# Chapter 1 - Continuous Time Signals

#### Continuous Time Signal
A continuous time signal is a function $x(t)$ that maps values from $\Reals$ to $\Reals$ or $\Complex$.

#### Causal Signals
A signal, $x(t)$, is said to be causal if it has value $0$ for $t<0$.

#### Non-causal Signals
A signal, $x(t)$, is said to be *non-causal*, or *not causal*, if $x(t)$ is not $0$ for all $t<0$.

#### Finite Support
A signal, $x(t)$, is said to have *finite support*, or *finite duration*, if there exists inputs $T_1$ and $T_2$ such that $x(t) = 0$ for $t < T_1$ and $t > T_2$.

#### Infinite Support
A signal, $x(t)$, is said to have *infinite support*, or *infinite duration*, if it does not have infinite support.

There are three types of inifinite support:
1. *Right-sided*: the signal has domain $(T_1, +\infty)$
1. *Left-sided*: the signal has domain $(-\infty, T_2)$
1. *Two-sided*: the signal has domain $(-\infty, +\infty)$

## 1.1 Basic Signal Operations

#### Signal Addition
Signal addition is of the form: $z(t) = x(t) + y(t)$, where the amplitude of the signal $z(t)$ is the net amplitude of $x(t)$ and $y(t)$.

#### Scalar Multiplication
Scalar multiplication is of the form: $z(t)=\alpha x(t)$. The amplitude of the output is proportional to $\alpha$.

#### Time Shift
A time shift is of the form: $z(t)=x(t-\tau)$. When $\tau > 0$, the time shift is said to be a *delay*. When $\tau < 0$, the time shift is said to be an *advance*.

#### Time Scale
A time scale is of the form: $z(t)=x(at)$. For $\lvert a \rvert > 1$, the scaling is said to be a compression. For $\lvert a \rvert < 1$, the scaling is said to be an expansion. For $a < 0$, a time reflection over $t=0$ occurs.

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##### Figure 1.1.1 - Signal Transformations
<img src="Images/Figure1.1.1.png" width=400>


##### Figure 1.1.2 - Signal Combinations
<img src="Images/Figure1.1.2.png" width=400>

## 1.2 Combinations of Operations

### Review of Reflections
Every $t$ becomes a $-t$. 

$x(t) = \begin{cases} t & 0 \leq t \leq 8 \\ 0 & otherwise \\ \end{cases}$

### Example 1
$x(t) = \begin{cases} t & 0 \leq t \leq 8 \\ 0 & otherwise \\ \end{cases}$

### How to find $y(t) = x(at - b)$
#### Method 1 (recommended)
1. Find $v(t) = x(t-b)$
1. Find $w(t) = v(\lvert a \rvert t) = x( \lvert a \rvert t - b)$
1. If $a \gt 0$, then $\lvert a \rvert = a$. $y(t) = w(t) = x(at - b)$
1. If $a \lt 0$, then $\lvert a \rvert = a$. $y(t) = w(-t) = x(-\lvert a \rvert t - b)$

#### Method 2
1. Find $v(t) = x(\lvert a \rvert t)$
1. Find $w(t) = v(t - \frac{b}{\lvert a \rvert}) = x(\lvert a \rvert \left(t - \frac{b}{\lvert a \rvert} \right))$

### Example
Find $x(3t-5)$

## Lecture 5

### The Impulse Function
$\delta(t) = \begin{cases} 0 & t \ne 0 \\ \infty & t = 0 \\ \end{cases}$

$\int_{-\infty}^{\infty} \delta(t) dt = 1$

$\delta(t - \alpha) = \begin{cases} \infty & t=\alpha \\ 0 & t \ne \alpha \\ \end{cases}$

$\delta(2t - 3) =  \begin{cases} \infty & 2t-3=0, t=3/2 \\ 0 & t \ne 3/2 \end{cases}$
#### Properties
$f(t) \delta(t- \alpha) = f(\alpha) \delta(t-\alpha)$

$\int_a^b f(t) \delta(t - \alpha)dt= \begin{cases} f(\alpha) & \alpha \in [a, b] \\ 0 & otherwise \end{cases}$

#### Unit step
$u(t) = \begin{cases} 1 & t \ge 0 \\ 0 & t \lt 0 \end{cases}$

$\delta(t) = \frac{du(t)}{dt}$

$u(t) = \int_{-\infty}^{t} \delta(\tau)d\tau = \begin{cases} 1 & t \ge 0 \\ 0 & t \lt 0 \end{cases}$

$u(t - \tau) = \begin{cases} 1 & t \ge \tau \\ 0 & t \lt \tau \end{cases}$

$u(-t + 5) = \begin{cases} 1 & t \le 5 \\ 0 & t \gt 5 \end{cases}$

The difference between $u(t)$ and $u(t-1)$ is a finite support pulse from 0 to 1.
#### Ramp
$r(t) = t u(t) = \begin{cases} t & t \ge 0 \\ 0 & t \lt 0 \end{cases}$

$\frac{dr(t)}{dt} = \frac{tdu(t)}{dt} + (1)u(t) = t\delta(t) + u(t) = u(t)$

#### Derivatives
1. $\cos(2 \pi t)\left[ u(t) - u(t-1) \right] = \begin{cases} 0 & t \lt 0 \\ \cos(2\pi t) & 0 \le t \lt 1 \\ 0 & t \ge 1 \end{cases}$

    Using the product rule:

    $\cos(2\pi t)\left[ \delta(t) - \delta(t-1) \right] + -\sin(2\pi t)(2\pi)\left[u(t)-u(t-1)\right]$

1. $u(t) - 2u(t-1) + u(t-2) = \begin{cases} 0 & t \lt 0 \\ 1 & 0 \le t \lt 1 \\ -1 & 1 \le t \lt 2 \\ 0 & t \ge 2 \end{cases}$

## Energy and Power of Signals

### Energy
The energy of a signal $x(t)$ could be finite or infinite.

Given a real or complex signal, $x(t)$, the energy, $E_x$, is defined as:

$$E_x = \int_{-\infty}^{\infty} \lvert x(t) \rvert^2 dt$$

1. If $x(t)$ has finite support, the domain is $[a, b]$, and therefore $E_x$ is a finite integral with bounds $a$ and $b$.

1. If $x(t)$ has infinite support, $E_x$ is an improper integral and may or may not have finite value.

1. If $x(t)$ is periodic, $E_x$ is infinite.

### Power
The power, $P_x$, of an aperiodic signal is defined as:

$$P_x = \lim_{T \to \infty} \frac{1}{2T} \int_{-T}^{T} \lvert x(t) \rvert^2 dt$$

The power, $P_x$, of a periodic signal with period, $T_0$, $x(t)$ is defined as:

$$P_x = \frac{1}{T_0}\int_{t_0}^{t_0 + T_0} \lvert x(t) \rvert^2 dt$$

The most convenient starting times, $t_0$ are $-\frac{T_0}{2}$ and $0$. The bounds of integration will be $\left[ -\frac{T_0}{2}, \frac{T_0}{2} \right]$ and $\left[ 0, T_0 \right]$ respectively.

For a periodic signal, the power is the energy of one period normalized by the length of the period.

**FACT:** A finite energy aperiodic signal has zero power.

$$P_x = \lim_{T \to \infty} \frac{1}{2T} \int_{-T}^{T} \lvert x(t) \rvert^2 dt$$

$$=\lim_{T \to \infty} \frac{1}{2T} [N]$$
Where $N$ is some finite number.

$$\lim_{T \to \infty} \frac{N}{2T} = 0$$

### Procedure
1. Determine if $x(t)$ is finite support or infinite support.
    - If finite support: $E_x \lt \infty$, $P_x = 0$
1. If $x(t)$ is infinite support, determine periodicity of $x(t)$
    - If aperiodic, calculate $E_x$, $P_x$
    - If periodic: $E_x = \infty$, calculate $P_x$

### Facts
If:
$$x(t) = A\cos(\omega_0 t + \theta)$$
Then:
$$E_x = \infty$$
$$P_x = \frac{A^2}{2}$$

If:
$$x(t) = \sum_k A_k\cos(\omega_k t + \theta)$$
Then:
$$E_x = \infty$$
$$P_x = \sum_k \frac{A_k^2}{2}$$