Tektronix TDS 320 Oscilloscope, 1000' RG-58/U coaxial cable (conveniently still in the box), High-Pass filter Circuit on bread-board, and Tektronix CFG 253 function generator)
An 11V amplitude square wave produced in the function generator was sent through a high-pass filter (a differentiator circuit):
The square wave also was used as a trigger for the oscilloscope on channel 2 - when the square wave went high (was positively sloped), the scope would look for incoming data on channel 1. The pulse, produced by the differentiator circuit, traveled down the inner conductor of the coaxial cable, as well as directly to the scope.
If the varying load resistor, R-load, were absent (i.e., infinite resistance), the pulse would reflect and travel back through the inner conductor. This reflection could be seen on the scope as a peak of diminished amplitude and a broader half-width than the pulse seen directly out of the differentiator circuit. (Click here to see a sample pulse)
From the Electromagnetic Theory, we know that when a signal encounters a change in impedance, some of the signal can be reflected. The ratio of the voltage reflected to the input voltage is (VR/VI) = (RL-Z) / (RL+Z), where RL is R-load and Z is the characteristic impedance of this coaxial cable (Z = 53.5 ohms). From this, we can see that the magnitude of the reflected peaks will not be as great as the input pulse's peak. The reflected peaks can, however, be negated. It can be seen form the above equation that the reflected peaks can be inverted (as when RL<Z). (Click here to see a sample pulse)
In determining the magnitude and sign of the reflected peaks which will be observed on the scope, it can be seen from the circuit diagram and knowledge of correct use of the (VR/VI) equation that one must consider that the reflected pulse will "see" the resistor in the differentiator circuit (R=217.9 ohms) as being another R-load. However, because 217.9 > 53.5, the pulse will not be inverted a second time (which would counteracting any inversion from the case of RL<Z and invert any pulse reflected when RL>Z).
If RL = Z, you might think no reflection would appear on the oscilloscope, right? Check it out!
If you crank up the volts/division on the oscilloscope (to 0.5 mV/div), set the seconds/div to 1 µs/div and set the square waves to a frequency of 9 kHz, you can see multiple reflections. Each one has a broader half-width and a smaller amplitude - as could be assumed from our knowledge of the ratio of (VR/VI). (Here it is.)
To learn of how we imported data from the oscilloscope to Mathematica (where the plots from above were generated), please click here.
Back to the table of contents, please! Discuss the conclusions