more, so the lights dim. Or, you can be even more intuitive, and note that it's obvious that when you put a coil of wire (= motor = inductor) across the two sides of your house's electrical wiring, the coil of wire will pull the + and - conductors together in voltage at first. Until the magnetic field sets up to provide reactance, the coil of wire is basically a short circuit. And finally, a capacitor is just the dual of the inductor.
- psycovic23 15y ago
  His method is a bit more complex, but you eventually need to describe how different frequencies change the characteristics of a system. Initially, it's nice to see impedance in terms of a + bi, but that's not easy to manipulate when you eventually want to know a circuit's behavior at various frequencies. This is because (a + bi) does not inherently describe the sinusoidal behavior of AC. By changing things into the frequency domain (where writing a phasor implies a driving rotational force at some variable frequency), we can now see how the impedance of an entire system varies as frequency changes. http://betterexplained.com/articles/intuitive-arithmetic-with-complex-numbers/ http://betterexplained.com/articles/intuitive-arithmetic-wit... http://en.wikipedia.org/wiki/Phasor http://en.wikipedia.org/wiki/Phasor It's really just a math abstraction. Maybe this analogy is a stretch, but you interface with stacks using push and pop, and there's an inherent size to the stack that you keep in the back of your mind. You don't care about the individual memory addresses and their values, you just want to manipulate the stack such that it doesn't overflow/underflow. Just like AC - you don't care where in the complex plane an impedance is at a given frequency (which would be in terms of a+bi), you'd rather just see how its behavior changes as frequencies vary. And why do you care about impedances for a whole range of frequencies? One example: http://www.ecoustics.com/electronics/products/articles/131062.html http://www.ecoustics.com/electronics/products/articles/13106...
jwcacces 15y ago
If you take the term "put simply" and then add the term "Laplace transform" you're doing something wrong.

haberman 15y ago

The only place where I've encountered the term "impedance" is in relation to the hi-z and lo-z instrument cables that you use to connect eg. electric guitars to amplifiers. You'd sometimes see these hi-z to lo-z converters. Is this the same thing? The video's explanation would lead me to believe that impedance only applies to AC circuits. But instrument cables don't run AC, do they?

sparky 15y ago
Yes, the 'z' in hi-z and lo-z cables is the standard variable name for impedance (like 'v' for voltage or 'i' for current). 'AC' in this context means 'not DC', which is to say 'varies over time'. Instrument cables certainly qualify under that definition, though it is definitely confusing to call it 'alternating' (which we normally think implies 'goes back and forth between 2 states'), where perhaps 'varying' or 'time-varying' would be more appropriate.
tkok 15y ago
Audio signals are alternating. The voltage corresponds to the sound pressure picked up by the microphone, which typically looks more or less like a bunch of sinusoids.
- kwantam 15y ago
  It's not just typically a bunch of sinusoids: Fourier tells us that it is always the sum of a bunch of sinusoids. [1] http://en.wikipedia.org/wiki/Fourier_transform http://en.wikipedia.org/wiki/Fourier_transform
kwantam 15y ago
In the case you're talking about, impedance means "characteristic impedance", i.e., one parameter which (along with speed of light in the given medium and length) defines the behavior of a transmission line. [1] It is the same thing in the sense that mathematically both are a complex (i.e., Real plus Imaginary) relationship between voltage and current. However, a transmission line is a two-port network while a capacitor is a one-port network, so these two senses of impedance correspond to different physical phenomena. In a capacitor, the impedance defines the relationship between voltage and current at the port (i.e., between the two terminals of the cap; this is sometimes called the driving point impedance). For a transmission line, the characteristic impedance defines, among other things, the equivalent impedance seen at one end of the line when the line is either infinitely long or an integral number of wavelengths long and terminated at the other end by an impedance of the same value as the characteristic impedance. Instrument cables do indeed carry AC signals, inasmuch as the signal on the line corresponds to the frequency of the note(s) you're playing. AC is just "non-zero frequency"---any signal which is not a constant, unchanging (DC) value is considered AC. So, 50/60Hz line power is an AC signal, the 2.4GHz signals used by your 802.11 network card are AC, and middle C (261.6 Hz) is AC. [1] http://en.wikipedia.org/wiki/Characteristic_impedance http://en.wikipedia.org/wiki/Characteristic_impedance

xd 15y ago

Part 2: http://benkrasnow.blogspot.com/2011/06/tutorial-electrical-impedance-made-easy_19.html http://benkrasnow.blogspot.com/2011/06/tutorial-electrical-i...

mrothe 15y ago

I don't know about the rules in America, but this circuit design wouldn't be approved for use here in Europe. Circuits are not allowed to draw what is called 'reactive power', because it stresses the grid. Therefore you have to compensate it (in the simplest case using an inductor or a capacitor).