Machine-translated by Google. Questions are authored in English.
Top

Cathode Ray Oscilloscope Questions and Answers

21.

The rise time of an oscilloscope and its bandwidth are related approximately by the product

Answer: A

For a response approximating a single time constant, the rise time in seconds multiplied by the bandwidth in hertz is about 0.35. A hundred megahertz instrument therefore has a rise time of about 3.5 nanoseconds, which limits the edges it can resolve.

Please log in or register to report this question.

Answer: B

The series resistance in the probe raises the input impedance tenfold and reduces the effective input capacitance, so a high impedance circuit is disturbed far less. The price is a tenfold loss of signal, which must be allowed for in reading the screen.

Please log in or register to report this question.

Answer: C

The divider must attenuate by ten for direct voltage and for the fastest edges alike, which requires the capacitive division to match the resistive one. A square wave of known shape is displayed and the trimmer set until the top of the wave is flat.

Please log in or register to report this question.

Answer: D

Too much capacitance in the probe passes the high frequencies too readily, so the edges are exaggerated and each transition overshoots before settling. Under-compensation gives the opposite, a rounded and slowly rising top.

Please log in or register to report this question.

Answer: A

The capacitor blocks any steady component, so a small alternating signal riding on a large direct voltage can be examined at high sensitivity. The penalty is that the true zero level is lost and low frequency square waves are distorted.

Please log in or register to report this question.

Answer: B

On direct coupling the trace would be driven far off the screen at any sensitivity able to show a few volts of ripple. Blocking the steady component brings the ripple to the centre of the screen where it can be measured at high sensitivity.

Please log in or register to report this question.

Answer: C

The horizontal plates are then driven by a second input signal instead of by the sawtooth, so the display becomes a plot of one signal against the other. Lissajous figures, hysteresis loops and device characteristics are all obtained this way.

Please log in or register to report this question.

Answer: D

An electronic switch shares one gun between the two channels, which is far cheaper than a tube with two guns. The compromise is that neither channel is displayed continuously, so a genuinely simultaneous view of two single events needs a dual beam tube.

Please log in or register to report this question.

Answer: A

Channel one is drawn for a whole sweep, then channel two for the next, and so on. At fast sweep speeds the eye blends them into two steady traces, so the mode suits high frequency work; at slow speeds the traces are seen to flicker alternately.

Please log in or register to report this question.

Answer: B

Chopping switches between the channels many times within one sweep, so both appear on every sweep and neither flickers at slow speeds. At fast speeds the chopping rate becomes comparable with the signal and the traces break into visible dashes.

Please log in or register to report this question.

Loading…
Tags: Cathode Ray Oscilloscope Questions and Answers || Cathode Ray Oscilloscope MCQ Questions and Answers || Cathode Ray Oscilloscope GK Questions and Answers || Cathode Ray Oscilloscope GK MCQ Questions || Cathode Ray Oscilloscope Multiple Choice Questions and Answers || Cathode Ray Oscilloscope GK || GK on Cathode Ray Oscilloscope || Electrical Measurements Questions and Answers || Electrical Measurements MCQ Questions and Answers || Electrical Measurements GK Questions and Answers || GK on Electrical Measurements