Why is the impedance of a resistor the same in time and frequency domains if the Laplace transform of a constant is not the same constant?


Solution 1

Impedance is usually done in Fourier analysis, not Laplace transform, but that's a quibble. The reality is that you transform the signal, not the impedance. The impedance is an operator that modifies the signal, and it's the effect that operator has on the signal in the transformed domain that we deal with.

To review, the basic linear circuit elements are:

  1. resistor: $V=I\,R$ ($R$ real),
  2. capacitor: $V = \frac{1}{C}\int I \operatorname{d}t$, and
  3. inductor: $V = - L \frac{\operatorname{d} I}{\operatorname{d} t}$.

Putting them all on the same footing, we can write them all as operators that have the form $V(t) = \int A(t,t')\, I(t') \operatorname{d}t'$. On doing so the kernels of the operators ($A$) are given by:

  1. $R\, \delta(t-t')$,
  2. $\frac{1}{C}\, \Theta(t'-t)$, and
  3. $\frac{\partial}{\partial t} \delta(t-t')$,

respectively. Fourier transforming the time domain kernel with respect to $t$ and inverse Fourier transforming with respect to $t'$ gives the frequency space representation of a kernel, $V(\omega) = \int A(\omega,\omega')\, I(\omega') \operatorname{d}\omega'$. For the three linear elements their angular frequency domain kernels are:

  1. $R\, \delta(\omega-\omega')$,
  2. $\frac{1}{i\omega C}\, \delta(\omega-\omega')$, and
  3. $i\omega L\, \delta(\omega-\omega')$.

The short version usually given is: Fourier/Laplace transforms are linear, so multiplying one version of the signal is the same as multiplying the other by the same constant; derivatives become multiplication by frequency; and integrals become division by frequency.

Solution 2

how come the impedance of a resistor is the same in the time domain as in the frequency domain

If two time domain quantities are proportional, then their Laplace domain representations are proportional:

$$v(t) = Ri(t) \Leftrightarrow V(s) = RI(s)$$

where $R$ is a constant. This is a property of the Laplace transform due to linearity of integration.

But it isn't quite correct to call $R$ the impedance in the time domain; the concept of impedance (generally) assumes AC steady state (sinusoidal excitation in which all transients have decayed to insignificance).

More to the point, the impedance of a circuit element is the ratio of the phasor voltage to phasor current:

A phasor is represented by a constant complex number, usually expressed in exponential form, representing the complex amplitude (magnitude and phase) of a sinusoidal function of time. Phasors are used by electrical engineers to simplify computations involving sinusoids, where they can often reduce a differential equation problem to an algebraic one.

The impedance of a circuit element can be defined as the ratio of the phasor voltage across the element to the phasor current through the element, as determined by the relative amplitudes and phases of the voltage and current.


Related videos on Youtube

Author by


Updated on June 12, 2020


  • userManyNumbers
    userManyNumbers over 3 years

    I'm sure this is a silly question, but if the Laplace transform of a constant is not a constant, e.g.

    $$\mathfrak{L}[1] = \frac{1}{s}$$

    then how come the impedance of a resistor is the same in the time domain as in the frequency domain? That is, why, for a resistor of resistance R,

    $$\mathfrak{L}[R] = R~$$

    Couldn't you just take out the factor of $R$ and have $R \cdot \mathfrak{L}[1] = R \cdot \frac{1}{s}$?