https://community.wolfram.com RSS Feed for Wolfram Community showing any discussions tagged with Signal Processing with no replies sorted by most replies. https://community.wolfram.com/groups/-/m/t/3589102 &[Wolfram Notebook][1] [1]: https://www.wolframcloud.com/obj/9f23858c-9e7c-4883-8a89-dc7e65f09a5d Guhan Thiagarajan 2025-12-11T00:20:39Z https://community.wolfram.com/groups/-/m/t/3580921 &[Wolfram Notebook][1] [1]: https://www.wolframcloud.com/obj/c6388a9d-cdfc-4d41-9d79-caccd5ba5c68 Adael Reyes 2025-11-23T22:34:43Z https://community.wolfram.com/groups/-/m/t/3366063 This publication was done with the collaboration of ***Mauro Miguel de Jesús García Enríquez*** **Introduction** In past posts, we explored the behavior of the sine function with natural and real numbers in the argument, and now we're taking it to the next level! *What would result from combining random behavior with numbers distributed randomly?* This was the question Mauro asked when we discussed a certain wave type and the previous post I wrote. In this post, we want to explore a particular type of numbers: Prime numbers. The initial step is to examine the behavior of the sine wave, we have developed a series of functions that generate the sine function with prime numbers as parameters. These functions have been plotted with varying numbers of terms and domains to have a better understanding of the behavior where the argument has natural and real parameters. After obtaining a visual representation, the behavior is analyzed numerically. A table of the decimal behavior when the parameter of the sine function is an integer and a real number has been created. This table enables the examination of the behavior up to five decimal places. It also facilitates the determination of the values taken by the sine function at a certain point and the observation of the differences between them. In the final visual analysis, a comparison is made between the binary representation of the decimal numbers of the sine function, where the argument is natural and real, and where n ranges from 1 to 100. This allows for a comprehensive examination of the function's behavior. &[Wolfram Notebook][1] [1]: https://www.wolframcloud.com/obj/c80ea9c7-6d54-4d3a-822f-1228efaff2ea Jesús Adrián Montesinos Correa 2025-01-28T22:03:55Z https://community.wolfram.com/groups/-/m/t/3178170 Periodogram seams to be one data point short on the frequency axis. Thus the frequencies are incorrect. With an input vector of length n, I think the periodogram should have n / 2 +1 points. The example I have should have a peak at frequency 0.2. Instead the peak is at 0.203. The input data is 30 cycles over a length of 150. Scaling the frequency axis by DataRange -> {0, (n / 2 -1) / n} seams to give the correct frequency. Compare with Fourier, plotting 0 to 1/ 2, which is n / 2 + 1 points. &[Wolfram Notebook][1] [1]: https://www.wolframcloud.com/obj/80be1c5e-7847-43a3-b6de-62a033aed0dd Philip Isaacson 2024-05-17T15:56:37Z https://community.wolfram.com/groups/-/m/t/3120650 &[Wolfram Notebook][1] [1]: https://www.wolframcloud.com/obj/984f22b4-0b73-453c-863a-077e341a61ab Housam Binous 2024-02-11T15:15:09Z https://community.wolfram.com/groups/-/m/t/3119889 &[Wolfram Notebook][1] [1]: https://www.wolframcloud.com/obj/a6b8af18-f813-4eba-a473-c23e29c06ffc Housam Binous 2024-02-11T12:35:40Z https://community.wolfram.com/groups/-/m/t/3089190 The InverseFourierSequenceTransform of DiracDelta is found, but not the more correct DiracComb: ![enter image description here][1] Doing the inverse does not work with constant r: ![enter image description here][2] Is this a bug or are there any extra parameters to be given to get correct output? &[Wolfram Notebook][3] [1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot_2023-12-21_22-16-34.png&userId=340012 [2]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot_2023-12-21_22-18-02.png&userId=340012 [3]: https://www.wolframcloud.com/obj/6f18fbf6-6bea-4e71-a91e-64bcdf306d55 Bert Aerts 2023-12-21T21:21:08Z https://community.wolfram.com/groups/-/m/t/2973753 Hoping for some insight from the signal processing folks. &[Wolfram Notebook][1] [1]: https://www.wolframcloud.com/obj/3529fe82-5014-46c3-bc0d-4e624746cc4e Gareth Russell 2023-07-21T17:23:33Z https://community.wolfram.com/groups/-/m/t/2951051 Hello all, I am trying to find a built-in function in Mathematica for the room impulse response (RIR), Is there one that does the job? Thanks, Jana Rusrus 2023-07-03T20:39:43Z https://community.wolfram.com/groups/-/m/t/2508076 Hi all, I did the ShortTimeFourier(stf) Transformation of the following signal. ![enter image description here][1] And got the results as shown in this Spectrogram. ![enter image description here][2] Then I wanted to find the peak frequency of this each turn using PeakDetect as: Qpeak = Table[PeakDetect[GaussianFilter[Abs[stf[[b]]["Data"][[turn]]], \[Sigma]], \[Sigma]/ 2, 0, qthreshold], {b, numpeaks}]; And plot it using MatrixPlot as below. ![enter image description here][3] As you can see, the right side of the plot is the mirror image of the left side. So my question is that I couldn't figure out which side (left or right) shows the imaginary side of the amplitude and which side shows the real side? Could you please help me with this? I appreciate any help you can provide. [1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=f1.png&userId=2506303 [2]: https://community.wolfram.com//c/portal/getImageAttachment?filename=6429f2.png&userId=2506303 [3]: https://community.wolfram.com//c/portal/getImageAttachment?filename=f3.png&userId=2506303 S Wije 2022-04-11T18:11:33Z https://community.wolfram.com/groups/-/m/t/2338835 Hello! According to the documentation, "it is also known as autocorrelation or cross-correlation function (ACF or CCF)" . Consider two series: don2 = {1.4700335660303452`, 3.938502200833952`, -2.3409981833059916`, -5.206682610803625`, \ -1.5119300878777509`, -3.877572858901246`, -0.54429724639661`, \ -2.71863911017752`, -4.720466309132462`, -4.9525515879098805`, 0.7471418147389777`, -0.7476753577979416`, -2.0373110008808197`, \ -2.2646947237306243`, 5.119048907352052`, 7.431707238981527`, 6.210342150215659`, -3.432918073555193`, -1.2724627838605187`, \ -2.5942814891801476`, 1.6379652872900294`, 1.8672442571174344`, -2.178087718675407`, 0.511717534220395`, -0.12794872714807326`, -3.493540043669867`, \ -2.076520676407341`, -0.020149221736978973`, -0.23566065497338684`, \ -0.9865291935808429`, -0.7274734194559436`, -0.08506759516156492`, \ -0.5359663222662654`, -2.372956774837462`, -1.5096330893537484`, 0.8687632738817399`, -1.8208987088858706`, -1.6012141611738784`, 1.7474049593318028`, 2.0618021152562562`, -0.3578432908793836`, 5.4607600668245055`, 2.831853971353688`, 0.9958723615545395`, -3.973835080221228`, 1.7249884984446382`, -0.9668510202178453`, 0.025070997030766762`, 1.3498754178110701`, 0.3039176050052319`, -1.5328993929300694`, -1.342354373782186`, 0.014071932471644288`, 1.1270691117867102`, 4.927260371512951`, 9.091993757162534`, 0.6464858344433324`, 2.598030745145927`, 6.84962621619767`}; don1 = {871.4281994478697`, 2358.9651127499164`, 497.0197383100749`, -940.3646592912628`, -739.5054930991755`, \ -1492.3180092337025`, -872.4843915093261`, -807.4571701991008`, \ -1051.7738241220868`, -2054.0353238452117`, 1845.0678906068024`, 462.55566003359127`, -329.8765141915137`, -1707.3552259982353`, 1660.4867031116046`, 1017.803797790395`, 711.0912693464415`, -269.40038456466937`, -952.7865411939238`, \ -514.1978049977151`, 680.9316719242015`, 859.6573038472379`, -780.5624685756338`, 59.709725844848435`, 84.38908186389745`, -1188.6162931806248`, 136.07855613612628`, 409.79006512267733`, 257.78393246168343`, -272.9260069138363`, -723.128703503171`, \ -377.5180831523867`, -155.24777831126994`, -432.05064088310246`, \ -67.65773696268661`, 193.24578222735`, -495.23616890942867`, -885.320612323234`, 812.5870090051285`, 868.6291241513609`, 560.2177206343404`, 967.5225791749835`, 566.0707853469139`, 2780.472100597479`, 26.222127021410415`, 1628.19479341435`, 113.34966858870757`, 292.3958115184505`, 1151.8488521254912`, -535.5242493191618`, -1240.155476238113`, \ -694.4329951865366`, 3.9872914187872084`, 535.0210633722745`, 1872.2482295022128`, 1744.9986746091017`, 555.1551401387915`, 293.87713749144314`, 947.3013172212959`}; and build a 2-d vector time series Don = TemporalData[{Transpose@{Range@Length@don1, Standardize@don1}, Transpose@{Range@Length@don1, Standardize@don2}}] DateListPlot[%, PlotRange -> All] I thought the list resulting from CorrelationFunction[Don, {30}] corresponds to the empirical cross-covariance between the variables. But while In[10]:= Correlation[don1, don2] Out[10]= 0.740266 we have that : In[13]:= QuantityMagnitude@ Last@Transpose@First@CorrelationFunction[Don, {30}]["Paths"] DateListPlot[CorrelationFunction[Don, {30}], PlotRange -> All] (* Croisé *) Out[13]= {1., 0.337196, 0.0717911, 0.0248392, 0.0893278, -0.00237899, \ -0.0203818, -0.128466, -0.170179, -0.0440539, 0.00504939, -0.114143, \ -0.0579385, 0.0900887, 0.135494, 0.0894005, 0.0555025, 0.0411966, \ -0.0582309, 0.0361019, 0.0501823, -0.0542087, -0.0280744, -0.0237793, \ -0.0048104, 0.0363897, 0.0376791, 0.109378, 0.0411235, -0.0551814, \ -0.141162} But with lag 0, the correlation should be 0.74! So, it seems that the function above is the autocovariance of don1*don2, which is different from their cross-covariance. Am I right? Thanks in advance, Claude Claude Mante 2021-08-10T16:50:23Z https://community.wolfram.com/groups/-/m/t/2321377 Hello everyone. I have performed the [Karhunen Loève decomposition][1] of some data using the function [KarhunenLoeveDecomposition][2]. The code I have used is as follows: R1 = Flatten[Import["Realization1.txt", "Table"]] {bT, mT} = KarhunenLoeveDecomposition[{R1, R2, R3, R4, R5}]; being R2 to R5 loaded the same way as R1. The size of all Ri is `{144}` if I calculate it as `Dimensions[R3]`. I have problems understanding the output. As far as I understand, mT should be the eigenvectors of the covariance matrix, however, it's dimensions are {5,5}. As far as I know, there should be 5 vectors of length 144. Is this wrong or am I doing the KL decomposition badly? Best regards. Jaime. [1]: https://en.wikipedia.org/wiki/Karhunen%E2%80%93Lo%C3%A8ve_theorem [2]: https://reference.wolfram.com/language/ref/KarhunenLoeveDecomposition.html Jaime de la Mota 2021-07-17T19:01:52Z https://community.wolfram.com/groups/-/m/t/2304987 Ok, for those who are begging like me, the fft of Matlab is equivalent in Wolfram to Fourier[list,FourierParameters->{1,-1}] this is because the Mathematica model used by Mathematica and MatLab are different, the difference is the normalization of Fourier DiscreteTransform used, the link below has more information about this, good luck. https://math.stackexchange.com/questions/3233050/fft-from-scilab-is-different-than-wolfram-alpha Jairo Smith Quilumbaquin Lanchimba 2021-07-04T00:16:19Z https://community.wolfram.com/groups/-/m/t/2239081 The Diffie Hellman key exchange protocol is a method for two parties to share secrets over a non-secure channel. Strength of cryptographic protection follows from the difficulty of the discrete logarithm problem (see: [Guillevic & Morain, Discrete Logarithms][1]). Industry standard implementations make use of elliptic curve addition rules, but here we will consider the simple tent map instead: It[x_] := 1 - Abs[1 - 2 x]; (* tent map, x in [0,1] *) ItN[x_, n_] := 1 - Abs[1 - 2 Mod[2^(n - 1) *x, 1]]; (* tent map nth power *) After some analysis it is clear that the following equality is **True**: Nest[It,x,n]==ItN[x, n] ; If we write $\phi(x)=\phi^{1}(x)$ for the tent map, and $\phi^n(x)=\phi(\phi^{n-1}( x ) )$ as the nested tent map, we can immediately see a time savings by computing $\phi^n(x)=\phi(x,n)$ when using the alternate, function. Since we can compute $\phi(x,n)$ relatively quickly, we can in principle use it in a discrete logarithm problem where $\phi(x,n+m)=\phi(\phi(x,n),m)=\phi(\phi(x,m),n)$. For chosen generator $g$, private key $n$ determines a public key $\phi(g,n)$. Assuming that $g=2/p$ with $p$ a safe prime (cf. [A005385][2]), then $k = p \; \phi(g,n)$ is an integer satisfying either $k = (2)^n \mod p$ or $k = - (2)^n \mod p$. The classical complexity of solving for $n$ is at best $\mathcal{O}(\sqrt{n})$ (see also: [Factoring Discrete Logarithms][3]). First we choose a prime denominator, say $p_0=2943167$, and observe that the tent map cycle of generator $g = 2/p_0$ visits $(p_0-1)/2 = 1471583$ unique values, as is proven by a brute force check: AbsoluteTiming[ TentCycleGraph[pr1_] := Graph[DirectedEdge[2 #/pr1, It[2 #/pr1]] & /@ Range[(pr1 - 1)/2]]; pr = Prime[2^17 + 2^16]; dat = ConnectedComponents[TentCycleGraph[pr]]; Length /@ dat] which returns the correct number in about 10 seconds. The integer $1471583$ is also the size of the key space, while the 10 second time figure gives some idea how fast the inverse function could possibly be computed. In all practical applications, a much larger keyspace would be necessary. Before moving on to key-exchange, let us take a look at the distribution of $k$ vs $n$ : AbsoluteTiming[ invdat = MapIndexed[{pr #1, #2[[1]]} &, NestList[It, 2/pr, len - 1]];] Histogram3D[invdat, Boxed -> False] ![dist][4] The distribution is almost normal, which indicates good dispersion of public key values, and also reaffirms the difficulty of finding inverse values. Now for the easy part. In secret, Alice and Bob both choose integer-valued keys $n$, and generate public keys using $\phi(2/p,n)$: AbsoluteTiming[{ KeyA = RandomInteger[{10^5, 10^6}], PubA = pr*ItN[2/pr, KeyA]}] AbsoluteTiming[{ KeyB = RandomInteger[{10^5, 10^6}], PubB = pr*ItN[2/pr, KeyB]}] For example, Alice has $n=392085 \rightarrow \phi = 1105606$ while bob has $n=956058 \rightarrow \phi = 2735806$. Both $\phi$ keys are passed through an unencrypted public channel. A third party, an eavesdropper name Eve, intercepts both $1105606$ and $2735806$. Meanwhile, secret values $392085$ or $956058$ do not enter the public channel, so Eve can not find either value without inverting $\phi$ (easy enough with a small keyspace, but difficult in general). Alice and Bob can then compute $p \; \phi(1105606/p,956058) = p \; \phi(2735806/p,392085) = 2257734$, a shared secret key. In practice, Alice and Bob both own and operate small, black box mail servers, which are inconspicuously addressed online. After symmetric key exchange, the friends can then use SMTP to send encrypted messages. There is no problem with contemplative, inquisitive Eve, who deals with exclusion by going back to nature and writing poetry. While Eve is pacifistic, we can readily expect the existence of many more aggressive adversaries who even hate nature poety. So it really is important for Alice and Bob (and Eve) to invest time and effort in securing their devices. [1]: https://hal.inria.fr/hal-01420485v2 [2]: https://oeis.org/A005385 [3]: https://crypto.stanford.edu/pbc/notes/crypto/factoring.html [4]: https://community.wolfram.com//c/portal/getImageAttachment?filename=histogram.png&userId=234448 Brad Klee 2021-04-07T17:00:56Z https://community.wolfram.com/groups/-/m/t/2148941 Hi everyone! I am trying to develop modeling in Mathematica that presuppose the use of Fourier analysis and synthesis applied to financial historical series (eg stocks, financial indices, currencies, and so on). So I wanted to ask you if anyone knows any textbooks, papers, or any other references on these topics for my purposes (even better if with examples of applications on Mathematica). The purpose of my research is to apply Fourier analysis and synthesis on financial time series (as already mentioned), in order to extrapolate from them the relative market cycles, both to be able to carry out appropriate analyzes and to be able to make forecast analyzes. The output should be to obtain a graph similar to the one attached, where above there is a "candlestick" graph and below the sinusoidal trend of the market cycle reconstructed through the application of Fourier. This kind of analysis would also make it possible to create a sort of trading system to identify entry points (sell/buy) to the market, exploiting the phase-lag of the various cycles, and their crossover (the idea in fact is to exploit cycles with different periods- eg. daily, weekly, monthly, ergo with amplitude and different frequency) This in turn implicitly requires a strategy to optimize the cycle obtained as a function of the volatility of the underlying chart, as well as the optimal choice of parameters to be used in the Fourier transform (and this is another point I wanted to ask). Furthermore, another question (secondary, but equally important for me to understand) is that by doing some research, I could see how the Hilbert-Huang transform could be a more efficient strategy in this sense, since financial time series are non-series. stationary, so I was also wondering if it was appropriate to model everything using this type of transform. Many thanks in advance. Giuseppe Vonella 2020-12-27T11:43:54Z https://community.wolfram.com/groups/-/m/t/2143838 I have a very short script to import a piece of `.wav` audio and use `SpeechRecognize` to transcribe it, but the usual `Export` command doesn’t work to save the resulting text. How do I go about saving the text produced from `SpeechRecognize` to a text file? Steven Buehler 2020-12-21T17:58:04Z https://community.wolfram.com/groups/-/m/t/2117912 hello, I'm trying to model a transfer function by specifying the zeros, poles and gain. Following the document that says: TransferFunctionModel[{z,p,g},s] specifies the zeros z, poles p, and gain g of a transfer-function model. I do this in Mathematica: In[43]:= tmf = TransferFunctionModel[{2 - 3 I, 0.5 + 0.1 I, 3}, s] Out[43]= TransferFunctionModel[{{{ Complex[2, -3], Complex[0.5, 0.1], 3}}, 1}, s] which is ***incorrect*** because of the following: In[44]:= TransferFunctionZeros[tmf] Out[44]= {{{}, {}, {}}} please kindly advise the correct syntax of zeros-poles-gain specification with an example. Thanks! Thomas Tse 2020-11-18T13:25:37Z https://community.wolfram.com/groups/-/m/t/2114280 First, thanks for read. I'm a math teacher in highschool. If you connect Arduino Uno to Mathematica and press the switch, the value read by DeviceRead is changed to '0' or '1'. When the switch is pressed using the EmitSound, Sound, and SoundNote functions, the sound corresponding to '1' is output. And in the end, I want to make an 88-key piano. However, an error code appears stating that the value of '0' or '1' output by v=Dynamic[DeviceRead["Arduino",11]] is not applicable in EmitSound[Sound[SoundNote[v]]]. What am I missing? If there is a difference in time that occurs from the moment the switch is pressed and connected to the EmitSound function of Mathematica through the Arduino, how can we overcome it? Thanks, again. wkdrpqkfk Jang 2020-11-13T06:10:26Z https://community.wolfram.com/groups/-/m/t/2002949 Hello everyone. Some time ago I wrote a code that adds some values to the beggining of a vector of data (X, f(X)=Y) which is later used for Shannon interpolation as follows: data = sample; data = SortBy[data, First];(* just in case time data is not properly sorted *) {xmin, leftY} = First@data; (* saving data min *) {xmax, rightY} = Last@data ;(* saving data max *) dDelta = (xmax - xmin)/ 20; (* step *) leftF = BSplineFunction[{{xmin, leftY}, {xmin - dDelta, leftY - dDelta (rightY - leftY)/(xmax - xmin)}, {xmin - 2 dDelta, 0}, {xmin - 3 dDelta, 0}}]; Now, I need to do a similar interpolation, but in 2D, meaning that now the dimensions of data are n*3 (X, Y, f(X,Y)) instead of n*2 and I don't know how to adapt the code for leftF to work with 2D data. Jaime de la Mota 2020-06-12T12:27:58Z https://community.wolfram.com/groups/-/m/t/1989401 Hallo, I'm trying to use Analyze Dynamic Sensitivity new features in Mathematica 12. All the functions have an input model in its graphical representation. My question is if I have a PDE with its initial/boundary conditions, How can I formulate it as a graphical representation model to use these functions? Maha Youssef 2020-05-28T19:08:33Z