# Re: problem with ofdm bandwidth and sampling rate **URL:** \ **Category:** GNU Radio\ **Created:** [June 3, 2015, 5:30pm UTC](https://www.ruby-forum.com/t/re-problem-with-ofdm-bandwidth-and-sampling-rate/242194 "2015-06-03T17:30:53Z")\ **Posts on this page:** 8\ **Page:** 1
**Author:** ![musicdenotation](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/m/ea666f/32.png) [@musicdenotation](https://www.ruby-forum.com/u/musicdenotation)\ **Post date:** [June 3, 2015, 5:30pm UTC](https://www.ruby-forum.com/t/re-problem-with-ofdm-bandwidth-and-sampling-rate/242194/1 "2015-06-03T17:30:53Z")
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**Author:** ![unknown](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/u/4bbf92/32.png) [@unknown](https://www.ruby-forum.com/u/unknown)\ **Post date:** [June 5, 2015, 8:41am UTC](https://www.ruby-forum.com/t/re-problem-with-ofdm-bandwidth-and-sampling-rate/242194/2 "2015-06-05T08:41:41Z")
On 03.06.2015 17:29, [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid) 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: [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid) www.cel.kit.edu KIT – University of the State of Baden-W??rttemberg and National Laboratory of the Helmholtz Association ---
**Author:** ![unknown](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/u/4bbf92/32.png) [@unknown](https://www.ruby-forum.com/u/unknown)\ **Post date:** [June 5, 2015, 10:16am UTC](https://www.ruby-forum.com/t/re-problem-with-ofdm-bandwidth-and-sampling-rate/242194/3 "2015-06-05T10:16:55Z")
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**Author:** ![unknown](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/u/4bbf92/32.png) [@unknown](https://www.ruby-forum.com/u/unknown)\ **Post date:** [June 5, 2015, 10:29am UTC](https://www.ruby-forum.com/t/re-problem-with-ofdm-bandwidth-and-sampling-rate/242194/4 "2015-06-05T10:29:13Z")
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, [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid) 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:removed\_email\_address@domain.invalid > > 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: [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid) www.cel.kit.edu KIT – University of the State of Baden-W??rttemberg and National Laboratory of the Helmholtz Association ---
**Author:** ![unknown](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/u/4bbf92/32.png) [@unknown](https://www.ruby-forum.com/u/unknown)\ **Post date:** [June 5, 2015, 11:24am UTC](https://www.ruby-forum.com/t/re-problem-with-ofdm-bandwidth-and-sampling-rate/242194/5 "2015-06-05T11:24:07Z")
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**Author:** ![unknown](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/u/4bbf92/32.png) [@unknown](https://www.ruby-forum.com/u/unknown)\ **Post date:** [June 5, 2015, 11:03am UTC](https://www.ruby-forum.com/t/re-problem-with-ofdm-bandwidth-and-sampling-rate/242194/6 "2015-06-05T11:03:41Z")
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 ---
**Author:** ![unknown](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/u/4bbf92/32.png) [@unknown](https://www.ruby-forum.com/u/unknown)\ **Post date:** [June 5, 2015, 11:36am UTC](https://www.ruby-forum.com/t/re-problem-with-ofdm-bandwidth-and-sampling-rate/242194/7 "2015-06-05T11:36:59Z")
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**Author:** ![unknown](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/u/4bbf92/32.png) [@unknown](https://www.ruby-forum.com/u/unknown)\ **Post date:** [June 5, 2015, 11:48am UTC](https://www.ruby-forum.com/t/re-problem-with-ofdm-bandwidth-and-sampling-rate/242194/8 "2015-06-05T11:48:05Z")
Hi Alphonso, On 06/05/2015 11:22 AM, [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid) 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