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I read the following sentence. "In wideband systems, the transmission bandwidth of a single channel is much larger than the coherence bandwidth of the channel." What is 'coherence bandwidth of the channel'?

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  • $\begingroup$ Sometimes, it feels a bit like your questions are asked when you read something, but hesitate to read the next page in the same book/paper... I don't mind this site having a lot of interesting questions, but coherence bandwidth is something that google can answer. $\endgroup$ – Marcus Müller Jun 5 '16 at 16:27
  • $\begingroup$ @MarcusMüller: I generally agree with that, but in this case at least the corresponding wikipedia article is a bit meagre. $\endgroup$ – Matt L. Jun 5 '16 at 16:34
  • $\begingroup$ Yeah, we might need to fix that one. $\endgroup$ – Marcus Müller Jun 5 '16 at 19:21
  • $\begingroup$ @Marcus Müller:No, that is not true. I google and read a lot. But some things are not clear to me even after I do that, and the questions keep nagging me. Like the one on the transients in frequency domain. This is a good forum to raise the questions. Some times I get convincing answers. Sometimes I know what the opinion of others is on these questions. In this case, I read previously about delay spread but not about coherence bandwidth. $\endgroup$ – Seetha Rama Raju Sanapala Jun 5 '16 at 20:40
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If you have a multipath channel with impulse response

$$h(t)=\sum_{k=0}^{m-1}a_k\delta(t-\tau_k)\tag{1}$$

the corresponding frequency response is

$$H(j\omega)=\sum_{k=0}^{m-1}a_ke^{-j\omega\tau_k}=e^{-j\omega\tau_0}\sum_{k=0}^{m-1}a_ke^{-j\omega(\tau_k-\tau_0)}\tag{2}$$

Assume that the delays are ordered such that $\tau_0<\tau_1<\ldots <\tau_{m-1}$. Note from $(2)$ that the periods of the complex exponentials are given by $1/(\tau_k-\tau_0)$, and the smallest period is $1/(\tau_{m-1}-\tau_0)$, which is simply the inverse of the largest delay difference. This largest delay difference is called the channel's delay spread. Its inverse, i.e., the smallest period of the complex exponentials in the frequency domain is referred to as the coherence bandwidth.

The importance of the coherence bandwidth is the following: in a frequency range which is considerably smaller than the coherence bandwidth, the channel's frequency response can be considered flat (i.e., it doesn't change much). So if your signal's bandwidth is small compared to the channel's coherence bandwidth, it will not be distorted very much. If the signal bandwidth is larger than the coherence bandwidth, the signal will experience frequency-selective fading.

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  • $\begingroup$ Matt L: So, coherence bandwidth of the channel is dependent on the multipath environment, not just on the channel under consideration, isn't? Same physical channel can have different coherence bandwidths under different multipath environments. $\endgroup$ – Seetha Rama Raju Sanapala Jun 5 '16 at 20:44
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    $\begingroup$ @SeethaRamaRajuSanapala: The "multipath environment" is the channel. $\endgroup$ – Matt L. Jun 5 '16 at 20:46
  • $\begingroup$ Matt L:Thanks for the clarification. You have a lot of patience. Your comment on multipath environment, I can take that as convention - as - agreed upon language, But in the normal sense, channel is independent of its multipath environment, The physical channel - say the space between two points remains the same, but the conditions in the same space can change and so the multipath environ, As I said, it is also possible to include the multipath in the channel definition and use it so. Thanks. These are the fine points I miss sometimes and get confused. $\endgroup$ – Seetha Rama Raju Sanapala Jun 5 '16 at 20:57
  • $\begingroup$ @SeethaRamaRajuSanapala in DSP, a channel is all the things that affect a signal from emitter to receiver; that's the normal sense. No other sense makes sense :) $\endgroup$ – Marcus Müller Jun 5 '16 at 21:17
  • $\begingroup$ @Marcus Müller: OK, Thanks Marcus Muller. I need like you and Matt L who make things crystal clear - to remove some such confusions in my head - . $\endgroup$ – Seetha Rama Raju Sanapala Jun 5 '16 at 21:21

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