This is the Q function for matlab, assignment help

Anonymous
timer Asked: Sep 15th, 2016
account_balance_wallet $50

Question description

I will attach the list of questions and the required reading materials that have references in the questions. One of the questions asks for a Matlab code and one asks for a C function and the rest are just calculations. I will also attach a Q function for Matlab that might help to solve the first question.



0XOWLSDWK   Multipath Introduction 7KHUH DUH WZR FKDUDFWHULVWLFV RI UDGLR FKDQQHOV WKDW SUHVHQW VHULRXV GLIILFXOWLHV IRU WHOHFRPPXQLFDWLRQ V\VWHPV 2QH LV WKH WUHPHQGRXV G\QDPLF UDQJH WKDW PXVW EH DFFRPPRGDWHG GXH WR WKH ODUJH FKDQJH LQ SDWK ORVV ZLWK GLVWDQFH $QRWKHU LV WKH SUHVHQFH RI PXOWLSOH SURSDJDWLRQ SDWKVEHWZHHQWUDQVPLWWHUDQGUHFHLYHUWKDWSURGXFHHFKRHVRIWKHWUDQVPLWWHGVLJQDO LQ WKH UHFHLYHG VLJQDO 7KLV SKHQRPHQRQ LV FDOOHG VLJQDOVFDQUHVXOWLQ multipath ,QWHUIHUHQFH EHWZHHQ PXOWLSDWK fading:H¶OOFRYHUIDGLQJLQWKHQH[WOHVVRQ Theory &RQVLGHUDQHQYLURQPHQWOLNHWKDWGLDJUDPPHGLQ)LJXUHLQZKLFKWKHUHH[LVWREMHFWVFDSDEOH RIUHIOHFWLQJUDGLRZDYHV  Figure 7.1. In the presence of reflecting objects there can be many paths between two radios, each having a different propagation distance. The situation shown might represent a top view of two radios on either side of a street lined with buildings. 0D[ZHOO¶V HTXDWLRQV DUH OLQHDU ,I QRWKLQJ LQ WKH HQYLURQPHQW LV PRYLQJ RU RWKHUZLVH FKDQJLQJ ZLWK WLPH WKHQ ZH KDYH D OLQHDU WLPHLQYDULDQW V\VWHP )RU VXFK V\VWHPV WKH LPSXOVH UHVSRQVH DQG WUDQVIHU IXQFWLRQ FRQFHSWV DUH DSSOLFDEOH DQG JHQHUDOO\ YHU\ XVHIXO /HW¶V WDNH RXU V\VWHP LQSXWWREHWKHWUDQVPLWWHGVLJQDODQGWKHRXWSXWWREHWKHUHFHLYHGVLJQDO&RQFHSWXDOO\ZHFRXOG WUDQVPLW DQ HOHFWURPDJQHWLF LPSXOVH 7KH LPSXOVH ZRXOG WUDYHO DORQJ WKH YDULRXV SDWKV IURP WUDQVPLWWHU WR UHFHLYHU 7KH '63 k  SDWK ZRXOG SURGXFH D UHFHLYHG LPSXOVH DIWHU D WLPH GHOD\ tk 'XH WR WKH GLVWDQFH WUDYHOHG DQG WKH LQWHUDFWLRQ ZLWK UHIOHFWLQJ REMHFWV LW ZLOO KDYH VRPH DPSOLWXGH ak 7KHUHIRUH WKH WRWDO UHFHLYHG ILHOG ZKLFK LV WKH impulse response RI WKH WUDQVPLWWHUUHFHLYHU V\VWHPZLOOEH N−  h t = ∑ ak δ t − tk  k=      "!#!%$ &'( *) @ A +*' '*,-$.)/0"12& )3 '45+'6&'4 78* 9:;)/ ' <   =>?(="?(=  0XOWLSDWK   7KH WUDQVIHU IXQFWLRQ RU IUHTXHQF\ UHVSRQVH RI WKH V\VWHP LV WKH /DSODFH WUDQVIRUP RI WKLV HYDOXDWHGDW s = jω = j πf  ∞ H ω = ∫ h t e − j C  N− B = ∑ ak e D D −j πft dt    πt k f C k= 7KLV LV D VXP RI SKDVRUV )RU FHUWDLQ IUHTXHQFLHV WKH SKDVRUV PLJKW EH LQ SKDVH DQG DGG XS WR D ODUJH UHVSRQVH )RU RWKHU IUHTXHQFLHV WKH\ PLJKW DGG RXW RI SKDVH DQG JLYH D ORZ UHVSRQVH :H VHH WKDW WKH HIIHFW RI PXOLWSDWK LV WR FUHDWH D UHVSRQVH WKDW YDULHV ZLWK IUHTXHQF\ ± D W\SH RI IUHTXHQF\VHOHFWLYHILOWHU Multipath Statistics $OWKRXJK LQ SULQFLSOH WKH LPSXOVH UHVSRQVH LV FRPSOHWHO\ GHWHUPLQHG E\ WKH SURSDJDWLRQ HQYLURQPHQWLQSUDFWLFHWKLVHQYLURQPHQWLVVRFRPSOH[WKDWZHDUHLQFOLQHGWRWUHDWWKHLPSXOVH UHVSRQVH DV D UDQGRP SURFHVV 5DQGRP SURFHVVHV DUH W\SLFDOO\ FKDUDFWHUL]HG E\ WKHLU moments 7KHILUVWDQGVHFRQGPRPHQWVDUH E N− t = k= E N− E N− t G = k k G ∑ ak k=  G ∑F a t F  G G ∑F a t k= E N− k k ∑F a k=   G k %HFDXVH WKH a¶V UHSUHVHQW YROWDJHV ZH VTXDUH WKHP WR JHW SRZHU  7KH ILUVW PRPHQW t  LV MXVW WKHDYHUDJHWLPHRIDUULYDO7KHILUVWDQGVHFRQGPRPHQWVFDQEHFRPELQHGWRJHWWKHroot-meansquare delay spread σh =  t H − t H    ,I ZH FRQWLQXHG WR FDOFXODWH KLJKHURUGHU PRPHQWV ZH ZRXOG JHW DQ LQFUHDVLQJO\ DFFXUDWH GHVFULSWLRQ RI WKH LPSXOVH UHVSRQVH +RZHYHU HYHQ WZR PRPHQWV SURYLGHV XV ZLWK XVHIXO LQIRUPDWLRQ QDPHO\ WKH DYHUDJH WLPH RI DUULYDO DQG WKH WLPH VSUHDG RI WKH PXOWLSDWKV 2QH IXQFWLRQ WKDW LV FRPSOHWHO\ GHWHUPLQHG E\ LWV ILUVW WZR PRPHQWV LV WKH *DXVVLDQ 6R E\ VWRSSLQJ DW WKH ILUVW WZR PRPHQWV RI PHDQ t DQGYDULDQFH I h t  ZH DUH HIIHFWLYHO\ PRGHOLQJ LW DV DQ HTXLYDOHQW *DXVVLDQ ZLWK σ h       "!#!%$ &'( *) +*' '*,-$.)/0"12& )3 '45+'6&'4 78* 9:;)/ ' < =>?(="?(=  0XOWLSDWK   I a t  ZLWK YDULDQFH σ h  DQG OHW¶V FKRRVH D WLPH RULJLQ VR WKDW WKH &RQVLGHU D *DXVVLDQ IXQFWLRQ PHDQLV]HUR7KHVSHFWUXPRID*DXVVLDQLVDOVR*DXVVLDQ n a t =    πσ h e  t   − m   σh  l ⇔ A f = e − e− m πl σ l f l h    p πo σ o f o =     6ROYLQJ IRU WKH IUHTXHQF\ ZH f ≈   σ h  :H FDQ WDNH WKLV DV WKH EDQGZLGWK RI WKH *DXVVLDQ :H ZLOO XVH WKLV WR GHILQH WKHcoherence bandwidthRIWKHLPSXOVHUHVSRQVH 7KH VSHFWUXP LV GRZQ WR ò LWV SHDN YDOXH ZKHQ h ILQG Bc =   σh     7KH FRKHUHQFH EDQGZLGWK VSHFLILHV WKH PLQLPXP IUHTXHQF\ GHYLDWLRQ RYHU ZKLFK WKH WUDQVIHU IXQFWLRQFDQFKDQJHVLJQLILFDQWO\,QRWKHUZRUGVLIWZRVLJQDOVGLIIHULQIUHTXHQF\E\PXFKOHVV WKDQ Bc WKHQ WKH\ H[SHULHQFH HVVHQWLDOO\ WKH VDPH UHVSRQVH WKURXJK WKH SK\VLFDO FKDQQHO ,I WZR BcWKHQWKH\PD\H[SHULHQFHFRPSOHWHO\GLIIHUHQWUHVSRQVHV VLJQDOVGLIIHUE\PXFKPRUHWKDQ 7KH UHODWLRQ EHWZHHQ WKH VLJQDO EDQGZLGWK FKDQQHO B DQG WKH FRKHUHQFH EDQGZLGWK RI WKH SK\VLFDO Bc LV LPSRUWDQW ,I B << Bc  WKHQ DOO IUHTXHQF\ FRPSRQHQWV LQ WKH VLJQDO DUH HVVHQWLDOO\ PRGLILHG LQ WKH VDPH PDQQHU DQG WKH UHFHLYHG VLJQDO LV VLPSO\ D WLPHGHOD\HG DQG DWWHQXDWHG YHUVLRQRIWKHWUDQVPLWWHGVLJQDO,QHIIHFWWKHFKDQQHOLVVLPSO\DQDWWHQXDWRU7KLVLVUHIHUUHGWR DV flat fading 2Q WKH RWKHU KDQG LI Bc << B  WKHQ GLIIHUHQW IUHTXHQF\ FRPSRQHQWV DUH PRGLILHG GLIIHUHQWO\ E\ WKH FKDQQHO DQG WKH UHFHLYHG VLJQDO FDQ EH VHULRXVO\ GLVWRUWHG ,Q HIIHFW WKH FKDQQHO LV WKHQ D ILOWHU ZLWK VRPH FRPSOLFDWHG WUDQVIHU IXQFWLRQ 7KLV LV UHIHUUHG WR DV frequency selective fading. ,Q WKLV FDVH ZH RIWHQ PXVW WDNH VWHSV WR XQGR WKH GLVWRUWLRQ LQWURGXFHG E\ WKH SK\VLFDOFKDQQHO6RPHZD\VWRGRWKLVDUHWKURXJKWKHXVHRI equalizersRULQ&'0$V\VWHPV rake filters%XWPRUHDERXWWKDWODWHU Example 7.1 N− $VVXPH q h t = ∑ ak δ t − tk  k= as =  t s = µs  ZLWK r au =  t t = µs  av =  t v = µs 7KHQ w t = t VR σh = w  w = w +   +  w w w  +  +      w w w w  +   +  w w w  +  +  w µs = µs  w w µs = µs w  x −  = µs DQG Bc = MHz  JJKLM N O PJQRS QTTUS QRVWUXTY TZW"[#[%\QS Z ]^(S WQ*_ `*ZW^ ^*a-\b_/WQc"d2] _eS QR^4W5Q`^6]^4T f8Q*S gT:U;_/S ^ h ijk(i"Mk(iK  0XOWLSDWK   Simulation )LJXUHVKRZVD0DWKFDGSURJUDPWKDWVLPXODWHVDFKDQQHOLQZKLFKWKHUHLVDGLUHFWSDWKDQG IRXU RWKHU PXOWLSDWKV ZLWK UDQGRP WLPHV RI DUULYDO VSUHDG RYHU DERXW  QV DQG UDQGRP DPSOLWXGHV
Homework 3 3.1 Due Friday, Sept. 16, 2015 Write a Matlab program that will read in impulse response data in the form of a two column table with the first column being time and the second column voltage and then calculate t and σ h . On the same plot it should then graph the data and the function V (t ) = V0e − 1 t− t  2  σh the data, i.e., 3.2 1/2    2 where V0 is chosen so that this function has the same total energy as ∫ V (t ) 2 dt = ∑ Vk 2 where Vk is the kth impulse response datum. Apply your program from Problem 3.1 to the following data: t V 1 1.00 2 0.75 4 0.50 7 0.25 10 0.10 3.3 For the data shown in Fig. 7.5 of the course notes, what is the maximum channel bandwidth for which we would have flat fading? Interpret “<<” as one order of magnitude and “eyeball” your answer from the plots, i.e., don’t try and numerically calculate values. 3.4 Write a C function called rayrand() that returns a value of type double drawn from a Rayleigh distribution. Make use of the C function rand() that is part of stdlib.h and generates uniformly-distributed integers between 0 and RAND_MAX (defined in stdlib.h). The following C code will generate a number of type double drawn from a distribution that is uniform in the interval [0,1) : ((double)rand())/RAND_MAX 3.5 Cars drive down a highway at 90 km/hr toward a base station 3 km away operating at 1900 MHz. The average received signal level at the base station is –85dBm. The system uses digital modulation with a bit rate of Rb bits per second. Bit errors occur when the received signal fades below –110 dBm. The average fade cannot last longer than half the bit time for the system to operate reliably. Assuming Rayleigh fading, what is the maximum allowable bit rate? EE432: RF Engineering for Telecommunications Washington State University 09/09/16 Homework 3 Due Friday, Sept. 16, 2015 2/2 3.6 A base station uses identical antennas for transmit and receive and a mobile uses a single antenna for transmit and receive. The base station transmits 40 dBm and the mobile receives –90 dBm. The mobile transmits 27 dBm. The minimum acceptable power level is –110 dBm. If you want to employ switched diversity reception at the base station, how many receiving antennas do you need so that the probability of the base station receiving –110 dBm or more from the mobile is at least as high as the probability that the mobile receives –110 dBm from the base station? 3.7 A transmitter has 3 dB antenna gain, its antenna is 2 m above the ground, and it transmits 100 mW of power. A receiver has 3 dB antenna gain, its antenna is 5 m above the ground, and it requires a signal of at least –100 dBm to operate. Assume a model of the form (3.9) for received power plus Rayleigh fading. In other words, (3.9) gives the average received power. How far can the transmitter be from the receiver so that the link works 95% of the time? If the receiver adds an identical antenna and uses N = 2 diversity reception, how far away can the transmitter be? EE432: RF Engineering for Telecommunications Washington State University 09/09/16
Ground Reflections 3.1 Ground Reflections Introduction When dealing with radio waves in a terrestrial environment we almost never observe free-space propagation of the type we talked about in the last lesson. This is because the waves interact with the environment: ground, trees, buildings, etc. The result is that we typically have several reflected, refracted, diffracted waves interfering with one another at any given point. (We’ll talk about diffraction later.) This complicates things quite a bit. In this lesson we’ll consider the simple model of a transmitter-receiver pair above an infinite “ground plane.” We don’t live on an infinite flat plane, but this simple model does have some explanatory power in the real world, and it will motivate some empirical models that we’ll learn about later. Geometry Assume a transmitter is a height ht above a flat, smooth ground and a receiver is at a height hr while the ground distance between them is r. This is illustrated in Figure 3.1. Let the reflection coefficient of the ground be Γ. In our model we will assume this is a constant. In fact, as you know from a course like EE351, this is really a function of the field polarization and the angle of incidence. TX r1 ht RX r2 θ θ hr r Figure 3.1: Geometry for the ground reflection problem (the “tworay model”). There are two paths for radio waves to take from transmitter to receiver: a direct path and a reflected path. Call the length of the direct path r1 and that of the reflected path r2. The law of reflection requires the angle of incidence to equal the angle of reflection (the angles θ in the figure). The result is that the reflected field appears to come from a “mirror image” of the source at a distance ht below the ground. Basic geometry gives us the distances EE432: RF Engineering for Telecommunications Scott Hudson, Washington State University 08/25/08 Ground Reflections 3.2 r1 = r 2 + (ht − hr ) 2 (3.1) r2 = r 2 + (ht + hr ) 2 We’ll be interested in cases where r >> ht , hr . For example r might be hundreds or thousands of meters while the h’s are only a few or maybe a few tens of meters. Then the following approximations are good (recall that for small x, 1 + x ≈ 1 + x / 2 ): (ht − hr ) 2 2r (h + hr ) 2 r2 ≈ r + t 2r r1 ≈ r + (3.2) The path difference is ∆r = r2 − r1 ≈2 (3.3) ht hr r As we’d expect from Figure 3.1, this goes to zero at large distances. Theory The receiver will “see” two transmitters – the real transmitter above the ground and a virtual, or mirror image, transmitter below the ground. The intensity of each of these fields in the absence of the other would be given by the Friis equation (2.6). When both are present we cannot simply add the powers because it is the EM fields that add, and the fields have both amplitude and phase. The amplitude is proportional to the square root of the intensity and the phase is 2π times the distance traveled in wavelengths. The amplitude of the reflected field is also multiplied by Γ, the reflection coefficient of the ground. The total power is proportional to the magnitude squared of this total field, or PR = PT −j 2π r1 λ GT GR λ e ( 4 π) 2 r1 2 +Γ e −j 2π 2 r2 λ r2 (3.4) Note that if there is no ground reflection ( Γ = 0 ) we simply get (2.6), as we should expect. For large r, (3.2) tells us that r1 and r2 both approach r, so the phase terms in (3.4) will approach one another in both phase and amplitude. EE432: RF Engineering for Telecommunications Scott Hudson, Washington State University 08/25/08 Ground Reflections 3.3 If Γ = −1 , which would correspond to a perfectly conducting ground with a certain polarization, the two terms will tend to cancel each other and the received power will drop very rapidly. We have in this special case e −j 2π r1 λ r1 +Γ e −j 2π 2 r2 λ r2 2π 2π 1 − j r1 − j r2 ≈ 2 e λ −e λ r 2π − j ∆r 1 = 2 1− e λ r π 2 (3.5) π − j ∆r 1 j ∆r = 2 e λ −e λ r = 2 2 π∆r 4 sin 2 2 r λ so PR ≈ PT GT GR λ2 2πht hr sin 2 2 (2πr ) λr (3.6) The sine term oscillates between 0 and 1 while the “envelope” PT GT GR λ2 /(2πr ) 2 has the 1 / r 2 behavior of free-space propagation. The approximation sin 2 1 / x = 1 / x 2 is good to 1.5 dB for x ≥ 1 . So, for large enough r, we can write PR ≈ PT GT GR 2 ht hr r4 2 (3.7) Here the received power is falling off as 1 / r 4 , much more rapidly than it would in free space. It is also interesting to note that there is no wavelength dependence in this expression. The envelope of (3.6) and (3.7) are equal when r = 2πht hr / λ . If we neglect the sine oscillations in (3.6) we can create a composite propagation model as follows: PR = EE432: RF Engineering for Telecommunications λ2 ( 2 πr ) 2 2 2 ht hr PT GT GR 4 r PT GT GR 2 πht hr λ 2 πht hr ;r ≥ λ ;r < Scott Hudson, Washington State University (3.8) 08/25/08 Ground Reflections 3.4 This is an example of a breakpoint model, the breakpoint being r = 2πht hr / λ where the model changes from one type of behavior to another. If we use the breakpoint as the reference distance r0 then PR ,dBm r r0 = r P0,dBm − 40 log r0 P0,dBm − 20 log ; r < r0 ; r ≥ r0 r0 = 2πht hr / λ (3.9) P0,dBm = PT ,dBm + GT ,dB + GR ,dB + 20 log λ2 (2π) 2 ht hr (Keep in mind that everything we’ve done is valid only for the case Γ = −1 .) The important point here is that the presence of the ground can cause the field to decay very differently from the 1 / r 2 behavior of free space. It is instructive to ask what happens to (3.9) if we change the height of one or both of the antennas. First, notice that the model depends only on the product ht hr . Let’s say we change the heights so that the new value of this product is (ht hr )' = aht hr . For example, if we double the height of one of the antennas, then a = 2 . The new breakpoint will be r0 ' = ar0 . Since there is no change in the transmitted power or antenna gains P0′,dBm = P0,dBm − 20 log λ2 λ2 + 20 log (2π) 2 ht hr (2π) 2 aht hr (3.10) = P0,dBm − 20 log a If we are at a small distance where r < r0 , r0 ' , then the new model is PR′,dBm = P0,dBm − 20 log a − 20 log = P0,dBm − 20 log = PR ,dBm r r0 r ar0 (3.11) In other words, for distances less than either breakpoint, there is no change in the received power. This makes sense because (3.9) models the close-in field as being the same as in freespace. That is, for small distances the ground is assumed to have no effect, hence, the antenna heights shouldn’t have any effect. On the other hand, at a large distance r > r0 , r0 ' EE432: RF Engineering for Telecommunications Scott Hudson, Washington State University 08/25/08 Ground Reflections 3.5 PR′, dBm = P0, dBm − 20 log a − 40 log r ar0 = P0, dBm − 20 log a + 40 log a − 40 log = PR , dBm + 20 log a r r0 (3.12) and received power is changed by the constant amount 20 log a . For example, if you increase either antenna height by a factor of 2, then received power at large distances increases by 6 dB. Figure 3.2 shows an example in which ht hr is increased by a factor of 10. 0 20 Pr (dBm) 40 60 80 100 1 0.5 0 0.5 log(r/75m) 1 1.5 2 Figure 3.2: Effect of changing antenna height. Solid curve shows PR for rt = rr = 2m , GT = 10dB , GR = 3dB and PT = 1W . Dashed curve is the case where we change to rt = 20m . The effect is to extend the breakpoint to a larger distance. Simulation Figure 3.3 shows a numerical (Mathcad) calculation of (3.4) for the case Γ = −1 and ht = 10λ, hr = 3λ . Also shown are 1 / r 2 and 1 / r 4 behaviors characteristic of small and large r values. EE432: RF Engineering for Telecommunications Scott Hudson, Washington State University 08/25/08 Ground Reflections 3.6 0 relative Pwr (dB) 50 100 150 200 2 1 two ray n=2 n=4 0 1 2 3 log(r/r0) Figure 3.3: Ground reflection model (3.4) (solid curve), envelope of (3.6) (dotted line), and (3.7) (dashed line) for ht=10λ, hr=3λ. The breakpoint is at 188λ. When received power falls off as 1 / r n , we say that the propagation constant is n. For free space the propagation constant is n = 2 . At large distances the ground-reflection model with Γ = −1 has a propagation constant of n = 4 . Thus we see that it is quite possible, even in a relatively simple geometry, to get a propagation constant different than that of free space. Example 3.1 Assume a propagation model of the form (3.9). Suppose a cellular base station transmits 30dBm (1W) of power and has an antenna with gain of 7dB and height 10m. A roof-mounted car antenna is 1.5m high and has 3dB gain. The frequency is 1900MHz. How far away can the car be and still receive –90dBm of signal? The wavelength is 300/1900 or 0.158m. (3.9) then gives r0 = 597m, P0 = −47.5dBm. We then solve − 90 = −47.5 − 40 log(r / 597 m) to get r = 6894m. So the car can go about 7km from the base station. Experiment A 915MHz transmitter was placed 2.5ft above a sidewalk. Field strength measurements were made at various distances and at a height of 3ft. The observed data are show in Figure 3.4 along with (3.4) and the free-space model. A reference distance of r0 = 5 ft was used. The groundreflection model does capture the general behavior of the data. In Figure 3.5 we plot the observed EE432: RF Engineering for Telecommunications Scott Hudson, Washington State University 08/25/08 Ground Reflections 3.7 data along with 1 / r n models for n = 2,3,4 . This illustrates the breakpoint concept. The theoretical break point of 2πht hr / λ is about 48 feet in this case. Since log(48 / 5) = 0.98 we would expect a breakpoint near 1 on the horizontal axis. This is indeed what we observe. The data are fairly well described by n = 2 for small r and by n = 3 for large r. 30 40 Pr (dBm) 50 60 70 80 0 0.2 0.4 two-ray observed free-space 0.6 0.8 1 log(r/r0) 1.2 1.4 1.6 1.8 Figure 3.4: Ground reflection measurements above a sidewalk at 915MHz. Reference distance r0 is 5ft. Circles are observed data. Solid curve is eqn. (3.4) with a Γ of −0.6. Dashed curve is freespace model. EE432: RF Engineering for Telecommunications Scott Hudson, Washington State University 08/25/08 Ground Reflections 3.8 30 Pr(dBm) 40 50 60 70 80 0 0.2 0.4 n=2 n=3 n=4 observed 0.6 0.8 1 log(r/r0) 1.2 1.4 1.6 1.8 Figure 3.5: Data of Figure 3.2 compared to various 1 / r n models. n = 2 gives a descent description for small r while n = 3 gives a descent description of large r. References 1. Rappaport, T. S., Wireless Communications: Principles and Practice, Prentice Hall, 1996, ISBN 0-13-375536-3. 2. Ishimaru, A., Electromagnetic Wave Propagation, Radiation, and Scattering, Prentice Hall, 1996, ISBN 0-13-249053-6. EE432: RF Engineering for Telecommunications Scott Hudson, Washington State University 08/25/08
Ground Reflections 3.1 Ground Reflections Introduction When dealing with radio waves in a terrestrial environment we almost never observe free-space propagation of the type we talked about in the last lesson. This is because the waves interact with the environment: ground, trees, buildings, etc. The result is that we typically have several reflected, refracted, diffracted waves interfering with one another at any given point. (We’ll talk about diffraction later.) This complicates things quite a bit. In this lesson we’ll consider the simple model of a transmitter-receiver pair above an infinite “ground plane.” We don’t live on an infinite flat plane, but this simple model does have some explanatory power in the real world, and it will motivate some empirical models that we’ll learn about later. Geometry Assume a transmitter is a height ht above a flat, smooth ground and a receiver is at a height hr while the ground distance between them is r. This is illustrated in Figure 3.1. Let the reflection coefficient of the ground be Γ. In our model we will assume this is a constant. In fact, as you know from a course like EE351, this is really a function of the field polarization and the angle of incidence. TX r1 ht RX r2 θ θ hr r Figure 3.1: Geometry for the ground reflection problem (the “tworay model”). There are two paths for radio waves to take from transmitter to receiver: a direct path and a reflected path. Call the length of the direct path r1 and that of the reflected path r2. The law of reflection requires the angle of incidence to equal the angle of reflection (the angles θ in the figure). The result is that the reflected field appears to come from a “mirror image” of the source at a distance ht below the ground. Basic geometry gives us the distances EE432: RF Engineering for Telecommunications Scott Hudson, Washington State University 08/25/08 Ground Reflections 3.2 r1 = r 2 + (ht − hr ) 2 (3.1) r2 = r 2 + (ht + hr ) 2 We’ll be interested in cases where r >> ht , hr . For example r might be hundreds or thousands of meters while the h’s are only a few or maybe a few tens of meters. Then the following approximations are good (recall that for small x, 1 + x ≈ 1 + x / 2 ): (ht − hr ) 2 2r (h + hr ) 2 r2 ≈ r + t 2r r1 ≈ r + (3.2) The path difference is ∆r = r2 − r1 ≈2 (3.3) ht hr r As we’d expect from Figure 3.1, this goes to zero at large distances. Theory The receiver will “see” two transmitters – the real transmitter above the ground and a virtual, or mirror image, transmitter below the ground. The intensity of each of these fields in the absence of the other would be given by the Friis equation (2.6). When both are present we cannot simply add the powers because it is the EM fields that add, and the fields have both amplitude and phase. The amplitude is proportional to the square root of the intensity and the phase is 2π times the distance traveled in wavelengths. The amplitude of the reflected field is also multiplied by Γ, the reflection coefficient of the ground. The total power is proportional to the magnitude squared of this total field, or PR = PT −j 2π r1 λ GT GR λ e ( 4 π) 2 r1 2 +Γ e −j 2π 2 r2 λ r2 (3.4) Note that if there is no ground reflection ( Γ = 0 ) we simply get (2.6), as we should expect. For large r, (3.2) tells us that r1 and r2 both approach r, so the phase terms in (3.4) will approach one another in both phase and amplitude. EE432: RF Engineering for Telecommunications Scott Hudson, Washington State University 08/25/08 Ground Reflections 3.3 If Γ = −1 , which would correspond to a perfectly conducting ground with a certain polarization, the two terms will tend to cancel each other and the received power will drop very rapidly. We have in this special case e −j 2π r1 λ r1 +Γ e −j 2π 2 r2 λ r2 2π 2π 1 − j r1 − j r2 ≈ 2 e λ −e λ r 2π − j ∆r 1 = 2 1− e λ r π 2 (3.5) π − j ∆r 1 j ∆r = 2 e λ −e λ r = 2 2 π∆r 4 sin 2 2 r λ so PR ≈ PT GT GR λ2 2πht hr sin 2 2 (2πr ) λr (3.6) The sine term oscillates between 0 and 1 while the “envelope” PT GT GR λ2 /(2πr ) 2 has the 1 / r 2 behavior of free-space propagation. The approximation sin 2 1 / x = 1 / x 2 is good to 1.5 dB for x ≥ 1 . So, for large enough r, we can write PR ≈ PT GT GR 2 ht hr r4 2 (3.7) Here the received power is falling off as 1 / r 4 , much more rapidly than it would in free space. It is also interesting to note that there is no wavelength dependence in this expression. The envelope of (3.6) and (3.7) are equal when r = 2πht hr / λ . If we neglect the sine oscillations in (3.6) we can create a composite propagation model as follows: PR = EE432: RF Engineering for Telecommunications λ2 ( 2 πr ) 2 2 2 ht hr PT GT GR 4 r PT GT GR 2 πht hr λ 2 πht hr ;r ≥ λ ;r < Scott Hudson, Washington State University (3.8) 08/25/08 Ground Reflections 3.4 This is an example of a breakpoint model, the breakpoint being r = 2πht hr / λ where the model changes from one type of behavior to another. If we use the breakpoint as the reference distance r0 then PR ,dBm r r0 = r P0,dBm − 40 log r0 P0,dBm − 20 log ; r < r0 ; r ≥ r0 r0 = 2πht hr / λ (3.9) P0,dBm = PT ,dBm + GT ,dB + GR ,dB + 20 log λ2 (2π) 2 ht hr (Keep in mind that everything we’ve done is valid only for the case Γ = −1 .) The important point here is that the presence of the ground can cause the field to decay very differently from the 1 / r 2 behavior of free space. It is instructive to ask what happens to (3.9) if we change the height of one or both of the antennas. First, notice that the model depends only on the product ht hr . Let’s say we change the heights so that the new value of this product is (ht hr )' = aht hr . For example, if we double the height of one of the antennas, then a = 2 . The new breakpoint will be r0 ' = ar0 . Since there is no change in the transmitted power or antenna gains P0′,dBm = P0,dBm − 20 log λ2 λ2 + 20 log (2π) 2 ht hr (2π) 2 aht hr (3.10) = P0,dBm − 20 log a If we are at a small distance where r < r0 , r0 ' , then the new model is PR′,dBm = P0,dBm − 20 log a − 20 log = P0,dBm − 20 log = PR ,dBm r r0 r ar0 (3.11) In other words, for distances less than either breakpoint, there is no change in the received power. This makes sense because (3.9) models the close-in field as being the same as in freespace. That is, for small distances the ground is assumed to have no effect, hence, the antenna heights shouldn’t have any effect. On the other hand, at a large distance r > r0 , r0 ' EE432: RF Engineering for Telecommunications Scott Hudson, Washington State University 08/25/08 Ground Reflections 3.5 PR′, dBm = P0, dBm − 20 log a − 40 log r ar0 = P0, dBm − 20 log a + 40 log a − 40 log = PR , dBm + 20 log a r r0 (3.12) and received power is changed by the constant amount 20 log a . For example, if you increase either antenna height by a factor of 2, then received power at large distances increases by 6 dB. Figure 3.2 shows an example in which ht hr is increased by a factor of 10. 0 20 Pr (dBm) 40 60 80 100 1 0.5 0 0.5 log(r/75m) 1 1.5 2 Figure 3.2: Effect of changing antenna height. Solid curve shows PR for rt = rr = 2m , GT = 10dB , GR = 3dB and PT = 1W . Dashed curve is the case where we change to rt = 20m . The effect is to extend the breakpoint to a larger distance. Simulation Figure 3.3 shows a numerical (Mathcad) calculation of (3.4) for the case Γ = −1 and ht = 10λ, hr = 3λ . Also shown are 1 / r 2 and 1 / r 4 behaviors characteristic of small and large r values. EE432: RF Engineering for Telecommunications Scott Hudson, Washington State University 08/25/08 Ground Reflections 3.6 0 relative Pwr (dB) 50 100 150 200 2 1 two ray n=2 n=4 0 1 2 3 log(r/r0) Figure 3.3: Ground reflection model (3.4) (solid curve), envelope of (3.6) (dotted line), and (3.7) (dashed line) for ht=10λ, hr=3λ. The breakpoint is at 188λ. When received power falls off as 1 / r n , we say that the propagation constant is n. For free space the propagation constant is n = 2 . At large distances the ground-reflection model with Γ = −1 has a propagation constant of n = 4 . Thus we see that it is quite possible, even in a relatively simple geometry, to get a propagation constant different than that of free space. Example 3.1 Assume a propagation model of the form (3.9). Suppose a cellular base station transmits 30dBm (1W) of power and has an antenna with gain of 7dB and height 10m. A roof-mounted car antenna is 1.5m high and has 3dB gain. The frequency is 1900MHz. How far away can the car be and still receive –90dBm of signal? The wavelength is 300/1900 or 0.158m. (3.9) then gives r0 = 597m, P0 = −47.5dBm. We then solve − 90 = −47.5 − 40 log(r / 597 m) to get r = 6894m. So the car can go about 7km from the base station. Experiment A 915MHz transmitter was placed 2.5ft above a sidewalk. Field strength measurements were made at various distances and at a height of 3ft. The observed data are show in Figure 3.4 along with (3.4) and the free-space model. A reference distance of r0 = 5 ft was used. The groundreflection model does capture the general behavior of the data. In Figure 3.5 we plot the observed EE432: RF Engineering for Telecommunications Scott Hudson, Washington State University 08/25/08 Ground Reflections 3.7 data along with 1 / r n models for n = 2,3,4 . This illustrates the breakpoint concept. The theoretical break point of 2πht hr / λ is about 48 feet in this case. Since log(48 / 5) = 0.98 we would expect a breakpoint near 1 on the horizontal axis. This is indeed what we observe. The data are fairly well described by n = 2 for small r and by n = 3 for large r. 30 40 Pr (dBm) 50 60 70 80 0 0.2 0.4 two-ray observed free-space 0.6 0.8 1 log(r/r0) 1.2 1.4 1.6 1.8 Figure 3.4: Ground reflection measurements above a sidewalk at 915MHz. Reference distance r0 is 5ft. Circles are observed data. Solid curve is eqn. (3.4) with a Γ of −0.6. Dashed curve is freespace model. EE432: RF Engineering for Telecommunications Scott Hudson, Washington State University 08/25/08 Ground Reflections 3.8 30 Pr(dBm) 40 50 60 70 80 0 0.2 0.4 n=2 n=3 n=4 observed 0.6 0.8 1 log(r/r0) 1.2 1.4 1.6 1.8 Figure 3.5: Data of Figure 3.2 compared to various 1 / r n models. n = 2 gives a descent description for small r while n = 3 gives a descent description of large r. References 1. Rappaport, T. S., Wireless Communications: Principles and Practice, Prentice Hall, 1996, ISBN 0-13-375536-3. 2. Ishimaru, A., Electromagnetic Wave Propagation, Radiation, and Scattering, Prentice Hall, 1996, ISBN 0-13-249053-6. EE432: RF Engineering for Telecommunications Scott Hudson, Washington State University 08/25/08

Tutor Answer

afzalahmad
School: Carnegie Mellon University

flag Report DMCA
Review

Anonymous
Totally impressed with results!! :-)

Similar Questions
Hot Questions
Related Tags

Brown University





1271 Tutors

California Institute of Technology




2131 Tutors

Carnegie Mellon University




982 Tutors

Columbia University





1256 Tutors

Dartmouth University





2113 Tutors

Emory University





2279 Tutors

Harvard University





599 Tutors

Massachusetts Institute of Technology



2319 Tutors

New York University





1645 Tutors

Notre Dam University





1911 Tutors

Oklahoma University





2122 Tutors

Pennsylvania State University





932 Tutors

Princeton University





1211 Tutors

Stanford University





983 Tutors

University of California





1282 Tutors

Oxford University





123 Tutors

Yale University





2325 Tutors