From the DSO image,
Number of peak to peak division are 4 full and one is one fifth of ome division.
So total 4.2 division and volt/ division scale is at 5 m volt/ div.
So voltage will be # Of division * volt/ div
4.2* 5= 21 m V ( millivolt)
Similarly, the wave repeats after approximately 2.6 division ( on Y axis~ time axis)
And time per division scale is at 0.1 m second.
So time = 0.1 *2.6= 0.26 ms( milliseconds)
And frequency is 1/Time period
f= 1/.26 ms= 1000/ .26= 3846.15 Hz.
Compute the voltage, Period, and the frequency from the images of the waveform, voltage scale and...
Using images 6.6 – 6.8, determine the amplitude of the
calibration waveform. Be sure to include proper units:
[Image 6.6 – Calibration Waveform Display]
[Image 6.7 – Calibration Waveform Vertical (VOLT/DIV)
Setting]
[Image 6.8 – Calibration Waveform Horizontal (TIME/DIV)
Setting]
# of vertical divisions
VOLT/DIV
Amplitude
Determine the period and calculate the frequency:
# of horizontal divisions
TIME/DIV
Period
Frequency
The calibration waveform has a specified amplitude of 2.0
VPP at a frequency of 1.0 Khz. How accurate is the...
The amplitude, period and frequency of the voltage waveform shown in the graph below is:
Sine wave measurement:
Images 6.9 – 6.12 show a sine wave. Note that Image 6.9 is a
close up of the function generator’s frequency display.
Using images 6.10 – 6.12, determine the amplitude of the sine
wave. Be sure to include proper units:
# of vertical divisions
VOLT/DIV
Amplitude
Determine the period and calculate the frequency:
# of horizontal divisions
TIME/DIV
Period
Frequency
Assuming that the frequency displayed in Image 6.9 is more
accurate than the oscilloscope, what is the...
Q1:a) For the output waveform as shown below from oscilloscope determine the output voltages for inductive and capacitive connected in series with source and mention that the lagging and leading? The parameter (volt div=2, the input voltage Vp=5 and the 0=0) (2 degree) www.it Volt div-2 VL 4
If the period of a waveform is doubled, what happens to the frequency of that waveform? The frequency remains the same. The frequency doubles. The frequency becomes one-fourth as large. The frequency becomes four times as large. The frequency becomes half as large.
QUESTION 1 Consider a sinusoidal waveform that has a frequency of 1 MHz. What is the period of this waveform? If the TIME/DIV knob on the scope is set at 0.2 microseconds and the scope screen is divided into 10 major gridlines on the time axis, how many complete periods can be displayed on the oscilloscope screen?
From the following waveform, find the following: 30 Sine wave 20 10 Voltage 20ws 40W GOs 80s -10 -20 Time -30 A- The period T (2.5 Points) B- The frequency f (2.5 Points) C- The Angular frequency (2.5 Points) D- The voltage expression Vit) (7.5 points)
How to find RMS Voltage of this waveform? Periodic waveform has period of 14ms. The cycle starts at t=0 with voltage: V=1.6 x (10^6)(t^2.3) volts
Help fill in table
The CH2 waveform vr(t) is shown in red. 3. Now, double the function generator frequency (to f=1250Hz about 1250Hz) to display exactly two complete cycles in 8 horizontal divisions. Do not change the amplitude of e(t). Again, both waveforms should be centered vertically on the scope grid. Adjust the vertical POSITION knobs if necessary. Carefully sketch and label both waveforms on the grid. Use the same settings (VOLTS/DIV AND SEC/DIV from part A4.) as with the...
2. Given the voltage waveform in Figure. a). What is the frequency of the signal? b).What is the phase shift of the waveform? c) What is the maximum amplitude of the AC signal? d). What is the sine function that describes the waveform? e). What is the cosine function that describes the waveform? * Plot of a sinusoidal signal Magnitude -15 --20 -25 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 Time (ms) Figure P4.1 AC Signal to be Analyzed...