Answer to Question #200857 in Electrical Engineering for Bahawal tahir

Question #200857

a. Determine the amplitude, frequency and phase difference between the two waveforms 

illustrated in below figure:

b. Determine the pulse amplitude, frequency, rise time, and fall time of the waveform illustrated 

in below figure:

d. Determine the displayed rise time when a pulse waveform with a rise time of 21 ns is applied to 

an oscilloscope that has an upper cutoff frequency of (a) 20 MHz and (b) 50MHz.


1
Expert's answer
2021-06-02T06:34:03-0400

Part a

Amplitude for waveform A= "\\frac{V_A}{2}=\\frac{6*200}{2}=600mV"

Amplitude for waveform B= "\\frac{V_B}{2}=\\frac{2.4*200}{2}=240mV"

Frequency for waveform A= "\\frac{1}{T_A}=\\frac{1}{6*0.1}=1666.667 Hz"

Frequency for waveform B= "\\frac{1}{T_B}=\\frac{1}{6*0.1}=1666.667 Hz"

Phase difference="360*ft= 3600 * 1666.667*0.1=60^o"

Part b

Pulse amplitude=4 major divisions ="4*0.1=0.4 mV"

Frequency "f=\\frac{1}{T}=\\frac{1}{1.2}=0.893 MHz"

Rise time= "3 *0.4=0.12 \\mu sec"

Fall time="2*0.04=0.08 \\mu sec"

Part d

Rise time of the pulse waveform=21 ns

The 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 rising greater than the 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.


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