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    What happens when you change to square wave on an oscilloscope? I remember the only the peaks were shown on the screen but I'm not sure if an RMS value is produced. Thanks
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    (Original post by Mulkeen)
    What happens when you change to square wave on an oscilloscope? I remember the only the peaks were shown on the screen but I'm not sure if an RMS value is produced. Thanks
    The R.M.S. (Root Mean Squared) value of a continuously varying waveform is a way of describing the relationship between the peak value and the area enclosed by the waveform.

    Thus all cyclic waveforms have an R.M.S. value including square, sine, triangular, saw, pulse and any combination/summation of these etc. All of these R.M.S. values will be different.

    The description of a 'perfect' square wave is one that transitions between voltage levels at infinite speed and remains perfectly stable (constant) once the new level is achieved.

    In practice, the actual shape of the square wave will depend on a number of factors which include how fast the leading and trailing edges of the wave transition between voltages.

    The square wave may not be perfect because the circuit that produced it contains components (capacitors and inductors) that store then release energy resulting in oscillation artefacts etc. This becomes more apparent as the frequency of the waveform increases.

    Some examples:


    IDEAL:


    NOT SO IDEAL:



    POOR SQUARE WAVE:

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    You're an absolute life saver, the 'ideal' example looked like what we found on our screen too. So the peak of that is the original peak value (without square wave enabled) divided by root2?
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    (Original post by Mulkeen)
    You're an absolute life saver, the 'ideal' example looked like what we found on our screen too. So the peak of that is the original peak value (without square wave enabled) divided by root2?
    I don't know what you mean by 'without square wave enabled'?

    The R.M.S. value of a pulse waveform shape (function) is given by:

    V_{rms} = V_p(\sqrt(\frac{t_1}{T}))

    where t1 / T is the waveform duty cycle.

    In the case of the square wave t1 / T = 1/2

    Hence the R.M.S value of a perfectly symmetrical square wave is

    V_{rms} = V_{peak}(\sqrt(\frac{1}{2}))

    V_{rms} = 0.707(V_{peak})

    or as you correctly say:

    V_{rms} = \frac{V_{peak}}{\sqrt2}

    V_{rms} = \frac{V_{peak}}{1.414} = 0.707(V_{peak})
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    (Original post by uberteknik)
    I don't know what you mean by 'without square wave enabled'?
    Okay I've had a task to do, and imagine if we have a oscillator connected to an oscilloscope, then the lab script says "what happen if the signal is changed to a square wave". When you uploaded a photo of the 'ideal' diagram with the lines, that is exactly what we saw. What does this mean? What is its function?
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    (Original post by Mulkeen)
    Okay I've had a task to do, and imagine if we have a oscillator connected to an oscilloscope, then the lab script says "what happen if the signal is changed to a square wave". When you uploaded a photo of the 'ideal' diagram with the lines, that is exactly what we saw. What does this mean? What is its function?
    Ahhh ok.

    The question wants you to compare the waveform shapes and one such measure is the rms value which will be different for different waveform shapes.

    My previous answer still stands.
 
 
 
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