Cathode Ray Oscilloscope Questions and Answers

Answer: C

Correct answer: 30 degrees

For an ellipse traced by two signals of equal frequency, \[ \sin\phi = \frac{y_{intercept}}{y_{max}} = \frac{1.5}{3} = 0.5, \] so \( \phi = 30^{\circ} \). The method needs no calibration of either axis, since it depends only on a ratio of two measurements made on the same axis - which is what made it the standard phase measurement before digital instruments.

Answer: D

The screen shows \( 3 \times 0.2 = 0.6 \) V, and the probe has divided the signal by ten, so the tip voltage is 6 V. Many instruments detect the probe and rescale the display automatically, in which case no multiplication is needed.

Answer: A

Correct answer: 10 ms

\[ \frac{10^{6}\ \mathrm{samples}}{100\times10^{6}\ \mathrm{samples/s}} = 10^{-2}\ \mathrm{s} = 10\ \mathrm{ms}. \] Memory depth divided by sample rate is the record length, and the trade it implies governs digital scope use: to capture a longer window the instrument must reduce the sample rate, which reduces the resolution of the fine detail within it.

Answer: B

Sampling below twice the signal frequency folds it about half the sample rate, so it appears at \( 100 - 60 = 40 \) Hz. The display looks like a perfectly respectable 40 Hz sinusoid, which is exactly why aliasing is so treacherous.

Answer: C

Full scale is eight volts across 256 levels, so each level is \( 8/256 = 0.031 \) V. Reducing the volts per division setting reduces the step in proportion, which is why a small signal is best examined at high sensitivity rather than by magnifying the stored record.

Answer: D

The screen shows instantaneous values, so what is measured is a peak or a peak to peak height. Any root mean square figure has to be computed from it with a factor depending on the shape of the wave, and that factor is only 0.707 of the peak for a sinusoid.

Answer: A

A megohm in parallel with twenty or thirty picofarads is the usual input, and the capacitance dominates above a few kilohertz. At one megahertz that capacitance is only a few kilohms, which is why a compensated probe matters so much in high impedance circuits.

Answer: B

Samples are written continuously into a circular buffer, so at the moment of the trigger a record of the recent past is already in memory. The trigger simply marks the position in that record, and the operator can display as much before it as after.

Answer: C

A faster electron spends less time between the plates and is deflected less, so sensitivity falls as the accelerating voltage rises. Brightness does improve, which is the conflict that post-deflection acceleration was devised to resolve.

Answer: D

Without the delay the edge that started the sweep would already have gone by the time the beam began to move, so the very transition of interest would be missing. Delaying the signal a little longer than the circuits take brings that edge onto the screen.