KaTeX demo

KaTeX demo Math formulae to demo LaTeX to MathML rendering 2020-04-15 demo math Given two discrete-time systems A and B connected in cascade to form a new system C = x \mapsto B(A(x)), we examine the following properties: ## Linearity If A and B are linear, i.e. for all signals x_i and scalars a_i,

\begin{aligned}
  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{aligned}

then C is also linear

\begin{aligned}
  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{aligned}

## Time Invariance If A and B are time invariant, i.e. for all signals x and integers k,

\begin{aligned}
  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{aligned}

then C is also time invariant

\begin{aligned}
  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{aligned}

## 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. ## Causality If A and B are causal, i.e. for all signals x, y and any choise of integer k,

\begin{aligned}
  \forall n < k, x[n] = y[n]\quad
  \Longrightarrow &\;\begin{cases}
  \forall n < k, A(x)[n] = A(y)[n]\\
  \forall n < k, B(x)[n] = B(y)[n]
  \end{cases}\\
  \Longrightarrow &\;\forall n < k, B(A(x))[n] = B(A(y))[n]\\
  \Longleftrightarrow &\;\forall n < k, C(x)[n] = C(y)[n]
\end{aligned}

then C is also causal. ## BIBO Stability If A and B are stable, i.e. there exists a signal x and scalars a and b that for all integers n,

\begin{aligned}
  |x[n]| < a &\Longrightarrow |A(x)[n]| < b\\
  |x[n]| < a &\Longrightarrow |B(x)[n]| < b
\end{aligned}

then C is also stable, i.e. there exists a signal x and scalars a, b and c that for all integers n,

\begin{aligned}
  |x[n]| < a\quad
  \Longrightarrow &\;|A(x)[n]| < b\\
  \Longrightarrow &\;|B(A(x))[n]| < c\\
  \Longleftrightarrow &\;|C(x)[n]| < c
\end{aligned}