C

#### [email protected]

- Jan 1, 1970

- 0

creating

sine waves with DACs.

First question: Assuming a perfect DAC with perfect analog

components,

obviously a DAC with fewer bits will create a sinewave with larger

steps,

and thus will have a higher percentage of THD+N, but how do I

calculate

the exact percentage?

Second question: I know about the Nyquist limit and it seems to me

that

as the Nyquist limit is approached the sinewave will have bigger

steps

no matter how many bits it has and thus will have a higher percentage

of THD+N, but how do I calculate the exact percentage?

Third question; In the real world I wouldn't have perfect analog

components; in fact I would purposely introduce a lowpass filter

at the output of the DAC to attenuate the switching noise. How

much would that change the answers to the questions above?

BACKGROUND:

We need to replace an old system that generates 20 Hz to 20 kHz sine

waves with a 12 bit DAC that puts out a 4096-step sine wave -- the

same number of steps whether it is putting out 20 Hz or 20 kHz.

A variable oscillator changes the clock rate of a counter that

gets the values from an EPROM lookup table.

We were discussing replacing the above with a modern DAC -- either

16 bits at 44.1 ksps or 24 bits at 96 ksps. The objection was

raised that at 20Khz we are putting out 4096 x 20,000 sps, or

81.92 Msps. I am guessing that 96 ksps with a added filter at

the DAC output is good enough. The final power stage starts

slew-rate limiting at 30-40 kHz with large signals and the small-

signal response is 3dB down at 50Khz and way down in the mud at

100KHz. I just don't see how it needs over 80 megasamples per

second to keep the THD+N reasonably low. Am I right?