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High-Speed DACs

Category: Datasheet/Specs
Product Number: ad9081 ad9082


I have a question about DAC bandwidth of AD9081/2.

In the datasheet,  it indicates "Useable analog bandwidth to 8 GHz" .

But on Table-1 of UG-1578, the "Max Tx Channel Bandwidth (GHz)" of AD9081 is 1.6. I can't figure out the relationship between these two bandwidth.

Does the "Useable analog bandwidth" represents the device bandwidth with CHANNELIZER PATH and MAIN DATA PATH bypass? The "Channel Bandwidth" means using one or both of the PATH?

I want to generate a QPSK intermediate frequency (IF) signal with 2Gsps sample rate, 4GHz IF, 2.5GHz bandwidth, which means the roll-off factor is 0.25. That's no problem if refer to the "Useable analog bandwidth", but it cannot be implemented if refer to the "Channel Bandwidth". Can I generate such a signal with AD9081 or do I have a better device choice?

  • Hello,

    The table is misleading.  The "Channel Bandwidth" limit of 1.2 GHz only applies when using the Channelizer Path that feeds into the Main Data Path since the maximum digital hand-off rate between two paths is 1.5 GSPS...........with the "Channel Bandwidth" being 80% of this data rate due to digital filter passband response.  For higher signal bandwidth applications, one can bypass the Channelizer Path and go directly into the Main Data path with no limitation (other than selecting proper JESD mode configuration to support data throughput rate of the total number of IQ data pairs while selecting DAC interpolation factor such that DAC clock rate  remains below 12 GHz.

    In your particular case...........a 2.5 GHz BW signal will require a higher IQ data rate than 2 GSPS since the IQ rate must be greater than 2.5 GHz/ 0.8 (or 3.125 GSPS) due to the digital filter passband bandwidth being 80% of the IQ rate. If your application could live with IQ rate of 3 GSPS (supporting 2.4 GHz of signal BW............than this would be more practical since it would allow Main Data Path to be configured for 4x interpolation (with digital NCO set to 4 GHz) such that the DAC would operate at 12 GSPS.

    Looking at the DAC output would see desired 2.4 GHz signal centered about 4 GHz IF output along with a DAC image having same bandwidth (due to discrete time sampling theory) residing at 8 GHz.

    Is this what your are hoping to achieve with a wideband QPSK signal?  Also...what is the symbol rate of the QPSK signal since its BW will be determined by its symbol rate along with RRC roll-off factor? 

  • Thanks for you help. I understand your explanation. And the DAC output spectrum you describe is what I want to get. The symbol rate of the QPSK signal is 2GHz, so 3GSPS IQ rate can't support enough oversampling factor for signal shaping despite sufficient bandwidth.

    Can I bypass both of Channelizer Path and Main Data Path, using 12GSPS DAC sample rate to generete the desired 2.5 GHz signal centered about 4 GHz IF without interpolation?  "Useable analog bandwidth to 8 GHz"  refers to the DAC output BW with 1X total interpolation (bypass these two data path) ?