I assume you mean "long" in regards to wavelengths, and also the wavelength is simply 30 cm at 1 GHz.
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Calculating accident in infection lines is a complicated task. You"ll have both resistive losses in the conductor(s) themselves, to add dielectric losses in the bordering materials.
One point you"ll have to include to your model is the self-inductance of the wire, which is another method of saying the you need to recognize its properties impedance. Losses will be least if the wire is terminated at both ends by its properties impedance, i m sorry will get rid of standing waves on the wire.
If you have actually standing waves, the peak current in the wire will certainly rise, boosting the I2R losses, and the top voltage will also rise, enhancing the dielectric losses.
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For a time-varying voltage used to a resistor, the instantaneous strength dissipated is just the present through the resistor times the voltage throughout the resistor. According to Ohm"s theory, the voltage is equal to I*R therefore the instantaneous power dissipation is $i^2(t)R$. Incorporate that over the period of the signal, divide by the period, and also you have actually the mean power. Now, to discover the present you need to know the load capacitance at the finish of the wire, and also calculate $i(t) = C dv(t)/dt$ for her 3.3V, 1 GHz signal. If the capacitance the the cable is far-reaching when contrasted to the load capacitance then the difficulty gets harder, and also at that allude I would suggest taking her netlist of parasitic R and also C values into SPICE.
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