I have a question regarding impedances. While it is intuitive that \$ R \$-element has impedance \$ R \$, it's not so easy for me to figure out impedances of \$ L \$ and \$ C \$.
Here is my, probably incorrect, reasoning. For element \$ C \$ following differential equation is satisfied: \$ i(t) = C \frac{\mbox{d} u}{\mbox{d} t}\$. Our input is \$u\$, and utput is \$i\$. By transforming this equation we have
$$ \frac{U(s)}{I(s)} = G(s) = \frac{1}{sC} $$
Since we cosider only sinusoidal inputs we know, that output will also be sinusoidal of the same frequency, but different magnitude and phase, so it's fine to use Fourier instead of Laplace, then let \$s:=j \omega\$, then
$$G(s)=\frac{1}{j \omega C} = -j \frac{1}{\omega C} $$, which is indeed equal to impedance of element \$C\$.

Now - is that a coincidence, or the actual reasoning behind impedances and whole method of solving circuits based on them? If so, why take \$u\$ as an input, and \$i\$ as an output? Is it to make analogy to well known \$R=\frac{U}{I} \$?

If I tried to use not necessarily sinusoidal voltage and current, would I use the more general form: \$\frac{1}{sC}\$? Would it work?

  • \$\begingroup\$ The R element has resistance, L and C have impedance \$\endgroup\$ Commented Feb 1, 2017 at 17:23
  • \$\begingroup\$ One could say that impedance is a complex number which consints of resistance - $$Re(Z)$$ and reactance $$Im(Z)$$. \$\endgroup\$
    – gabe
    Commented Feb 1, 2017 at 17:37
  • \$\begingroup\$ Impedance is defined as V/I. So the result is no coincidence. It appears that you are studying this in a formal setting (university). As you continue on, I think everything will make more sense. \$\endgroup\$
    – user57037
    Commented Feb 1, 2017 at 21:17
  • \$\begingroup\$ Yes, the Laplace form (s domain) can be used for transient analysis as well as steady state analysis. At least that is what I remember from my university days. \$\endgroup\$
    – user57037
    Commented Feb 1, 2017 at 21:20
  • \$\begingroup\$ Impedance of a pure inductance or capacitance has no resistance, its only when you combine inductance or capacitance or both with resistance that you get a complex number with real (resistance component) and an 'imaginary' (+/- j) component. \$\endgroup\$ Commented Feb 2, 2017 at 12:57

3 Answers 3


Your intuitions are correct. The "j" can be viewed in many ways. Many people see "j" as a unit vector along the "unreal" axis in the complex plane (traditionally called the Real and Imaginary axes). The "j" axis is rotated 90 degrees from the real axis. So "j" can really be thought of as a operator like square root or exponential that rotates a real number by 90 degrees. This is a natural mathematical operation for electronic circuits because the currents and voltages in and across an inductor or capacitor are always 90 degrees out of phase. So plotting the current or voltage associated with a L or C in the complex plane always results in the current and voltage leading or lagging the other by 90 degrees. This explains the power of complex math as applied to RLC circuits when driven by sinusoidal functions. jcos(wt) = cos(wt+90 degrees)! jsin(wt)=sin(wt+90 degrees)! The whole jw and 1/jw thing is just the realization that when using sine and cosine functions that jw*Function equals the time derivative of the function, and (1/jw)*function equal the time integral of the function. If you use this definition and apply it to the voltage and current associated with an L or C you will indeed find that v(t)/i(t) really does end up being jwL for an inductor and 1/jwC for the capacitor! This notion of "AC resistance" or, "impedance" only works for sinusoidal sources and responses, but there is a cute escape clause: Any function (square, triangle, or just periodic wiggle) can be expressed as a sum of sine waves of different frequency and phase. Since you can solve your circuit using any one frequency of a source you can always solve the problems for an arbitrary waveform by adding up the solutions using each frequency component in the source that makes up the waveform. If the waveform can be transformed then you can find this solution sum using the coefficients of the transformed waveform.


Intuitive answer - the whole point of impedance of inductor /capacitor is their reaction to frequency response. Your mathematics is correct but you also need to look at the impedance (resistance) frequency based. For example : Xc=1/wc is the capcitor resistance and it has low resistance on high frequency and high resistance on low frequency. Filters are being design using this concept for example. Another example is impedance matching, which the C/L impedance is being match to a resistor impedance (generally speaking)


I'm not entirely sure I understand your question, but I can point you in a direction where you may just grok the whole thing on your own.

Firstly, when you say "input" and "output" of some equation, that may make sense in a purely arithmetic sense, but it's a bit confusing to make such statements in electronics or mechanics, since the terms are used to identify sources and destinations of voltages or currents being injected or measured somewhere. I prefer to say that the equation \$i(t) = C \frac{du}{dt}\$ describes the relationship between current \$i\$ through and voltage \$u\$ across a capacitor \$C\$.

This is just a relationship, and it's not a "system" per se, with an input signal and output signal.

It's not yet clear what "impedance" means. Resistance is just a relationship, between current and voltage, but may only be applied to components where that relationship is time-invariant, which is not the case for capacitors and inductors. So, to be glib, I ask why can't resistance change with time? If we write the capacitor equation again, but rearrange it to reveal the equivalent "resistance" of the capacitor, as it changes with time, you'll see what I mean.

Start with Ohm's law, which I stress is merely a statement about some property we made up, which we call "resistance", and which is nothing more than a single value which embodies some relationship between current and voltage:

$$ R = \frac{V}{I} $$

This is valid for all values of \$I\$, \$V\$ and \$R\$, at all times, but we usually associate R with a constant value, and call the relationship "Ohm's law". There's nothing wrong with extending this relationship to include time, still with a constant resistance \$R\$, but \$I\$ and \$V\$ are usually expected to change over time:

$$ R = \frac{V(t)}{I(t)} $$

What, though, if \$R\$ is not constant? If \$R\$ changes with time too, can we extend Ohm's law to include components that behave this way? You're damn right we can, because the meaning of resistance is only "the ratio of voltage across a thing to the current through it, at some instant in time". The only difference is that since resistance can now change with time, it might be better to call it "equivalent resistance at some instant in time". The official name for that is "impedance", usually denoted by \$Z\$:

$$ R(t) = Z = \frac{V(t)}{I(t)} $$

By extension, we can use the capacitor equation to calculate a capacitor's "equivalent resistance at time t", just as we would for a regular resistor, using Ohm's law:

$$ R_C(t) = Z_C = \frac{V(t)}{I(t)} = \frac{V(t)}{C\frac{dV(t)}{dt}} $$

At this point you could stop and quite rightly say that you have all the information you need to solve any network of resistors and capacitors, simply because Kirchhoff's current and voltage laws, and Ohm's law can still be applied. We haven't lost any information, and those laws don't break just because there are derivatives in the equations. For example, a potential divider consisting of a resistor and capacitor can be analysed using \$Z_C\$ and \$R\$ in exactly the same way you would deal with two constant resistances:


simulate this circuit – Schematic created using CircuitLab

$$ \begin{aligned} V_{OUT1} &= V_{IN1}\frac{R_2}{R_1 + R_2} \\ \\ \\ V_{OUT2}(t) &= V_{IN2}(t)\frac{Z_{C1}}{R_3 + Z_{C1}} \\ \\ &= V_{IN2}(t)\frac{\frac{V_{IN2}(t)}{C_1\frac{dV_{IN2}(t)}{dt}}}{R_3 + \frac{V_{IN2}(t)}{C_1\frac{dV_{IN2}(t)}{dt}}} \end{aligned} $$

That equation for \$V_{OUT2}(t)\$ is complete and correct, but it makes my eyes bleed. To solve this equation, instead of leaving those derivatives in their natural form, we can use laplace transformations to derive a much simpler problem to solve with inverse Laplace transforms.

Before we actually write that last equation, stop at the step prior, and rearrange things to obtain the input-to-output relationship, which is called "gain", where \$Gain = \frac{Output}{Input}\$. This step doesn't change the truth or completeness of the equation in any way, but it is very useful when applying control theory to cascaded stages:

$$ \begin{aligned} Gain = g(t) = \frac{V_{OUT2}(t)}{V_{IN2}(t)} &= \frac{Z_{C1}}{R_3 + Z_{C1}} \\ \\ \end{aligned} $$

Now take the laplace transform of both sides, to obtain the "transfer function", in the s domain:

$$ \begin{aligned} G(s) &= \frac{\frac{1}{sC_1}}{R_3 + \frac{1}{sC_1}} \\ \\ &= \frac{1}{1+sR_3C_1} \end{aligned} $$

All of that is probably familiar to you, but the point to take home is this: This last equation is as valid, true and complete as the horrible equation with derivatives that we obtained before. No information has been lost, and any information you could extract from the derivative version can be obtained from this one. Time \$t\$ has disappeared, making it seem like something's missing, but don't forget that \$t\$ will return if you applied the inverse Laplace transform to get back into the time domain.

The second point I wish to make clear is that there's no mention of \$V_{IN2}\$ or \$V_{OUT2}\$, giving the impression that information has been lost, but again, if you applied the inverse Laplace transform, you will get it straight back. In other words, there's no difference between the Laplacian form of the gain expression and the original natural derivative form, except the domain in which it's expressed, \$t\$ or \$s\$.

The third point is that the expression on the right hand side makes no mention of \$V_{IN2}\$ or \$V_{OUT2}\$. In other words, it is valid for any function \$V_{IN2}(t)\$ or \$V_{OUT2}(t)\$ whether they be sinusoids, deltas, steps, triangles or whatever you throw at it.

Now let's address the Fourier transform part of your observations. It's important to note that as long as the gain function \$ g(t) = \frac{V_{OUT}}{V_{IN}}\$ is linear, the following is true:

$$ g(p + q + r) = g(p) + g(q) + g(r) $$

Whatever form \$V_{IN}\$ takes, it can be considered to be the sum of any number of sub-signals, p(t), q(t), r(t) etc. Consequently, if you have three identical systems with the same gain expression, fed from these three independent signal sources, and then you summed their outputs, the result would be identical to the output of one of the systems fed with the sum of p, q and r.

As long as the components that make up some network are linear (which resistors, capacitors and inductors are), then the transfer function of that system is also linear, and this linearity has interesting consequences for the harmonics that make up any arbitrary waveform. Those harmonics are all treated the same way by the system, regardless of whether they are all present in the signal simulataneously, or presented in isolation, as a single pure sinusoid. This gives meaning to the concept of frequency response; you can measure the response of the system to a single sinusoid, and that response will be the same for any sinusoid of that frequency present in any signal, regardless of the presence or absence of any other frequency components accompanying it.

Comparing the Laplace transform with the Fourier transform, the difference is subtle:

$$ \begin{aligned} \mathcal{L}\left[{f(t)}\right] &= {\int_{-\infty}^{\infty}{e^{-st}} \cdot f(t) \cdot dt} \\ \\ \mathcal{F}\left[{f(t)}\right] &= {\int_{-\infty}^{\infty}{e^{-j \omega t}} \cdot f(t) \cdot dt} \\ \\ \end{aligned} $$

The only change is the variable substitution \$s\$ for \$j\omega\$, which has a couple of important consequences. The first is that the Laplace transform is more general, in that its domain is the entire complex \$s\$-plane, whereas the Fourier transform's domain is constrained to a single axis of that plane. The Laplace transform is more suited to the solution of differential equations, since the restricted domain of the Fourier transform severely limits the functions for which a transform exists.

I believe you may plug in \$j\omega \$ in the place of \$s\$ because signals in the real world are not complex (and won't require the transform to operate over the entire \$s\$-plane), and since the function being transformed is a linear combination of simple sinusoids and their derivatives, a Fourier transformation is always possible.

There's one more thing I want to point out, to do with transfer functions in the \$s\$ domain. Remember that they keep all the information present in the raw differential equations? Well, they keep that data in a different form, where each sinusoid is represented as a vector on the complex \$s\$-plane. Complex numbers have magnitude and direction, which is in general far more useful to know than the actual time-domain behaviour (the shape) of a sinusoid. From those vectors we can derive amplitude and phase information, and we rarely need more than that. I can't remember the last time I actually had to employ the inverse Laplace transform to solve a differential equation of a network, in the time domain. Most filter and control system design and analysis is based on amplitude and phase, and the whole process is greatly simplified by using tools like "poles" and "zeros" which emerge naturally from the \$s\$-domain representation of a gain function.


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