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Determine the universal minimum distance between adjacent signal points, dmin.i = d for all i, and compute the average symbol energy; Es, in terms of a.b) Assuming ...

Question

Determine the universal minimum distance between adjacent signal points, dmin.i = d for all i, and compute the average symbol energy; Es, in terms of a.b) Assuming that nearest neighbor symbol errors dominate , show that Ps for the M-ary decision problem can be upper bounded byPs < 4QRNExpress the upper bound on Ps in terms of the average Eb No and compute the value needed to achieve Ps 10-5 . How does this compare with binary antipodal signaling?

Determine the universal minimum distance between adjacent signal points, dmin.i = d for all i, and compute the average symbol energy; Es, in terms of a. b) Assuming that nearest neighbor symbol errors dominate , show that Ps for the M-ary decision problem can be upper bounded by Ps < 4Q RN Express the upper bound on Ps in terms of the average Eb No and compute the value needed to achieve Ps 10-5 . How does this compare with binary antipodal signaling?



Answers

Derive an expression for the location of the centers of the dark bands in Young's experiment. Give your answer in terms of $\Delta r=r_{1}$ $-r_{2}$ and $m^{\prime}=0,1,2, \ldots$ Explain your answer. [Hint: The first minima on either side of the central maximum occur when $\mathrm{m}^{\prime}=0 .$ ]

Were given a log. Normal distribution with parameters mu equals 3.5 and Sigma equals 1.2 for a were asked to find the mean and standard deviation of the distribution. Now, for a log normal distribution, the mean is given by the following formula. So if we plug in our values, our perimeter values, we have e t 3.5 plus 1.2 squared, divided by two and this comes out to 68.3 The variance is given by this formula and again plugging in our parameter values. We get a variance of 14,000 907.17 which gives us a standard deviation of 122.9 So that answer is part A. Now for part B were asked for the probability that X is between 50 and 250. So here we have we've taken the log of X and then standardized it to the standard normal distribution because we know it's a log normal. Therefore, the log of X is a normally distributed random variable 0.954 minus 0.634 and this comes out to 0.3 to 0. The probability of 0.3 to 0. That X is between 50 and 2 50 and next. For part C, we're asked for the probability that X is less than its main value. Its mean value from Part A is 68.3 and this comes out to 0.726 This is slightly different than the answer in the back of the textbook. There is an error in the back of the textbook. It seems as though they forgot to divide by 1.2. You here now, the reason why the probability that X is less than the mean isn't exactly 50% is because the log normal distribution is right skewed.

But this problem has a the path difference in most of the difference in the path difference that's gonna be rt of minus our one. And according to the picture, that's just our ah, distance between the slits times the sign of data. Okay, um, now if they'd, uh, we weren't given data, but we can find they did by saying that the tangent. Well, we could even say the sign of data science data is also equal. Teoh. Why? Which is our opposite sign? We departed about the square root of y square close elsewhere for the hotness or the or the length of the ray in the middle. I guess maybe the from the origin to the point. Um, so that was per day part being just wants us to write this expression in terms of land. So we also have, uh, this phase difference is gonna be or the the path difference Eyes also m times lambda. Okay. And then subpart c. It just says Are is it a maximum or minimum? And so we just need to find out what is RM values that an integer Or is it a 0.5 or somewhere in between So let's go and calculate this. So we're gonna have on Dean, which is 0.15 millimeters. So in meters, it's just times 10 to the negative. Third. There were to multiply that by Why? No, but why buy? It was 1.8 centimeters, which is times 10 to the negative second meters. Divided by a square root of that 1.8 times 10 to the negative second square. Uh, us, our length squared, which is one when 40 centimeters, which is 1.4 meters. Yeah, it's a squared. That's gonna equal 1.92 a, uh, times 10 to the negative six power. Okay, if that is equal to em. Uh, time 6 43 terms. 10 to the negative Ninth. We just have to divide by that 6 43 times 10 to the negative night. That's RM value. It's gonna be equal. Teoh 2.9999 which is just basically equal to three. And so because it's a nice, solid integer, this is going to be a maximum in intensity for maximum condition. OK, that's basically just because we saw for em and got that the M value would be an an integer so they would be in phase and totally constructive interference. Thank you very much.

We will use the local linear approximation of F at the point p to approximate the value of FX que. We will then compare the error in our approximation but the distance between the points p and Q. So for part A to find the local linear approximation of F at P, let's go ahead and compute the partial derivatives of F. So we get that dad left. Del X, which is a partial derivative of F with respect to X, is equal to one over y plus Z because in our function we treat X as the only variable. So we can break this apart as X over y plus z plus why over y plus e And when we take the partial derivative with respect to X, why over y plus z is just zero and one over y plus z is a coefficient two x, and that's why we get this as the partial derivative of F with respect to X. Next, we have del f del y partial derivative of F with respect to why, and this time we treat why is the only variable. But we have y in our denominator and our numerator so to get around that well, not really get around that, but to compute the derivative of this function the partial derivative with respect to y well, the quotient room. So this is equal to the quantity Why plus z times Thebe Rive A tive of x plus Why? Which is just one? Because we're taking the partial derivative of F with respect to why so x is a constant minus the quantity x plus Why times the derivative of the denominator Which is why plus Z which is again one. Because this is a partial derivative of f with respect to why and this is all over a quantity why plus z squared and then next we have dlf del Z and we get that this is minus X plus why times the quantity y plus z to the minus two and again that's taking the partial derivative of F this time with respect to Z and using power rule. So from here we want to evaluate thes first order partial derivatives at the point P so dlf del X at the point P is equal to one half because that's 1/1 plus one. Dlf del y at the point p we get that this is one plus one times one, which is to minus negative one plus one times one which is zero. So we get that this is too all over one plus one squared, which is to over four. So that's also one half and then dlf del Z that's equal to minus, Let's see, minus one plus one, which is zero all over one plus one squared, which is just zero. So we get that our local linear approximation of F at the point P is equal to F evaluated at P, which is minus one plus one all over one plus one, which is zero plus one half times x plus one plus one half times the quantity. Why minus one plus zero times the quantity Z minus one. Of course, this is zero and this is zero. So we get that our local linear approximation is one half times the quantity X plus one plus one half times a quantity. Why minus one and then from here in part B, we want to take the absolute value of our local linear approximation evaluated at Kew, which is one half times the quantity negative 0.99 minus negative one, which is positive. 0.1 plus one half times of quantity. 0.99 minus one, which is negative 0.1 minus our function evaluated at Kew, which is negative 0.99 plus 0.99 all over 0.99 plus 1.1 So that's just zero. And this is all over the distance between P and Q, which is the square root as we have 0.1 squared plus negative 0.1 squared plus 1.1 minus one squared, which is 0.1 squared and then above. This is just zero. Those two cancel each other out, so that zero So we get zero minus zero and we get that this is equal to zero, and that completes the problem.

We're going to find a number and of serve intervals for Bar A, the midpoint approximation em Sovann party The trap Seidel approximation TN and Parsi Simpson's rule approximation as n. In order to ensure that the absolute error when approximating the integral in exercise five will be less than 10 to 94 we will use inequalities 12 13 and 14 in theory, um, 7.7 point two on page 540 of the textbook. So remember here that in exercise five Yeah, we were approximating Dean to grow from 1 to 3 off E to the negative two times six. But so the function we got to consider here is e to the negative two x defined on the interval off integration 13 closed interview. The other values we need here are the maximum over the interval 13 of the absolute values of thesis econ de and forth derivative off if and and that was calculated in exercise 11. So in exercise 11. Yeah, way found that que two is equal to the maximum value for X over 13 off. The absolute value of the second derivative of F is equal to for over he square. Okay. And cave for which is the maximum value over 13 of the absolute value of the fourth derivative off have is 16 over he square. Okay, those values k two before are used or inequalities 12 13 and 14. Right. We started then for a calculated an for the midpoint approximation. And I m seven. And here we use inequality 12 which, which read the absolute error in this method is less than or equal then b minuses. Uh, do you want to say to the third over 24 square times k two? Uh huh. And now we replace the values A is one b three. So we get three minus one to the third over 24 square times. And remember K two year, his value is for over his square. We can simplify this expression here to get for over Ah, three East square in square. So now what we do is to require that thes inequality or this'll last bank be less than 10 2 and 84 required these inequality to be true. Because if this inequality is true, it is automatically true that the absolute values lessons into an 84 Because the trans activity off the inequalities Yeah, eso we gotta solve this'll last inequality for an we write Did fact here if or over three e square and square is less than 10 to 94. Yeah, then, yeah, automatically, Thea, Absolute error in the mid point method will be less intend to the negative four. So we gotta solve what we said. He's inequality for n So starting from this inequality don't yet that end square is greater than four times then to the fourth over three e square. Okay, remember all the quantities that appear in thes inequality of positive So we can take terms from one side to the other of inequality without changing its sense. So we get this equivalent inequality and now we can take both eyes were route because all terms air positive. And remember that square root of thin squares and because it's positive so from here yet and greater than square root of thes expression car and is can be simplified if you want to. 200 over square to three e. Yeah, and this is approximately equal to 42 but point 479 So the first integer that satisfies thes inequalities 43. So we can take all right animals for three for midpoint approximation have we go for? Hmm? The trapezoidal approximation t n And he will use inequality 13 which says that the absolute error in this method is less than or equal to B minus eight to the third over 12 in square times, K two, we replaced values and it is equal to three minus one to the third over 12 and square times que two we remember here is four over his square. Yeah, well, and these simplifies to eight over three square in square. Yeah. Yeah. Okay, so now we required that this last bound be less than 10 to 94. Because if that is true, then it will be also be true that the absolute error will be less intense Trinity for And that is because trans activity of the inequalities. So if 8/3 is square and square is less than 10 to the 94 then the absolute error in the trapezoidal approximation will be less than 10 to 94 also, So we gotta solve thes inequality for n. So we start from there? Uh huh. Mhm. So from this inequality we get right, that and square is greater than eight times 10 to the fourth over three E square. Remember, everything in this inequality is positive. So we can take terms from one side to the other of the inequality without changing it sends. Ah, so we get this equivalent inequality and taking into account that everything here in this new inequality is positive and that squared of end squares and we can take root both eyes, which maintains thean equality because its operatives increasing for positive numbers. Uh, so we did that end Screwed, then square root of all these 8000, which is eight times 10 to the fourth over three E square if we want. If you want to notice, is equal to 200 ovary times square it two thirds okay and is approximately equal to 60 point serious. Seven four. So the first interview that satisfies this inequality is 61 so we can take mhm. Yeah. 61 7 travels for the Travis Idol approximation in our party we regard mhm seems since rule okay is in mhm. And here we use inequality. 14 which says that the absolute error in this method is less than or equal to B minus eight to the fifth over, 180 to the fourth times K four well, and then putting values. This is equal to three minus one to the fifth over. 180 into the fourth times K four, Remember K four. Here it is folly here, 16 over the square. So I put that in here times 16 over. But over his square it was, and he simplifies to 100 28 over 45 s Square and to the fourth. So now we require that this Let's, uh, pound be less than 10 2084. Because if that's true, then it will be also true that thes after error will be less in 10 to 24. Because off the transit Vitti trans activity of the inequalities. So if 180 I'm sorry 128 over 45 square, enter the fourth. It's less than 10 to the negative four than automatically the absolute error. And since his rule approximation will be less then 10 to 94. So we get to solve thes inequality for N. So we start from here Is inequality. Yeah, you want to solve for N? And from here we get into the fourth. Where, then what? A 100 28 times? 10 to the fourth over? Yeah, 45 e square. And remember, all terms that appear here are positive. So we can pass terms from one side to the other of the inequality without changing the sense of the inequality. So we get this inequality here that is equivalent to this one to the first one. So now we can take if, fourth a rude, both sides of the inequalities because all terms air positive. And we remember that the fourth, a route off enter the fourth is in because it's positive. So this implies that and it's greater than the fourth. Root off all these expression here. Right? Mhm. And okay, okay. Re calculate is if you want Thio have a reference, and our reference is equal to 20. Both rude off eight over 45 e square. If you want a reference of an equivalent expression and it is a routine illegal waas 7.8 768 Okay. Uh huh. So the first interview that sacrifices in what is eight. And we remember Simpson's rule required to use and even integer for the numbers of intervals. And in this case, n equals for satisfies that requirement so we can take indeed and equals eight. Uh huh, which is on even integer, right that's required Okay by Simpson school. Yeah, So we have found that the numbers of intervals needed to have a curiously tend to 94 for Simpson rule is eight is much more less than the numbers of intervals needed for Travis Oil, which is 61 midpoint approximation 43 to have the same accuracy tense tend to 94. So we expected that because we saw, in theory that more accurate of the three method for the same security is, it seems, his rule. So we have found that only eight of intervals are needed of on a security of 10 Trinity, for in the case of Simpson's rule,


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