Graphing a digit oscillating signal

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To graph a digital oscillating signal, place time on the horizontal axis and voltage on the vertical axis, with the maximum voltage reflecting the logic level used (e.g., 3.3V or 5V). The pulse width can be determined using the duty cycle, which is the ratio of the pulse width to the total period; the period is the time from one rising edge to the next. For a signal with a 25% duty cycle and a 1ms period, the high time lasts 250 microseconds, followed by a low time of 750 microseconds. The resulting graph will show a rectangular waveform, with the shape varying based on the duty cycle. Understanding these concepts allows for accurate representation of digital signals.
Mr. Hiyasaki
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How would I go about graphing a digtal oscillating signal? I don't quite understand what a pulse width is, or rather how to find it given only the oscillation in Hz and a duty cycle. I understand that the duty cycle is the ratio of the pulse width over the total period, but i don't understand what the pulse width or the period is for a digital signal.
 
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Mr. Hiyasaki said:
How would I go about graphing a digtal oscillating signal? I don't quite understand what a pulse width is, or rather how to find it given only the oscillation in Hz and a duty cycle. I understand that the duty cycle is the ratio of the pulse width over the total period, but i don't understand what the pulse width or the period is for a digital signal.
On your graph, put time on the horizontal axis, and voltage on the vertical axis. If your digital signal is based on 3.3V logic, the top of the vertical axis will be about 4V. If it is 5V logic, use 5V as the top of the vertical axis. Then graph your digital signal versus time.
You end up with a rectangular waveform, or a square wave if the duty cycle is 50%. If the duty cycle is less than 50%, then the high time is less than the low time in the rectangular waveform. The period T is the time from one rising edge to the next rising edge, or from falling edge to next falling edge. The frequency is f = 1/T.
So let's say you are asked to graph a digital 3.3V signal with a period of 1ms and a duty cycle of 25%. You start at the origin of your graph with a rising edge from 0V to 3.3V, draw to the right from (0,3.3V) to (250us,3.3V), draw a falling edge to (250us,0V), and then draw the low part of the signal to (1ms,0V). Then you can repeat if you want to draw more than one cycle.
Makes sense?
 
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