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Lets say I have a 1km long transmission line with a characteristic impedance of 50ohm. With a pulse generator I inject a really long pulse (30µs) the pulse generators termination is also 50 ohm. The response will look like the image below

Tenter image description here

So the amplitude U/2 of the reflections is because of the voltage divider formed by the two 50 ohm impedances. When using a signal generator with a known resistance (impedance) we should be able to calculate the impedance of any cable by the amplitudes of the reflections. Is my assumption correct?

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When using a signal generator with a known resistance (impedance) we should be able to calculate the impedance of any cable by the amplitudes of the reflections. Is my assumption correct?

It's a little bit easier than that - you don't need to wait for the reflection; at the point you apply the signal (50 Ω source), it is the voltage instantly developed at the cable that determines the impedance. For instance, if your pulse is 1 volt and has a source impedance of 50 Ω and, it connects to a 100 Ω cable, the voltage seen at the cable is: -

$$V_{CABLE} = 1\text{ volt}\cdot \dfrac{100}{100 + 50} = 0.6667\text{ volts}$$

If the cable were 50 Ω impedance then the peak voltage is 0.5 volts.

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  • \$\begingroup\$ But this assumes that the impedance of the cable is a real value? \$\endgroup\$
    – Yoomo
    Commented Dec 20, 2020 at 12:22
  • \$\begingroup\$ @Yoomo If you are using pulses then this is going to be the case. The characteristic impedance will be \$\sqrt{L/C}\$ and that will be resistive above 1 Mohm. \$\endgroup\$
    – Andy aka
    Commented Dec 20, 2020 at 12:24
  • \$\begingroup\$ So with your formula I can calculate the impedance of the cable, when I am using either a pulse or a DC voltage? \$\endgroup\$
    – Yoomo
    Commented Dec 20, 2020 at 12:26
  • \$\begingroup\$ @Yoomo use a pulse and pretty much after applying that pulse (via a resistor) look at the magnitude at the cable end where you apply the pulse. If the cable is short then you might get a reflection coming back too quickly and it screws up the measurement of course. \$\endgroup\$
    – Andy aka
    Commented Dec 20, 2020 at 12:29
  • \$\begingroup\$ I am a bit confused. When I use a long square wave pulse like above the impedance will be a real value and can be calculated like this Vcable=1 volt⋅(100/150)? \$\endgroup\$
    – Yoomo
    Commented Dec 20, 2020 at 12:40

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