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PGA970EVM: Why are the sign of DEMOD_PH1 and DEMOD_PH2 value inverted?

Takuto_Yoshioka
Intellectual 360 points

Part Number: PGA970EVM

Hello,

I am checking values of DEMOD_PH1 register and DEMOD_PH2 register.

Now primary waveform (voltage waveform of P1 pin) and secondary waveforms (voltage waveform of S1P and S2P) are below.

And relevant register values (DAC_SIN_NDSs, WAVEFORM_TABLE_LEN, DEMOD_PHx_DATAxs) are below.

And GUI configurations are below.

 

Refer to the primary waveform and secondary waveforms, I expected that DEMOD1_PH1_DATA register  value is less than zero.

Because the blue line is greater then green line, therefore  the  S1P pin voltage minus S2P pin voltage is less than zero.

But DEMOD1_PH1_DATA register value is +1455193 that is greater than zero.

I guess the sign of S1P pin voltage minus S2P pin voltage is inverted. 

Is that true?

Assuming that's the case, why is the sign inverted?

I appreciate if you give me your advice.

Thank you and best regards,

Takuto Yoshioka