怎么使用Agilent 33220A任意波形发生器进行FM调制

您好我认为您可能会混淆FM频率和FM偏差。
提到的20kHz限制是针对FM频率而不是FM偏差。
FM偏差可以满量程。
因此，如果你有一个100kHz的正弦波，你可以让它在内部偏离整个100kHz，所以它基本上从0Hz到300kHz。
FM频率就是频率偏差的应用速率。
如果我正确阅读下面的消息，您希望能够偏离100kHz正确吗？
如果是的话，你应该可以毫无问题地做到这一点。
目前尚不清楚为什么要使用两个函数发生器？
外部调制输入限制在20kHz，就像FM频率一样。
但我再次认为你可能会将FM频率与FM偏差混淆。



     以下为原文

  Hello I think you may be confusing the FM Frequency and the FM Deviation.  The 20kHz limitation mentioned is for the FM frequency, not the FM deviation.  The FM deviation can go full scale.  So if you had a 100kHz sine wave, you could have it deviate by the full 100kHz internally, so it would essentially go from 0Hz to 300kHz.

The FM frequency is just the rate at which the Frequency Deviation is applied.

If I read the message below correctly, you want to be able to deviate 100kHz correct?   If yes you should be able to do this with no problems.

It is not clear why you are using two function generators?  The External modulation input is limited to 20kHz, just like the FM frequency.  But again I think you may be confusing FM frequency with FM Deviation.

嗨Ihornburg，我非常了解fm频率和频率偏差。
我需要测试我的FM解调器电路。
我使用两个函数发生器的原因是内部FM频率限制在20 kHz。
我想扩展调制频率范围。
使用外部FM选项，可以使用“Modulation In”输入作为调制信号源，这样我就可以将其频率更改为超出有限的内部20 kHz情况。
问题是现在我需要注意第二个信号发生器的幅度，其输出连接到第一个信号发生器。
{载波信号属性（幅度，频率）和FM选项（内部或外部和频率偏差）从第一个信号发生器调整。
因为调制信号Am的幅度也影响FM信号的频率偏差，即Df = kf * Am，其中kf是以Hz / V给出的FM灵敏度。
在我应用具有100 kHz偏差的外部FM的第一个线程中，需要应用Am = 10Vpk-pk = + 5V的幅度。
这意味着Df = 100kHz = kf * Am = kf * 5 ==>> kf = 20kHz / V对吗？
我问的问题与后一个问题相同：为了得到特定的Df = 40kHz频率偏差，我需要从第一个信号发生器应用什么：频率偏差：Df = 40kHz和第二个信号发生器：调制信号
幅度Am = 10Vpk-pk？
如果我应用频率偏差：Df = 100kHz来自第一个信号发生器和调制信号幅度Am = 4Vpk-pk我最终会得到一个实际频率偏差Df = kf * Am = 20 * 2 = 40kHz或者我是否需要申请
Df = 40kHz，Am = 10Vpk-pk？
还请再看看我以前的问题。
谢谢HIDIR



     以下为原文

  Hi Ihornburg,
I do understand the fm frequency and the frequency deviation very well.
I need to test my FM demodulator circuit.
The reason why I use two function generators is that the internal FM frequency is limited to 20 kHz. I wanted to extend the modulating frequency range. Using the external FM option one can use the "Modulation In" input as a modulating signal source, so that I can change its frequency beyond the limited internal 20 kHz case. The problem is that now I need to be careful about the amplitude of the second signal generator whose output is connected to the first signal generator. { The carrier signal properties (amplitude, frequency) and the FM options (internal or external and frequency deviation) are adjusted from the first signal generator. }
Because the amplitude of the modulating signal Am also effects the frequency deviation of the FM signal which is Df=kf*Am, where kf is the FM sensitivity given in Hz/V. In my first thread for applying an external FM with 100 kHz deviation one needs to apply Am=10Vpk-pk=+5V in amplitude. Which means Df=100kHz=kf*Am=kf*5 ==> kf=20kHz/V right ?
I am asking the same questions as in my latter thread :
In order to have a specific Df=40kHz frequency deviation what do I need to apply from 
              the 1st signal generator : frequency deviation : Df=40kHz
and from the 2nd signal generator : modulating signal amplitude Am=10Vpk-pk ?

If I apply frequency deviation : Df=100kHz from the 1st signal generator and modulating signal amplitude Am=4Vpk-pk will I end up with an actual frequency deviation of Df=kf*Am=20*2=40kHz OR do I need to apply Df= 40kHz and Am=10Vpk-pk ?
Please also again take a look to my previous questions.
Thanks 
HIDIR

这里有三种不同的频率： - 有载波频率Fc。
让我们假设这是1 MHz信号 - 频率偏差为Fd。
让我们假装这是80 kHz - 有调制信号的频率，Fm。
让我们假装这是20 kHz，这是规格中指出的最大频率。
无论您使用内部还是外部调制源，载波频率都将更改为最大1.08 MHz和最小0.92 MHz。
输出从0.92 MHz变为1.08 MHz并返回1.08 MHz的速度由调制信号的频率决定。
最快的指定速率是20 kHz - 无论您使用内部还是外部信号。
如果您使用外部信号，+ 5 V将为您提供最大Fd。
在我的例子中，它将给你80 kHz，发电机的输出将是1.08 MHz。
这将是16kHz / V.
如果我将频率偏差设置更改为500 kHz，那么它将为100kHz / V.
在+5 V时，我将获得1.5 MHz。
在-5 V时，我将获得0.5 MHz。
我可以多快地从0.5 MHz改为1.5 MHz再回到0.5 MHz，受到20 kHz规格的限制。
如果使用的外部信号小于+/- 5V，则需要增加偏差。
例如，如果要在外部信号为+4 V时达到1.08 MHz，则需要将仪器设置调整为100 kHz偏差。
在+5 V时，仪器将输出1.10 MHz，但在+4 V时，仪器将输出1.08 MHz。
使用FM的外部信号不会以任何方式提高33220A的指定性能。
它只是允许您动态地改变输出频率以响应外部信号。
最诚挚的问候，谢丽尔



     以下为原文

  There are three different frequencies in play here:
- There is the carrier frequency, Fc.  Let's pretend that this is a 1 MHz signal
- There is the frequency deviation, Fd.  Let's pretend this is 80 kHz
- There is the frequency of the modulating signal, Fm.  Let's pretend this is 20 kHz, the maximum frequency as noted in the specifications.

Whether you are using an internal or external modulation source, the frequency of the carrier will be changed to a maximum of 1.08 MHz and a minimum of 0.92 MHz.  How quickly the output changes from 0.92 MHz to 1.08 MHz and back to 1.08 MHz is determined by the frequency of the modulating signal.  The fastest specified rate is 20 kHz -- whether you use an internal or an external signal.

If you are using an external signal, +5 V will give you the maximum Fd.  In my example, it will give you 80 kHz and the output of the generator wll be 1.08 MHz.  This would be 16kHz/V.  If I change my frequency deviation setting to 500 kHz, then it would be 100kHz/V.  at +5 V, I will get 1.5 MHz.  At -5 V I will get 0.5 MHz.  How quickly I can change from 0.5 MHz to 1.5 MHz and back to 0.5 MHz is limited by the specification of 20 kHz.  

If you are using an external signal that is less than +/- 5V, then you will need to increase your deviation.  For example, if you want to reach 1.08 MHz when the external signal is +4 V, you need to adjust the instrument settings to a 100 kHz deviation.  At +5 V, the instrument will output 1.10 MHz, but at +4 V the instrument will output 1.08 MHz.  

Using an external signal for FM does not increase the specified performance of the 33220A in any way.  It simply allows you to dynamically change the output frequency in response to an external signal.

Best Regards,
Cheryl

谢丽尔先生，谢谢你的回复。
我知道+ 5V给出了从信号发生器中选择的全部偏差。
它解决了我关于频率偏差的问题。
您说“最快的指定速率是20 kHz - 无论您使用内部信号还是外部信号。”
但是我可以将调制信号频率改变到20KHz以上，这决定了我可以在最大值之间扫描的速度。
（+）和分钟。
（ - ）偏差。
我看到它是信号发生器本身的限制。
这是否意味着虽然我改变了调制频率（来自第二个信号发生器，第一个是“调制输入”）但是有20 KHz的限制？
它仅限于内部调制情况吗？
最诚挚的问候，HIDIRE日期：HIDIR于2013年2月22日下午1:14



     以下为原文

  Hi Mr. Cheryl,

Thank you for your reply.
I understood that +5V gives the full deviation chosen from the signal generator . It solves my problems about hte frequency deviation.

You said that " The fastest specified rate is 20 kHz -- whether you use an internal or an external signal.". However I can change the modulating signal frequency beyond 20KHz, which determines how fast i can sweep between the max. (+) and min. (-) deviation. I see that it is a limitation of the signal generator itself. Does it mean that although i change the the modulating frequency (from the second signal generator which is "Modulation In" for the first) there is a limit of 20 KHz? Is it a limitation only for the internal modulation case?

Best regards,
HIDIR

Edited by: HIDIR on Feb 22, 2013 1:14 PM

33220A的输入具有有限的带宽，无法处理无限频率。
使用内部或外部调制信号时，性能仅为+指定+高达20 kHz。
使用内部调制信号时，设置受仪器固件限制为最大20 kHz。
使用外部调制信号时，不会为20 kHz以上的频率指定仪器的性能。
如果继续将输入频率增加到20 kHz以上，则在信号衰减或边缘减慢之前，您可能会获得更高的性能。
在使用20 kHz以上的信号时，您必须对特定单元进行特征化（测试），以确定您的应用是否具有可接受的性能。
可能还存在仪器到仪器的可变性。
使用一个单元的功能在第二个单元上可能无法正常工作。
我希望有所帮助！
最诚挚的问候，谢丽尔



     以下为原文

  The input of the 33220A has a finite bandwidth and cannot handle infinite frequencies.  The performance is only +specified+ up to 20 kHz when using either an internal or external modulation signal.   

When using an internal modulation signal, the setting is limited by the instrument firmware to a maximum of 20 kHz.  

When using an external modulation signal, the performance of the instrument is not specified for frequencies above 20 kHz.  If you continue to increase the input frequency beyond 20 kHz, you will likely obtain some higher performance before the signal is attenuated or the edges are slowed. You will have to characterize (test) your particular unit to determine if you get acceptable performance for your application when using signals above 20 kHz.  There may also be instrument-to-instrument variability.  What works using one unit may not work as well on a second unit.

I hope that helps!

Best Regards,
Cheryl

非常感谢你。
最好的问候，HIDIR



     以下为原文

  Thank you very much.
Best regards,
HIDIR