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I am reading about high speed PCB layout design from TI (Texas pdf)
This document mentioned the software AppCAD from Agilent to calculate impedance. Here is the snapshot.
My question here is whether this Z0 a complex number or not.
If it omits imaginary part, why will we use this software instead of calculator? Does H(Height) and dielectric \$\varepsilon_r\$ matter in this scenario?
\$Z_0\$ is the characteristic impedance of a transmission line is given as:$$Z_{0}=\sqrt{\frac{R+j\omega L}{G+j\omega C}}$$ \$R\$ and \$L\$ are the series resistance and inductance per unit length. \$G\$ and \$C\$ are the parallel conductance and capacitance per unit length across the transmission line.
If \$R\$ and/or \$G\$ are significant then energy is lost to heating resulting in what is called a lossy line.
If these two are insignificant, then they are approximated as zero, reducing the expression to:$$Z_{0}=\sqrt{\frac{L}{C}}$$ which is purely resistive.
If it omits imaginary part...
Actually it doesn't. The linear resistivity of the copper and the conductivity of the fibreglass is sufficiently small that the \$j\omega\$ factors cancel out from the numerator and the denominator.
The real parts are the ones being ignored, or put in a better way are treated as being insignificant.
Addition:
why will we use this software instead of calculator? Does H(Height) and dielectric \$\epsilon_r\$ matter in this scenario?
The IPC-2141 standard provides a calculation for L and C. Cadence has some good information. I repeat the IPC-2141 microstrip equations to show the need for \$H\$ and \$\epsilon_r\$: $$Z=\frac{87}{\sqrt{\epsilon_r + 1.41}}\text{ln}\left(\frac{5.98H}{0.8W+T}\right)$$ $$C=\frac{0.67(\epsilon_r + 1.41)}{\text{ln}\left(\frac{5.98H}{0.8W+T}\right)}\text{pF/in}$$ $$L=5.071\text{ln}\left(\frac{5.98H}{0.8W+T}\right)$$
Clearly all the parameters including the thickness \$T\$ are needed. Either the software or a hand calculator can be chosen to calculate.
If the line is lossless, then it has a real characteristic impedance.
If the characteristic impedance has an imaginary component, then the line is lossy.
We often model our lines as lossless even when they aren't actually perfectly lossless. This tool is providing an approximate model of a lossless line and the Z0 it reports is real.
In the design of transmission lines Z0 is usually real, at least the desired Z0 is. That's because, normally the source and load terminations are considered to be real, or have been tuned to be real.
In the provided screen shot the design of a microstrip line is presented, and the height and dielectric constant are elements of the equation for the characteristic impedance such a transmission line, so they are necessary.
If by the "imaginary part" of the transmission line you are referring to the delay of the line that is a function of the length of the line and the speed of light in the line, that is different from the characteristic impedance. The speed of light in a material is also a function of the dielectric constant.
So there are two complex numbers that have been discussed.
My question here is whether this Z0 a complex number or not
To find \$Z_0\$ start with a representation of a short section of transmission line (t-line): -
R, L, C and G can be thought of as "per metre" values but, as will be suggested further down, they can be based on "per mm" or per micron all the way to zero length.
The model in your question uses only L and C hence, the section becomes this: -
The section has an input impedance \$Z_{IN}\$ on the left that we want to find. On the right, more sections are connected. They all have the same input impedance.
We can now solve for \$Z_{IN}\$ in the left: - $$$$ $$Z_{IN}\hspace{0.5cm} =\hspace{0.5cm} sL + \dfrac{1}{sC}||Z_{IN}\hspace{0.5cm} =\hspace{0.5cm} sL + \dfrac{Z_{IN}}{1 +sC Z_{IN}} $$
$$Z_{IN}\cdot (1 + sCZ_{IN})\hspace{0.5cm} =\hspace{0.5cm}sL + s^2LCZ_{IN} + Z_{IN}$$
$$sCZ_{IN}^2\hspace{0.5cm} =\hspace{0.5cm} sL + s^2LCZ_{IN} $$
$$Z_{IN}^2 \hspace{0.5cm}=\hspace{0.5cm}\dfrac{L}{C} + sL Z_{IN}$$
That looks about as far as it goes but, if we imagine our section of t-line shrunk towards zero length, \$sL\cdot Z_{IN}\$ becomes insignificant compared to \$\frac{L}{C}\$ hence: -
$$Z_{IN} = \sqrt{\dfrac{L}{C}}$$
And clearly that is purely resistive. However, we have ignored the effects of R and G meaning that the simplified formula only applies at frequencies greater than typically 100 kHz. If you considered R, L, C and G you'd derive this: -
$$Z_{IN} = Z_0 = \sqrt{\dfrac{R + j\omega L}{G + j\omega C}}$$
Derivation of Characteristic Impedance of R, L, G and C section gives you the full math (similar to the above and, without the complication of using the telegraphers equations).
For normal cable constructions the impedance exhibits a complex impedance at around the mid audio frequencies. This is well-known in telephony of course.
The image below is from a spreadsheet I made and also presented in this answer. It's of a typical telecom cable and, around 1 kHz, it has an impedance magnitude of 600 Ω.
The formula for impedance magnitude is not too hard to prove: -
$$|Z_0| = \sqrt{\sqrt{\dfrac{R^2+\omega^2 L^2}{G^2+\omega^2 C^2}}}$$
The formula for the impedance angle is a tad harder to prove: -
$$\text{Angle} = \dfrac{\arctan\left[{\dfrac{\omega\cdot (LG-CR)}{RG+\omega^2 LC}}\right]}{2}$$
If it omits imaginary part, why will we use this software instead of calculator? Does H(Height) and dielectric εr matter in this scenario?
The calculator helps you establish the inductance and capacitance per metre based on physical distances and, that's pretty useful even though your calculator doesn't actually display those values. \$H\$ and \$\epsilon_r\$ totally matter in this respect.
