characteristic-impedanceresistance
Why is there voltage drop across resistors?
Characteristic impedance is the ratio between voltage and current in a transmission line, but resistors also have impedance that translates to a voltage drop. Why does power dissipate across a resistor? How can a transmission line be lossless?
The characteristic impedance of a cable is this mathematically: -
\$Z_0 = \sqrt{\dfrac{R + jwL}{G + jwC}}\$
Where
It has nothing to do with the overall length of the cable. It has everything to do with the ratios of the parameters listed in the equation above.
Simplifications exist - at above about 1MHz you can say that inductive reactance dominates series resistance and capacitive effects dominate the leakage conductance. This makes the equation: -
\$Z_0 = \sqrt{\dfrac{L}{C}}\$
If you are testing at 100 Hz you won't get any sensible indication of characteristic impedance. In fact at 100 Hz it makes no sense to terminate the cable at all.
If you want to know how characteristic impedance is determined try a few experiments using this website: -
If you are intent on understanding the attenuation at 100 Hz then use only R and C and, because standing waves will be virtually non-existent, you can consider R and C as lumped parameters equivalent to the whole length of the cable.
For instance, find a coax spec and look-up the capacitance per unit length and resistance per unit length and convert to equivalent values for your cable length. Basically you get an RC low pass filter with maybe 2 ohm and 1nF (numbers off the top of my head for 10m of cable).
Ask your self how much attenuation is going to happen at 100 Hz. Not much is my impression because the basic cut-off frequency is 79 MHz but, for a cable that is 1 km long, R might be 200 ohms and C might be 200nF and the cut-off frequency will be more like 2 kHz and you will start to see 100 Hz sinewaves being attenuated a little bit.
A confusing article. If you don't already know what you're looking for, then it doesn't really help. What you're missing is that the IC pin is more or less insignificant compared to the PCB trace, for most systems.
If you follow the strategy in figure 7, then you will get good results, as long as your receiver can cope with half the swing that the transmitter puts out.
Figure 7 shows source and sink termination, the 100\$\Omega\$ resistors matching the line, and assuming no contribution or degradation from the IC pins at all. This is pretty much valid for most rise times and most 'low speed' systems, say below 500MHz, even for a few pF of silicon and pin capacitance, as long as the PCB transmission line is taken care of.
The alternative matching scheme for a) where you want more receiver swing and b) you have point to point connection is to use series source-only line termination. This means that you leave out the shunt R at the receiver. The transmitter launches a half-height signal into the line, it is reflected to full height at the receiver (a process it sounds like you know about), and then the reflection is absorbed in the source resistor. This half to full height reflection is why it can only be used point to point, not with multi-drop connections.
Obviously, for matching 50\$\Omega\$ lines, you'd use 50\$\Omega\$ resistors, though you may want to reduce the value of the source resistor to take account of the output resistance of the driver. Some drivers might have such a high source resistance that no additional source resistor needs to be used.
Make sure than your drivers have the capacity to drive 50\$\Omega\$ lines, that is a 100\$\Omega\$ load. TI used 100\$\Omega\$ lines as an example as that's kinder to drivers, and they take less width on a PCB.
Where TI's calculation of pin inductance and capacitance does become useful is when the chip has an internal termination, or frequencies have gone so high that it does become significant. I figure that if you're designing GHz serial links, you will not be scratching around on stack exchange looking for help, or you will be following the reference design board faithfully.
Best Answer
Because that's the definition of resistance, \$R=\frac UI\$.
If a device doesn't have a voltage drop if current flows through it, it can't be a resistor.
Because one is a characteristic impedance, and the other a resistance – you should probably just read on a few pages in your transmission line theory book, and will understand the difference. They have the same unit, because they're a voltage per current, but they don't describe the same phenomenon.
In a resistor, a charge passing through it loses energy (energy per charge = voltage). In a transmission line, we can't speak of charges flowing (there's no net current of electrons in direction of wave propagation), but have electrical and magnetic Field Strengths, given in Volt per Meter, and Ampere per Meter, respectively. The ratio of these fields give you the wave impedance.