Language/OS - Multiplatform Resource Library

home *** CD-ROM | disk | FTP | other *** search

/ Language/OS - Multiplatform Resource Library / LANGUAGE OS.iso / dsp / dspgroup / 00index.arc / CISDSP.ARC / CREST.TXT < prev next >

Wrap

Text File | 1988-07-28 | 12.9 KB | 263 lines

Crest Factor Evaluation of Multitone Waveforms under SSB Modulation Franklin Antonio, N6NKF, 4/19/88 5am edition Summary: Crest Factor of a multitone waveform under SSB modulation is discussed. Equations for crest factor of the transmitted waveform are developed. Crest Factor causes a loss in transmit power, which is calculated as Waveform Xmit Power Loss -------- --------------- 1-tone 0 db 2-tones 3 db (any phases, any frequencies) 3-tones depends on frequencies and phases optimum unknown. Probably >2 db Discussion: Barry McLarnon has suggested the use of (n,m)-ary FSK modulation for an amateur radio HF modem. During each modulation interval, the modulator would generate n out of m possible audio frequency sinusoids add them, and send them into the microphone input of an HF SSB transceiver. An important characteristic of such multi-tone waveforms is their "crest factor". This is the ratio of the waveform peak value to the waveform's RMS value. Simple sine waves have a crest factor of sqrt(2), or 3db. The power amplifier in the HF transceiver is voltage limited, and we must reduce the power level until the amplifer can handle the peaks. (or at least nearly so) For example, if a PA is rated at 100W PEP (Peak Envelope Power), and we feed it a waveform with a crest factor = CF, then we must reduce average power by 2/(CF^2). If CF = 10, we must reduce our 100W rig to 2W, a 17db loss! There has been some discussion of designing multi-tone waveforms specifically so that they will have a low crest factor. One example in the literature is "Multitone Signals with Low Crest Factor", by Stephen Boyd, IEEE Ckts & Syst, Oct 1986, pp1018-1022, and "Comments on Multitone...", Ouderaa, Schoukens, Renneboog, IEEE C&S, Sept 1987, pp1125-1127. They looked at waveforms of the form.. s(t) = cos(W*t+P1) + cos(2*W*t+P2) + cos(3*W*t+P3) + ... + cos(n*W+Pn) (1) They then describe methods for choosing the phase offsets P1...Pn to minimize (or nearly minimize) the crest factor of s(t). These guys have great sounding names, and they solve an interesting problem, but it isn't our problem. Important is the fact that we really care about the crest factor of the waveform AFTER SSB MODULATION. (because that's where the power amp lives) SSB modulation changes the crest factor of a waveform. To understand why, we have to do some math, which follows. Boyd, Ouderra, etc (BOSR) define crest-factor in a manner which doesn't directly help us, because they didn't deal with SSB modulation, yet we must. You cannot derive the crest-factor after SSB modulation by knowing only the crest-factor before modulation. If you minimize the crest factor of a signal by varying some parameter of the signal (such as the relative phases of two sinusoidal components) then you haven't necessarily minimized the crest factor after SSB modulation. Gobldegook, but it gets clearer later... What is SSB modulation anyway? It's really just a frequency shift, but that's easier to say than to write mathematically. Consider a signal s(t). Now what does that turn into when it's SSB modulated? ssb(s(t)) = cos(Wc*t+Pc)*s(t) - sin(Wc*t+Pc)*H(s(t)) (2) where, Wc and Pc are the frequency and phase of the local oscillator, and H() represents a Hilbert Transform. The Hilbert Transform of a signal is simply the signal with every frequency component phase shifted 90 degrees. Now let's try it with an example signal. How about a square wave? [Square wave example was Barry's idea] A square wave has the marvelous property that it's crest factor = 1 !!! Can't do better than that. Would it be a reasonable waveform to stick into an SSB modulator? Remember a description of a square wave from some class on Fourier theory... s(t) = sin(t) + (1/3)sin(3t) + (1/5)sin(5t) + ... (3) Because it's represented here as a sum of sin()'s, we immediately know how to compute it's Hilbert transform too. Shift a sin() 90 degrees, and you just get a -cos().. H(s(t)) = - cos(t) - (1/3)cos(3t) - (1/5)cos(5t) - ... (4) Now note that while this example s(t) is very well behaved (ie CF=1), it's Hilbert Transform is very badly behaved. It's crest factor is infinite! In particular, at t=0, we have - 1 - (1/3) - (1/5) - ... which is a series that doesn't converge, ie is infinite. (minus infinite actually) Right here we have a hint that something that wasn't immediately obvious is going on. Substituting (3) and (4) into (2)... ssb(s(t)) = cos(Wc*t+Pc) * [ sin(t)+(1/3)sin(3t)+(1/5)sin(5t)... ] - sin(Wc*t+Pc) * [ cos(t)+(1/3)cos(3t)+(1/5)cos(5t)... ] (5) We can decide what waveform we generate and send to the SSB modulator, and might, in some systems, be able to control the LO frequency Wc, but typically we will be entirely unable to control the LO phase Pc. If we could choose Wc and Pc precisely, then we could make the large peaks of the cos(t)+... term occur when sin(Wc*t+Pc) was exactly zero. We get to choose the signal we generate, but the devil can choose Pc, so to "rotate" our waveform thru any phase angle he chooses. In the example, this means that ssb(s(t)) can take on very large (infinite actually) values. All the devil has to do is choose Pc to make Wc*t+Pc be nonzero at time t=0, and the resulting ssb(t) blows up. Actually, because the frequency of the carrier, Wc, is large (maybe 14 MHz) relative to the frequency components of the input signal (audio frequencies), the carrier is whipping around so fast (ie Wc*t+Pc is changing so rapidly) that the carrier samples all the good and bad parts of both s(t) and H(s(t)). BOSR only deals with the case where Wc=0, Pc=0, hence they ignore H(s(t)). We took a waveform with Crest Factor = 1, and put it thru an SSB modulator. What came out had Crest Factor = Infinity. This is not good. Now it turns out that luckily most waveforms are more well behaved than square waves under SSB modulation. Most waveforms don't blow up like this, but the lesson is that we must look for the maximum value of the crest factor of ssb(t) under all possible local oscillator phases. Lets define a new measure, call it.. Peak/Avg Envelope Ratio (PAR): PAR(s(t)) = MAX [ CrestFactor( ssb(s(t)) ) ] / sqrt(2) (6) maximum taken over time and all possible values of Pc. PAR() is the measure we should care about. The sqrt(2) divisor makes PAR=1 (ie 0db) for a single sine wave. When we look at other signals, PAR tells us how far we have to turn down the signal power at the power-amplifer RELATIVE to the power level we could handle for a single sine wave. A simpler definition of PAR() follows. The math will get easier, and the sqrt(2) will go away. Fortunately, we don't have to compute ssb(s(t)) for all possible Pc to figure out PAR(s(t)). We can do it algebraically. I prefer to do the algebra with complex exponentials instead of sin()'s and cos()'s. Sometimes it's simpler. We use, exp(j*(W1*t+P1)) (7) to represent a sinusoid. Here, j = sqrt(-1). This seems gnarly to those who aren't used to it. Remember in the above calculations, i had to evaluate the Hilbert Transform of signals. You may think about the complex exponential as mathematical trick that carries the Hilbert Transform around with every signal, so that i don't have to calculate it as i go. This is sometimes called the complex signal model. Such complex signals are also sometimes called phasors. I will try to use capital leters for phasors, and lowercase for scalar (ie ordinary) signals. Also, re[], and im[] mean the real and imaginary part of a phasor. For example, re[ Exp(j*W1*t+P1) ] = cos(W1*t+P1) (8) im[ Exp(j*W1*t+P1) ] = sin(W1*t+P1) In this notation, the formula for SSB modulation is simpler than before... ssb(S(t)) = re[ exp(j*(Wc*t+Pc)) * S(t) ] (9) SSB modulation simply means multiply the signal phasor by the carrier phasor, then take the real part. We can also now redefine Peak-to-Average Envelope Ratio now in phasor terms. It becomes... PAR(S(t)) = MAX [ Mag(S(t)) ] / RMS[ Mag(S(t)) ] (10) Now i only need take the MAX over time, whereas before the MAX was also over all possible LO phases. The phasors carry around with them all the info i need to evaluate all phases simultaniously. This definition is equivalent to the the previous one. Phasors on stun! ... Let's immediately jump to some useful waveforms, ie (n,m)-ary FSK, for which each waveform of interest is a sum of n equal amplitude sinusoids of specified phase, and compute PAR. Start with (2,m)-ary FSK. If i generate a signal by adding two equal amplitude tones, can i cleverly pick the relative phases of the two tones to minimize the Peak-to-Average Envelope ratio? Here's such a two-tone signal, represented as a phasor. The sqrt(2) normalizes S(t) so that it's RMS value is 1. S(t) = (1/sqrt(2)) * [ Exp(j*W1*t+P1) + Exp(j*W2*t+P2) ] (11) Now, using the following identities... Exp(j*x) = cos(x) + j*sin(x) sin(a) + sin(b) = 2 * sin((1/2)*(a+b)) * cos((1/2)*(a-b)) cos(a) + cos(b) = 2 * cos((1/2)*(a+b)) * cos((1/2)*(a-b)) we can put S(t) into a more form which makes it's properties more intuitive.. S(t) = sqrt(2) * cos[ ((W1-W2)*t+(P1-P2))/2] * Exp[j*((W1+W2)*t+(P1+P2))/2] (12) Ok, so at first it doesn't look very intuitive. Ignore the stuff inside the parenthesis for a moment. The Exp[] is a complex exponential which always has magnitude = 1. The cos[] has it's maximum value = 1. So, by immediate observation, we see that PAR(S(t)) = sqrt(2) (13) You can choose the relative phases of the two sinusoids, P1 & P2, but you won't affect the Peak-to-Average Envelope! It's always sqrt(2). Ie you have to turn down the power by 3db to xmit the sum of two sinusoids. How about that for a no-hope answer! Interestingly, it doesn't generalize... What happens for a sum of 3 sinusoids? (ie for (3,m)-ary FSK) S(t) = (1/sqrt(3)) * [ Exp(j*W1*t+P1) + Exp(j*W2*t+P2) + Exp(j*W3*t+P3) ] (14) The trig identity gods aren't with us this time, so the thing doesn't pop out magically into something obvious. (at least not under my hand) But we can get an interesting result. We want to put this thing into the form S(t) = a(t) * Exp( b(t) ) (15) where a(t) and b(t) are real, just as before. Then we'll look at the maximum of a(t), and ignore b(t). I'll spare you the intermediate trig doodles... a(t) = sqrt[ 1 + (2/3)*cos((W1-W2)*t+(P1-P2)) (16) + (2/3)*cos((W2-W3)*t+(P2-P3)) + (2/3)*cos((W3-W1)*t+(P3-P1)) ] Now, because we started with the goal of doing (3,m)-ary FSK, we know the frequencies of the tones we want to generate, so W1,W2,W3 are known and fixed. We now get to pick the relative phases RP1=P1-P2, and RP2=P2-P3. For the case W1,W2,W3 = 1,2,3, the phases RP1=0, RP2=pi produce PAR(S(t))=1.29, (ie 2.2 db) which is probably near optimum for this choice of frequencies. Other frequency triplets will have other optimum phases. So for every set of (W1,W2,W3) in the (n,m)-ary FSK, we can optimize RP1,RP2 to minimize PAR(S(t)). Because minimizing sqrt(x) is the same as minimizing x, we can drop the sqrt[] function and minimize what's inside. Interestingly, this looks a lot like what BOSR were doing, but it isn't the same. In the case of a 3-tone signal, they chose phases to minimize abs [ cos(W1*t+P1) + cos(W2*t+P2) + cos(W3*t+P3) ] (17) and instead, we choose phases to minimize 1 + (2/3)*cos((W1-W2)*t+(P1-P2)) + (2/3)*cos((W2-W3)*t+(P2-P3)) + (2/3)*cos((W3-W1)*t+(P3-P1)) (18) which, amazingly, is a nearly equivalent form. Unfortunately, the phase values that minimize one do not necessarily minimize the other. The equations above for the 3-tone case can be easily generalized for the n-tone case. I haven't been able to solve for the optimal values of PAR or the phases producing same directly, but a computer program could be easily written to evaluate (18) for every frequency triple (W1,W2,W3), trying phases RP1,RP2 each in [0,2pi], on a 64x64 grid, for each calculation varying t in [0,lcm periods 1/W1,1/W2,1/W3]. Franklin Antonio, N6NKF