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System Cascade Connection
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\documentclass[a4paper,12pt]{article} | |
\usepackage[english,vietnamese]{babel} | |
\usepackage{amsmath} | |
\usepackage{enumerate} | |
\usepackage{lmodern} | |
\title{System Cascade Connection} | |
\author{{\selectlanguage{vietnamese}Nguyễn Gia Phong}} | |
\begin{document} | |
\selectlanguage{english}\maketitle | |
Given two discrete-time systems $A$ and $B$ connected in cascade to form | |
a new system $C = x \mapsto B(A(x))$. | |
\section{Linearity} | |
If $A$ and $B$ are linear, i.e. for all signals $x_i$ and scalars $a_i$, | |
\begin{align*} | |
A\left(n \mapsto \sum_i a_i x_i[n]\right) &= n \mapsto \sum_i a_i A(x_i)[n]\\ | |
B\left(n \mapsto \sum_i a_i x_i[n]\right) &= n \mapsto \sum_i a_i B(x_i)[n] | |
\end{align*} | |
then $C$ is also linear | |
\begin{align*} | |
C\left(n \mapsto \sum_i a_i x_i[n]\right) | |
&= B\left(A\left(n \mapsto \sum_i a_i x_i[n]\right)\right)\\ | |
&= B\left(n \mapsto \sum_i a_i A(x_i)[n]\right)\\ | |
&= n \mapsto \sum_i a_i B(A(x_i))[n]\\ | |
&= n \mapsto \sum_i a_i C(x_i)[n] | |
\end{align*} | |
\section{Time Invariance} | |
If $A$ and $B$ are time invariant, i.e. for all signals $x$ and integers $k$, | |
\begin{align*} | |
A(n \mapsto x[n - k]) &= n \mapsto A(x)[n - k]\\ | |
B(n \mapsto x[n - k]) &= n \mapsto B(x)[n - k]\\ | |
\end{align*} | |
then $C$ is also time invariant | |
\begin{align*} | |
C(n \mapsto x[n - k]) | |
&= B(A(n \mapsto x[n - k]))\\ | |
&= B(n \mapsto A(x)[n - k])\\ | |
&= n \mapsto B(A(x))[n - k]\\ | |
&= n \mapsto C(x)[n - k] | |
\end{align*} | |
\section{LTI Ordering} | |
If $A$ and $B$ are linear and time-invariant, there exists signals $g$ and $h$ | |
such that for all signals $x$, $A = x \mapsto x * g$ and $B = x \mapsto x * h$, | |
thus \[B(A(x)) = B(x * g) = x * g * h = x * h * g = A(x * h) = A(B(x))\] | |
or interchanging $A$ and $B$ order does not change $C$. | |
\section{Causality} | |
If $A$ and $B$ are causal, i.e. for all signals $x$, $y$ and integers $k$, | |
\begin{multline*} | |
x[n] = y[n]\quad\forall n < k | |
\Longrightarrow\begin{cases} | |
A(x)[n] = A(y)[n]\quad\forall n < k\\ | |
B(x)[n] = B(y)[n]\quad\forall n < k | |
\end{cases}\\ | |
\Longrightarrow B(A(x))[n] = B(A(y))[n]\quad\forall n < k | |
\iff C(x)[n] = C(y)[n]\quad\forall n < k | |
\end{multline*} | |
then $C$ is also causal. | |
\section{BIBO Stability} | |
If $A$ and $B$ are stable, i.e. there exists a signal $x$ | |
and scalars $a$, $b$ that | |
\begin{align*} | |
|x[n]| < a\quad\forall n \in \mathbb Z | |
&\Longrightarrow |A(x)[n]| < b\quad\forall n \in \mathbb Z\\ | |
|x[n]| < a\quad\forall n \in \mathbb Z | |
&\Longrightarrow |B(x)[n]| < b\quad\forall n \in \mathbb Z | |
\end{align*} | |
then $C$ is also stable, i.e. there exists a signal $x$ | |
and scalars $a$, $b$, $c$ that | |
\begin{align*} | |
|x[n]| < a\;\forall n \in \mathbb Z | |
&\Longrightarrow |A(x)[n]| < b\;\forall n \in \mathbb Z\\ | |
&\Longrightarrow |B(A(x))[n]| < c\;\forall n \in \mathbb Z | |
\iff |C(x)[n]| < c\;\forall n \in \mathbb Z | |
\end{align*} | |
\end{document} |
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