Re: problem with ofdm bandwidth and sampling rate


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On 03.06.2015 17:29, [email protected] wrote:

used_bw = occupied_carriers * df
df = used_bw / occupied_carriers = 10khz / 228
df = 43.86hz --> tsym = 22.8ms
it’s very confusing why i have to adjust my used bw by changing the
sampling rate. when i follow the equations and the standard i choose
one mode (RM A, SO 3). that means i have to use 228 carriers
and have a symbol duration with 24ms which leads to 41.66hz carrier
spacing. so the used bw is 228 * 41.66hz = 9.5khz (approx. 10khz).
this bw should be fix and should not change by the sampling rate.
sorry for the maybe silly questions but i want to keep on that stuff.
Well, the measurable on-air bandwidth is always defined by the
speed/rate at which your soundcard processes your digital samples. If
you remember that B ~ 1/T and that the rate of your soundcard controls T
it becomes obvious that changing the sample rate changes the bandwidth.
If you don’t want that, you could use a resampler that adapts its
resampling ratio to the sampling rate of the soundcard.

Felix

@Marcus
you mean if there were pilots for a channel estimation in the
transmitted signal?
alphonso


Karlsruhe Institute of Technology (KIT)
Communications Engineering Lab (CEL)

Felix W., M. Sc.
Research Associate

Kaiserstra??e 12
Building 05.01
76131 Karlsruhe

Phone: +49 721 608-46276
Fax: +49 721 608-46071
E-Mail: [email protected]

www.cel.kit.edu

KIT – University of the State of Baden-W??rttemberg and
National Laboratory of the Helmholtz Association


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Hi Alphonso,

keep in mind that if you don’t resample on RX side your signal
processing has to deal with the oversampled signal. If you use the same
chain to demodulate your signal I can see why it only works when you
resample on RX side.

– Felix

On 05.06.2015 10:15, [email protected] wrote:

Gesendet: Donnerstag, 04. Juni 2015 um 23:40 Uhr
hope it works the next time.
the sampling rate. when i follow the equations and the standard i
T it becomes obvious that changing the sample rate changes the

Phone: +49 721 608-46276
Fax: +49 721 608-46071
E-Mail:[email protected]

www.cel.kit.edu

KIT – University of the State of Baden-W??rttemberg and
National Laboratory of the Helmholtz Association


Karlsruhe Institute of Technology (KIT)
Communications Engineering Lab (CEL)

Felix W., M. Sc.
Research Associate

Kaiserstra??e 12
Building 05.01
76131 Karlsruhe

Phone: +49 721 608-46276
Fax: +49 721 608-46071
E-Mail: [email protected]

www.cel.kit.edu

KIT – University of the State of Baden-W??rttemberg and
National Laboratory of the Helmholtz Association


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Hi Alphonso,

well, your OFDM mod block uses the whole nyquist bandwidth of your
sampled signal, ie. at the output of the OFDM mod block, all frequencies
in [-pi;pi] (relative frequencies, as sampling rate doesn’t really exist
in this world) are used. After the resampler, only frequencies in
[-(11/48)pi;(11/48)pi] are occupied. You feed that signal directly into
the OFDM demod block. How is that block supposed to know to only look
into that sub band? You will have to resample exactly inversely to your
TX resampler, before trying to demodulate that.

You still have to move the upsampled signal from 0 Hz to your desired
center frequency, and then you’ll have to convert to a real signal,
before being able to feed things to a sound card. On the receive side,
you’d need to do the inverse.

Best regards,
Marcus


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Hi Alphonso,

On 06/05/2015 11:22 AM, [email protected] wrote:

i think i have understand the concept now: the sampling rate is valid
global for the hole graph.if there is a previous/following block or
source with a different rate, i have to resample it with such a
resampler block. i am right?
No. There is no sampling rate in GNU Radio, really. It’s just a
numerical concept for each block. For example, look at the standard
“signal source”. It takes a frequency and a sampling rate argument. If
you use f=1000 and sampling rate=4000, the signal source (float mode)
will just calculate how to scale the values that it puts into the sine
function. You will get exactly the same signal (really, no difference
anywhere) if you used f=0.5, rate=2, or f=1e9 and rate=4e9.
So, each block has only “sample” as a means to measure time, and
1/sample as a means to measure frequency. You need to start thinking
relative to Nyquist frequency, i.e. to say “this signal source generates
a signal at 0.25 of the sample rate, which is equivalent to 0.5 nyquist
frequency, because this is real sampling”.
You’ll quickly see that the OFDM modulator and demodulator always uses
the whole possible bandwidth (complex sampling, hence nyquist frequency
== sampling rate), and your resampler generates a signal where only a
part of the bandwidth is occupied.

Best regards,
Marcus

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