text:
\begin{document}

\input{acronym.tex}

\title{Wavelet Classification for Over-the-Air Non-Orthogonal Waveforms }
\author{\IEEEauthorblockN{Tongyang Xu and Izzat Darwazeh}
 \IEEEauthorblockA{Department of Electronic and Electrical Engineering,
University College London, London, UK\\
 Email: {tongyang.xu.11@ucl.ac.uk, i.darwazeh@ucl.ac.uk}}}

\maketitle

\begin{abstract}

Non-cooperative communications using non-orthogonal multicarrier signals are challenging since self-created inter carrier interference (ICI) exists, which would prevent successful signal classification. Deep learning (DL) can deal with the classification task without domain-knowledge at the cost of training complexity since neural network hyperparameters have to be extensively tuned. Previous work showed that a tremendously trained convolutional neural network (CNN) classifier can efficiently identify feature-diversity dominant signals while it failed when feature-similarity dominates. Therefore, a pre-processing strategy, which can amplify signal feature diversity is of great importance. This work applies single-level wavelet transform to manually extract time-frequency features from non-orthogonal signals. Composite statistical features are investigated and the wavelet enabled two-dimensional time-frequency feature grid is further simplified into a one-dimensional feature vector via proper statistical transform. The dimensionality reduced features are fed to an error-correcting output codes (ECOC) model, consisting of multiple binary support vector machine (SVM) learners, for multiclass signal classification. Low-cost experiments reveal 100\% classification accuracy for feature-diversity dominant signals and 90\% for feature-similarity dominant signals, which is nearly 28\% accuracy improvement when compared with the CNN classification results.

\end{abstract}

\begin{IEEEkeywords}
Signal classification, wavelet, machine learning, SVM, non-cooperative, non-orthogonal, SEFDM, waveform, experiment, software defined radio.
\end{IEEEkeywords}

\section{Introduction}

In traditional communication systems, cooperation between a transmitter and a receiver is the default configuration to ensure reliable signal recovery at the receiver. Therefore, signal format is the important side information that has to be mutually known to both transmitters and receivers. The side information is delivered from a transmitter to notify a receiver and such an overhead would additionally occupy either time, frequency or space resources. Furthermore, wireless channels are time-variant and the side information of signal format would be out of date when a signal reaches a receiver after a time delay, which would subsequently cause inaccurate signal detection. Therefore, a more reliable solution is required, which avoids transmitting side information and lets the receiver timely extract signal format information from received signals.

An intelligent receiver can automatically identify signal formats based on data training. Deep learning (DL) is initially proposed to deal with image processing since it can automatically and efficiently extract features from two-dimensional images. The representative deep learning strategy is \ac{CNN}, which employs multiple convolutional layers for feature extractions. CNN has been successfully applied in single carrier modulation classification \cite{OShea_classification_2018} and multicarrier \ac{OFDM} modulation classification \cite{OFDM_classification_Zhou2019ARM, OFDM_classification_access2019}. The classification for non-orthogonal \ac{SEFDM} signals \cite{TongyangTVT2017} has been theoretically and practically investigated in work \cite{tongyang_VTC2020_DL_classification}, in which a trained \ac{CNN} classifier can efficiently identify feature-diversity dominant signals while it cannot accurately classify feature-similarity dominant signals. Although the deep learning classifier is trained to automatically extract signal features without domain-knowledge, the tremendous fine-tuning for optimal neural network hyperparameters is time consuming and inefficient. Therefore, manually extracting signal features, based on expert knowledge and traditional machine learning (ML), would be more efficient and convincing.

This work will firstly study different statistical features in \ac{SVM} for non-orthogonal signal classification. Modelling results reveal that either time-domain or frequency-domain statistical features are unable to train accurate classifiers for non-orthogonal signals. Therefore, a wavelet transform \cite{wavelet_tour_Mallat_2008, wavelet_transform_TIT_1990} based time-frequency feature extraction approach is applied in this work. Previous work has explored multilevel structured wavelet decomposition \cite{wavelet_classification_TPAMI_1997, wavelet_classification_TITB_2007} and wavelet scattering \cite{wavelet_scattering_TSP_2014} for feature extraction and classification. This work focuses on a single-level wavelet filtering (WF) strategy. Results indicate that the time-frequency feature with statistical dimensionality reduction can assist \ac{SVM} to identify signals at high accuracy. In addition, this work evaluates classifier accuracy for signals at different Es/N0. Finally, a low-cost experiment is set up to verify the trained classifiers using over-the-air signals.

The main contributions of this work are as the following.
\begin{itemize}
\item{ Statistical features are investigated in \ac{SVM} for non-orthogonal signal classification.  }
\item{ Two-dimensional time-frequency features are evaluated via single-level wavelet transform. Various time-frequency feature dimensionality reduction methods are studied to simplify the features and further improve the classification accuracy.  }
\item{ Low-cost over-the-air experiment is designed for non-orthogonal signal classification. Practical results verify the robustness of the wavelet classification.}
\end{itemize}

\section{SEFDM Waveform}

The time-domain SEFDM signals are illustrated in Fig. \ref{Fig:ML_feature_combined_G_I_G_II} where two types of signal patterns are presented. It is inferred that the classification of Type-I signal pattern is easier than that of Type-II signal pattern since the signal features in Type-I are more distinguishable.

\begin{figure}[t!]
\begin{center}
\includegraphics[scale=0.35]{ML_feature_combined_G_I_G_II.eps}
\end{center}
\caption{Signal feature diversity and similarity visualization by modulating the same QPSK data. (a) Type-I signal pattern. (b) Type-II signal pattern.}
\label{Fig:ML_feature_combined_G_I_G_II}
\end{figure}

The discrete format of one time-domain SEFDM symbol is defined as
\begin{equation}
X_k=\frac{1}{\sqrt{N}}\sum_{n=0}^{N-1}s_{n}\exp\left(\frac{j2{\pi}nk\alpha}N\right),\label{eq:SEFDM_discrete_signal}\end{equation}
where the expression is very similar to that of OFDM except the bandwidth compression factor $\alpha=\Delta{f}\cdotp{T}$, in which $\Delta{f}$ is the sub-carrier spacing and $T$ is the time period of one SEFDM symbol. The signal spectral bandwidth in \eqref{eq:SEFDM_discrete_signal} is compressed when $\alpha<1$ and is equivalent to that of OFDM when $\alpha=1$. The number of sub-carriers is determined by $N$. $s_{n}$ is the $n^{th}$ single-carrier symbol within one SEFDM symbol and $X_k$ is the $k^{th}$ time sample with $k=0,1,...,N-1$. The instantaneous power for one SEFDM symbol is computed in the following
\begin{equation}\label{eq:SEFDM_square_signal}
\begin{split}
|X_k|^2&=\frac{1}{N}\sum_{n=0}^{N-1}\sum_{m=0}^{N-1}s_{n}s^{*}_{m}\exp\left(\frac{j2{\pi}(n-m)k\alpha}N\right)\\
&=\frac{1}{N}\sum_{n=0}^{N-1}|s_{n}|^2+\\
&\underbrace{\frac{1}{N}\sum_{n=0}^{N-1}\sum_{m\neq{n},m=0}^{N-1}s_{n}s^{*}_{m}\exp\left(\frac{j2{\pi}(n-m)k\alpha}N\right)}_{ICI}.
\end{split}
\end{equation}

It is clearly shown that the \ac{ICI} term in \eqref{eq:SEFDM_square_signal}, which is related to the value of $\alpha$, determines the possibility of identifying different SEFDM signals. It is inferred that when SEFDM signals have similar values of $\alpha$, the ICI term will become similar and would complicate signal classification.

\section{Classification Strategies}

\subsection{CNN Classification}

A multi-layer CNN classifier is trained in a recent work \cite{tongyang_VTC2020_DL_classification} to automatically extract signal features in either time-domain or frequency-domain. Based on extracted features, classification results are compared in Fig. \ref{Fig:CNN_accuracy_performance_Type_I_Type_II_wavelet}, in which the time-domain classifier achieves higher accuracy than its frequency-domain counterpart. Classification accuracy can reach 95\% when considering limited number of non-orthogonal signals in Type-I. However, the accuracy drops greatly when adding more similar signals in Type-II.

\begin{figure}[t!]
\begin{center}
\includegraphics[scale=0.43]{CNN_accuracy_performance_Type_I_Type_II_wavelet.eps}
\end{center}
\caption{CNN classification accuracy for SEFDM signals considering either time-domain or frequency-domain features.}
\label{Fig:CNN_accuracy_performance_Type_I_Type_II_wavelet}
\end{figure}

\subsection{SVM Classification}

The limitation of the previous work \cite{tongyang_VTC2020_DL_classification} is obvious and the motivation for this work is to accurately classify Type-II signals. The training of a multi-layer CNN classifier is time-consuming since it requires extensive hyperparameter tuning and iterative back propagation optimization. Therefore, it would be more efficient to use traditional machine learning strategies with manual feature extractions. The SVM classifier, based on domain-knowledge dependent features, is applied in this work. Firstly, the training is fast since features are obtained in advance rather than time-consuming data training. Secondly, the methodology of machine learning is deterministic and its working principle can be well explained. Since there are multiple signal classes in Type-I and Type-II, therefore a multiclass \ac{ECOC} model \cite{ECOC_Dietterich_1995} is applied here. A one-versus-one \cite{ECOC_2014_codingDesign} coding strategy is implemented for separating different classes, which simplifies the multiclass classification task into multiple binary class classification tasks. Thus, multiple binary SVM learners, with a polynomial kernel of order two, are used for the multiclass classification.

\section{Feature Selection}

This section will firstly explore the impacts of different one-dimensional statistical features and their combinations either in time-domain or frequency-domain. The second part will investigate the impact of two-dimensional time-frequency features via the single-level wavelet transform.

\subsection{Statistical Features}\label{subsec:statistical_features}

The commonly used statistical feature is arithmetic mean, which computes the average value of a dataset. Variance is used to measure the variations of a dataset. Small variance indicates that the values of dataset elements are closer to the arithmetic mean while large variance indicates that the dataset elements are spread out away from the mean. Skewness \cite{Feature_skewness_2011} is a way to measure data distribution characteristics. Negative skewness indicates that a dataset distributes more data to the left side relative to its mean; positive skewness indicates that data is more distributed to the right side of the mean. The ratio between the maximum value and the minimum value is also studied here and the MaxMin ratio can tell the fluctuations of a dataset. Interquartile range (IQR) \cite{book_IQR_1996} is a way to measure data dispersion, which equals the difference between the 25th percentile and the 75th percentile.

\subsection{Time-Frequency Features}

The previous work \cite{tongyang_VTC2020_DL_classification} revealed that independent time-domain features or frequency-domain features cannot efficiently identify Type-II signals. Therefore, the joint analysis of time-frequency signal features is important since feature diversity would be enhanced by considering two domains. This section applies wavelet transform \cite{wavelet_tour_Mallat_2008} to manually extract hidden signal features in time-frequency dimensions.

\begin{figure}[t!]
\begin{center}
\includegraphics[scale=0.25]{wavelet_spectrogram_1_08_07_v3.eps}
\end{center}
\caption{Spectrogram of OFDM and SEFDM signals after wavelet transform. For the purpose of illustration, the signals are simply generated with $N$=12 sub-carriers at a data rate $R_s$=1 bit/s.}
\label{Fig:wavelet_spectrogram_1_08_07}
\end{figure}

There are two types of wavelet transform for time-frequency analysis, namely \ac{CWT} and \ac{DWT}. \ac{CWT} provides a detailed representation for signals by using fine scale factors. It therefore leads to high-resolution signal analysis and can capture crucial signal features. However, the obvious disadvantage of \ac{CWT} is its higher computational complexity over \ac{DWT}. A large time-frequency spectrogram grid would be obtained with the fine representation of scales. In this work, we would like to explore the accurate signal transient localization via detailed time-frequency analysis. Therefore, the high-resolution wavelet transform \ac{CWT} is used rather than its coarse wavelet transform \ac{DWT}.

\begin{figure}[t!]
\begin{center}
\includegraphics[scale=0.41]{time_frequency_feature_grid_transform_v3.eps}
\end{center}
\caption{ One-dimensional wavelet feature generation based on wavelet filtering and statistical feature dimensionality reduction.}
\label{Fig:time_frequency_feature_grid_transform}
\end{figure}

There are several wavelet candidates for wavelet transform. This work employs the widely used Morse wavelet and the effects of different wavelets are not taken into account. The \ac{CWT} time-frequency analysis for OFDM and SEFDM signals using Morse wavelet is illustrated in Fig. \ref{Fig:wavelet_spectrogram_1_08_07}. It is clearly shown that with the reduction of alpha, the frequency scales for SEFDM shrink to show the effect of bandwidth compression while its time scales are stretched to show the time-domain sample characteristics. Typical artificial intelligent solutions are to feed the time-frequency grid as an image to a deep learning neural network such as \ac{CNN}. However, this would cause extra training complexity since the optimal neural network hyperparameters have to be tuned based on iterative attempts. Therefore, pre-processing is required to simplify the two-dimensional time-frequency feature representation into a one-dimensional feature vector as illustrated in Fig. \ref{Fig:time_frequency_feature_grid_transform}. The strategy is to maintain the fine frequency scales of \ac{CWT} while reducing time samples dimensionality using the statistical knowledge explained in Section \ref{subsec:statistical_features}.

\section{Classifier Training and Testing}

To have a realistic training scenario, channel/hardware impairments have to be considered. The wireless channel \ac{PDP} and hardware impairments are defined in \cite{tongyang_VTC2020_DL_classification} and are reused in this work. Signals are generated according to Table \ref{tab:table_signal_specifications} where 2048 time samples are produced at the transmitter for each OFDM/SEFDM symbol. There is no synchronization mechanism between the transmitter and the receiver. Therefore, the receiver would capture 2048 time samples and randomly truncate 1024 samples for training. At the training stage, 2,000 OFDM/SEFDM symbols are generated for each class (i.e. each $\alpha$) following the data augmentation principle in \cite{tongyang_VTC2020_DL_classification}. In this case, there are overall 8,000 symbols for the Type-I signal pattern and 14,000 symbols for the Type-II signal pattern. For testing, there are overall 3,200 OFDM/SEFDM symbols for Type-I and 5,600 symbols for Type-II.

\begin{table}[t!]
\caption{Signal specifications}
\centering
\begin{tabular}{ll}
\hline \hline
$\mathbf{Parameter}$ & $\mathbf{Signal}$  \\[0.5pt] \hline
Sampling frequency (kHz) & 200 \\
IFFT sample length & 2048 \\
Oversampling factor & 8  \\
No. of data sub-carriers & 256 \\
Bandwidth compression factor $\alpha$ & 1,0.95,0.9,0.85,0.8,0.75,0.7\\
Modulation scheme & QPSK  \\ \hline \hline
\label{tab:table_signal_specifications}
\end{tabular}
\end{table}

\begin{figure}[t!]
\begin{center}
\includegraphics[scale=0.375]{statistics_SVM_20dB.eps}
\end{center}
\caption{Statistical feature based SVM classification accuracy trained and tested at Es/N0=20 dB. }
\label{Fig:statistic_test_singleEsN0}
\end{figure}

At first, we assume a simple training and testing scenario, in which both the training data and testing data are contaminated by \ac{AWGN} at a single Es/N0=20 dB. Multiple time-domain statistical features are extracted from the training dataset, which are labelled as `T-Statistics-SVM'. Joint statistical features are investigated by combining each statistical feature. In addition, the raw data without any manual feature extractions is also evaluated. Results in Fig. \ref{Fig:statistic_test_singleEsN0} show that all the statistical features cannot properly classify Type-I signals. It should be noted that even the joint feature cannot improve the accuracy. Similar results are obtained as well for the Type-II signals which have more challenging signal feature-similarity issues. The same feature extraction and training operations are repeated to the frequency-domain dataset, which are labelled as `F-Statistics-SVM'. The same conclusion is obtained in Fig. \ref{Fig:statistic_test_singleEsN0} that single domain statistical features cannot classify signals even in the frequency-domain.

The above results naturally lead to the joint time-frequency analysis, which would enhance the feature extraction efficiency. Wavelet transform will create a two-dimensional time-frequency feature grid. The scale range of the Morse wavelet is configured to have 7 octaves and 10 scales per octave. Therefore, considering both real and imaginary part of a signal, there are overall 140 frequency scales. In terms of time scale, following the signal specifications in Table \ref{tab:table_signal_specifications} and the 50\% random symbol truncation mechanism, 1024 time sample scales will be reserved. Therefore, \ac{CWT} will generate a pair of two-dimensional 70$\times$1024 time-frequency analysis matrices.

\begin{figure}[t!]
\begin{center}
\includegraphics[scale=0.34]{wavelet_SVM_20dB.eps}
\end{center}
\caption{Wavelet based SVM classification accuracy trained and tested at Es/N0=20 dB. }
\label{Fig:wavelet_test_singleEsN0}
\end{figure}

There are many ways to reduce the time-frequency feature dimensionality. This work applies statistical transform to reduce the amount of time samples. Thus, the two-dimensional 70$\times$1024 time-frequency grid is simplified into a 70$\times$1 frequency-scale vector following the dimensionality reduction method in Fig. \ref{Fig:time_frequency_feature_grid_transform}. Different statistical transform methods are evaluated at each frequency scale and results are shown in Fig. \ref{Fig:wavelet_test_singleEsN0}. It is clearly seen that the IQR and variance features enable higher classification accuracy than other features, which can even classify the feature-similarity dominant Type-II signals. The following classifier training will be based on those two statistical features.

\begin{figure*}[ht]
\begin{center}
\includegraphics[scale=0.48]{wavelet_combined_single_mixed_EsN0.eps}
\end{center}
\caption{Wavelet classifier accuracy tested at Es/N0 ranging from -20 dB to 50 dB. The classifier is trained at (a) Es/N0=20 dB. (b) Es/N0=20, 30, 40 dB. (c) Es/N0=0, 10, 20 dB. (d) Es/N0=0, 10, 20, 30, 40 dB. (e) Es/N0=-20, -10, 0, 10, 20, 30, 40, 50 dB. (f) Es/N0=-20, -10, 0, 10, 20, 30, 40, 50 dB. Note: var indicates variance, iqr indicates interquartile range and vi indicates variance-interquartile range.  }
\label{Fig:wavelet_combined_single_mixed_EsN0}
\end{figure*}

A wavelet classifier is firstly trained using data at a fixed Es/N0=20 dB and tested at various Es/N0 with accuracy results shown in Fig. \ref{Fig:wavelet_combined_single_mixed_EsN0}(a). It clearly shows that all the curves reach the peak accuracy at 20 dB. However, for other Es/N0 values, accuracy drops significantly. It indicates that training data at a fixed Es/N0 is not robust to train a classifier that can classify signals at a wide range of Es/N0.

To train a robust classifier, a dataset covering different Es/N0 (20, 30, 40 dB) is generated. The classification results are shown in Fig. \ref{Fig:wavelet_combined_single_mixed_EsN0}(b), in which better accuracy is reached at high Es/N0 for both Type-I and Type-II signals. However, the accuracy at low Es/N0 still needs improvement.

To enhance the classification sensitivity at low Es/N0, a dataset, covering low Es/N0 (0, 10, 20 dBs), is trained with results shown in Fig. \ref{Fig:wavelet_combined_single_mixed_EsN0}(c). All the curves are raised to achieve higher accuracy at low Es/N0. It should be noted that the variance feature enabled wavelet classifier can identify signals even below noise power and it achieves 78\% classification accuracy when Es/N0=0 dB. However, its performance drops obviously at high Es/N0, especially those beyond Es/N0=20 dB. For the IQR feature trained classifiers, both Type-I and Type-II curves are stable at high Es/N0. It should be noted that the IQR feature trained Type-I classifier outperforms the variance feature trained model at high Es/N0. It is concluded from the figure that the variance trained model is robust at low Es/N0 while the IQR trained model is robust at high Es/N0.

Based on the above results, it is inferred that classifiers trained at high Es/N0 would enable high testing accuracy merely at high Es/N0 while classifiers trained at low Es/N0 would lead to high testing accuracy at low Es/N0. This indicates that a wider Es/N0 range has to be considered for the training data. In Fig. \ref{Fig:wavelet_combined_single_mixed_EsN0}(d), classifiers are trained with data covering an Es/N0 range from 0 dB to 40 dB with an increment step of 10 dB, which basically combines the two Es/N0 ranges in Fig. \ref{Fig:wavelet_combined_single_mixed_EsN0}(b) and Fig. \ref{Fig:wavelet_combined_single_mixed_EsN0}(c). It clearly shows accuracy improvement for all the curves at both low and high Es/N0. In Fig. \ref{Fig:wavelet_combined_single_mixed_EsN0}(e), a wider Es/N0 range between -20 dB and 50 dB is considered. The variance feature trained classifier shows apparent accuracy improvement for classifying Type-I signals at high Es/N0 while all other curves have no obvious improvement. However, there is still a minor performance degradation for the variance feature based classifier at high Es/N0 when compared with the IQR trained classifier. The robust feature performance of variance at low Es/N0 and IQR at high Es/N0 inspires to combine the two features for a more reliable classifier.

The composite classifiers, trained by joint variance and IQR features, can reach high classification accuracy for both Type-I and Type-II signals at both low and high Es/N0 ranges in Fig. \ref{Fig:wavelet_combined_single_mixed_EsN0}(f). Therefore, the composite classifiers will be used in the following over-the-air experiments.

\section{Low-Cost Experiment and Results}

The experiment is operated indoor in an open space, in which facilities would cause signal reflections and further result in frequency selective channel impairments. In addition, people movement in the space would cause Doppler spread and therefore dynamic spectral fluctuations. This work will use a pair of low-cost Analog Devices \ac{SDR}
PLUTO \cite{PlutoSDR} to practically transmit and classify over-the-air signals. The signals are designed according to Table \ref{tab:table_signal_specifications} and transmitted at a free-licensed 900 MHz (33-centimeter band) carrier frequency.

\begin{figure}[t!]
\begin{center}
\includegraphics[scale=0.4]{experiment_setup_wavelet_classification_PLUTO_SDR_v2.eps}
\end{center}
\caption{Low-cost experiment setup for the wavelet classifier training and testing. Four data collection points are labelled as `L-a', `L-b', `L-c' and `L-d'.}
\label{Fig:SVM_wavelet_classification_testbed}
\end{figure}

The experiment setup, shown in Fig. \ref{Fig:SVM_wavelet_classification_testbed}, is low cost since a laptop and two PLUTO devices are sufficient to realize signal generation, over-the-air transmission, signal reception and classifier training. In order to collect diversified data from an indoor environment, we fix the position of the transmitter side SDR device and place the receiver side SDR device at different locations. In this case, a number of training datasets, impaired by channel multipath fading, power degradation and Doppler effect, are collected. Unlike the CNN classifier where a large number of training symbols are required for feature extractions, the wavelet classifier can manually extract features based on a limited dataset. Therefore, in this experiment, at each location, 400 symbols are collected for the Type-I signal pattern and 700 symbols for the Type-II signal pattern. There are four data collections considering four different locations of the receiver. Therefore, the overall collected training symbols for Type-I and Type-II are 1,600 and 2,800, respectively. For testing, the same process is repeated with four data collections. To have a fair comparison with the previous work \cite{tongyang_VTC2020_DL_classification}, the number of testing symbols per class is fixed at 800.

\begin{figure}[t!]
\begin{center}
\includegraphics[scale=0.67]{confusion_matrix_wavelet_classification_Type_I_Type_II.eps}
\end{center}
\caption{Confusion matrix visualization for wavelet classification.}
\label{Fig:confusion_matrix_wavelet_classification_Type_I_Type_II}
\end{figure}

The collected data will be used to train wavelet classifiers off-line using Matlab. Once a wavelet classifier is trained, the model will be saved. Therefore, SDR devices will reuse the saved model for online signal classification and there is no need to re-train classifiers. Thus, the off-line training is a one-time operation. The confusion matrices are presented in Fig. \ref{Fig:confusion_matrix_wavelet_classification_Type_I_Type_II}. The classification accuracy for the Type-I signal pattern is nearly 100\%. For Type-II signals, the accuracy is 90\%, which is much higher than the 70.75\% in \cite{tongyang_VTC2020_DL_classification} where a transfer learning enabled \ac{CNN} classifier is applied.

\balance

\section{Conclusion}

This work aims to explore typical machine learning (ML) algorithms for non-orthogonal signal classification in non-cooperative communications. Multiple statistical approaches are tested for feature extractions in either time-domain or frequency-domain but showing unreliable classification accuracy. Wavelet transform is therefore applied to extract two-dimensional time-frequency features, which are further converted to a one-dimensional feature vector using statistical transform. Simulation results discovered that Es/N0 has great impacts on classification accuracy at the training stage. Results show increased classification accuracy over a wide range of training Es/N0. Classifiers are trained and tested with results showing that variance and IQR are the most efficient features. The combination of variance and IQR, associated with wavelet transform, enables classification accuracy up to 100\%. Furthermore, the wavelet classifier can even identify signals when the signal power is below its noise power. Results show that the variance feature enabled wavelet classifier achieves 78\% classification accuracy when Es/N0=0 dB. A low-cost experiment is set up using one laptop and two SDR devices. Practical results verify the efficacy of the wavelet enabled time-frequency features. Confusion matrices are obtained to show nearly 100\% classification accuracy for the Type-I signal pattern and 90\% accuracy for Type-II.

\bibliographystyle{IEEEtran}
\bibliography{Tongyang_Ref}

\end{document}


text:
{\bf Problem.} If the digit represented by $\triangle$ satisfies the following addition problem $$ \begin{array}{c@{}c@{\;}c@{}c@{}c@{}c} & & 3 & 2 & 1 & \triangle_6\\ & & & \triangle & 4 & 0_6\\ &+ & & & \triangle & 2_6\\ \cline{2-6} & & 4 & 2 & \triangle & 1_6\\ \end{array} $$what is the value of $\triangle$?
{\bf Level.} Level 4
{\bf Type.} Number Theory
{\bf Solution.} We notice that in the units column, it's impossible for $\triangle+2=1_6$. So, it must be the case that $\triangle+2=11_6=7$. That would mean $\triangle=7-2=\boxed{5}$. We can check that our answer works by plugging our value for triangle into the original problem: $$ \begin{array}{c@{}c@{\;}c@{}c@{}c@{}c} & & 3 & 2 & 1 & 5_6\\ & & & 5 & 4 & 0_6\\ &+ & & & 5 & 2_6\\ \cline{2-6} & & 4 & 2 & 5 & 1_6.\\ \end{array} $$The addition problem still works, so our answer is correct.

set_name: MATH


TITLE: Let $X_t$ be an Ornstein Uhlenbeck process. Is $X_t - \frac{1}{2} \int_o^{t} X_u du$ a Brownian motion?
QUESTION [1 upvotes]: Let $X_t=e^{-t}B_{e^{2t}}$ be an Ornstein Uhlenbeck process with $\mathbb{E}[X_tX_s]=e^{-|t-s|}$. Is $Y_t = X_t - \frac{1}{2} \int_o^{t}X_u du$ a Brownian motion? I managed to prove that $Y_t$ is a centered Gaussian process. The only thing I need to show is that it has $\mathbb{E}[Y_sY_t]=min(s,t)$, but I could not figure this out. Any helps would be appreciated.

REPLY [2 votes]: Hint: $EX_tX_s=e^{-t-s} (e^{2t}\wedge e^{2s})$. $EX_t\int_0^{s}X_u du=\int_0^{s} E(X_tX_u)du$ and $E\int_0^{t} X_u du \int_0^{s}X_vdv=\int_0^{t}\int_0^{s} E(X_uX_v)dudv$. All these are easy to compute.


TITLE: Time dilation, mass-energy equivalence, and implications for the passage of time
QUESTION [0 upvotes]: I'm an amateur physics enthusiast, and there is a question that's been in the back of my mind for some time that I haven't been able to answer on my own. First, I want to define a couple of terms and make sure my understanding isn't breaking down there before diving into my question on the implication.
First is the concept of time dilation. When an object / person / particle is traveling at an extremely high velocity or is near a massive gravitational object, they experience time dilation relative to an outside observer. E.g., me and my twin brother are on earth, I travel in a rocket at the speed of light for one light-year, and return to earth at the same velocity, I experienced the trip as having been instant, but my twin brother is now two years older. Is this generally the correct understanding?
Next is the concept of mass-energy equivalence expressed in Einstein's famous equation, that mass is highly concentrated energy. So effectively, any piece of matter (an object with mass), is electrical in nature and simply condensed down so far that it appears solid to us as an observer in spacetime. Again, please correct me here if I'm wrong.
If both of these definitions are correct, it seems to me that all of us as observers are really just highly condensed balls of electrical energy, traveling at light speed at an atomic or subatomic level. Does that mean that, as physical objects, we are not experiencing time, but rather just feel we are experiencing time as higher level observers?
When I google "Time is an illusion", I get a lot about Carlo Rovelli, of which I've read a decent amount (although not his book The Order of Time, just a synopsis so far), but I don't really get a definitive answer. I would love any input at all to help me better understand this quandary. Thank you!

REPLY [1 votes]: When an object / person / particle is traveling at an extremely high velocity or is near a massive gravitational object, they experience time dilation relative to an outside observer. [...] Is this generally the correct understanding?

It is commonly described that way, but it isn't a very good way of describing it. It's better to just say that elapsed time = length of worldline. Your movements over the course of your life can be drawn as a curve in spacetime, and the length of the curve is the length of your life. If you meet up with someone else at two points in your life, different amounts of time may have passed for the two of you between those points because the curves had different shapes between those points. The longer your worldline, the more room there is in it for the atomic processes that cause clocks to tick off the seconds, and human skin to accumulate wrinkles, and all of the other things that we associate with the passage of time. We don't notice differences of length most of the time because our worldlines are all almost straight and parallel to each other (we don't ever change speed by very much, on the scale of the speed of light).
The twin paradox is like the fact that if you drive in a straight line from A to B, while someone else takes a detour, their odometer will show a larger total distance traveled. The odometer that shows the larger distance will always be that of the car that changed direction at some point on the trip, simply because a straight line is the shortest distance between two points. In spacetime, because of the funny minus sign in the distance formula, it ends up being the other way around, and the straight line is longer instead of shorter, meaning that that person ends up older. Otherwise, every aspect of your intuition about the driving example should carry over to the worldline example.

Next is the concept of mass-energy equivalence expressed in Einstein's famous equation, that mass is highly concentrated energy. So effectively, any piece of matter (an object with mass), is electrical in nature and simply condensed down so far that it appears solid to us as an observer in spacetime.

Mass-energy equivalence means that mass and energy are two names for the same thing. So mass isn't highly concentrated energy, it's just energy.
This is complicated by the fact that the words "mass" and "energy" are used in several different ways, that aren't all equivalent. For example "rest mass" and "rest energy" are the same as each other but not the same as "total energy." And "total energy" and "relativistic mass" are the same thing, even though they don't use the same word as a modifier, but most physicists avoid the term "relativistic mass" because it's considered confusing. This is just the English language being a mess as usual, and isn't really related to the underlying physical concepts.
I don't understand why you say "electrical" since electricity isn't related to anything else in this question.

it seems to me that all of us as observers are really just highly condensed balls of [...] energy, traveling at light speed at an atomic or subatomic level.

This is sort of true. In modern physics, particles/fields (they are the same thing in quantum mechanics) that have mass can be understood as two separate massless fields that are coupled to each other. In the particle picture, you can to some extent imagine this as a particle that moves at the speed of light at every instant, but constantly reverses direction so that over longer time periods its average speed is less than the speed of light.

Does that mean that, as physical objects, we are not experiencing time, but rather just feel we are experiencing time as higher level observers?

We really experience time.
To put it another way, if we don't experience time, then nothing does. Then there would be no use for the phrase "experiencing time" and instead you'd have to use "feel as though experiencing time" everywhere instead. Then we'd probably drop the "feel as though" part for brevity (which wouldn't create ambiguity since we weren't using "experiencing time" by itself any more), which would put us back where we started. There's no meaningful difference between something really being true and merely seeming in every way to be true.
The exact nature of our experience of time comes down to the ways in which the various fundamental particles we're made of interact with each other. This is very complicated in detail, but the key point is that it's the same everywhere (throughout the visible universe and all the way back to shortly after the big bang, as far as we can tell). So it's complicated but in a consistent way. That's why you can talk about the relative aging of two twins, for example: even though they're in different places between their two meeting points, the particle interactions that make them older are the same, and in the one with the longer worldline there's room for more of those interactions.


TITLE: Removing zero-counts in exponential decay measurement
QUESTION [1 upvotes]: I'm doing an experiment where I'm measuring the mean-lifetime of muons. I have a set of data points for the number of decays against time, resembling an exponential  distribution (of course). But in the experiment, there weren't a lot of counts measured (approximately 500) so for some time-values I got zero counts where it doesn't really fit to the exponential distribution.
I was wondering if it is justifiable to remove these points and what the reasoning behind that is. It's a much better fit if I do remove them compared to the experimental value of the muon lifetime given by e.g.
Source- M. Tanabashi et al. (Particle Data Group),
Review of Particle Physics, Phys. Rev. D 98, 030001 (2018)

REPLY [2 votes]: From your description of the data, it appears you choose some time interval $T_k$, and then count the number of decays, $N_k$, observed during that interval, and then repeat for many different intervals: $k \in [1, 2, ..., n]$.
Then you posit:
$$ N_k \propto e^{T_k/\tau} $$
This is not correct (see others' comments).
Given a uniform random decay rate, the time between decays will be exponentially distributed and the number of decays in a fixed time interval with be Poisson distributed:
$$P(N_k) = \frac{\lambda_k^{N_k}e^{-\lambda_k}}{N_k!}$$
where:
$$ \lambda_k \equiv \frac{T_k}{\tau}$$
where $\tau$ is the lifetime.
At this point, there is nothing to plot (unless all $T_k$ are equal, then you should histogram $N_k$ and get a Poisson distribution that can be fit). If you have many different $T_k$, and:
$$ \sum_{k=1}^n{N_k} \approx 500 $$
then simultaneously fitting all $T_k$ may not be the best approach. Here you want to use maximum-likelihood. Note that in this case, the $N_k=0$ data are extremely important, as:
$$ P(0) = e^{-T_k/\tau} $$
is a valid, and expected, measurement.


TITLE: Show that the Jordan measure is zero
QUESTION [2 upvotes]: I want to show that if C is a Jordan measurable and a measure 0 set then $\int_C 1 =0$ (this is, its Jordan Measure is zero)
My approach is to show that $L(X_C,P)=0$ for all partition $P$, But I am not sure in how to proceed in this. Thanks a lot for your help.
Note:
My definition of measure 0 is that there exist rectangles $R_1,R_2,\ldots$ such that: $C \subset \bigcup R_i$ and the the sum of the volumes $\sum v(R_i) < \epsilon$.
Jordan measurable for me is a set tha is bounded and its boundary has measure zero.

REPLY [0 votes]: If there is some non trivial rectangle $R$ such that
$\inf_{x \in R} 1_C(x) \neq 0$ then you must have $R \subset C$, in which
case $C$ cannot have measure zero. Hence $\inf_{x \in R} 1_C(x) = 0$ for
all rectangles $R$, and so
$L(1_C, P) = 0$.


TITLE: Measurable Maps and Continuous Functions
QUESTION [2 upvotes]: I'm having a bit of trouble understanding the connection between measurable maps and continuous functions.  Namely, I know the following: Assume $f:(\Omega_{1}, \mathcal{B}_{1}) \rightarrow (\Omega_{2}, \mathcal{B}_{2})$

$E \subset \Omega_{2} \quad \mbox{open} \Rightarrow f^{-1}(E) \quad \mbox{open} $ by continuity
$f$ is a measurable map $\Rightarrow$ $( A \in \mathcal{B}_{2} \Rightarrow f^{-1}(A) \in \mathcal{B} _{1})$

What I'm uncertain of is the connection between the two.  Say, for instance, that $\Omega_{2} = \mathbb{R}$ and $\mathcal{B}_{2} = \mathcal{B}$, the $\sigma$-algebra generated by the collection of open intervals of $\mathbb{R}$ and $\Omega_{2} = \mathbb{R}^{d}$ and $\mathcal{B}_{2} = \mathcal{B}^{d}$, the same for $\mathbb{R}^{d}$.  Does knowledge of how $\mathcal{B}_{1}$ and $\mathcal{B}_{2}$ were generated help?  And if so, how does it lead the ability to talk about the whole of $\mathcal{B}_{1}$?
On the other hand, I've heard it mentioned that I can use compliments and limits of sets to somehow say that $f^{-1}(A)$ is measurable for all intervals $A \subset \mathbb{R}$, regardless of whether it's open, closed, or half-open, just by knowing that 1. from above.  How is that so?

REPLY [5 votes]: If $\Omega_1$ and $\Omega_2$ are topological spaces and $\mathcal{B}_1$ and $\mathcal{B}_2$ are their Borel sigma-algebras, then every continuous function is measurable (but, in general, not every measurable function is continuous).
Proof. Let $u:\Omega_1\to\Omega_2$ be a continuous function. That $u$ is $\mathcal{B}_1/\mathcal{B}_2$ measurable is equivalent to the assertion that $\mathcal{C}=\mathcal{B}_2$, where
$$
\mathcal{C}=\{B\in\mathcal{B}_2\mid u^{-1}(B)\in\mathcal{B}_1\}.
$$
Thus, one has to show that $\mathcal{C}=\mathcal{B}_2$. This is done in two steps.

First, $u$ is continuous hence one already knows that $\mathcal{T}_2\subseteq\mathcal{C}$, where $\mathcal{T}_2$ is the topology of $\Omega_2$ (in other words, $\mathcal{T}_2$ is the collection of its open sets).
Since $\mathcal{C}\subseteq\mathcal{B}_2$ by definition, one gets the double inclusion
$$
\mathcal{T}_2\subseteq\mathcal{C}\subseteq\mathcal{B}_2.
$$
By definition of Borel sigma-algebras, $\mathcal{B}_2=\sigma(\mathcal{T}_2)$, that is, $\mathcal{B}_2$ is the sigma-algebra generated by $\mathcal{T}_2$ (or equivalently, the smallest sigma-algebra which contains $\mathcal{T}_2$). Hence, the double inclusion above implies
$$\mathcal{B}_2=\sigma(\mathcal{T}_2)\subseteq\sigma(\mathcal{C})\subseteq\sigma(\mathcal{B}_2)=\mathcal{B}_2,
$$
that is, $\sigma(\mathcal{C})=\mathcal{B}_2$.
Second, and this will be enough to conclude, one wants to show that $\mathcal{C}=\sigma(\mathcal{C})$, or equivalently, that $\mathcal{C}$ is a sigma-algebra. It happens that $\mathcal{C}$ fulfills all the axioms defining a sigma-algebra and that it is easy to show, so I will be lazy and leave you this part of the proof.


TITLE: How to count degrees of freedom
QUESTION [0 upvotes]: I am able to visualize and see that $y=A \sin{x}$ has 1 degree of freedom, because $z=0$ and $y$ depends on $x$. However, its plot looks like a 2D plane, even though according to the DOF it is a 1D motion. Or am I just missing some point in it? Also if any of $x$, $y$ or $z$ depends on another one of them, then it is not a freedom for that particle to move freely in it, right?

REPLY [1 votes]: If a particle is constrained to move along the curve you described, it is one DOF because you only need one coordinate to describe the configuration of the particle. The easiest way to see this is to use the arc length along the curve as the parameter for locating the particle. Give me the arc length $s$ and I can tell you where the particle is:

Another way to think about it is if you give me $x$, I can also tell where the particle is. It is at $(x,A \sin(x), 0)$, as another author explained.
A particle moving along a helix is also 1DOF motion, despite the helix being a space curve that occupies 3D space. You gave me an example involving a plane curve.


text:
\appendix

\section{An estimate for integrals over a neighborhood of the boundary}

The following lemma is a slight modification of~\cite[Lemma~5.1]{PSu:2012}.
\begin{lem}
\label{lem: Norm in the neighborhood}Let⋅$\Sigma$ be a uniformly
weakly Lipschitz domain in~$\R^{d}$. Then for each⋅$q\in[1,\infty)$
and any~$\varepsilon>0$
\begin{equation}
\norm u_{q,\rbr{\partial\Sigma}_{\varepsilon}\cap\Sigma}\lesssim\varepsilon^{1/q}\norm u_{1,q,\Sigma}^{1/q}\norm u_{q,\Sigma}^{1/q^{\dual}},\qquad u\in C_{c}^{\infty}\rbr{\wb{\Sigma}}.\label{est: Norm in the neighborhood}
\end{equation}
The⋅constant in the inequality depends only on⋅$q$,⋅$d$ and~$\Sigma$.
\end{lem}
\begin{proof}
Recall that $B$ denotes the open unit ball centered at the origin
and $B_{+}$ denotes the open unit half-ball with~$x_{d}\in\rbr{0,1}$.
Let $S_{t}$ be the cross-section of⋅$B$ at⋅$x_{d}=t$ and $P_{t}$
be the piece of⋅$B_{+}$ with~$x_{d}\in\rbr{0,t}$. If⋅$\rbr{W_{k},\omega_{k}}$
are local boundary coordinate patches, then $\omega_{k}\rbr{W_{k}\cap\Sigma}=B_{+}$
and⋅$\omega_{k}\rbr{W_{k}\cap\partial\Sigma}=S_{0}$, and for any⋅$y\in\omega_{k}\rbr{W_{k}\cap\Sigma}$
\[
\dist\rbr{y,S_{0}}\le L_{\Sigma}\dist\rbr{x,W_{k}\cap\partial\Sigma},
\]
where~$x=\omega_{k}^{-1}\rbr y$ and~$L_{\Sigma}=\sup_{k}\seminorm{\omega_{k}}_{C^{0,1}}$.
It⋅follows that $\omega_{k}\rbr{W_{k}\cap\rbr{\partial\Sigma}_{\varepsilon}\cap\Sigma}\subset P_{\varepsilon/\varepsilon_{1}}$
with~$\varepsilon_{1}\cellradius=L_{\Sigma}^{-1}$. On⋅the⋅other
hand, we know that the cover is sufficiently tight in the sense that
the union of⋅$\omega_{k}^{-1}\rbr{B_{+}}$ contains $\rbr{\partial\Sigma}_{\delta}\cap\Sigma$
for some~$\delta>0$. Therefore, taking $\varepsilon_{0}=\varepsilon_{1}\meet\delta$,
we can insure that $\rbr{\partial\Sigma}_{\varepsilon}\cap\Sigma$
is covered by⋅$\cbr{W_{k}}$ for any~$\varepsilon\le\varepsilon_{0}$.

Now, using a partition of unity~$\cbr{\varphi_{k}}$ subordinate
to⋅$\cbr{W_{k}}$ (see⋅Section~\ref{sec: Notation}) and making a
change of variables to flatten out the boundary, we reduce⋅(\ref{est: Norm in the neighborhood})
to proving that, for any $\varepsilon\le\varepsilon_{0}$ and any
smooth function~$u$ on⋅$B_{+}$ vanishing near the boundary of⋅$B$,
it holds that
\begin{equation}
\norm u_{q,P_{\varepsilon/\varepsilon_{0}}}\lesssim\varepsilon^{1/q}\norm u_{1,q,B_{+}}^{1/q}\norm u_{q,B_{+}}^{1-1/q}.\label{est: Norm in the neighborhood for half-ball}
\end{equation}
By⋅the⋅divergence theorem, for any⋅$t\in\rbr{0,1}$ we have
\[
\int_{S_{t}}\abs{u\rbr{x^{\prime},t}}^{q}\dd x^{\prime}=-\int_{B_{+}\!\setminus P_{t}}\partial_{x_{d}}\abs{u\rbr x}^{q}\dd x,
\]
and hence
\[
\begin{aligned}\int_{S_{t}}\abs{u\rbr{x^{\prime},t}}^{q}\dd x^{\prime} & \le q\int_{B_{+}\!\setminus P_{t}}\abs{\partial_{x_{d}}u\rbr x}\abs{u\rbr x}^{q-1}\dd x\\
 & \le q\biggrbr{\int_{B_{+}}\abs{\partial_{x_{d}}u\rbr x}^{q}\dd x}^{1/q}\biggrbr{\int_{B_{+}}\abs{u\rbr x}^{q}\dd x}^{1-1/q}.
\end{aligned}
\]
Integrating in $t$ from $0$ to $\varepsilon/\varepsilon_{0}$ now
gives~(\ref{est: Norm in the neighborhood for half-ball}).
\end{proof}

config: arxiv


TITLE: Finding the equation of an ellipse using eccentricity
QUESTION [0 upvotes]: The ellipse $E$ has eccentricity $\frac{1}{2}$, focus $(0,0)$ with the line $x=−1$ as the corresponding directrix. Find an equation for $E$.
The equation I get is $(x+\frac{1}{3})^2 + y^2=\frac{4}{9}$ which is a circle, radius $\frac{2}{3}$. Is this wrong? I feel it must be because I then have to find the other focus and directrix.

REPLY [0 votes]: Let's do this step by step.
\begin{align*}
\lvert PF\rvert &= e\,\lvert PL\rvert \\
\lvert PF\rvert^2 &= e^2\,\lvert PL\rvert^2 \\
x^2+y^2 &= \left(\tfrac12\right)^2(x+1)^2 \\
4x^2+4y^2 &= x^2+2x+1 \\
3x^2-2x+4y^2 &= 1 \\
3\left(x^2-2x\cdot\tfrac13+\tfrac19\right) + 4y^2 &= 1+3\cdot\tfrac19 \\
3\left(x-\tfrac13\right)^2 + 4y^2 &= \tfrac43 \\
\tfrac94\left(x-\tfrac13\right)^2 + 3y^2 &= 1 \\
\left(\frac{x-\tfrac13}{\tfrac23}\right)^2 +
\left(\frac{y-0}{\sqrt{\tfrac13}}\right)^2 &= 1
\end{align*}
So the center would be at $(\tfrac13,0)$, the length of the semiaxis in $x$ direction would be $\tfrac23$ and the length of the semiaxis in $y$ direction would be $\sqrt{\tfrac13}$. The grid in the following illustration has distance $\tfrac13$ between the lines.

Note that the above computation heavily relies on the fact that the directrix is vertical, which makes the whole ellipse axis-aligned. Otherwise we'd have to account for rotation, which would complicate things. See this post of mine for reading geometric properties of a generic equation. In the case of a focus and directrix description, it would be easier to first transform coordinates to axis-align the directrix, though.
I'm not sure how you obtained your equation, so I can't tell you where you went wrong.


TITLE: Simultaneously diagonalizable linear operators on a space of square matrices
QUESTION [0 upvotes]: Let $V$ be the space of $n×n$ matrices over the field $F$. For a fixed $n×n$ matrix $A$ over $F$, let $T_A$ be the linear operator on $V$ defined by $T_A(B) = AB −BA$. Consider the family of linear operators obtained by letting $A$ vary over all diagonal matrices. Prove that the operators in this family are simultaneously diagonalizable.

Consider any two linear operators $T_{A_1}$ and $T_{A_2}$, then $T_{A_2}\left( T_{A_1}(B) \right) = T_{A_2}(A_1B-BA_1) = $ $ A_2(A_1B-BA_1)-(A_1B-BA_1)A_2 = A_2A_1B-A_2BA_1-A_1BA_2+BA_1A_2 = $ $A_1(A_2B-BA_2)-(A_2B-BA_2)A_1 = T_{A_1}\left( T_{A_2}(B) \right)$. Hence $T_{A_2}$ and $T_{A_1} $ commute. Now it is enough to show that $T_{A_2}$ and $T_{A_1} $ are diagonalizable, since operators are simultaneously diagonalizable if and only if they commute and are diagonalizable.  But how do i show that $T_{A_2}$ and $T_{A_1} $ are diagonalizable?

REPLY [1 votes]: The effect of the (additive) commutator $T_A$ with a diagonal matrix$~A$ on an elementary matrix $E_{k,l}=(\delta_{i,k}\delta_{j,l})_{i,j=1}^n$ is scalar multiplication by $a_i-a_j$, where $a_i$ is the diagonal entry of$~A$ at position $i,i$. Therefore all such operators $T_A$ diagonalise on the basis $\{E_{k,l}\mid k=1,\ldots,n; l=1,\ldots,n\,\}$ of$~V$.


text:
\begin{document}

\subsection{Stochastic Automata Networks}
\label{subsection:stochastic_automata_networks}

For a continuous-time Markov chain of interacting processes the discrete state space factorizes in a natural way into the state spaces of the individual processes.
Each process is itself a Markov chain over its own state space $\Space_{\mu}$ and is called a \emph{stochastic automaton} $\A_{\mu}$.
The set $\{\A_1, \dots, \A_d\}$ of $d$ stochastic automata is called a \emph{Stochastic Automata Network}, cf.~\cite{Plateau2000}.
The full state space $\Space$ consists of all possible combinations of states in each automaton, i.e., for $n_{\mu} = \vert \Space_{\mu} \vert$ and $n = \max_{\mu} n_{\mu}$ it is given by
\begin{align}
\label{eq:state_space}
	\Space = \bigtimes\limits_{\mu = 1}^d \Space_{\mu} \qquad \text{with} \qquad \vert \Space \vert = \prod\limits_{\mu = 1}^d n_{\mu} = \mathcal{O}\left(n^d\right).
\end{align}
Each state $x \in S$ specifies its state $x[\mu] \in \Space_{\mu}$ in each automaton $\A_{\mu}$.

We now consider transitions between states in the full state space $\Space$, which are given by one or more transitions between states within the individual state spaces $\Space_{\mu}$.
Such transitions are called \emph{functional}/\emph{constant} and \emph{synchronized}/\emph{local} according to the following definitions.
\begin{itemize}
	\item[(a)] 	A transition is called functional if the transition rate is a function of the current state in $\Space$. Otherwise it is called constant.
	\item[(b)] A transition in a state space $\Space_{\mu}$ may force or prevent transitions in another state space $\Space_{\nu}$ at the same time. Such transitions in $\Space$ are called synchronized. Otherwise they are called local.
\end{itemize}
It is known that the infinitesimal generator $\Q$ of a Stochastic Automata Network with $d$ automata can be represented as a sum of $d$ Kronecker products, cf.~\cite{Plateau2000}.
To this end,  the infinitesimal generator is split into two parts describing local and synchronizing transitions, respectively, i.e., $\Q = \Q_{\text{local}} + \Q_{\text{sync}} $.
Since the local transitions only change the state in their corresponding automaton, we represent the local part as
\begin{align}
\label{eq:local_transition_part}
	\Q_{\text{local}}  = \sum\limits_{\nu = 1}^d ~ \bigotimes\limits_{\mu < \nu} \Id_{\Space_\mu} \otimes ~\Q_{\text{local}} ^{(\nu)} \otimes \bigotimes\limits_{\mu > \nu} \Id_{\Space_\mu}
\end{align}
where $\Id_{\Space_{\mu}}$ denotes the identity in $\Space_{\mu}$ and $\Q_{\text{local}} ^{(\nu)}$ describes the local transition in $\Space_{\nu}$.
The synchronized transitions are represented similarly. Instead of one term of Kronecker products, two terms are needed for each synchronized transition, cf.~\cite{Plateau2000}.
The whole infinitesimal generator for $d$ automata and $s$ synchronized transitions is given by
\begin{align}
\label{eq:general_cp_transiton_matrix}
	\Q = \Q_{\text{local}} + \Q_{\text{sync}} = \sum\limits_{\nu = 1}^{2 s + d} ~ \bigotimes\limits_{\mu = 1}^d \Q_{\nu}^{(\mu)}.
\end{align}
Instead of $\vert \Space \vert \cdot \vert \Space \vert = \mathcal{O}\left(n^{2d}\right)$ entries, one only has to store $\left(2 s + d\right) d n^2$ entries.
Following the current state of the art in tumor progression models we only consider local transitions from now on.

\end{document}