For a given RF transformer, the Insertion Loss @ 10 MHz is 0.5 dB, and the Return Loss @ 10 MHz is 25 dB, with impedances of 50 Ohm. Let's say I put in a 1 V, 10 MHz sine wave, what happens?
Return loss tells you how much of the input signal is reflected. Return loss is the ratio between the reflected power and input power:
$$ \mathrm{RL}=\frac{P_{ref}}{P_{in}} $$
If the input signal is 0 dBm and there is 25 dB return loss, then the component will create a reflected wave of -25 dBm back toward the generator.
In your example, I assume you mean a 1 V rms signal (as opposed to 1 V amplitude or 1 V peak-peak). This is +13 dBm. With 25 dBm return loss the reflected wave has -12 dBm power or 56 mV rms amplitude.
The insertion loss tells you how much power is lost in the signal passing through the component. Insertion loss is the ratio between output power and input power:
$$\mathrm{IL}=\frac{P_{out}}{P_{in}}$$
If the input signal is 0 dBm and there is 0.5 dB insertion loss, the transmitted signal (continuing towards the final load) is -0.5 dBm. In your example, +13 dBm - 0.5 dB gives +12.5 dBm power or 943 mV rms amplitude.
If the windings are 1:1, are these losses the same if I run the signal from the secondary to the primary?
In and ideal world, yes. This is because of the reciprocity theorem. In the real world there might be slight differences in the measured characteristics due to differences between the connectors on each side, etc.
If this characteristic is important for your application you can look for a transformer with a "reverse return loss" and "reverse insertion loss" specifications. If the vendor offers S-parameter characteristics of the part, you can look at the S12 (reverse transmission) and S22 (reverse reflection) characteristics. If they are the same as the S21 and S11, then your device is symettric.
Does this change how impedance is transformed across windings?
If the turns ratio is 1:1 there won't be any impedance transformation.
What if the winding impedances aren't the same?
In rf, if things are done right, the impedance you see looking into the primary depends more on how the secondary is loaded than on characteristics of the transformer itself. If you want to transform impedances you will choose the turns ratio so that, for example, a 75 ohm load can be driven by the secondary, while the primary looks like a 50 ohm load to the generator.
So if I understand correctly, Insertion is the efficiency from one winding to the next, and Return is the reflected portion of the original signal?
Insertion loss is the power loss from input to output. It applies to many kinds of rf devices, not just transformers.
when you say loss, you mean the ratio between input and output, not the difference, correct?
Yes, a ratio in watts is a difference in dBm.
They can be the same thing. A transmission line will attenuate an amount (based on distance) whilst maintaining impedances so, it could be regarded as causing "insertion loss".
Here's what wiki says about cable attenuation: -
Insertion Loss
Insertion loss, also referred to as attenuation, refers to the loss of
signal strength at the far end of a line compared to the signal that
was introduced into the line. This loss is due to the electrical
impedance of the copper cable, the loss of energy through the cable
insulation and the impedance caused by the connectors. Insertion loss
is usually expressed in decibels dB with a minus sign. Insertion loss
increases with distance and frequency. For every 3 dB of loss, the
original signal will be half the original power (\$\sqrt2\$ of
amplitude).
Taken from here
Best Answer
An accepted formula for calculating the power needed by a receiver operating at a specific data rate is this: -
Power (dBm) = -154 dBm + 10\$log_{10}(\text{data rate})\$
Formula from this excellent book by Chistopher Haslett: -
The module in question has a data rate throughput as high as 300 kbps so, in that case the receiver input power needed would be: -
Power (dBm) = -154 dBm + 54.77 dB = 99.28 dBm
OK, if your data rate is only low (maybe 1 kbps) the receiver power needed might only be -124 dBm and, if the data rate is only 100 bits per second, the receiver will need a power of -134 dBm.
So, you decide based on your data rate.
Formulas and other link budget information from here.
Don't forget to add another 20 db loss for fade margin.