12.5: Transformaciones teóricas numéricas para convolución
- Page ID
- 87087
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Aquí observamos las condiciones colocadas en una transformada lineal general para que pueda soportar la convolución cíclica. La forma de una transformación lineal de una secuencia longitud-N de número viene dada por
\[X(k)=\sum_{n=0}^{N-1}t(n,k)x(n) \nonumber \]
para\(k=0,1,...,(N-1)\). La definición de convolución cíclica de dos secuencias viene dada por
\[y(n)=\sum_{m=0}^{N-1}x(m)h(n-m) \nonumber \]
para\(n=0,1,...,(N-1)\) y todos los índices evaluados módulo\(N\). Nos gustaría encontrar las propiedades de la transformación tales que soportará la convolución cíclica. Esto significa que si\(X(k),\; H(k),\; Y(k)\) son las transformaciones de\(x(n),\; h(n),\; y(n)\) respectivamente,
\[Y(k)=X(k)H(k) \nonumber \]
Las condiciones se derivan tomando la transformada definida en las ecuaciones anteriores de ambos lados de la ecuación que da
\[Y(k)=\sum_{n=0}^{N-1}t(n,k)\sum_{m=0}^{N-1}x(m)h(n-m) \nonumber \]
\[Y(k)=\sum_{m=0}^{N-1}\sum_{n=0}^{N-1}x(m)h(n-m)t(n,k) \nonumber \]
Haciendo el cambio de las variables de índice,\(l=n-m\) da
\[Y(k)=\sum_{m=0}^{N-1}\sum_{l=0}^{N-1}x(m)h(l)t(l+m,k) \nonumber \]
Pero a partir de la ecuación, esto debe ser
\[Y(k)=\sum_{n=0}^{N-1}x(n)t(n,k)\sum_{m=0}^{N-1}x(m)t(m,k) \nonumber \]
\[Y(k)=\sum_{m=0}^{N-1}\sum_{l=0}^{N-1}x(m)h(l)t(n,k)t(l,k) \nonumber \]
Esto debe ser cierto para todos\(\(x(n),\; h(n)\) y\(k\), por lo tanto de las ecuaciones anteriores tenemos
\[t(m+l,k)=t(m,k)t(l,k) \nonumber \]
Para\(l=0\) nosotros tenemos
\[t(m,k)=t(m,k)t(0,k) \nonumber \]
Por lo tanto,\(t(0,k)=1\). Para\(l=m\) nosotros tenemos
\[t(2m,k)=t(m,k)t(m,k)=t^2(m,k) \nonumber \]
Porque de igual manera\(l=pm\) tenemos
\[t(pm,k)=t^p(m,k) \nonumber \]
Por lo tanto,
\[t^N(m,k)=t(Nm,k)=t(0,k)=1 \nonumber \]
Pero
\[t(m,k)=t^m(1,k)=t^k(m,1) \nonumber \]
Por lo tanto,
\[t(m,k)=t^{mk}(1,1) \nonumber \]
Definir\(t(1,1)=\alpha\) da la forma para nuestra ecuación de transformación lineal general como
\[X(k)=\sum_{n=0}^{N-1}\alpha ^{nk}x(n) \nonumber \]
donde\(\alpha\) es una raíz de orden\(N\)