PXA复杂功率测量

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 -

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....

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



以下为原文

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

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 -

是的，我正在使用前面板，我和Q都已连接。
问候



以下为原文

Yes, I'm using front panel and BOTH I and Q are connected.

Regards

嗨 - 我做了I / Q输入的实验。
当我生成W-CDMA信号时，当使用I + jQ输入路径时，当我移除I或Q输入时，集成功率下降3dB。
相反，当我在同一相位上产生具有两个音调的多音调信号时，当我移除I输入时，信号从显示器消失。
当我移除Q输入时，看不到功率差异。
当我将第二个音调的相位偏移90度时，当我移除I或Q时，功率下降3dB。
当我将第二个音调偏移180度时，当我移除Q输入时，信号从显示屏上消失（所有电源都在Q通道中）。
现在，当我比较I + jQ输入之间的积分功率时，我只输入相同角度的两个音调的情况，对于仅I路径，功率高6 dB。
当我比较I + jQ输入之间的积分功率和I仅在相位角偏移90度时输入，那么我看到仅I路径的结果高出3 dB。
因此，当电源仅包含在I或Q输入中的一个时，当您将输入设置为I + jQ而不是使用单个通道（仅I或仅Q）输入时，您将看到信号测量值降低6 dB
。
这是因为Asin的傅立叶变换（wt）= 1 / 2A u（W-w）+ 1 / 2A u（W + w）。
当I和Q通道的功率相等时，当您将输入设置为I + jQ而不是使用单通道（仅I或Q）输入时，您将看到信号测量值降低3 dB。
问候 -



以下为原文

Hi -

I did an experiment with the I/Q inputs. When I generate a W-CDMA signal, the integrated power when I remove either the I or Q input when using I+jQ input path drops by 3 dB. Conversely, when I generate a multi-tone signal with two tones at the same phase, then when I remove the I input, the signal disappears from the display. When I remove the Q input, no difference in power is seen. When I offset the phase of the second tone by 90 degrees, then when I remove either I or Q, the power drops by 3 dB. When I offset the 2nd tone by 180 degrees, then when I remove the Q input, the signal disappears from the display (all of the power is in the Q channel). Now, when I compare the integrated power between the I+jQ input and just I only input in the case of the two tones that are at the same phase angle, the power is 6 dB higher for the I only path. When I compare the integrated power between I+jQ input and I only input when the phase angles are offset by 90 degrees, then I see a 3 dB higher result for the I only path. Therefore, when the power is contained in just one of the I or Q inputs, you will see the signal measure 6 dB lower when you set the input to I+jQ as opposed to using a single channel (I only or Q only) input. This is because of the Fourier Transform of Asin(wt) = 1/2A u(W-w) + 1/2A u(W+w). When the power is equally in the I and Q channels, you will see the signal measure 3 dB lower when you set the input to I+jQ as opposed to using a single channel (I only or Q only) input.

Regards -