Discrete-time signals, energy and power of discrete signals, signal symmetries, block diagrams, classification, convolution.

Dubulicious
2024-03-25

$$x(n)=\sin\left(2\pi\frac{1}{8}n\right)\approx\\\{\dots~-1~-0.7~\underline{0}~0.7~1~0.7~\dots\}$$

The underline indicates `n=0`.

Average of three last values of `x(n)`:

$$y(n)=\frac{1}{3}x(n)+\frac{1}{3}x(n-1)+\frac{1}{3}x(n-2)$$

A stable system decaying at a rate of 0.9:

$$y(n)=0.9y(n-1)+x(n)$$

A unstable system diverging at a rate of 1.1:

$$y(n)=1.1y(n-1)+x(n)$$

The Dirac delta function is used for impulses.

$$\delta(n)=\begin{cases}1&\text{if }n=0\\0&\text{otherwise}\end{cases}$$

The Step function (Heaviside/unit).

$$u(n)=\begin{cases}1&\text{if }n\geq0\\0&\text{otherwise}\end{cases}$$

These two functions can be combined to make other signals.

The **energy** of a signal is defined as:

$$E\equiv\sum_{n=-\infty}^{\infty}|x(n)|^2$$

The **average power** of a signal is defined as:

$$P\equiv\lim\limits_{N\to\infty}\frac{1}{2N+1}\sum\limits_{n=-N}^{N}|x(n)|^2$$

Even if a signal has infinite energy, the average power could still be finite.

We can also express the energy as:

$$E\equiv\lim\limits_{N\to\infty}E_N$$

where

$$E_N\equiv\sum_{n=-N}^N|x(n)|^2$$

$$\text{energy signal}\iff 0 < E < \infty,P=0$$

$$\text{power signal}\iff 0 < P < \infty,E=\infty$$

A signal is **symmetric** or **even** if `x(n) = x(-n)`.

A signal is **antisymmetric** or **odd** if `x(n) = -x(-n)`.

**Any signal** `x(n)` can be written as a sum of an odd and even signal.

$$

\begin{split}

x(n) &= x_e(n) + x_o(n) \\

x_e(n) &= \frac{1}{2}(x(n)+x(-n)) \\

x_o(n) &= \frac{1}{2}(x(n) - x(-n))

\end{split}

$$

For block diagrams see lecture slides or chapter 2.2.2 in the book.

We call systems different things depending on their properties, a system is called:

**static:**if**without memory****dynamic:**if**without memory,****time-invariant:**where I/O characteristics**don't change with time,****time-variant:**where I/O characteristics**change with time,****linear system:**when it satisfies the**superposition principle,****non-linear system:**when it**doesn't**satisfy the**superposition principle,****causal:**where the**output**at any time**only depends**on the**current**and**past inputs,****bounded-input bounded-output (BIBO) stable:**where**bounded inputs**`|x(n)| \leq M_x` always produce**bounded outputs**`|y(n)| \leq M_y`.

Focus is on Linear Time-Invariant (LTI) systems in this course.

$$\delta(n) \rightarrow \boxed{\text{LTI system}} \rightarrow h(n)$$

The **impulse response** `h(n)` of an LTI system is the output signal when the input signal is a unit impulse `\delta(n)` at time `n=0`.

An arbitrary signal can be written as the sum of delayed impulses:

$$x(n) = \sum_{k=-\infty}^\infty x(k)\delta(n-k)$$

Since LTI systems are linear and time-invariant the output `y(n)` will be the sum of impulse responses:

$$y(n) = x(n)*h(n) = \sum_{k=-\infty}^\infty x(k)h(n-k)$$

**Unit impulse** is the **identity element** of convolution:

$$x(n)*\delta(n) = \sum_{k=-\infty}^\infty x(k)\delta(n-k) = x(n)$$

Convolution fulfills the:

**commutative law:**\(x*h = h*x\),**associative law:**\((x*h_1)*h_2 = x * (h_1*h_2)\),**distributive law:**\(x*(h_1+h_2) = x*h_1 + x*h_2\)

If we have two **finite-length signals** `x(n)` and `h(n)` with lengths `M` and `N` then the convolution \(y(n) = x(n) * h(n)\) will have length `M+N-1`

An LTI system is only causal if its impulse response fulfills: `h(n) = 0` for all `n < 0`.

If we take the example of a recursive system:

$$y(n) = ay(n-1)+x(n)$$

We can rewrite it in terms of the memory from `n=0`:

$$y(n) = a^{n+1}y(-1) + \sum_{k=0}^n a^kx(n-k)$$

Constant-coefficient difference equations describe **LTI** systems.

If the state of the system at `n=0` is zero (**relaxed**): e.g. `y(-1)=0` we get the **zero-state response**:

$$y_{zs}(n) = \sum_{k=0}^na^kx(n-k)$$

If the input signal is zero, `x(n) = 0`, we get the **zero-input response (natural response)**:

$$y_{zi}(n) = a^{n+1}y(-1)$$

The output signal is the sum: `y(n) = y_{zi}(n) + y_{zs}(n)`.