Answer to Question #200854 in Electrical Engineering for Bahawal tahir

Question

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.

Answer 1

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