Let me know one thing. Why is impedance represented by a complex number when considering loss?

Usually it is said that imaginary part is because of loss. Is it True? If it is true then does that imply that a line with characteristic impedance Z0=50 ohm has zero loss?

  • 8
    \$\begingroup\$ Huh? This question needs to be reworded. It is difficult to tell what is being asked here. \$\endgroup\$ – Olin Lathrop Sep 25 '12 at 18:31

I certainly hope I managed to decrypt what's being asked here. If not, please clarify so I can provide a more relevant answer.

Let's start with the definition of the lossy and lossless models:

In lossy model the resistance of the line itself matters for the calculation! Basically we're running such voltage through the line that even if we just took replaced the signal with a DC voltage equal to the RMS voltage of our signal, we'd still have losses.

Using the lossless model, if we replaced the signal with DC voltage equal to RMS voltage of the signal, we'd have no considerable losses. In this case, only reactive components matter and the resistance of the line itself isn't important.

This brings us to the basic equation for the transmission line characteristic impedance:

\$Z_0=\sqrt{ \frac {R + j \omega L} {G + j \omega C}}\$

Keep in mind that G is conductance in parallel with the capacitance, and that R is resistance in series with the inductance. In the lossless model, we don't have the two of them, so their values are taken as zero. We can then rewrite the equation as

\$Z_0=\sqrt{ \frac {0 + j \omega L} {0 + j \omega C}}\$

which is equivalent to

\$Z_0=\sqrt{ \frac {j \omega L} {j \omega C}}\$

Which we can rewrite as \$Z_0=\sqrt{ \frac {j \omega} {j \omega } \frac {L}{C}}\$

The \$j \omega\$ parts prodce one and we get in the end only

\$Z_0=\sqrt{\frac {L}{C}}\$

So the lossless model doesn't have the imaginary part because the imaginary units cancelled themselves out.


Impedance is represented as a complex number, regardless of whether you are considering loss or not, so you must understand this before you start thinking about lossless/lossy transmission lines.

In general terms, impedance is the ratio V/I (voltage over current). V and I are not necessarily constant in time, so they are represented by their frequency components. So impedance is now thought as the ratio of two sine waves of a certain amplitude and phase at a particular frequency. Such ratio will be another wave of certain amplitude and phase.

So you see, in its basic form, you need two numbers to represent the ratio of two sine waves: amplitude and phase. A complex number is a mathematical convenience to carry over those two values, although not directly amplitude and phase as such, but the x-y components of the related phasor or vector. You will frequently see Z expressed as a function of f or w, which is frequency, and this is just an expression that generalizes the ratio V/I for any frequency f (just replace with the desired frequency and you'll get the real and imaginary parts, which you can then translate to amplitude and phase of said ratio).


It is interesting to think about the meaning of the fact, rather than the mathematics involved. Of course mathematics are a necesary tool, but a tool to understand and express a meaning. When both E and H maintain a 90° phase difference (where phase refers to a time relationship), or in circuit terms, V and I are normal to each other (in time), the power delivered to the line to a given load is maximum. It can be seen in P = ExH or P = 1/2Re(VI*). If E and H travel perpendicular to the propagation direction, and additionally have a 90° TIME phase difference, we have the best case. Maximum power delivered. In this case, characteristic impedance is real

  • \$\begingroup\$ I think your answer is trying to get at something the other answers miss, which is that representing impedances as complex numbers allows things like Ohm's Law work with things other than resistors. It would be helpful, however, if you defined the letters you're using. \$\endgroup\$ – supercat Aug 22 '13 at 19:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.