1 GSW Multipath Channel Models

Transcription

1 In the general case, the moble rado channel s pretty unpleasant: there are a lot of echoes dstortng the receved sgnal, and the mpulse response keeps changng. Fortunately, there are some smplfyng assumptons that can often be used to make the mathematcal descrpton of the channel a bt easer, and whch allow the determnaton of few smple parameters that provde a good ndcaton the sort of problems any gven system s lkely to have. Ths chapter s about these assumptons, and a couple of these smple parameters that can be derved usng them that characterse the qualty of a channel: most notably the spread and the coherence bandwdth. There s also a bref ntroducton to fadng, but there s much more about tme-varyng channels n the next chapter on Tme-Varant Multpath Channels.. The Moble Rado Channel n General In the most general case, the moble rado channel can be charactersed ether by a tmedependent mpulse response h(, t) or a tme-dependent frequency response H(, t), where t s the tme. Ths s a smple extenson of the usual representaton of lnear tme-nvarant systems to the case where the system sn t tme-nvarant and the mpulse response changes wth tme. Wth tme-varant systems, the output of the system for any gven nput sgnal x(t) can be determned usng the tme-dependent mpulse response:, and n the general case, h(, t) can be anythng. y t h t x t d (0.) The sort of nformaton we d lke to know s over what range of tme wll the energy n a transmtted mpulse arrve at the recever, and how quckly the mpulse response changes wth tme. Long mpulse responses and fast-changng channels both cause problems for recevers... The Problem of Long Impulse Responses Long mpulse responses (long relatve to the length of a symbol) result n ntersymbol nterference. Consder a smple channel consstng of two rays, wth a between them of just over one symbol perod: power Frst ray between rays symbol n symbol n+ symbol n+ symbol n+3 symbol n+4 one symbol perod Second ray symbol n- symbol n symbol n+ symbol n+ symbol n+3 Fgure - Intersymbol Interference Caused by Multpath At all tmes, the recever, whch s recevng the sum of the sgnals from both rays n the mpulse response, s recevng energy from two dfferent symbols. The nterference caused by the echo s known as ntersymbol nterference, and makes the recever s task of workng out 007 Dave Pearce Page 8/08/008

2 what symbol was transmtted much more dffcult. (The part of the recever wth the task of removng the effects of multpath nterference s called the equalser. Ths s often the most dffcult task a recever has to perform.) However, reduce the to one tenth of a symbol perod (or alternatvely ncrease the length of the symbol to ten tmes the between the rays), and the ntersymbol nterference almost dsappears, as now the recever s, for almost all of the tme, only recevng energy from one symbol. power Frst ray between rays symbol n symbol n+ symbol n+ symbol n+3 symbol n+4 one symbol perod Second ray symbol n symbol n+ symbol n+ symbol n+3 symbol n+4 Fgure - Multpath Causng Less Intersymbol Interference The spread of a channel s a measure of the length of tme over whch the energy transmtted at one nstant arrves at the recever. It tells you how powerful an equalser you wll need when you re transmttng at a certan symbol rate; or alternatvely, what symbol rate you can use wthout the need for an equalser... The Problem of Tme-Varyng Channels Knowng how fast the channel s changng can be very useful for a number of reasons. An equalser has to work out what the mpulse response of the channel s, so t can undo the effects the multpath nterference n the receved sgnal, and ths task s much harder f the channel keeps changng. Systems usng equalsers typcally work by transmttng a known seres of bts (known as a tranng sequence or as plot symbols) that allow the recever to work out the mpulse response of the channel. The recever can then adapt the equalser based on ths channel mpulse response. The problem wth usng a tranng sequence s that the recever works out what the channel mpulse response was durng the tme when the tranng sequence s transmtted, not when the nformaton s beng transmtted; and the channel s changng all the tme. Know how fast the mpulse response of a channel s changng, and you know how often you need to transmt these tranng sequences so that the recever can keep ts equalser up-to-date...3 The Tme-Varant Impulse Response It s perhaps nterestng to pause and consder what h(, t) s. It s the mpulse response for an mpulse leavng the transmtter at tme t. It s not the mpulse response of the channel at tme t. If you took a photograph of the system at any tme t, and then tred to determne the mpulse response h(, t ) just by lookng at the photograph, you couldn t do t. Ths s perhaps an over-smplfcaton. For example, some desgns of equalser, once set-up usng a tranng sequence, can track changes n the channel usng the receved data tself. However the general prncple remans true: t s harder to desgn an accurate equalser for a rado channel that s changng quckly. 007 Dave Pearce Page 8/08/008

3 Tme t Tme t Tme t 3 Impulse Transmtter Recever Transmtter Recever Transmtter Recever Fgure -3 Path of an Impulse Through a Fast-Movng Channel For example, consder the case shown n the fgure above, where the drect path between the transmtter and the recever s obstructed by two large sheets of copper, each wth a hole n t. Both sheets are travellng upwards (very fast), and the holes are arranged so that an mpulse leavng the transmtter at tme t would go straght through both holes and arrve at the recever. However, at any gven tme, a photograph would show that the holes n the two sheets of copper are not n lne, and that there s never a lne-of-sght lnk between the transmtter and the recever. The photograph does not contan enough nformaton to work out h(, t), you need to know how fast thngs are movng n the envronment as well. In most cases, for the moble rado channels that we re nterested n, ths pont s largely academc, snce the speed of just about anythng n the envronment s a very small fracton of the speed of lght. Ths s the frst of the smplfyng assumpton we can make: the mpulse response does not change over the amount of tme taken for an mpulse to travel from the transmtter to a recever. Ths approxmaton s qute reasonable: consder a 6 km long rado channel (qute long for moble rado), so that an mpulse travellng at the speed of lght would take 0 s to get from the transmtter to the recever. Anythng travellng more than one mllmetre n that tme would be movng at more that 50 m/sec (80 km/hr), and anythng movng less than one mllmetre s unlkely to make a sgnfcant dfference to the channel mpulse response at the frequences used by moble phones (wth wavelengths typcally around 5 cm).. Clusters of Uncorrelated Scatterers Another assumpton that s often made s that the recever receves a fnte number of echoes of a transmtted mpulse, and as the objects move n the envronment of the transmtter and recever, the ampltudes and relatve phases of these echoes can change relatvely quckly, but ther s change relatvely slowly. There are two physcal models of scatterng that predct exactly ths. The frst assumes that there are groups of scatterers (objects whch reflect rado energy) spread all around the recever, sort of lke ths: The same thng s true for the tme-dependent frequency response as well. H(,t) s the gan of the component at frequency of an mpulse transmtted at tme t, and s the Fourer transform of the tme-dependent mpulse response. 007 Dave Pearce Page 3 8/08/008

4 Scatterers Set Scatterers Set Transmtter Recever Scatterers Set 3 Fgure -4 Model wth Groups of Dstrbuted Scatterers The second, and n many cases rather more plausble explanaton s that there are a lot of local scatterers around the recever, and a smaller number of large reflectors spaced all around the recever: Reflector Reflector Local Scatterers Recever Transmtter Reflector 3 Fgure -5 Model wth Local Scatterers and Dstant Reflectors Both models predct that there wll be groups of rays arrvng at the recever, wth each group arrvng at about the same tme. If the dfference n s between two rays from the same group s too small, then the recever cannot resolve these nto two separate rays 3, they effectvely merge nto one ray, wth an ampltude and phase gven by the sum of the vectors representng the ndvdual rays n the group. In both cases, the mpulse response then looks somethng lke the followng dagram, wth the ampltude and relatve phases of each of the groups of rays changng wth tme: 3 What ultmately determnes whether two rays arrvng at slghtly dfferent tmes are resolvable or not s the samplng rate and bandwdth of the recever. If nformaton about a change n the transmtted sgnal arrves va two dfferent rays less than a sample perod apart, then the recever wll not be able to dstngush the rays: n one sample nether has the new nformaton, the next tme the recever looks, they both have. Recevers typcally sample the ncomng sgnal at between two and four tmes the transmtted symbol rate, so any rays arrvng wthn a small fracton of the symbol rate wll not be resolved. Ths means that the number of resolvable multpath rays n a rado channel s a functon of the symbol rate: systems wth hgher transmtted symbol rates (and hence hgher samplng rates) have more resolvable multpath components n the channel mpulse response. 007 Dave Pearce Page 4 8/08/008

5 ampltude Net Ampltude of Group Net Ampltude of Group Net Ampltude of Group 3 Fgure -6 Impulse Response for Groups of Scatterers In order to change the phase of the receved sgnal from each group of scatterers, the recever (or possbly the scatterers themselves) have to move a sgnfcant fracton of a wavelength. At typcal moble phone frequences (about GHz), ths mples a movement of a few centmetres. That s all t takes to sgnfcantly change the relatve phase of the component rays n any of these group of rays, and hence the ampltude and phase of the resultant of these rays. However, to change the assocated wth ths ray (the tme between transmttng the sgnal and recevng the energy from ths group of scattered rays), the recever would have to move a sgnfcant dstance towards (or away from) the transmtter. Even assumng a fast symbol rate of 5 Mbaud, to make a sgnfcant dfference to the tself (say, one-tenth of a symbol perod) would requre the recever to move 60 meters. The dfference s over three orders of magntude: the phase and ampltudes of the resolvable rays arrvng at the recevers change much more rapdly than ther s. It s often helpful to plot the mpulse responses as a functon of tme. In the case of the three clusters of scatterers consdered here, we mght get somethng a bt lke ths:.5 ampltude 0.5 tme Fgure -7 Example Tme-Varant Three-Ray Impulse Response Note that whle the s of the three rays don t change very fast wth tme, the ampltudes of the rays do change. Ths s typcal of most moble rado channels. 007 Dave Pearce Page 5 8/08/008

6 .3 Multpath and the Power Delay Profle Although the ampltude of each of the resolvable multpath rays n the channel mpulse response s constantly changng, the power receved n each ray can usually be averaged over a tme durng whch the s of the rays themselves are reasonably constant (as we ve seen, the s usually change much less rapdly than the ampltudes and phases of these rays). Plottng ths average power aganst gves the power profle. A typcal power profle mght look somethng lke: mean power -85 db -88 db 0 db spread 3 db spread -95 db Fgure -8 Sample Power-Delay Profle (Remember that n ths fgure the short-term fadng has been averaged out over tme: ths s a plot of the mean power receved n each ray, and remans vald over the length of tme that a recever moves suffcently far to change the s of the rays.) The power- profle plot s useful, snce t provdes a smple vsual mage of the average range of tmes over whch the energy transmtted n one symbol arrves at the recever, and hence a measure of the ntersymbol nterference expected. Whlst a pcture s useful, what would be even more useful s a sngle number that can be used to descrbe ths range of tmes, and tell us how much ntersymbol nterference to expect from the channel. The most obvous way to come up wth a sngle number (the dfference n tme between the frst multpath ray arrvng and the last multpath ray arrvng) sn t very helpful, snce n theory at least the multpath rays decay away for ever 4. At some pont you have to say there s so lttle energy arrvng after ths tme I m just gong to gnore t. Two other possbltes are to defne the spread of the channel n terms of the range of s over whch rays arrve wthn 3 db or 0 db of the ray wth the maxmum power (the 3dB spread and 0 db spread, see the fgure above). These have the advantage that they are very easy to measure from a plot of the power profle, but the dsadvantage that they can gve msleadng answers. For example, consder the followng two power profles: 4 Imagne a recever on a cty street between two buldngs. In theory, some of the energy arrvng at the recever wll have bounced between the two buldngs thousands of tmes before endng up at the recever, and have a huge. Of course, the amount of energy that arrves after ths length of tme s neglgble. 007 Dave Pearce Page 6 8/08/008

7 mean power 0 db spread mean power 0 db spread -80 db -80 db -90 db -90 db Fgure -9 Two Power Delay Profles wth the Same 0 db Delay Spread Both of these profles have the same 0 db spread, but the second one actually has most of the receved power arrvng after a long ; t s just that none of these ndvdual rays qute manages to make t to wthn 0 db of the largest beam. Effectvely, most of the receved power s beng gnored. It s a bt of a slly example perhaps, and totally unrealstc, but t does llustrate the pont that the 3 db or 0 db spread fgure completely neglects all of the power arrvng outsde the range specfed, and there could be a consderable amount of ths power, spread out over a wde range of s, beng gnored. That power could cause a lot of ntersymbol nterference. The usual soluton s to use the rms spread (sometmes just called the spread snce t s so commonly used). Ths fgure ncludes all of the receved power, weghted by what effect t has on the amount of ntersymbol nterference t causes: exactly what we want to know..3. The rms Delay Spread If we normalse the power profle by ensurng that the area under the curve s one (or equvalently that the sum of all the powers n all the rays s one), then we have a probablty densty functon, descrbng the probablty that any random photon of energy arrvng at the recever has taken a gven length of tme to get from the transmtter to the recever. The standard devaton of ths probablty densty functon s the rms spread. It s a good measure of the range of s over whch most of the power arrves. In other words, the rms spread s the square root of the mean value of the square of the dfference between the experenced by each photon of energy and the average. It s probably easer to understand n mathematcal notaton: P P (0.) where s the rms spread, P s the receved power n the th ray, s the of the th ray, and s the mean gven by: P P (0.3) 007 Dave Pearce Page 7 8/08/008

8 Consder a smple example: the mpulse response shown n the fgure below: mean power Fgure -0 Smple Power-Delay Profle Ths conssts of just two rays: arrvng at tmes of and 3 unts after the mpulse was transmtted. The frst ray arrves wth a power of, and the second ray arrves wth exactly half ths power (t doesn t matter what the unts of power are, they cancel out). Then, the mean s: P P (0.4) and the rms spread s: P 7 / / 3 / /9 P (0.5) and n both cases the unts are whatever unts the s of the orgnal rays were specfed n 5. 5 There s another way to do ths sum whch s often easer, and that s to use the fact that the varance of a dstrbuton s the mean of the square of the values mnus the square of the mean of the values. Here, ths gves: and: P P P P Dave Pearce Page 8 8/08/008

9 .3. Delay Spread of Contnuous Power-Delay Profles In some cases the power- profle doesn t consst of a fnte number of dscrete rays, arrvng wth dfferent tmes, but t looks much more lke a contnuous curve, wth some power occurrng wth any value of. We mght get a power- profle that looks more lke ths: mean power densty Fgure - An Example Contnuous Power Delay Profle where the functon plotted s now a power densty: the amount of power that arrves wthn a small range of s d centred around a, dvded by d. Ths sn t a problem: we can calculate a spread n exactly the same way: normalse the power- profle so that t has a total receved power of one, and then calculate the standard devaton of the resultant probablty densty functon. The only dfference s that we now have to ntegrate over all possble s, rather than just summng over the ndvdual rays: P d P d (0.6) where: P d P d (0.7) See the problems for an example of ths..3.3 Problems wth the rms Delay Spread Sadly, the rms Delay Spread does have a problem when t s used n real lfe, and that s nose. If you try and measure a moble channel, you ll get a result that has some nose on t, for example: 007 Dave Pearce Page 9 8/08/008

10 mean power densty Real mpulse response mean power densty Measured mpulse response wth nose nose threshold Fgure - Measured Impulse Response wth Nose Wth the nose, t s not obvous where the mpulse response ends. If we nclude all the nose n the calculaton of the rms Delay Spread, we d end up wth an answer that s much too bg (n theory, nfnte). On the other hand, f we use some nose threshold (as shown n the fgure above) and gnore everythng below that lmt, we would be mssng some parts of the real mpulse response, and that would (usually) gve an answer that was too low. Wthout some knowledge of the shape of the mpulse response (for example, that t looks lke an exponental decay), t can be very dffcult to estmate a good value for spread n cases where there are a lot of very low energy rays arrvng at dfferent tmes (and hence a lot of energy arrvng below ths nose threshold)..4 The Coherence Bandwdth Any parameter or result n sgnals and systems theory expressed n the tme doman has a correspondng parameter or result n the frequency doman. In the case of the spread, the correspondng parameter n the frequency doman s the coherence bandwdth: the range of frequences over whch the gan of the channel remans about the same..4. A Smple Example of Coherence Bandwdth Perhaps a smple example mght help to llustrate the concept. About the smplest example I can thnk of s a channel mpulse response wth just two rays n t (one arrvng after a tme, the other after a of ), each wth the same ampltude A. Input a sngle frequency cosne wave wth unt ampltude cos(t) nto ths channel, and the recever wll receve the sum of these two ed rays:, cos cos r t A t A t (0.8) and usng the formula for the sum of two cosnes, ths can be expressed as: r, t Acos t cos (0.9) whch represents a cosne wave wth an ampltude of: Acos (0.0) and a phase relatve to the transmtted sgnal cos(t) of: 007 Dave Pearce Page 0 8/08/008

11 (0.) (at least when cos(( ) / ) s postve; t s greater than ths when cos(( ) / ) s negatve). If we plotted the mpulse response, and the ampltude of the output from ths channel as a functon of frequency, we d get somethng lke ths: ampltude A gan frequency Fgure -3 Impulse Response and Gan of Smple Channel Put a second cosne wave at a slghtly dfferent frequency + through the same channel, and the ampltude of the receved sgnal becomes: Acos (0.) The queston we need to answer s: how smlar s the response of ths channel at these two frequences? Clearly f = 0, then the channel has the same gan for both frequences, but what f s a small frequency dfference? The channel gan changes smoothly, so the gan would be smlar at two slghtly dfferent frequences, although very dfferent when the frequences are more wdely separated. Even ths very smple case turns out to be rather awkward to solve, mostly due to the fact that we re tryng to compare two oscllatons that can dffer n both ampltude and n phase. The usual mathematcal way of quantfyng how smlar two varables are s n terms of ther correlaton co-effcent, defned as the mean value of the product of the complex conjugate of one functon (mnus ts mean) wth the other functon (mnus ts mean), dvded by ther standard devatons 6 : x x y y E (0.3) x y 6 We have to take the complex conjugate of one of the functons n the case where the functons are complex. Dong so ensures that the correlaton coeffcent s one when the two functons are dentcal, snce one number multpled by ts complex conjugate s always real. 007 Dave Pearce Page 8/08/008

12 Ths gves a value of one f the x and y are perfectly correlated (n the sense that y s dentcal to x or at least a constant multple of x), and zero f the value of y s completely ndependent of the value of x. When the two quanttes can dffer n both ampltude and phase, the task of decdng how smlar they are has an added complexty. The usual technque s to use complex numbers to represent each component n the mpulse response of the channel: the ampltude and phase of each oscllaton beng represented by the ampltude and phase of a complex number. For example, the sgnal n the frst ray: cos Acos t A t (0.4) would be represented by a complex number wth ampltude A and phase angle : Aexp j (0.5) and the real sgnal n ths ray can be re-created by multplyng ths complex number by a complex oscllaton at the orgnal frequency, and then takng the real part of the result: Aexp j exp j t Aexp j t Acos t (0.6) Usng ths notaton, we can represent the sum of the two cosne waves n equaton (0.9) n complex form as:, cos cos Aexp j t Aexp j t Aexp j exp j exp j t r t A t A t (0.7) and then use the complex representaton of ths sgnal:, exp exp r t A j j E (0.8) Smlarly, we can use as complex representaton of the channel s output at the hgher frequency +, and use the complex representaton of the sum of the two rays at ths frequency:, exp exp r t A j j E (0.9) We then need to fnd the correlaton coeffcent between these two complex representatons. One bg advantage of ths technque s that the mean value of both of these sgnals s zero: a complex exponental oscllaton has a zero mean, and the mean of the sum of two varables s the sum of the means of each varable. Ths leaves us wth the problem of fndng the correlaton between r E (, t) and r E (, t) as a functon of the frequency dfference, averaged over all possble values of. The expectaton value of the product of the complex conjugate of r E (, t) and r E (, t), from equatons (0.8) and (0.9) gves: 007 Dave Pearce Page 8/08/008

13 E A exp j exp j exp j exp j (0.0) and multplyng out the brackets gves the four terms: E A j j j exp j exp j exp j exp exp exp (0.) The mddle two terms are functons of, and average out to zero, leavng just the frst and last terms when the expectaton value s taken: exp A exp j j (0.) Rather than calculate the standard devatons for the denomnator of the correlaton coeffcent, we can take a short cut: we know the correlaton coeffcent wll be one when = 0, and when = 0 ths expectaton value s A, so we can wrte the fnal correlaton coeffcent of the frequency response of the two channels as: j exp j exp (0.3) Ths s a complex correlaton coeffcent: f we wanted to know over what range the absolute magntude of the correlaton coeffcent had a value greater than, we d have to use: exp j exp j exp j exp j cos 4 cos 4 (0.4) so to determne the range of frequences over whch the channel gan has a correlaton wth a magntude of greater than, we d have to use: cos (0.5) In ths case (two equal powered rays arrvng after a of and respectvely), the spread s: so we could wrte: (0.6) 007 Dave Pearce Page 3 8/08/008

14 cos (0.7) whch s nterestng, snce t shows that the coherence bandwdth s nversely proportonal to the spread, the constant of proportonalty dependng on the correlaton coeffcent of the channel gans over the range of frequences (n other words, how close to the same you want the channel response to be at the two frequences consdered). Puttng some numbers nto ths: consder an mpulse response that looks lke ths: power 3 (s) Fgure -4 Smple Two-Ray Power Delay Profle Then, f we wanted to know what range of frequences could be used so that there was a correlaton wth a magntude of at least 0.9 between any two frequences, we could frst calculate the spread (0.5 s n ths case) and then: cos Mrad/s 44 khz 0.5 μs (0.8) but for a correlaton of at least 0.5 between frequences, the coherence bandwdth ncreases to: cos Mrad/s 333 khz 0.5μs (0.9) Ths s a very smple case (two equal powered rays), but the general concluson remans true for all other channels: the coherence bandwdth s nversely proportonal to the spread, the constant of proportonalty dependng on the power profle of the channel, and the maxmum correlaton coeffcent requred between the channel gan at dfferent frequences wthn the coherence bandwdth..4. Calculatng the Coherence Bandwdth n the General Case For the general case, a smple expresson for the coherence bandwdth can be determned from the mpulse response of the channel and some results from Fourer theory. Any channel wth a tme-dependent mpulse response of h(, t) has a tme-dependent frequency response of H(, t) obtaned by takng the Fourer transform of the mpulse response wth respect to the, so we can calculate the channel frequency response at any gven frequency usng the Fourer transform ntegral: 007 Dave Pearce Page 4 8/08/008

15 ,, exp H t h t j d (0.30) and the correlaton coeffcent between the frequency response of the channel at two dfferent frequences s then: H t H t H t H t E,,,, (0.3) H, t Now for all channels we re nterested n, the mean value of the frequency response s zero (the phase of the output from each ray constantly rotates n phase as the frequency changes, averagng out to zero over all frequences), so we can smplfy ths to: E H, t H, t (0.3) H, t and snce we know that the correlaton co-effcent must be one when s zero (the same trck as we used above to avod havng to work out the standard devaton), we can replace the denomnator: E H, t H, t (0.33) E H, t H, t The next step uses a verson of the Wener-Khnchn theorem from Fourer theory: the autocorrelaton of the frequency response s tmes the Fourer transform of the square of the modulus of the mpulse response (n other words, of the power profle) 7 : 7 Consder takng the Fourer transform of the autocorrelaton of the frequency response (that s the autocorrelaton functon, not the correlaton co-effcent), defned as: H H d Takng the nverse Fourer transform of ths autocorrelaton functon and reversng the order of ntegraton gves: [contnued on next page ] 007 Dave Pearce Page 5 8/08/008

16 H, t H, td h, t exp j d (0.34) Now the expectaton value of H (, t) H(, t) s the mean value of ths product over all possble frequences, whch could be expressed as:,, lm,, s E H t H t H t H t d s s (0.35) so we could express our normalsed correlaton coeffcent as: s H, t H, td E H, t H, t s s lm s E H, t H, t s H, t H, td s s s (0.36) whch when cancellng out the factors of s and usng the result from equaton (0.34) gves:, exp h t j d h, t d (0.37) exp j d H H d exp j d exp H H j d d H H exp j d exp j d exp H h j d exp h H j d h h h and f the nverse Fourer transform of the autocorrelaton functon s tmes the power profle, then the autocorrelaton functon must be the Fourer transform of tmes power profle. exp h j d 007 Dave Pearce Page 6 8/08/008

17 Ths provdes a smple way 8 to calculate the coherence bandwdth: all you need to do s take the Fourer transform of the power profle of the channel and then normalse t..4.3 An Example of Coherence Bandwdth Consder an example: the ampltude of the mpulse response, power profle, frequency response and frequency correlaton of a rado channel are shown n the fgure below: Impulse Response h(, t) (s) Power-Delay Profle h(, t) (s) Frequency Response H(, t) freq (MHz) Frequency Correlaton () freq_offset (MHz) Fgure -5 Correlaton Bandwdth of Example Channel (I ve plotted the frequency response of ths mpulse response from 5.5 to 6.5 MHz, rather than around zero, snce to a rado transmsson, the only nterestng part of the spectrum s around the carrer frequency, whch I ve assumed here s 6 MHz.) Of course, ths just provdes a plot of the coherence between two frequences. Agan, just as wth the spread, we d deally lke a sngle number gvng the range of frequences over whch the response of the channel s approxmately the same. Ths channel has a spread of about.7 s, and the channel changes sgnfcantly n gan n around 0. MHz (that s /5). It s clear, I thnk, that the channel has a completely dfferent (.e. totally uncorrelated) gan f you move n frequency by 0.58 MHz (/). If you want two frequences that have approxmately the same gan, you can t move more than about (/50) MHz; see the expanded fgure below: 8 I should apologse to any mathematcans readng ths who are probably spttng wth fury at ths pont. I ve taken a few short cuts wth ths dervaton, for example, the autocorrelaton as I ve defned t can take nfnte values, whch means you can t take the Fourer transform. However, I hope ths smplfed dervaton at least serves to ndcate where the result comes from. For more dscusson on ths pont, see the chapter on Fourer theory. 007 Dave Pearce Page 7 8/08/008

18 H(, t) 4 3 / /5 /50 freq (MHz) Fgure -6 Channel Frequency Response and Coherence Bandwdths Whch fgure s of the most use depends on what you want to use t for. If you want to know whether the channel looks flat, then you ll want a coherence bandwdth over whch the gan of the channel s approxmately the same, perhaps wth a correlaton coeffcent of greater than 0.9. If you want to know how by how much you have to change the frequency to be confdent that the gan of the channel has completely changed (perhaps the sgnal has faded, and you re thnkng of choosng a new carrer frequency to use), you ll want a correlaton coeffcent of close to zero. If you want to know whether you need to use an equalser or not, you ll want a correlaton coeffcent of about The exact relatonshp between the rms spread and the coherence bandwdth s dependent on the shape of the power profle, but n every case, the coherence bandwdth s nversely proportonal to the spread. Increase the spread, and you reduce the range of frequences over whch the channel appears to have a smlar gan. I ll fnsh ths chapter wth a summary of some useful approxmatons: Approxmaton Correlaton coeffcent B ~ 0 rms B ~ 0.5 B 3 5 rms 50 rms > 0.9 Notes The channel gan at frequency s almost entrely ndependent of the gan at. Changng frequency by results n a channel wth a completely dfferent channel gan. The channel gan at frequency s smlar to the gan at. If s greater than the symbol rate, an equalser may not be requred. The channel gan of frequency s almost exactly the same as the gan at 3. 9 Ths s a rough rule-of-thumb only. 007 Dave Pearce Page 8 8/08/008

19 .5 Problems ) A channel has the mpulse response shown n the fgure below: Power (pw) Delay (s) ) What s the mean of ths channel? ) What s the spread? ) Estmate the coherence bandwdth. Estmate the maxmum symbol rate that can be used through ths channel wthout the need for an equalser. (Assume an equalser s not requred f the spread s less than one-ffth of the symbol perod.) ) A rado channel has an mpulse response contanng two rays of equal power, one wth a of s, and the other wth a of 3 s. Calculate the rms spread for ths channel, and hence estmate the dfference n frequences for whch the correlaton between the gans s 0.5 by usng /5. Then calculate the exact dfference n frequences that gves an expected correlaton of 0.5. How accurate s the approxmate formula n ths case? What s the actual value of the correlaton between two frequences /5 apart n frequency? 3) Repeat queston for a rado channel wth an mpulse response of h() = exp(-a). How accurate s the approxmaton n ths case? 4) What coherence bandwdth gves a maxmum correlaton of 0.9 between the channel gans for the mpulse response n queston? 007 Dave Pearce Page 9 8/08/008