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?
Best Answer
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.