Hi all:

Thanks in advance.

Example:gnuradio/gr-digital/examples/ofdm/rx_ofdm.grc.

If I set the sample rate of usrp equal to 1Msps,then what is

the sample rate of the output of the FFT block?Thank you very much.

Best regards,

xd

The question is ill-formed, the concept of a ‘sampling rate’ does not

really apply here.

However, to an item rate, you can do the math yourself:

Assume N is the FFT length, and CP the length of cyclic prefix in

samples. r is the incoming sampling rate.

First, data goes through the HPD, which takes out the CP. The outgoing

rate of that block is thus

r2 = (N+CP)/N * r

Actually, it’s output is already OFDM symbols (“FFT-ready”). So, the

output item rate is

r3 = r2/N

Then, the packet header is removed. This further reduces the rate, but

let’s ignore that. It’s also dependent on the packet length.

Since the FFT doesn’t change the rate, what you want is most likely r3.

M

Hi Martin:

Thank you so much.I understand.

Best regards,

xd

Obviously I flipped the fraction. It’s N/(N+CP).

M

Hi Martin:

Thanks.But I’m confused about it.

```
First, data goes through the HPD, which takes out the CP.
```

The outgoing rate of that block is thus

r2 = (N+CP)/N * r

```
Now r2>r.But according to the downsampling,when the sample
```

become less than before,the sample rate become smaller than before.Now

the data goes through the HPD,the sample become less ,but the r2>r?Maybe

I’m wrong.Can you help me?Thanks.

I think maybe this:r2=r*N/(N+CP)

Best regards,

xd

Hi Martin:

Thanks.

If I add a block(Vector to Stream) after FFT block.Then

the sample rate become r4=r3*N.

Am I right?Thank you very much.

Best regards,

xd