How can I prove the Convolution property of Continuous Fourier Transform?
155410
Do a Google search for "Fourier Transform."  I'm sure that someone has a proof on the web.
163726
It is Parseval's theorem you seek. "The total energy contained in a waveform x(t) summed across all of time t is equal to the total energy of the waveform's Fourier Transform X(f) summed across all of its frequency components f."\n\nFor a periodic signal, take the energy in each frequency for one period.  Sum those and you have the total energy.  The maximum amplitude during the period would be if all of that energy occured at once (i.e. an  impulse).  Convert that to dBm or DBW and you have your answer in dB.\n\nLook at the last equation in the link in "sources".
248592
Yes. There is not a continuous one, though. It is tricky to come up with an explicit example, but showing that one exists isn't too bad. First show that R is in one-to-one correspondence with 2^N. Then since Nx{0,1) is equivalent to N, we get R^2 is equivalent to (2^N)^{0,1} which is equivalent to 2^N which is equivalent to R.\n\nIn general, if X is an infinite set, there is a bijection from X to X^2. The axiom of choice is needed for this, though.
823238
I wont give you the entire proof, just give you the key arguments.\n\nknowing the joint distibution of (X, Y) allows to compute the marginal densities of X and Y.\nIndeed, let f_X be the marginal density of X, then:\n\nf_X(x) = &#92;int f(x, y) d_y\n\nf_Y(y) = &#92;int f(x, y) d_x\n\nNow, you might compute E(Y| X = x) = &#92;int y f_{Y|x} (y) d_y\n\nwhere f_{Y| x} is the conditionnal density of Y knowing the event X = x\n\nrecall that f_{Y| x}  = f(x, y) / f_X(x), and you just have to compute the resulting integrals\n\nregards,\n\njfp
81441
Two reasons:\n1. Because it results in a finite spectrum result for a signal. Fourier Transform typically results in infinite spectrum result for a function (e.g if the function has C1 or C0 discontinuities e.g. square wave, triangular wave, ramp, sawtooth)\n\n2. FT assumes that signals are "stationary" and that their frequency-domain can be computed over infinite-time window and will not change. This is unrealistic - most real-world signals are not stationary in a window >10ms (e.g. speech, telecomms, EKG). WT computes its transform based on both a time- and frequency-window.\n\nAs to why you compute DWT rather than the direct Continuous (integral formulation) WT, it's for the same simple reason you generally use DFT instead of CFT - because it's mathematically tractable and lends itself to efficient digital computation (order(NlogN) time); also this formulation can be very efficiently parallelised whether in hardware or software esp. if N is a power of 2 e.g. 64,256,512,1024,... and people have developed these implementations very far (they're a standard toolbox feature or hardware IP).\n\nI don't know the theoretical links offhand but they should be trivial to find.
1111738
the best way to appr this problem is to apply the first principle i.e.\nf'(x)= Lt     [f(x+h)-f(x)]/h\n       h->0\nnow putting f(x)=e^x we have\nf'(x)= Lt    [e^(x+h)-e^x]/h\n       h->0\n     =Lt    [e^x(e^h-1)]/h\n       h->0\n\nnow (e^h-1)/h where h tends to 0 is 1\nhence f'(x) = e^x\ntherefore the derivative of e^x is e^x
348048
A sequence {x_n} is Cauchy if given e>0 we can find an integer N such that || x_m - x_n || < e for all m,n > N.  I want to prove a lemma first:\n\nLEMMA: Let f: A -> R^m be uniformly continuous.  If {x_n} is a Cauchy sequence, then so is { f(x_n) }.\n\nProof: Let e > 0 be given.  We must show that there exists an integer N such that || f(x_m) - f(x_n) || < e whenever m, n > N.  Since f is uniformly continuous, we can find d > 0 such that || f(x) - f(y) || < e whenever || x -y || < d.  Since d is given -- and only depends on e! -- there is an integer N such that || x_m - x_m || < d for all m, n > N.  This means || f(x_m) - f(x_n) || < e for all m, n > N.\n\nWe now prove the main problem with the following.\n\nPROPOSITION: A sequence {x_n} is Cauchy if and only if lim x_n exists.\n\nProof: First assume that L = lim x_n exists.  Given e > 0, there exists an integer N such that || x_n - L || < e/2 whenever n > N.  For m, n > N, we have, by the Triangle Inequality,\n\n|| x_m - x_n || <= || x_m - L || + || x_n - L || < (e/2) + (e/2) = e.\n\nConversely, say that {x_n} is a Cauchy sequence.  We give a Proof by Contradiction, so assume that lim x_n does *not* exist.  That means given any point L in R^m, real e > 0, and integer N we can find an integer m > N such that || x_m - L || > e.  Since e is given, we can find an integer N such that || x_m - x_n || < e for all m, n > N.  Choose n = N+1 and L = x_n.  Then there exists an integer m > N such that\ne < || x_m - L || = || x_m - x_n || < e\n\nwhich is a contradiction.  Hence lim x_n must exist.\n\nCOROLLARY: If {x_n} is Cauchy and f: A -> R^m is uniformly continuous, then lim f(x_n) exists.\n\nProof: If {x_n} is Cauchy and f: A -> R^m is uniformly continuous, then { f(x_n) } is also Cauchy.  The limit of a Cauchy sequence exists, so L = lim f(x_n) exists.
264987
WALSH FUNCTIONS \nThe Walsh functions form an orthogonal and complete set of functions representing a discretized function (Rao, 1983). As in the FFT, the discrete function must be padded before decomposition, so that the total number of samples reaches the nearest power of two. Figure a shows the first sixteen Walsh functions, which form a complete set describing sequences with sixteen samples. Two important characteristics of the Walsh functions are their compactness (representing the lower order functions requires fewer samples), and the simplicity and quickness of their construction.
972053
look here:\n\nhttp://en.wikipedia.org/wiki/Wavelet
744417
My answer assume that you have a good knowledge of integral calculus. If it's not the case, then use a soft like Matlab...\n\nNotations :  I will use SIGMA for the 1 to infinite sum and INTEGRAL for the integral between -T/2 and T/2.\n\nIf you consider a periodic signal x(t) with a T = 1/f0 period, its Fourier development is :\n\nx(t) = a0/2 + SYGMA[k=1 -> infinite]  (ak*cos(2*PI*f0*t)) +  SYGMA[k=1 -> infinite] (bk*sin*(2*PI*f0*t))\n\nwhere f0 = 1/T is the fundamental frequency  of the signal, a0/2 the mean value and ak, bk are the Fourier coefficients in cosinus and sinus.\n\nThe Fourier coefficients are calculated like this :\n\nak = 2/T * INTEGRAL[-T/2 -> T/2] (x(t)*cos(2*PI*k*f0*t) dt)    for k >= 0\n\nbk = 2/T * INTEGRAL[-T/2 -> T/2] (x(t)*sin(2*PI*k*f0*t) dt)     for k >= 1      \n\nAfter developpement, you will find the values  of all a and b coefficients in function of k.\n\nI hope this will help :)
71756
Here's a classic:\n\nUsing the following partial deriviatives:\n\nInt(f(x)*g(x))dx = f(x)*G(x) - Int(f'(x)*G(x))dx\n\nwhere G(x): the primitive function of g(x)\n     and f'(x): the derivative of f(x)\n\nNow watch the following 'proof':\n\n    Int(1/x^2 * 2x)dx  =  1/x^2 * x^2 - Int(-2/x^3 * x^2)dx\n           {just take f(x) = 1/x^2 and g(x) = 2x}\n\n    this yields:\n\n    Int(2/x)dx  =  1 - Int(-2/x)dx  = 1 + Int(2/x)dx    \n\n    substracting Int(2/x)dx on both sides yields:\n\n    0 = 1
776321
the question is unclear.\nDo you want an explicit formula for Y[n or do you just want to write out the convolution]
779791
(f(2) - f(0))/(2 - 0) = 2, but f'(x) < 0 for every x, except x=1 where it is undefined.\n\nThe function has to be continuous in [0,2] and differentiable in (0,2).  For this f(x), it is undefined at x=1, so the conditions of the MVT are not met.  Math is still safe!\n\nCheers!
669001
e^f(x) = x is the best answer i can give you.
74348
If you look at the standard epsilon/delta proof that G(x) = x^2 is continuous 0, it all works for F(x) too. \n\nThat is, given epsilon (e), you can find delta (d) such that /G(x) - G(0)/ < e when /x - 0/ < d.\n(Since for -1 < x < 1,  x^2 < |x|, just pick d = min(e, 1))\n\nNow looking at F(x), given d, pick that same e you would use with G(x). If x is rational, F(x) = G(x)\nso /G(x) - G(0)/ = /F(x) - F(0)/ < e.  If x is irrational, /G(x) - G(0)/ = /0 - 0/ < e.
873172
u can get them with the help of second and higher derivatives.\ninflection point is there where the second derivative is zero.
396592
A continuous spectrum of what?