Question

Please help with completing this table?

Using the O'scope as the standard measurement device, adjust the function generator for the indicated frequencies and voltage amplitudes, using peak to peak as a reference voltage. If the indicated voltage amplitude is not in peak to peak change it to peak to peak and adjust the function generator appropriately. Fill in each column with the required information: v/div, # of divisions, time/div, the number of divisions for one complete cycle, and period. t/DIV #OF DIV VERT FREQUENCY SIGNAL V/DIV. #OF DIV HORIZ PERIOD AMPLITUDE 2v p-to-p k Hz 1.414VAC 2k Hz. 3v p-to-p 150k Hz 10v p-to-p 24k Hz. 600mv p-to-p 40k Hz. v p-to-p 400 Hz. 1.25V pk. 100k Hz. 350mv pk. 10k Hz Sv pk. 500 Hz. 3.54VAC 400k Hz.

Answer 1

The completed table is given below ==>

FREQUENCY SIGNAL AMPLITUDE V/DIV. t/DIV. OF DIV. HORIZ PERIOD OF DIV. VERT. 2 Ik Hz. 2v p-to-p 1 msec 2k Hz. 1.414VAC 1 500 usec 500 usec 1 usec 150k Hz. 1 2 3 5 24k Hz. 10 usec 0.5 msec 6.67 usec 41.67 usec 25 usec 2.5 msec 40k Hz. 10 usec 0.2 0.5 6.67 4.17 25 25 1 5 3v p-to-p 107 P-to-p 600mvp-lo-p lv p-to-p 1.25V pk. 350my pk. Sv pk. 3.54VAC 400 Hz. 1 msec 10 usec 100k H. 10 usec 20 sec 0.1 10k Hz. 100 usec 1 msec 2 500 Hz. 400k Hz. 2 sec 25 usec I usec 25

Please note following points ==>

If signal amplitude = x peak, then signal amplitude (peak to peak) = 2x

If signal amplitude = x AC, then signal amplitude (peak to peak) = (2.1.414)x

Also, period, T = 1/frequency

Let's take first case now ==>

(i) frequency = 1 kHz ; Amplitude (peak to peak) = 2 V

Time period = 1/(1 kHz) = 1 msec

Set V/Div. = 1 V and t/Div. = 500 usec

Therefore, # vertical division = Amplitude(peak to peak) / V/Div. = 2/1 = 2

Therefore, # horizontal division = Time period / t/Div. = 1 msec / 500 usec = 2

Similarly, you can do other cases also.

NOTE - For any other help, please let me know in the comments.