# Impedance calculation from voltage and current (exponential) waveforms

If we want to find the impedance across two terminals (a port) at a frequency $$\\omega_0\$$ using the voltage waveform across $$\v(t)\$$) and current through ($$\i(t)\$$) the terminal, general intuition is to use discrete Fourier transform for both voltage and current waveforms and take the ratio $$Z(\omega)|_{\omega=\omega_0}=\frac{V(\omega)}{I(\omega)}|_{\omega=\omega_0}.$$ To my understanding, the answer should not depend on the shape of the waveform as long as there is a $$\\omega_0\$$ frequency component in the waveforms. - Please correct me if I am missing any point here. But this doesn't seem to be true when the waveforms have an imaginary frequency component (growing/decaying in time).

Example:

For example, let's take $$\i(t)=I_0 \sin(\omega_0t)e^{\alpha t}\$$ and suppose we have an ideal inductor $$\L\$$ to measure the impedance (in fact, an ideal inductor is chosen to explain my question). Now we have, $$v(t)=L{di(t) \over dt}=L I_0 e^{\alpha t} \left(\alpha \sin ( \omega_0 t)+ \omega_0 \cos ( \omega_0 t)\right),$$ and the Fourier transforms, $$I(\omega)=-i\sqrt{\frac{\pi }{2}} I_0\big( \delta (i \alpha -\omega +\omega_0)- \delta (-i \alpha +\omega +\omega_0)\big)$$ $$V(\omega)= \sqrt{\frac{\pi }{2}} I_0 L\big(i \alpha \delta (i \alpha -\omega +\omega_0)+ \omega_0 \delta (i \alpha -\omega +\omega_0)-i \alpha \delta (-i \alpha +\omega +\omega_0)+ \omega_0 \delta (-i \alpha +\omega +\omega_0)\big)$$

Obviously, we cannot find $$\V(\omega)/I(\omega)\$$ at $$\\omega =\omega_0\$$ but if we put $$\\omega =\omega_0+i \alpha\$$, and consider one-sided Fourier transform we have

$$\frac{V(\omega)}{I(\omega)}|_{\omega=\omega_0+i \alpha}=(i\omega_0+ \alpha)L$$

This is, in fact, equivalent to the impedance of an inductor when $$\\omega =\omega_0+i \alpha\$$. But here, I am interested in finding the impedance at $$\\omega_0\$$.

If we use Laplace transform, we will get $$\V(s)/I(s)=s L\$$, and then we can simply put $$\s=j\omega_0\$$ to obtain the impedance across the terminals (inductor in the example). However, to my knowledge, we can't use Laplace transform for the discrete set of numbers (i.e. voltage and current signals in the time domain).

Even if we use z-transform (which said to be more like discrete Laplace transform), we will still be ending up with $$\V(z)/I(z)=(i\omega_0+ \alpha)L\$$

My questions

1. What is the physical meaning of the real part $$\(\alpha L)\$$ observed in the Fourier transform in this example?

2. How can we obtain the equivalent impedance across a port using voltage and current waveforms when the waveforms are exponentially growing (or decaying)?

• You can't use the Laplace transform to calculate a time domain response, which requires discretisation. You absolutely can use the Laplace transform representation to find the impedance at a given frequency. – loudnoises Mar 20 at 8:08
• May be related – Unknown123 Mar 20 at 8:15
• Thanks, @loudnoises, That is exactly my question. What I have is voltage and current signals, thus cannot use Laplace transform directly. I am trying to use curve fitting to find good fitting function and then use Laplace transform-but will not be very accurate. Question here is how to find the equivalent impedance directly from the waveform. – Pojj Mar 20 at 8:37
• Is there any particular reason you're using exponential waveforms? Maybe you should investigate how an LCR meter works. – loudnoises Mar 20 at 8:48
• Since you're only interested in a single frequency, this is an ideal application for the Goertzel algorithm. But, for your purposes, the Goertzel is no better or worse than the DFT, it's just a more computationally efficient way of getting the same answer as the DFT. I would say your "real" question is what is the actual meaning of $Z(\omega)|_{\omega=\omega_0}$, especially since your stimulus waveform has a decidedly more complex frequency structure. – Mr. Snrub Mar 20 at 17:23

My solution is in accordance with the expected result:

From FOURIER TRANSFORMS:

$$\mathcal{F} \{ e^{-at}\sin(\omega_0t)u(t)\} = \frac{\omega_0}{(a+j\omega)^2+\omega_0^2}$$ $$\mathcal{F} \{ e^{-at}\cos(\omega_0t)u(t)\} = \frac{a+j\omega}{(a+j\omega)^2+\omega_0^2}$$

and the inductor current as:

$$i(t) = I_0e^{\alpha t}\sin(\omega_0t)u(t)$$

In frequency domain:

$$\boxed{I(\omega)= \frac{I_o\omega_0}{(-\alpha+j\omega)^2+\omega_0^2}}$$

Then the inductor voltage is:

$$v(t) = L\frac{\mathrm{d}i(t) }{\mathrm{d}t }$$

or

$$v(t)=LI_0 \left[\alpha e^{\alpha t}\sin(\omega_0t)+ \omega_0e^{\alpha t} \cos(\omega_0t) \right ]u(t)$$

Similarly, in frequency domain:

$$V(\omega)=LI_0 \left [ \frac{\alpha\omega_0}{(-\alpha+j\omega)^2+\omega_0^2} + \frac{\omega_0(-\alpha+j\omega)}{(-\alpha+j\omega)^2+\omega_0^2} \right ]$$

or

$$\boxed{V(\omega)=LI_0\omega_0 \left [ \frac{j\omega}{(-\alpha+j\omega)^2+\omega_0^2} \right ] }$$

The impedance is:

$$Z(\omega)=\frac{V(w)}{I(\omega)}$$

or

$$\boxed{Z(\omega)= j\omega L}$$

From LAPLACE TRANSFORMS, the same result can be achivied:

$$I(s) = \frac{I_0 \omega_0}{(-\alpha + s)^2+ \omega_0^2}$$

and

$$V(s)= LI_0\omega_0 \frac{s}{(-\alpha + s)^2+ \omega_0^2}$$

The impedance is:

$$Z(s)=\frac{V(s)}{I(s)} = sL$$

doing $$\ s = j\omega\$$

$$Z(j\omega)= j\omega L$$

EDIT:

The main goal of my answer was to point out an error in the calculation of a particular Fourier transform present in question. Although the OP does not explicitly make clear the fact that the INDUCTANCE IS UNKNOWN - it seems to me that more important is the determination of the INDUCTANCE rather than the IMPEDANCE of inductor from the measurements in time.

Note that, in this case, the OP will be faced with a linear system identification procedure (since in practice it will be a series RL circuit due to the inductor's DCR). There are many techniques already used by LCR meters. In other hand seems the OP intention is to do some inovative work using growing exponentials. One problem I realize that, in the question, would be difficult to generate a exponentally growing sine CURRENT. Maybe a VOLTAGE with a similar shape could be applied on RL circuit. Measurement in time suggest a easier form of curve fitting (expected versus measured) to determine the parameters L and R.

• Additionally, the expression for $V(\omega)$ also could be obtained directly from $I(\omega )$ through the differentiation property of Fourier Transform: $\frac{df(t)}{dt} \Rightarrow j\omega F(\omega )$. – Dirceu Rodrigues Jr Mar 21 at 19:35
• ...which would then mathematically demonstrate that the input waveform doesn't matter, because $Z( \omega ) = {V( \omega ) \over I ( \omega ) } = { L \cdot j \omega I(\omega) \over I(\omega)} = j \omega L$. Very cool! – Mr. Snrub Mar 21 at 20:12
• I agree, since the impedance in wider sense is $Z(s)=\frac{V(s)}{I(s)}$ and Laplace Transform accepts signals not "absolutely integrable" (not decaying in time - as step and ramp, for example). In other hand, this "Dirichlet condition" is necessary for Fourier Transform (exponentially decaying is OK). – Dirceu Rodrigues Jr Mar 21 at 20:45
• For example: If $i(t)=I_ou(t)$ (step current in a inductor), then $I(s) = \frac{I_0}{s}$. So, $v(t) = L\frac{\mathrm{d}i(t) }{\mathrm{d} t}=LI_0\delta (t)$ (a voltage impulse) and $V(s)=LI_0$. Finally, $Z(s)=\frac{V(s)}{I(s)}=sL$. Try yourself with a new example where $i(t)=I_0t u(t)$ (a current ramp)... or, alternativelly, here also could be used the property $\frac{\mathrm{d}f(t) }{\mathrm{d} t} \Rightarrow sF(s)$. – Dirceu Rodrigues Jr Mar 21 at 22:43
• Interesting, here I missed to take one-sided input function $u(t)$ (I directly used Mathematica to find Fourier Transform) - I think, that is where I missed it. BTW, can we still apply the Fourier Transform if the waveform is growing (i.e. $a<0$ in your answer) – Pojj Mar 22 at 7:59