Answer to Question #200857 in Electrical Engineering for Bahawal tahir

Question

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.

Answer 1

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.