Answer to Question #200854 in Electrical Engineering for Bahawal tahir

Question #200854


c. Determine the minimum time/division sensitivity for an oscilloscope that is to be used to 

investigate a 50 MHz waveform.Assume that the time base magnifier expands the horizontal 

display by a factor of 5.

e. A 1kHz triangular wave with a peak amplitude of 10 V is applied to the vertical deflecting 

plates of a CRT. A 1kHz sawtooth wave with a peak amplitude of 20 V is applied to the 

horizontal deflecting plates. The CRT has a vertical deflection sensitivity of 0.4cm/V,and a 

horizontal deflection sensitivity of 0.25cm/V.Assuming that the two inputs are synchronised, 

determine the waveform displayed on the screen.

f. A digital storage oscilloscope has a sampling rate of 100MS/s. Determine the number 

of samples taken during one cycle of a 3MHz sine wave, and during a 15µs pulse. Also, 

estimate the maximum time period of a glitch that might be missed by the sampling 

process.


1
Expert's answer
2021-06-01T03:46:04-0400

Part c

As horizontal deflection is adjusted to 5divisionss and frequency of 50mHz, it means the time period "2 \\mu sec=20n sec" . Hence the time/division selectors will be at unsec "\\implies 5*4 nsec =20nsec"

Part e



Part f

Rise time of the pulse waveform=21 ns


Frequency corresponding to this rise time=1/21 ns


=47.277 MHz


(a) If oscilloscope upper cutoff frequency=20 MHz<47.277 MHz then the pulse waveform will not be allowed to pass by the oscilloscope because of having a frequency of rise greater than upper cutoff frequency of oscilloscope.


(b) If oscilloscope upper cutoff frequency=50 MHz>47.277 MHz then the displayed rise time of the pulse will be its actual rise time which is 21 ns.


Q2


The sampling rate of digital storage oscilloscope =100 MS/s


Number of samples taken during a cycle of 3 MHz sine wave


=100/3


=33.33


~33


Number of samples taken during a 15 micro second pulse


=100 M × 15 u


=1500


Maximum time period of a glitch missed by the sampling process


=1/(100M) sec


=10^-8 sec


=10 ns or 0.01 microseconds


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