# Impulse Response of wide band wireless channel **URL:** \ **Category:** GNU Radio\ **Created:** [August 6, 2015, 1:18pm UTC](https://www.ruby-forum.com/t/impulse-response-of-wide-band-wireless-channel/242905 "2015-08-06T13:18:17Z")\ **Posts on this page:** 2\ **Page:** 1
**Author:** ![Rafa\_F](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/r/958977/32.png) [@Rafa\_F](https://www.ruby-forum.com/u/Rafa_F)\ **Post date:** [August 6, 2015, 1:18pm UTC](https://www.ruby-forum.com/t/impulse-response-of-wide-band-wireless-channel/242905/1 "2015-08-06T13:18:17Z")
Hii Is the Impulse Response of wide band wireless channel has only few significant components as compared to the channel delay spread ? This question is not related to gnu radio. I asked it, just to know the opinion of the people from this group who have good experiences in this field of communication. Thanks… ---
**Author:** ![monika\_bansal](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/m/7c8e57/32.png) [@monika\_bansal](https://www.ruby-forum.com/u/monika_bansal)\ **Post date:** [August 6, 2015, 9:13pm UTC](https://www.ruby-forum.com/t/impulse-response-of-wide-band-wireless-channel/242905/2 "2015-08-06T21:13:17Z")
Hi Monika, you’re right, this is a bit off-topic for this list; however, I kind of like the fact that there’s much that can be discussed with respect to actual GNU Radio implementations here, and hence, I find that your “discussion kickstart” /is/ interesting. I’d therefore like to give you /one/ short answer, and let you do your decision on your own 🙂 That depends on how you define your channel- “wide band wireless channel” says not very much about its specific properties: There’s the definition that you consider communication “wide band” when the channels’ coherence bandwidth, which is typically defined as $b\_c=\frac{2\pi}{d}$, $d$ being the delay spread, is smaller than the signal bandwidth $b\_s$. Now, this would unambigous if not a) the most interesting case would be when $b\_s \approx b\_c$ (because doing $b\_c \gg b\_s$ is “easy”), and b) people would completely agree what $d$ is – is it the mean delay spread (if $h(t)$ is the complex frequency response, $\overline{d}={\int {h(\tau)\tau d\tau}}\cdot{\left(\int{h(\tau)d\tau} \right)^{-1}}$), or is it the RMS delay spread ($d\_\text{RMS} = \sqrt{\int {h(\tau){\left(\overline{d} - \tau\right)}^2 d\tau}\cdot{\left( \int{h(\tau)d\tau} \right)^{-1}}}$), or is it maybe a measure based on 90% of the energy being contained within a time frame? You might want to think about why you want to consider the delay spread at all – typically, you care about whether your channel is inter-symbol-interference free; hence, if whatever measure you use for the delay spread (most of the time you either use the mean delay spread or the root mean square delay spread) is smaller than your symbol duration, you’d call that channel ISI-free, and having ISI otherwise. So maybe you just want to say “we consider channels for which our communication suffers ISI”. Another problem here is: If you model your channel as tapped delay line, the number of taps alone doesn’t say much about the phase response and hence, delay over that channel. If you make some assumptions on the nature / distribution of the coefficients, then you might come to the conclusion that a channel with a high delay spread is never flat, and you might then call it wideband channel (again, you’ll clearly need to define this for yourself). If you consider your channel to be as general as possible, it might as well just be an all-pass filter, which means it might have a flat power profile, but a very high delay spread. This would, for example, be no problem for e.g. an OFDM system where each subcarrier’s phase is independent of the others (imagine an OFDM signal with DPSK subcarriers) – since the amplitude response is flat, you don’t need to equalize the individual subchannels. However, if your OFDM system was to carry QAM, then suddenly, you will need to understand the phase effects for every single subcarrier. That’s when you start adding preambles and pilot tones all over your OFDM frames. It’s very much up to you to pick the right channel model. The trick here is figuring out what existing channel model describes your application sufficiently well (or can be slightly adapted), and just saying “adapting the channel model XYZ”, listing the parameters might be much better than trying to fit your channel model into any specific terminology - there’s always people who’ll want to understand this for yourself. If you look at the example above, you might have wondered how bad the problem of getting that channel state information is – especially, when not only there’s a change of channel influence over frequency, but also, if your channel starts exhibiting non-infinite coherence time (which goes with non-zero Doppler frequency). Like delay spread, there’s different ways people define the coherence time (Doppler freq), and different people define the statistic measures based on that (Doppler spread), and refering to a known and well-tested channel model makes it harder to argue against the “realism” of your observations. Now, coming back to GNU Radio (which [I hope] justifies posting this mail here): You’re doing digital signal processing, so your channels are digital. The integrals up there break down to finite sums. The fact alone that your signal, and hence, the operations you apply on it, have a specific rate and therefore, bandwidth. If you need to oversample your signal significantly to be able to reconstruct it means that the channel influence is large compared to your signal’s bandwidth, right? So, if I was in a situation where I was spontaneously asked whether I had a wide band channel, and I didn’t prepare for that question, I’d just have a look at how many samples per symbol I need to reconstruct my signal – if it’s \> 4, I’m pretty surely in wide band channels. As you noticed from my discussion above, things get a bit subjective when you use words like “wide”. It’s just often better to give actual relative measurements than to rely on “squishy” human terms; the interested audience will have no problem if you tell them that /you /consider the channel to be wide band, because $b\_c$ is only $0.95 b\_s$, because that is a valid opinion; saying “it’s not wide band, because $b\_s$ is but $1.05 b\_c$” is as much a valid opinion, if you ask me. Best regards, Marcus