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By referring to the http://www.analog.com/static/imported-files/data_sheets/AD9850.pdf, for page 9 and figure 5, why the quantized D/A converter affect the amplitude of the output waveform?

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The output of the D/A converter is a series of steps, in which the voltage moves directly from the previous value to the next. This is also known as "zero-order hold".

The output spectrum would be perfectly flat if the output were a series of infinitesimally-narrow impulses. The zero-order hold is equivalent to putting such a series of impulses through a filter whose impulse response is a rectangular pulse whose width is equal to the sample period. The frequency response of such a filter is the dashed sin(x)/x line shown in Figure 5, which is why the amplitudes of the harmonics in the output spectrum are affected in the manner shown.

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then why this attenuation only occur for highest frequency, but not on lower frequency? – user37970 Feb 28 '14 at 15:00

Below is a picture of a sine wave sampled 8 times in 3 cycles of the sine wave (i.e. fairly close to the Nyquist criteria). The sampled waveform does not look ideal but, its RMS level is exactly the same as the sine wave's RMS level i.e. energy content is the same. See conclusions below picture: -

enter image description here

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hi andy aka,beside the imperfect of DAC, do the decreasing of the amplitude relate to the phase of the signal? – user37970 Mar 4 '14 at 8:28
@user37970 Sorry dude I don't understand your question. – Andy aka Mar 4 '14 at 8:30

This is due to the Zero Order Hold effect of a DAC.

A DAC's input is a sequence in discrete time. It's output is continuous time. The DAC "fills in" between the input number values and output voltage by "holding" the output steady in between values. This is technically called a Zero-Order Hold which may be modeled as (1-exp(-sT))/(1 - z^-1). In the frequency domain this is a 1st order Sinc function with zeros at the sample rate. This Sinc function primarily attenuates at its Zeros but there is some roll off in the band of interest too.

Intuitively : i) we know sampling in the time domain generates replicas of the signal in the frequency domain at multiple of the sample rate frequency; ii) the sampled data is the signal at the DAC's input; iii) we know the output is continuous so the frequency replicas must have be removed (or at least significantly reduced). iv) the Sinc function of the ZOH performs this function by "notching" out the replicas.

Strictly speaking a DAC should be followed by a filter to "smooth out" the signal in the time domain thereby attenuating all replications in the frequency domain; but this is off topic for this question.

enter image description here

UPDATE: The diagram shows how the frequency of the input signal to the DAC is related to the magnitude of the output signal.

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thus it means that the rate of filling of the voltage = sinc function ?..and why this problem occurs when it is at highest frequency? – user37970 Feb 28 '14 at 14:50
I would say : the Sinc function is caused by the "filling of the voltage" . The reason the maximum attenuation (or zeros) occurs at multiple of the sample rate is because the "filled in" signal is no longer a sampled signal. – akellyirl Feb 28 '14 at 14:54
the "filled in" signal is no longer a sampled signal" is it mean it is an alias signal? – user37970 Feb 28 '14 at 14:59
No. It means it's continuous time. Another way to understand the notches is: imagine applying a sinewave of Frequency 1MHz to the DAC that is sampled at 1MHz. What would you see at the DACs output? Answer: just DC. Therefore its transfer function has a zero in the frequency domain at the sample rate. – akellyirl Feb 28 '14 at 15:02
then what is the output of DAC below and above 1 MHz?..then what is the relationship between transfer function,amplitude and frequency? as we known that, amplitude of output waveform follow sinc function, then how about the transfer function? – user37970 Feb 28 '14 at 15:20

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