Intuitive Guide to Principles of Communications By Charan Langton

Transcription

1 Intuitive Guide to Priniples of Communiations By Charan Langton Understanding Frequeny Modulation (FM), Frequeny Shift Keying (FSK), Sunde s FSK and MSK and some more The proess of modulation onsists of mapping the information on to an eletromagneti medium (a arrier). This mapping an be digital or it an be analog. The modulation takes plae by varying the three parameters of the sinusoid arrier. 1. Map the info into amplitude hanges of the arrier 2. Map the info into hanges in the phase of the arrier 3. Map the info into hanges in the frequeny of the arrier. The first method is known as amplitude modulation. The seond and third are both a from of angle modulation, with seond known as phase and third as frequeny modulation. Let s start with a sinusoid arrier given by its general equation t () = Aos(2 π ft+ φ ) 0 This wave has an amplitude A a starting phase of fð 0 and the arrier frequeny, f. The arrier in Figure 1 has amplitude of 1 v, with f of 4 Hz and starting phase of 45 degrees. Generally when we refer to amplitude, we are talking about the maximum amplitude, but amplitude also means any instantaneous amplitude at any time t, and so it is really a variable quantity depending on where you speify it. Figure 1 A sinusoid arrier of frequeny 4, starting phase of 45 degrees and amplitude of 1 volt. The amplitude modulation hanges the amplitude (instantaneous and maximum) in response to the information. Take the following two signals; one is analog and the other digital

2 . a. Analog message signal, m(t) b. Binary message signal, m(t) Fig 2 Two arbitrary message signals, m(t) About Amplitude Modulation (AM) The amplitude modulated wave is reated by multiplying the amplitude of a sinusoid arrier with the message signal. st () = mt () t () = A mt ()os(2 π ft+ φ ) 0 The modulated signal shown in Figure 3, is of arrier frequeny f but now the amplitude hanges in response to the information. We an see the analog information signal as the envelope of the modulated signal. Same is true for the digital signal. a. Amplitude modulated arrier shown with analog message signal as its envelope

3 . Amplitude modulated arrier shown with binary message signal as its envelope Figure 3 Amplitude modulated arrier, a. analog, b. digital Let s look at the argument of the arrier. What is it? The argument is an angle in radians. The argument of a osine funtion is always an angle we know that from our first lass in trigonometry. The seond term is what is generally alled the phase. In amplitude modulation only the amplitude of the arrier hanges as we an see above for both binary and analog messages. Phase and frequeny retain their initial values. Any modulation method that hanges the angle instead of the amplitude is alled angle modulation. The angle onsists of two parts, the phase and the frequeny part. The modulation that hanges the phase part is alled phase modulation (PM) and one that hanges the frequeny part is alled frequeny modulation (FM). How do you define frequeny? Frequeny is the number of 2 pð revolutions over a ertain time period. Mathematially, we an write the expression for average frequeny as f t φi( t+ t) φi( t) = 2π t 1 This equation says; the average frequeny is equal to the differene in the phase at time t + Dðt and time t, divided by 2pð (or 360 degrees if we are dealing in Hz.) Example: a signal hanges phase from 45 to 2700 degrees over 0.1 seond. What is its average frequeny?

4 = = Hz 360* 0.1 This is the average frequeny over time period t = 0.1 ses. Perhaps it will be different over 0.2 ses or some other time period or maybe not, we don t know. What is the instantaneous frequeny of this signal at any partiular moment in the 0.1 seond period? We don t really know given this information. The instantaneous frequeny is defined as the limit of the average frequeny as Dðt gets smaller and smaller and approahes 0. So we take limit of equation 1 to reate an expression for instantaneous frequeny, f i (t),. The f i (t) is the limit of f Dðt (t) as Dðt goes to 0. The phase hange over time Dðt is hanged to a differential to indiate hange from disrete to ontinuous. f () t = lim f () t i t 0 t φi( t+ t) φi( t) = lim t 0 2π t 1 dφi () t = 2π dt This last result is very important in developing understanding of both phase and frequeny modulation! The 2pð fator has been moved up front. The remaining is just the differential of the phase. Another way we an state this is by reognizing that radial frequeny wð is equal to ω = 2 π f ( t) i It is also equal to the rate of hange of phase, d( φi ( t)) ω = dt so again we get, 1 dφi () t ft () t = 2 2π dt Intuitively, it says; the frequeny of a signal is equal to its phase hange over time. When seen as a phasor, the signal phasor rotates in response to phase hange. The faster it spins (phase hange), the higher its frequeny. What does it mean, if I say: the phasor rotates for one yle and then hanges diretions, goes the oppo-

5 site way for one yle and then hanges diretion again? This is a representation of a phase modulation. Changing diretions means the signal has hanged its phase by 180 deg. We an do a simple minded phase modulation this way. Go N spins in lokwise diretions in response to a 1 and N spins in ounterlokwise diretions in response to a 0. Here N represents frequeny of the phasor. Figure 4 The arrier as a phasor, the faster it spins, the higher the frequeny. If frequeny is the rate of hange of phase, then what is phase in terms of frequeny? As we know from definition of frequeny that it is number of full 2 pð rotations in a time period. Given, a signal has traveled for 0.3 seonds, at a frequeny of 10 Hz with a starting phase of 0, what is its phase now? Phase now φ = 2π f t = 10 Hz 2 π.3 = 20 radians i This is an integration of the total number of radians overed by the signal in 0.3 ses. Now we write this as an integral,.3 φi() t = 2 π fi() t dt 0 and sine this is average frequeny, it is onstant over this time period, we get Phase now φ = 2π f t = 10 Hz 2 π.3 = 20 radians We note the phase and frequeny are related by Phase a Integral of frequeny Frequeny a Differential of phase i Phase modulation Let the phase be variable. Going bak to the original equation of the arrier, hange the phase (the underlined term only) from a onstant to a funtion of time. ( t) = A os( 2 π f t + φ( t) ) 3

6 We an phase modulate this arrier by hanging the phase in response to the message signal. φ () t = k m() t i p Now we an write the equation for the arrier with a hanging phase as st () = Aos(2 π ft + k mt ()) 4 p The fator k p is alled the phase sensitivity fator or the modulation index of the message signal. For analog modulation, this expression is alled the phase modulation. In phasor representation of an analog PM, the phasor slows down gradually to full stop and then again piking up speed in the opposite diretion gradually. In binary ase, the phasor does not slow down gradually but stops abruptly. This is easy to imagine. Take a look at equation 4. Replae the underlined terms by a 0 or a 1, or better yet, replae it by 180 oð if m(t) is 1 and -180 oð if it is a 0. Now, we have a binary PSK signal. The phase hanges from -180 oð to +180 o in response to a message bit. This is the main differene between analog and digital phase modulation in that in digital, the phase hanges are disrete and in analog, they are gradual and not obvious. For binary PSK, phase hange is mapped very simply as two disrete values of phase. st () = aos(2 π ft+ φ ) φ = 0 or π j j a. Phase modulation is response to a binary message. The phase hanges are abrupt. b. Phase modulation is response to an analog message. The phase hanges are smooth. Figure 5 Phase modulated arrier for both binary and analog messages.

7 Frequeny modulation FM is a variation of angle modulation where instead of phase, we hange the frequeny of the arrier in response to the message signal. Vary the frequeny by adding a time varying omponent to the arrier frequeny. f () t = f + k m() t t f where f is the frequeny of the unmodulated arrier, and k f a saling fator, and m(t), the message signal. The term k f m(t) an alled a deviation from the arrier frequeny. Example, a arrier with f = 100, k f = 8 and message bit rate = 1. Assume the message signal is polar so we have 1 for a 1 and -1 for 0. For m(t) = 1, we get f 1 = (1) = 108 Hz for m(t) = -1 f 2 = 100-8(1) = 92 Hz The phasor rotates at a frequeny of 108 Hz, as long as it has a signal of 1 and at 92 Hz for a signal indiating a 0 bit. Remember the equation relating phase and frequeny 1 dθi () t fi () t = 5 2π dt whih an also be written as t φi() t = 2 π fi() t dt 6 Now look at the arrier equation, st () = Aos(2 π ft+ φ ) 7 i We define the instantaneous frequeny of this signal as the sum of a onstant part, whih is the arrier frequeny, and a hanging part as show below. f () t = f + k m() t 8 t f The argument of the arrier is an angle. So we need to onvert this frequeny term to an angle. Do that by

8 taking the integral of this expression (invoke equation 6) t θ () t = 2π f dt + 2 πk m() t dt i f and sine f is a onstant, this is just equal to t t θi() t = 2π ft + 2 πkf m() t dt 9 Now we plug this as the argument of the arrier into equation 7 and we get st () = Aos(2π ft + 2 πkf mtdt () ) 10 t This represent the equation of a FM modulated signal. A deidedly unpleasant looking equation! This is about as far as you an get from intuitive. There is no resemblane at all to anything we an imagine. Analog FM is one of the more omplex ideas in ommuniations. It is very diffiult to develop an intuitive feeling for it. Bessel funtions rear their multi-heads here and things go from bad to worse. Thankfully this is where binary signals ome to the resue. The binary or digital form of Frequeny Modulation is very easy to understand. But before we do that, examine the following relationship between FM and PM. Let s write out both equations side by side so we an see what s going on. PM: st () = Aos(2 π ft + k mt ()) 11 p FM: st () = Aos(2π ft + 2 πkf mtdt () ) 12 t Note that in FM, we integrate the message signal before modulating. Both FM and PM modulated signals are oneptually idential, the only differene being in the first ase phase is modulated diretly by the message signal and in FM ase, the message signal is first integrated and then used in plae of the phase. Using a Phase modulator we an reate a FM signal by just integrating the message signal first. Similarly with a FM modulator we an reate a PM signal by differentiating the message signal before modulation.

9 Differentiator Derivative of m(t) Phase Modulator A FM signal m(t) Integrator Integral of m(t) Frequeny Modulator A PM signal Figure 6 Phase and Frequeny modulators are interhangeable by just hanging the form of the message signal. Differentiate the message signal and then feed to a PM modulator to get a FM signal and vie-versa for FM. Frequeny Shift Keying the digital form of FM The binary version of FM is alled the Frequeny Shift Keying or FSK. Here the frequeny does not keep hanging gradually over symbol time but hanges in disrete amounts in response to a message similar to binary PSK where phase hange is disrete. The modulated signal an be written very simply as onsisting of two different arriers. s () t = A os(2 π f t) 1 1 s () t = A os(2 π f t) 2 2 s 1 (t) in response to a 1 and s 2 (t) in response to a 0. Getting sophistiated, this an be written as a deviation from the arrier frequeny. s () t = A os(2 π ( f f)) t 1 s () t = A os(2 π ( f + f)) t 2 Here Dðf is alled the frequeny deviation. This is exursion of the signal above and below the arrier frequeny and indiates the quality of the signal suh in stereo FM reeption. Eah of these two frequenies f 1 and f 2, are an offset from the arrier frequeny, f. Let s all the higher of these f h and lower f l,. Now we an reate a very simple FM modulator as shown the Figure 7. We an use this table to modulate the inoming message signal. Amplitude of m(t) f h f l We send f l in response to a -1 and 0 in response to a +1. When we get a hange from a -1 to +1, we

10 swith frequeny, so that only one signal, either fh or fl is transmitted. The ombined signal an be written as, keeping mind that two terms are orthogonal in time and never our at the same time. s ( t) = A os(2 π f t+ φ ) + Aos(2 π f t+ φ ) 13 BFSK h h h l l l Figure 7 A FSK modulator In the modulator shown here, we are hanging the amplitude A l and A h whih are either 0 or 1 in response to the message signal, but never present at the same time. The above proess looks very muh like amplitude modulation. The only differene is that the signal polarity goes from 0 to 1(beause A h is either 0 or 1), rather than -1 to +1 as it does in BPSK. This is an important differene and we an try to understand FM better by making this lever substitution (Ref. 1). 1 1 A t = + A t ' Al() t = + Al() t 2 2 ' h() h() 14 These transformations turn A h and A l from 0,1 to -1, +1. Now we substitute these into equation 13 and get, s ( t) = os(2 π f t+ φ ) + os(2 π ft+ φ ) BFSK h h l l + A os(2 π f t+ φ ) + A os(2 π ft+ φ ) ' ' h h h l l l 15 The first two terms are just osines of frequenies f h and f l, so their spetrum ontribution is an impulse at frequenies f h and f l. The seond two terms are amplitude modulated arriers. These last two terms give rise to a sin x/x type of spetrum of the square pulse as shown below.

11 Figure 8 Sin x/ x spetrum of a square signal pulse The basi idea behind a FM spetrum is that it is a superimposition of impulses and sin x/x spetrum of the pulses. However, this is true only if the separation or deviation is quite large and the overlap is small. The ombinations are not linear and not easily predited unless of ourse you use a program like SPW, Matlab or Mathad as I do. The following figures show the spetrum of an FSK ase, where the arrier frequeny is 10, the message signal frequeny is 1, Dðf is equal 2. a. Binary message signal, m(t) = 2 b. Modulated signal, s(t) with f = 4

12 Figure 9- Spetrum of the modulated signal, Dðf is equal 2. Note the two impulses at frequenies 2 and 6 Hz. Here we see learly that there are two impulses representing the arrier frequenies, 2 and 6. If you ignore, these, the remaining are just two sin x/ x spetrums entered at frequenies 2 and 6, added together. We define a term alled the modulation index of a FM signal. m f f = 16 m where Dðf is what is alled the frequeny deviation, and f m is the frequeny of the message signal. f = f f This fator m, determines the oupied bandwidth and is a measure of the bandwidth of the signal. The FM broadasting in the US takes plae at arrier frequenies of app. 50 MHz. The message signal whih is musi, has frequeny ontent up to about 15 khz. The FCC allows deviation of 75 khz. For this, the m is f 75 m = = = 15 f 5 m m, is independent of the arrier frequeny and depends only on the message signal frequeny and the allowed deviation frequeny. This fator is not onstant for a partiular signal as is the AM modulation index. In FM broadasting m an go very high, sine the message signal has some pretty low frequenies, suh as 50 Hz. For FM it an vary any where from 1500 to 5 on the low end. FM signal is lassified in two ategories, Signals with m << 1 are alled Narrowband FM Signals with m >> 1 are alled Wideband FM

13 FM radio is an obviously a wideband signal by this definition. The smaller the deviation, the loser the main lobes lie and in fat overlap and larger the deviation, the further out the signal spreads. The following figures show the effet of inreasing modulation index on the signal spetrum and bandwidth. a. Binary message signal, m(t) = 1 b. Modulated signal f = 4, Dðf = 5. Spetrum shows two impulses at f; = 9 and 4-5 = -1

14 d. Modulated signal, f = 4, Dðf = 10 e. Spetrum shows two impulses at f; = 14 and 4-10 = -6 Note that the spetrum does look like a superposition of impulses and sin funtions. f. Modulated signal, f = 4, Dðf = 1

15 g. Spetrum shows two impulses at f; = 5 and 4-1 = 3 h. Modulated signal, f = 4, Dðf =.4 i. Spetrum shows two impulses at f; = 4.1 and = 3.9 Figure 10 Relationship of FM modulation index and its spetrum. As index gets large, the signal bandwidth inreases. In these figures, we an define the bandwidth as the spae between the main lobes or impulses. In fat as k f gets very small, the general rules of identifying the impulses do not apply and we have to

16 know about Bessel funtions in order to ompute the spetrum. There are two very speial ases of FM that bear disussing. Sundes FSK Sunde s FSK is a speial ase of the general form we have desribed above. Here the frequeny spaing of the two arriers, f h and f l is exatly equal to the symbol rate. or Dðf = f m m = 1 Two tones appear at frequenies whih are exatly the symbol rate and these help in the demodulation of this signal without external timing information. When you an extrat timing information from a signal, the detetion task is alled oherent and generally has better BER performane. a. Message signal with bit rate = 1.0 b. Modulated signal, f = 8, deviation = 1.0 k f = 1.0

17 Figure 11 Sunde s FSK spetrum with tones at = 9 and 7 Hz. The spetrum of Sunde s FSK is given by T s os( π ft s S( f) = δ( f.5 ) + δ( f +.5 ) Ts Ts π 4f Ts 1 17 The first two underlined terms are the two impulses (batman s ears in above spetrum) and the last term is the spetrum of the message signal, whih is not quite sin x/x as I had said earlier. Minimum Shift Keying - MSK Minimum Shift Keying, another speial from of FSK (also alled Continuous phase FSK or CPSFK) and a very important one. It is used widely in ellular systems.. MSK is also a speial from of PSK owing to the equivalene of phase and frequeny modulation whih we will disuss in more detail again. For now, let it suffie to say that MSK is a speial ase where or Dðf =.5 f m m = 0.5 The MSK spetrum has no disrete omponents unlike all the other FSKs inluding Sunde s above. The spetrum is just a bit wider than QPSK but its side lobes fall off muh more quikly than QPSK In wireless systems, it often offers better properties than QPSK due to its onstant envelope harateristis, as we an see in the omparison above.

18 a. MSK modulated signal, m(t) = 2.0, Dðf = 1 MSK spetrums does look a lot like a QPSK spetrum but is not the same. Figure 12 MSK signal and spetrum. Note that it has no Batman ears as in all other FSKs. The spetrum of MSK is given by 2 16T s os( 2π ft s S( f) = π 16 f Ts 1 18 (We will disuss MSK again from PSK point of view in the next tutorial.) M-ary FSK A M-ary FSK is just an extension of BFSK. Instead of two arriers, we have M. These arriers an be orthogonal or not, but the orthogonal ase would obviously give better BER. M-ary FSK requires onsiderably larger bandwidth than M-PSK but as M inreases, the BER goes down unlike M-PSK. Infat if number of frequenies are inreased, M-FSK beomes an OFDM like modulation.

19 Analog FM Response to a sinusoid not as simple as we would like I have side-skirted the issue of analog FM s ompliated spetrum. Binary or digital FSK allows us to delve into FM and is easy to understand but a majority of FM transmission suh for radio is analog. So we will touh on that aspet to give you the full understanding. Now we will look at what happens to a message signal that is a single sinusoid of frequeny f m when it is FM modulated. t st () = Aos(2π ft + 2 πkf mtdt () ) 19 Let s take a message signal as shown below. Figure 13 Just a lowly sinusoid that is about to be mangled to high-fidelity heights by a FM modulator. st ( ) = Aos(2π ft+ 2πk sin f t) f m Now set A = 1 and swith to radial frequeny for onveniene ( less typing.) st () = os( ω t+ k sin ω t) f m By the well-known trigonometri relationship, this equation is beomes st ( ) = os( ω t+ k sin ω t) f m = osω t os( k sin ω t) sinω tsin( k sin ω t) 20 f m f m Now the terms os( k sin ω t) an be expanded. How and why, we leave it to the mathematiians, and see that we again get a pretty hairy looking result. f m os( k sin ω t) = J ( k ) + 2 J ( k )os(2 ω t) + 2 J ( k )os(4ω 21 f m 0 f 2 f m 4 f

20 The sin( k sin ω t) is similarly written as f m sin( k sin ω t) = 2 J ( k )sin( ω t) + 2 J ( k )sin(3 ω t) + 2 J ( k f m 1 f m 3 f m 5 22 (Note k f = m) The underlined funtions are Bessell funtions and are funtion of the modulation index m or k f. They are seen many plaes where harmoni signals analysis and despite my bad-mouthing, they are quite harmless and benign. There are just so many of them. They ones here are of first kind and of order n. The terms of the osine expansion sine it is an even funtion, ontain only even harmonis of w m and sin expansion has only odd harmonis and this is obvious if you will examine the above equations even asually. Figure 14 shows the general shape of 4 Bessel funtions as the order is inreased. The x-axis is modulation index k f and the y-axis is the value of the Bessel funtion value. For example, for k f = 2, the values of the various Bessel funtions are J0 =.25 J1 =.65 J2 =.25 J3 =.18 The FM arriers take their amplitude values from these funtions as we will see below. Figure 14 Bessel funtion of the first kind of order n, shown for order = 0, 1, 2 and 4 X-axis is read as the modulation index k f, and y-axis is the value of the amplitude of the assoiated harmoni. Exept for the 0 order funtion, all others start at 0.0 and damp down with yle. Below we see a plot of what the os and sin expansions (equations 22, 21) ontaining the various Bessel funtions look like for just four terms.

21 expfos( 5, t ) t Figure 15 Plot of os( k sin t) = J0( k ) + 2 J2( k ) os(2 t) + 2 J4( k )os(4 ω ω ω f m f f m f expfsin( 5, t ) t Figure 16 Plot of sin( k sin ω t) = 2 J1( k )sin( ω t) + 2 J3( k )sin(3 ω t) + 2 J5( k f m f m f m Now we put it all together (plug equations 21, 22 into equation 20) And by trigonometri magi get, an expression for the FM modulated signal. st () = J( k )os( ω t) [ ω ω ω ω ] [ ω ω ω ω ] [ ω ω ω ω ] J ( k ) os( ) t os( + ) t + J ( k ) os( 2 ) t os( + 2 ) t J ( k ) os( 3 ) t os( + 3 ) t +... f f m m f m m f m m Remember this is a response to just one single solitary sinusoid. This signal ontains the arrier with

22 amplitude set by Bessel funtion of 0 order (the first underlined term), and sideband on eah side of the arrier at harmonially related separations of ωm,2 ωm,3 ωm,4 ω m,.... This is very different from AM, in that we know that in AM, a single sinusoid would give rise to just two sidebands (FFT of the sinusoid) on eah side of the arrier. So many sidebands, in fat an infinite number of them, for just one sinusoid! Eah of these omponents has a Bessel funtion as its amplitude value. For ertain values of k f, we an see that J0 funtion an be 0. In suh ase there would no arrier at all (MSK) and all power is in sidebands. For the ase of k f = 0, whih also means that there is no modulation, all power is in the arrier and all sideband Bessel funtion values are zero. FM is alled a onstant envelope modulation. As we know the power of a signal is funtion of its amplitude only. We note that the total power of the signal is onstant and not a funtion of the frequeny. For FM, the power gets distributed amongst the sidebands, the total power always remains equal to the square of the amplitude and is onstant.. Here we plot the above signal for k f = 2 and k f = 1. Figure 17 Modulated FM signal for k f = 2 and k f = 1, the red signal is moving faster indiated higher frequeny omponents. Examples of FM spetrums to a sinusoid. In the following signals, the only thing we are hanging is the modulation index k f. This inreases the frequeny separation (or deviation and inreases the bandwidth.)

23 k f = 0.5 k f = 1 k f = 2 k f = 4 Figure 18 Time domain FM signals for various k f values.

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25 Figure 19 Spetrum of FM signals for various k f values. Bandwidth of a FM signal The bandwidth of an FM signal is kind of slippery thing. Carson s rule states that is bandwidth of a FM signal is equal to Here Bandwidth. = 2( f + f ) est m f m 2 = T s or is equal to the baseband symbol rate, Rs. If the frequenies f1 and f2 hosen satisfy the following equation, the ross-orrelation between the arriers is zero, then this is an orthogonal set. T0 ρ = os(2 π ft) os(2 π ft) dt= The output of I is maximum just when output of Q hannel is zero. The deision at the reeiver is a simple matter of determining if the voltage is present. The BER of a BFSK system is given by E s Pe = 1 erf 2 2 N 0 If however, if the ross-orrelation is not zero, T0 ρ = os(2 π ft) os(2 π ft) dt then energy is present on both hannels at the same time and in presene of noise, symbol deision may be flawed. Expanding above equation, we get, sin( π( f2 f1) Ts) os( π( f2 f1) Ts) ρ = π ( f f ) T 2 1 s

26 Now set 4 f ( f2 f1) Ts = = 4k f m f sin(4 πkf ) os(4 πkf) ρ = 4π k f Figure 20 Cross-orrelation between I and Q FSK signals. The zero rossing points are the preferable points of operation. Plotting this equation against the 2 times modulation index k f, we get Figure 20. The y-axis whih is orrelation between I and Q hannels is 0 at speifi values of k f. These values are 1, 2, 3 and 4 and so on. The point with the lowest orrelation (negative value is onsidered the most effiient plae to operate an FSK. At first zero rossing, we note that the symbols are loated only half a symbol apart so, oherent detetion is not possible at this point. The next zero rossing, where kf = 1, is the Sunde s FSK. Coherent detetion is possible at this point beause symbols are at least ((f2-f1)ts = 1) one symbol apart. The 4 th zero rossing is MSK. Bandwidth of Sunde s FSK is given by 2 3 B= ( f2 f1) + = Hz T T s s And for MSK, 4 5 B= ( f2 f1) + = Hz T T s s

27 Sunde s FSK gives us the most spetrally effiient form of FSK that an be deteted inoherently, whereas MSK gives us a spetrum that is a lot like QPSK but rolls-off muh faster. Questions, orretions? Please ontat me, Charan Langton [email protected] Copyright Feb 2002, All rights reserved