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王慧

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[问答]

PXA复杂功率测量

当使用PXA测量复杂域中信号的功率时,我观察到IQ正弦波和IQ调制信号(例如WCDMA,DVB-H等)之间的3dB差异,即使它们具有相同(实际)功率

任何意见?

以上来自于谷歌翻译


     以下为原文

  When using the PXA to measure the power of a signal in the complex domain I observe a 3dB difference between an IQ sinewave and a IQ modulated signal (e.g. WCDMA, DVB-H etc.), even though they have the same (real) power. Any comments?  

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李欢

2019-7-8 16:32:13
嗨 - 调制信号时,其带宽会增加。
您需要确保整合信号带宽以测量整个信号功率。
因此,在频谱分析仪应用程序中,您可以增加RBW或使用通道功率测量,并确保集成/通道带宽足够大。
在IQ Analyzer模式下,您还可以增加集成带宽。
问候 -

以上来自于谷歌翻译


     以下为原文

  Hi -

When you modulate a signal, its bandwidth grows.  You need to make sure to integrate over the signal bandwidth to measure the entire signal power.  So, in the Spectrum Analyzer app you would increase your RBW or use the Channel Power measurement and make sure the integration / channel bandwidth is large enough.  In the IQ Analyzer mode, you can also increase your integration bandwidth.

Regards -
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王慧

2019-7-8 16:39:39
引用: 醉狼工作室 发表于 2019-7-8 17:35
嗨 - 调制信号时,其带宽会增加。
您需要确保整合信号带宽以测量整个信号功率。
因此,在频谱分析仪应用程序中,您可以增加RBW或使用通道功率测量,并确保集成/通道带宽足够大。

感谢您的答复。
当然,我们整合了整个信号带宽。
例如,当使用10MHz LTE(RF)信号时,我们在复杂IQ域中集成[-5MHz,5MHz]。
只是为了说明测量结果:我正在使用直接转换接收器来下转换RF信号并测量其在PXA中的功率。
我比较了在RF上使用CW(tome)和使用具有相同功率电平的调制信号(例如LTE,WCDMA,DVB-H)时的结果。
在频谱分析仪模式下测量调制信号在信号带宽/ 2上的积分功率,并将其与峰值或在I或Q信道中的基带中的CW音调的积分功率进行比较,结果匹配。
然而,当我改变为PXA I + jQ模式然后对于CW我看到与频谱分析仪模式相同的功率但是对于调制信号信号,我看到3dB较少(此时通过信号带宽积分以考虑负频率)。
这没有意义,所以我想知道FFT中是否存在用于我们缺少的I + jQ PXA测量的东西....

以上来自于谷歌翻译


     以下为原文

  Thank you for the reply.

Of course we integrate over the entire signal bandwidth. For example when using a 10MHz LTE (RF) signal, we integrate [-5MHz, 5MHz] in the complex IQ domain. 

Just to make clear on the measurement: I'm using a direct conversion receiver to down-convert an RF signal and measure its power in the PXA. I compare the results when using a CW (tome) at RF and when using a modulated signal (for example LTE, WCDMA, DVB-H) both with the SAME power level. When measuring in spectrum analyzer mode the integrated power of a modulated signal over Signal Bandwidth/2 and compare it either with peak or with the integrated power of the CW tone in baseband in either I or Q channel the results match. However when I change to PXA I+jQ mode then for a CW I see the same power as in spectrum analyzer mode but for modulated signal signal I see 3dB less (integrated over Signal Bandwidth this time to account for negative freq as well). This does not make sense, so I wonder if there is something in the FFT used for I+jQ PXA measurements that we are missing....
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王慧

2019-7-8 16:52:03
引用: asfszczgv 发表于 2019-7-8 17:42
感谢您的答复。
当然,我们整合了整个信号带宽。
例如,当使用10MHz LTE(RF)信号时,我们在复杂IQ域中集成[-5MHz,5MHz]。

更新:看起来这与PXA相关的是log-pwr avg而不是RMS。
你可否确认?

以上来自于谷歌翻译


     以下为原文

  Update: It looks like this is related to PXA is log-pwr avg instead of RMS. Can you confirm?
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李欢

2019-7-8 17:11:28
引用: asfszczgv 发表于 2019-7-8 17:54
更新:看起来这与PXA相关的是log-pwr avg而不是RMS。
你可否确认?

嗨 - 您似乎正在使用前面板I / Q输入。
它是否正确?
当您执行此测量时:“然而,当我更改为PXA I + jQ模式然后对于CW我看到与频谱分析仪模式相同的功率但是对于调制信号信号我看到3dB更少(这次集成在信号带宽上)
负频率也是如此。“当你执行I = jQ测量时,你是否将来自DUT的I和Q信号连接到PXA前面板I和Q输入?
如果不这样做,这可能是导致3 db差异的原因。
问候 -

以上来自于谷歌翻译


     以下为原文

  Hi -

It appears you are using the front panel I/Q inputs.  Is this correct?

When you performed this measurement:
“However when I change to PXA I+jQ mode then for a CW I see the same power as in spectrum analyzer mode but for modulated signal signal I see 3dB less (integrated over Signal Bandwidth this time to account for negative freq as well).”

Did you connect BOTH the I and Q signals from the DUT to the PXA front panel I and Q inputs when you performed the I=jQ measurement?
If you did not, this is the likely cause of the 3 db difference.

Regards -
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