System Cascade Connection

Linearity

If $A$ and $B$ are linear, i.e. for all signals $x_i$ and scalars $a_i$,

$$\{\begin{array}{l}A\left(\underset{i}{\Sigma }{a}_i{x}_i\right[\cdot \left]\right)=\underset{i}{\Sigma }{a}_{i}A\left(x_i\right)[\cdot ]\\ B\left(\underset{i}{\Sigma }{a}_i{x}_i\right[\cdot \left]\right)=\underset{i}{\Sigma }{a}_{i}B\left(x_i\right)[\cdot ]\end{array}(null)$$

then $C$ is also linear

$$\begin{array}{cc}\hfill C\left(\underset{i}{\Sigma }{a}_i{x}_i\right[\cdot \left]\right)& =B\left(A\right(\underset{i}{\Sigma }{a}_i{x}_i[\cdot ]\left)\right)\hfill \\ \hfill x& =B\left(\underset{i}{\Sigma }{a}_{i}A\right(x_i\left)\right[\cdot \left]\right)\hfill \\ \hfill x& =\underset{i}{\Sigma }{a}_{i}B\left(A\right(x_i\left)\right)[\cdot ]\hfill \end{array}$$

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}]