coaxialRFtransmission line
When feeding an RF signal to a coaxial transmission line, when do you have to take into consideration the capacitance of the cable? I am working with a 36 MHz signal. The total cable length is ~3 ft (RG316 with SMA connectors).
The source and load impedances are not equal. The source impedance is -130\$j \Omega\$ and the load impedance is 1 k\$\Omega\$.
I want to make sure I am calculating the transfer function correctly.
I understand you asking these sort of questions and I've tried to answer in the format that you presented the questions. Please forgive if I have misconstrued something.
Q1 - cable length when transmitting high data rate: -
(1) Impedance of the transmission line is independent of cable length
(2) Source impedance is independent of cable length
(3) Termination impedance is independent of cable length
Q2 - data rate for a fixed length cable: -
(1) Impedance of the transmission line is generally higher for better data success
(2) Source impedance can be zero for improved results providing the receiver has good terminator
(3) Termination impedance needs to be the correct value for the cable to minimize reflections corrupting the data
This should be true for any cable length and any wavelength.
No, this is not true for any wavelength. At low frequencies (as in telephony/audio) the characteristic impedance is dominated by R and C: -
It approximates to \$\sqrt{\frac{R}{jwc}}\$ i.e. complex
At dc it is \$\sqrt{\frac{R}{G}}\$ i.e. resistive
And at RF frequencies it is \$\sqrt{\frac{jwL}{jwC}} = \sqrt{\frac{L}{C}}\$ i.e. resistive
However, in a typical datasheet of any coaxial cable the capacitance of the cable per unit length is usually given. I don't understand the effect of this capacitance on the input impedance.
Data sheets do tend to give the capacitance per unit length (without mentioning L/m) and if you know the characteristic impedance you can calculate what L per metre is: -
\${Z_o}^2 = \frac{L}{C}\$ therefore \$L= C\times{Z_o}^2\$ = 100\$e^{-12}\times 50\times 50 = 0.25\$ uH per metre.
What would be the input impedance?
The input impedance of RG-58 at RF frequencies will be 50\$\Omega\$ resistive because there are inductive and capacitive components that are in ratio as per the formulas above. This assumes you are correctly terminating the cable in 50\$\Omega\$
EDIT This is about where the turning points are between audio (complex) impedances and HF resistive impedances. For a start, here is a good spec for RG-58. Below are the salient points: -
Notice the bottom two data highlighted in red - this is the inner and outer DC resistance per 1000ft - a total of 54\$\Omega\$ per 1000ft loop (304.8m). This equates to 0.1772\$\Omega\$ per metre.
For |jwL| to equal 0.1772, the frequency will be \$\frac{0.1772}{2\Pi L}\$ and if L = 0.25uH then F = 113kHz. Ten times higher in frequency (1.13MHz) and Zo pretty much approximates to \$\sqrt{\frac{L}{C}}\$ i.e. is 50\$\Omega\$ resistive.
For higher frequencies, Zo is a reliable resistive quantity, for frequencies down between 10kHz and 1MHz it's a mish-mash and at audio frequencies below 10kHz it becomes what is telephonically known as a "complex impedance" where the impedance is largely determined by series resistance and parallel capacitance and the impedance phase angle is about 45ยบ because \$\sqrt{-j}\$ is 45 degrees.
Best Answer
50 ohm air spaced cable has a capacitance of 66.7pF/m, dielectric spaced cables with velocity factors around 0.66 have around 100pF/m. This is a useful number, every time you connect two items together in the lab with a typical 1m coax cable, at low frequencies it is like adding 100pF across the signal. A 2m cable is like 200pF.
\$C=\frac{1}{Z_0 v}\$ where \$v=\frac{c}{\sqrt{\epsilon_r}}\$ is the velocity of propagation
As your 1m cable is around 20% of a wavelength at 36MHz you should really treat it as a transmission line.
Just looking at the capacitance, at low frequencies it looks like about 100pF, which at 36MHz has a reactance of about 44ohms. It is clearly going to influence your circuit where your impedance is around 1k.
Assuming that you are after the simple voltage transfer function from a source with 130k impedance to the 1k load, with no cable you have -42.3dB. Treating the cable as a simple 100pF shunt gives -69.4dB, treating it as a 5ns length of 50 ohm cable gives -67.4dB (from a simple SPICE sim).
Assuming that you are after the simple voltage transfer function from a source with -130j impedance (= 34pF cap at 36MHz) to the 1k load, with no cable you have -0.1dB. Treating the cable as a simple 100pF shunt gives -11.9dB, treating it as a 5ns length of 50 ohm cable gives -8.9dB (from a simple SPICE sim). Note that this can change quite dramatically with the length of the cable, use a 2m cable and you get -2.8dB, increasing your signal by over 6dB.