Lecture 5 - Notes

January 13, 2016

Stable Systems

definition: A system is BIBO (Bounded Input Bounded Output) Stable if and only if every bounded input produces a bounded output.3

If,

$$\begin{align} \left| x[n] \right| \lt B_x \end{align}$$

and,

$$\begin{align} \left| y[n] \right| \lt B_y \end{align}$$

for all $n$, and for some $B_x, B_y \in \mathbb R$ then the system is BIBO Stable.

Causal Systems

definition: A system is causal if its output does not depend on future input.

Examples

The system for the forward difference,

$$y[n] = x[n+1] - x[n]$$

is not causal because it depends on future values of $x[n]$, namely $x[n+1]$. The system for the backward difference,

$$y[n] = x[n] - x[n-1]$$

is causal because it only depends on past values of $x[n]$.

Causal Signals

definition: A signal is causal if,

$$x[n] = 0 ~\forall~ n \lt 0$$

Testing System Properties

Example

Determine the system given by,

$$T(x[n]) = \left(\cos{\pi n}\right)x[n]$$

is,

• Stable
• Causal
• Linear
• Time-Invariant

Solution

We can reduce the system to,

$$\begin{align} y[n] &= \left(\cos{\pi n}\right)x[n] \newline &= (-1)^n x[n] \end{align}$$

Stability

If $\left|x[n]\right| \lt B_x$, then since $(-1)^n$ is bounded for all $n$,

$$\begin{align} \left|y[n] \right| \lt B_x \end{align}$$

and the system is bounded.

Causality

By inspections we can see that $y[n]$ does not really on any values other than $x[n]$, therefor it is causal.

Linearity

Let,

$$x[n] = a_1 x_1[n] + a_2 x_2[n]$$

so,

$$\begin{align} y[n] &= (1)^n x[n] \newline &= (1)^n \left(a_1 x_1[n] + a_2 x_2[n]\right) \newline &= (-1)^n a_1 x_1[n] + (-1)^n a_2 x_2[n] \newline &= a_1 \left[(-1)^n x_1[n]\right] + a_2 \left[(-1)^n x_2[n]\right] \newline &= a_1 y_1[n] + a_2 y_2[n] \newline \end{align}$$

and the system is linear.

Time-Invariant

Given,

$$y[n] = (-1)^n x[n]$$

If we test we can find,

$$\begin{align} y[n] &= x[n] \newline y[n-1] &\neq x[n-1] \end{align}$$

so the system is not time-invariant.

Linear Time-Invariant Systems

definition: A Linear Time-Invariant (LTI) system is characterized by its impulse response $h[n]$, i.e., $y[n] = h[n]$ when $x[n] = \delta[n]$.

Example - Ideal Delay

Suppose,

$$\begin{align} y[n] = x[n - n_d] \end{align}$$

where $-\infty \lt n \lt \infty$. So,

$$\begin{align} h[n] &= \delta[n - n_d] \end{align}$$

Example - Moving Average

Suppose,

$$\begin{align} y[n] = \frac{1}{M_1 + M_2 + 1} \sum_{k = -M_1}^{M_2} x[n-k] \end{align}$$

So,

$$\begin{align} h[n] &= \frac{1}{M_1 + M_2 + 1} \sum_{k = -M_1}^{M_2} \delta[n-k] \newline &= \frac{1}{M_1 + M_2 + 1} \sum_{k = -M_1}^{M_2} \delta[n-k] \newline \end{align}$$

note this system is not causal.

Example - Accumulator

Suppose,

$$\begin{align} y[n] = \sum_{k = -\infty}^{n} x[k] \end{align}$$

then,

$$\begin{align} h[n] &= \sum_{k = -\infty}^{n} \delta[k] \newline &= \sum_{m = 0}^{\infty} \delta[n - m] \newline &= u[n] \newline \end{align}$$

Example - Downsampler

Suppose,

$$\begin{align} y[n] = x[Mn] \end{align}$$

in this case the system is not time-invariant. So we can't compute the unit response.

Example - Forward Difference

Suppose,

$$\begin{align} y[n] = x[n+1] - x[n] \end{align}$$

then,

$$\begin{align} h[n] = \delta[n+1] - \delta[n] \end{align}$$

Example - Backward Difference

Suppose,

$$\begin{align} y[n] = x[n] - x[n - 1] \end{align}$$

then,

$$\begin{align} h[n] = \delta[n] - \delta[n - 1] \end{align}$$

we can see that the backward difference is simply the forward difference time shifted by 1.