Oscilloscope Sensitivity and Voltage Analysis

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SUMMARY

The discussion focuses on calculating the amplitude and frequency of a sinusoidal voltage displayed on an oscilloscope with a vertical sensitivity of 5mV/cm and a horizontal sensitivity of 0.5μs/cm. The amplitude is determined by multiplying the vertical sensitivity by the height of the waveform in centimeters, leading to a peak amplitude of approximately 16mV. For frequency, the period is calculated based on the horizontal divisions, with a period of 3.2 cm translating to a frequency of 312.5 Hz. The discussion clarifies common misconceptions regarding voltage units and emphasizes the importance of understanding oscilloscope readings.

PREREQUISITES
  • Understanding of oscilloscope operation and sensitivity settings
  • Familiarity with voltage measurement units (mV vs. MV)
  • Knowledge of sinusoidal wave properties (amplitude, frequency, period)
  • Basic skills in unit conversion and calculations
NEXT STEPS
  • Learn how to interpret oscilloscope waveforms and measurements
  • Study the relationship between period and frequency in sinusoidal signals
  • Explore advanced oscilloscope features, such as triggering and persistence
  • Investigate common pitfalls in voltage unit conversions and calculations
USEFUL FOR

Electronics students, engineers, and technicians who work with oscilloscopes and need to analyze voltage waveforms accurately.

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Homework Statement


In the oscilloscope display above, the vertical sensitivity is 5mV/cm, and the horizontal sensitivity is 0.5μs/cm.
A) find the amplitude of the sinusoidal voltage
B) find the frequency of the sinusoidal voltagehttp://imgur.com/fnkXliz

Homework Equations


im not sure which equations to use for this question, if any are needed.

The Attempt at a Solution


I know you have to convert the units in this question I am assuming. It also says in the question that each block is 1cm by 1 cm, and the top of the amplitude is about 3.2 cm, but that isn't the correct answer.
 
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Ask yourself, what does it mean when it says 5 mV/cm? And so, what does it mean if you have a wave 3.2 cm tall on the screen?

Ask yourself, what does it mean when it says 0.5μs/cm? And how long (in cm) is one wave of the signal on the screen? And so, what does that mean?
 
i think it means (and correct me if I am wrong) that there is 5 megavolts (5000000 volts) in every centimeter on the y-axis. So if the wave is 3.2 cm tall would you multiple 5mV by 3.2 to get the amplitude?
 
reconrusty said:
i think it means (and correct me if I am wrong) that there is 5 megavolts (5000000 volts) in every centimeter on the y-axis. So if the wave is 3.2 cm tall would you multiple 5mV by 3.2 to get the amplitude?

First of all, mV is milliVolts, not MegaVolts... :-)

Next, I see the vertical amplitude as a little over 6 divisions tall. So the peak-to-peak amplitude will be a bit over 6 * 5mVpp. It's hard to tell if the question is asking for the peak or peak-to-peak "amplitude". Can you clarify that?

And on the horizontal axis, how many divisions between the positive peaks of the waveform (or alternately use the positive-going zero crossings to find the waveform's period)
 
reconrusty said:
http://imgur.com/fnkXliz
this is the graph posted in the question aswell

Better to show the picture:
fnkXliz.gif
 

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Okay so i got the first question right, it was asking for the peak not the peak to peak. So for the second part about finding the frequency, I counted roughly 3.2 blocks as the period, so the frequency is just 1/period right? I am not sure what to put in for units though, i thought frequency was measured in Hz, but when i try to put the answer in online it says the units are wrong... any help?
 
3.2 cm is how much time ?
 
Problem solved! thank you guys so much for the help :)
 

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