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Homework Help: Some subtleties regarding Impedance

  1. Jul 14, 2009 #1

    1. The problem statement, all variables and given/known data

    I am taking a first year electricity and magnetism course.
    In class, the following was taught:

    a. What is the impedance of an inductor?
    b. What is the total impedance of 2 components connected serially?

    But only at home I realized that I don't understand some of the derivation.

    2. Relevant equations

    Vp - Voltage at peak.
    L - Inductance.
    w - angular frequency
    theta - phase

    3. The attempt at a solution

    I was able to derive the impedance of a resistor and a capacitor using phasors.
    I had a problem with the derviation for an inductor:

    dI/dt = (Vp/L) * exp(j * (wt - theta))
    At this point the professor integrated both sides, but did not add a constant of integration. As much as I understand, this is equivalent to the assumption that when the voltage is zero, the current is also zero. Is this assumption really necessary and what is its justification?

    Regarding 2 components connected serially:

    We where told this is analogous to resistance and resistors. Trying to understand this, I wrote:

    V1 + V2 = V = Vp * exp(j * (wt - theta))
    I1 = I2 = I

    Therefore: Vp * exp(j * (wt - theta)) = V = Z1*I + Z2*I = (Z1 + Z2) * I
    which implies that the total impedance really is Z1+Z2.
    But the whole concept of impedance was developed assuming that the voltage across the component looks like: Vp * sin(wt - theta). But here, all the information I have is that the sum of the voltages is in that form. I don't see why each of them is of that form, and therefore I don't understand the justification for this derivation.

    Thanks in advance for your help,
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jul 14, 2009 #2


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    I'll post again, addressing your problem directly, in a minute. But first I want to point out that the impedance of an inductor can be derived in such a way that does not assume anything about the initial conditions other than that the current is sinusoidal. If we have:

    [tex] v_L(t) = L \frac{d}{dt}[i_L(t)] [/tex]​

    And these are both arbitrary functions of time, then we can define some other function of time zL(t) such that:

    [tex] z_L(t) = \frac{v_L(t)}{i_L(t)} = L \frac{ \frac{d}{dt}[i_L(t)]}{i_L(t)} [/tex]​

    At this point there are two equivalent things you can do:

    1. You can say that we are dealing with AC circuits and therefore the current is of the form [itex] i_L(t) = I_0 e^{i (\omega t + \phi)} [/itex]. Here I have used "i" instead of "j", because "j" (the electrical engineering convention) bugs me.

    2. You can take the Fourier transform of both sides of the equation, so that you are going from looking at signals in the time domain to looking at them in the frequency domain (i.e. looking at the complex impedance Z(ω), as well as I(ω) and V(ω), all of which are phasors).

    Approaches 1 and 2 are equivalent, because what Fourier analysis says is that any signal (i.e. any function of time) can be constructed as a combination of sinusoids (i.e. complex exponentials) of varying frequencies at varying amplitudes and phases. Therefore, without any loss of generality, you can say that the input has this sinusoidal form, and ask the question, what does the output look like as a function of an input at this frequency ω? A good way of thinking about it is that you are computing the frequency response of the system, i.e. how the system responds to an input at a given frequency. If you input a certain signal v(t) at a given frequency ω, the frequency response tells you by how much the output i(t) has been phase shifted and attenuated (or amplified) by the system. In this case, "the system" is just an inductor. However, this approach applies to more complicated linear circuits, which can be characterized by their frequency response (also sometimes called a transfer function), H(ω), defined as output/input (in the frequency domain). So, in this case, the output is being taken to be the current across the inductor, and the input is being taken to be the voltage, making H(ω) = 1/Z(ω).

    Since these two approaches are equivalent (either one will tell you how the component attentuates and phase shifts signals of frequency ω as a function of ω), we can make the substitution:

    [tex] Z_L = L \frac{ \frac{d}{dt}[I_0 e^{i (\omega t + \phi)}]}{I_0 e^{i (\omega t + \phi)}} [/tex]

    [tex] Z_L = L \frac{i \omega I_0 e^{i (\omega t + \phi)}}{I_0 e^{i (\omega t + \phi)}} [/tex]

    [tex] Z_L = i \omega L[/tex]​

    An inductor phase shifts current by π/2 (which comes from the factor of i) relative to the voltage across it and attenuates higher frequency signals more than lower ones. This is the frequency response of an inductor.
    Last edited: Jul 14, 2009
  4. Jul 14, 2009 #3


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    To address your problem directly:

    Omitting the constant of integration doesn't assume that i = 0 when v = 0. What it assumes is that when t = 0, the current is given by the same initial phase as the voltage + π/2, and its amplitude is attenuated from that of the voltage by the expected amount. In other words, you are assuming certain initial conditions for the current.

    You could certainly add in a constant of integration, C. However, since these are complex-valued functions, C is an arbitrary complex number. Therefore, there is nothing preventing you from writing:

    C = A/(iωL), where A is some other arbitrary complex constant. Doing so means that the whole expression is still divided by iωL, and we get the result we expect.

    Part of the problem here is that the functions we are using here as basis functions in the Fourier synthesis, namely sinusoids (or complex exponentials), are defined so that they are present for all time (they always have been oscillating and always will). If you want to know precisely how the current in an inductor will evolve with time given that a sinusoidal voltage is suddenly turned on at t = 0, then Laplace transform methods are more convenient than Fourier transform methods.
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