5th semester files

5th-Semester-Fall-2023/Signals-and-Systems/Notes/Chapter1.md

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+# 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.
+
+<br>
+
+##### 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:
+$$E_x = \infty$$
+$$P_x = \sum_k \frac{A_k^2}{2}$$

5th-Semester-Fall-2023/Signals-and-Systems/Notes/Chapter2.md

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+# Chapter 2 - Systems
+A *system*, $S$, takes in an input signal, $x(t)$, and outputs a signal, $y(t)$.
+$$x(t) \rightarrow \boxed{S} \rightarrow y(t)$$
+$$y(t) = S[x(t)]$$
+
+## Simple System Properties
+#### Linear Systems
+$S$ is said to be linear if it satisfies both hoogeneity and superposition.
+
+#### Homogeneity
+Scaling an input by a factor, $a$, scales the output by the same factor.
+$$ay(t) = S[ax(t)]$$
+
+#### Superposition
+Adding two inputs lead to the corresponding outputs being added.
+
+If
+$$y_1(t) = S[x_1(t)]$$
+$$y_2(t) = S[x_2(t)]$$
+Then
+$$y_1(t) + y_2(t) = S[x_1(t) + x_2(t)]$$
+
+#### Time Invariance
+$S$ is said to be time-invariant if delaying or advancing the input gives the same delay or advance in the output.
+$$y(t - \alpha) = S[x(t - \alpha)]$$
+
+#### Causality
+$S$ is said to be causal if the output does not depend on any future inputs. Theoretically, noncausal systems cannot be realized.
+
+#### Signal Boundedness
+$x(t)$ is said to be bounded if $\lvert x(t) \rvert \le B_x \lt \infty$
+
+Bounded example:
+$$x(t) = e^{-t} u(t)$$
+Unbounded example:
+$$x(t) = e^t u(t)$$
+
+#### Bounded-Input, Bounded-Output Stability
+$S$ is said to be bounded-input, bounded-output (BIBO) stable if every bounded input gives rise to a bounded output.
+
+Fact: The system is BIBO unstable if just one bounded input gives rise to an unbounded output.
+
+Every bounded $x(t)$ gives rise to a bounded $y(t)$.
+
+### Example 2.1
+$$y(t) = S[x(t)] = x^2(t)$$
+
+##### Homogeneity:
+$$S[ax(t)] = [ax(t)]^2 = a^2x^2(t)$$
+$$ay(t) = ax^2(t)$$
+Since $S[ax(t)] \ne ay(t)$, $S$ is not homogeneous, and therefore not linear.
+
+##### Time invariance:
+$$S[x(t-\alpha)] = [x(t-\alpha)]^2 = x^2(t-\alpha) =y(t-\alpha)$$
+Since $S[x(t-\alpha)] = y(t-\alpha)$, $S$ is causal
+
+##### BIBO stability:
+$$\lvert x(t) \rvert \le B_x \lt \infty$$
+$$\lvert y(t) \rvert = \lvert x^2(t) \rvert \le B_x^2 \lt \infty$$
+
+### Example 2.2
+$$y(t) = S[x(t)] = \cos(x(t))$$
+
+##### Homogeneity:
+$$S[ax(t)] = \cos(ax(t))$$
+$$ay(t) = a\cos(x(t))$$
+Since $S[ax(t)] \ne ay(t)$, $S$ is not homogeneous and therefore not linear.
+
+##### Time invariance:
+$$S[x(t-\alpha)] = \cos(x(t-\alpha)) = y(t-\alpha)$$
+
+##### BIBO stability:
+$$\lvert x(t) \rvert \le B_x \lt \infty$$
+$$\lvert y(t) \rvert \le 1$$
+
+### Example 2.3
+$$y(t) = S[x(t)] = \lvert x(t) \rvert$$
+
+##### Homogeneity:
+$$S[ax(t)] = \lvert ax(t) \rvert$$
+$$ay(t) = a \lvert x(t) \rvert$$
+Since $\lvert a \rvert \ne a$, $S$ is not homogeneous and therefore not linear.
+
+##### Time invariance:
+$$S[x(t-\alpha)] = \lvert x(t-\alpha)\rvert$$
+$$y(t-\alpha) = \lvert x(t-\alpha)\rvert$$
+Since $S[x(t-\alpha)] = y(t-\alpha)$, $S$ is time invariant.
+
+##### BIBO stability
+$$\lvert x(t) \rvert \le B_x \lt \infty$$
+$$\lvert y(t) \rvert = \lvert x(t) \rvert \le B_x \lt \infty$$
+
+### Example 2.4
+$$y(t) = S[x(t)] = mx(t) + b$$
+
+##### Homogeneity:
+$$S[ax(t)] = max(t) + b$$
+$$ay(t) = amx(t) + ab$$
+
+##### Superposition:
+$$y_1(t) = mx_1(t) + b$$
+$$y_2(t) = mx_2(t) + b$$
+$$y_1(t) + y_2(t) = m(x_1(t) + x_2(t)) + 2b$$
+
+For $b=0$, $S$ is linear, otherwise, neither homogeneity nor superposition are satisfied.
+
+##### Time invariance:
+$$S[x(t-\alpha)] = mx(t-\alpha) + b$$
+$$y(t-\alpha) = mx(t-\alpha) + b$$
+
+Since $y(t-\alpha) = x(t-\alpha)$, $S$ is time invariant.
+
+##### BIBO stability
+$$\lvert y(t) \rvert = \lvert mx(t) + b \rvert \le \lvert m \rvert \lvert x(t) \rvert + \lvert b \rvert \le \lvert m \rvert B_x \lt \infty$$
+
+### Example 2.5
+$$y(t) = S[x(t)] = tx(t+3)$$
+
+##### Homogeneity:
+$$S[ax(t)] = atx(t+3)$$
+$$ay(t) = atx(t+3)$$
+
+##### Superposition:
+$$y_1(t) = tx_1(t+3)$$
+$$y_2(t) = tx_2(t+3)$$
+$$S[x_1(t) + x_2(t)] = t[x_1(t+3) + x_2(t+3)]$$
+$$y_1 + y_2 = tx_1(t+3) + tx_2(t+3)$$
+
+Since both homogeneity and superposition are satisfied, $S$ is a linear system.
+
+##### Time invariance:
+$$S[x(t-\alpha)] = tx(t+3-\alpha)$$
+$$y(t-\alpha) = (t-\alpha)\cdot x (t+3-\alpha)$$
+Not time invariant.
+
+##### Causality:
+No
+
+##### BIBO stability:
+$$\lvert y(t) \rvert = \lvert tx(t+3) \rvert = \lvert t \rvert \lvert x(t+3)\rvert$$
+
+Since $\lvert t \rvert$ is not bounded, ther is no $B_y$ such that:
+$$\lvert y(t) \rvert \le B_y \lt \infty$$
+
+### Example 2.6 - Expansion
+$$y(t) = S[x(t)] = x({t \over 10})$$
+
+##### Homogeneity
+$$S[ax(t)] = ax({t \over 10}) = ay(t)$$
+
+##### Superposition
+$$y_1(t) = x_1({t \over 10})$$
+$$y_2(t) = x_2({t \over 10})$$
+$$S[x_1(t) + x_2(t)] = x_1({t \over 10}) + x_2({t \over 10})$$
+
+##### Time Invariance
+$$S[x(t - \alpha)] = x({t \over 10} - \alpha)$$
+$$y(t - \alpha) = x({t - \alpha \over 10})$$
+
+##### Causal
+$$y(-10) = x(-1)$$
+
+##### BIBO Stability
+Expansion affects input space, not output space
+
+### Example 2.7
+$$y(t) = S[x(t)] = {1 \over T} \int\limits_{t-T}^t x(\tau)d\tau + B$$
+
+##### Homogeneity
+$$S[ax(t)] = {1 \over T} \int\limits_{t-T}^t ax(\tau)d\tau + B$$
+$$ay(t) = {a \over T} \int\limits_{t-T}^t x(\tau)d\tau + B$$
+Not homogeneous unless $B = 0$.
+
+##### Superposition
+$$y_1(t) = {1 \over T} \int\limits_{t-T}^t x_1(\tau)d\tau + B$$
+$$y_2(t) = {1 \over T} \int\limits_{t-T}^t x_2(\tau)d\tau + B$$
+$$S[x_1(t) + x_2(t)] = {1 \over T} \int\limits_{t-T}^t [x_1(\tau) + x_2(\tau)]d\tau + B$$
+$$y_1(t) + y_2(t) = {1 \over T} \int\limits_{t-T}^t x_1(\tau)d\tau + {1 \over T} \int\limits_{t-T}^t x_2(\tau)d\tau + 2B$$
+
+##### Time Invariance
+$$S[x(t - \alpha)] = {1 \over T} \int\limits_{t-T}^t x(\tau - \alpha)d\tau + B$$
+Substitute $v = \tau - \alpha$:
+$$S[x(t - \alpha)] = {1 \over T} \int\limits_{t-T-\alpha}^{t-\alpha} x(v)dv + B$$
+$$y(t - \alpha) = {1 \over T} \int\limits_{t-T-\alpha}^{t-\alpha} x(\tau)d\tau + B$$
+
+##### Causality
+$y(t)$ depends on $x(t)$ which uses inputs from $t-T$ to $t$, and no future inputs. Therefore, the system is causal.
+
+If instead, inputs ranged from $t-T$ to $t+T$, the system would rely on future inputs and would not be causal.
+
+##### BIBO Stability
+$$\lvert x(t) \rvert \le B_x < \infty$$
+$$|y(t)| = \left| {1 \over T} \int\limits_{t-T}^t x(\tau)d\tau + B \right|$$
+$$|y(t)| \le \left| {1 \over T} \int\limits_{t-T}^t x(\tau)d\tau \right| + |B|$$
+$$|y(t)| \le {1 \over T} \int\limits_{t-T}^t |x(\tau)|d\tau + |B|$$
+$$|y(t)| \le {1 \over T} \int\limits_{t-T}^t B_x d\tau + |B|$$
+$$|y(t)| \le {1 \over T} \left[\tau B_x\Big|_{t-T}^{t}\right] + |B|$$
+$$|y(t)| \le {1 \over T} T B_x + |B|$$
+$$|y(t)| \le B_x + |B|$$
+
+### Example 2.8 - AM
+$$y(t) = S[m(t)] = m(t)\cos(\omega_c t)$$
+
+##### Homogeneity
+$$S[am(t)] = am(t)\cos(\omega_c t) ay(t)$$
+
+##### Superposition
+$$y_1(t) = m_1(t)\cos(\omega_c t)$$
+$$y_2(t) = m_2(t)\cos(\omega_c t)$$
+$$S[m_1(t) + m_2(t)] = [m_1(t) + m_2(t)]\cos(\omega_c t)$$
+$$y_1(t) + y_2(t) = [m_1(t) + m_2(t)]\cos(\omega_c t)$$
+
+##### Time Invariance
+$$S[m(t-\alpha)] = m(t-\alpha)\cos(\omega_c t)$$
+$$y(t - \alpha) = m(t-\alpha)\cos(\omega_c (t -\alpha))$$
+
+##### Causality
+Yes
+
+##### BIBO Stability
+$$|m(t)| \le B_x$$
+$$|y(t)| = |m(t)\cos(\omega_c t)| \le |m(t)| \le B_x$$
+
+### Example 2.9 - FM
+$$y(t) = S[m(t)] = \cos\left(\omega_c t + \int\limits_{-\infty}^{t} m(\tau) d\tau \right)$$
+
+##### Homogeneity
+$$S[am(t)] = \cos\left(\omega_c t + \int\limits_{-\infty}^{t} am(\tau) d\tau \right)$$
+$$ay(t) = a\cos\left(\omega_c t + \int\limits_{-\infty}^{t} m(\tau) d\tau \right)$$
+
+##### Time Invariance
+$$S[m(t-\alpha)] = \cos\left(\omega_c t + \int\limits_{-\infty}^{t} m(\tau - \alpha) d\tau \right)$$ 
+$$y(t-\alpha) = \cos\left(\omega_c (t - \alpha) + \int\limits_{-\infty}^{t} m(\tau) d\tau \right)$$ 
+##### BIBO Stability
+$$|y(t)| \le 1$$
+
+## Linear, Time-Invariant Systems
+The implulse response $h(t)$ of an LTI system is the output of the system when
+
+$$\delta(t) \rightarrow \boxed{\text{LTI}} \rightarrow h(t)$$
+
+Causal if $h(t) = 0$ for $t<0$.
+
+BIBO stable if:
+$$\int |h(t)|dt < \infty$$
+
+If $h(t)$ has finite supports, it is *always* BIBO stable. If $h(t)$ has infinite support, BIBO stability must be checked.
+
+A linear, constant coefficient ODE with input $x(t)$ and output $y(t)$ is LTI under zero initial conditions and when $x(t)$ is a causal input.
+
+### Circuit Application
+An RLC circuit with no initial voltage across the capacitor and no initial current through the inductor is LTI.
+
+#### Without Initial Condition
+For a capacitor with no initial voltage:
+$$V(t) = S[I(t)] = {1\over C} \int\limits_0^t I(\tau) d\tau$$
+
+##### Homogeneity 
+$$S[aI(t)] = {1\over C}\int\limits_0^t aI(\tau) d\tau = {a\over C}\int\limits_0^t I(\tau) d\tau = aV(t)$$
+
+##### Superposition
+$$V_1(t) = {1\over C}\int\limits_0^t I_1(\tau) d\tau$$
+$$V_2(t) = {1\over C}\int\limits_0^t I_2(\tau) d\tau$$
+$$S[I_1(t) + I_2(t)] = {1\over C} \int\limits_0^t [I_1(\tau) + I_2(\tau)]d\tau = V_1(t) + V_2(t)$$
+
+##### Time Invariance
+$$S[I(t - \alpha)] = {1\over C}\int\limits_{-\alpha}^{t-\alpha} I(\tau - \alpha) d\tau$$
+$$S[I(t - \alpha)] = {1\over C}\int\limits_{-\alpha}^{t-\alpha} I(p) dp$$
+If $I(t)$ is causal:
+$$S[I(t - \alpha)] = {1\over C}\int\limits_{0}^{t-\alpha} I(p) dp$$
+$$V(t - \alpha) = {1\over C}\int\limits_0^{t-\alpha} I(\tau) d\tau$$
+
+
+#### With Initial Condition
+For a capacitor with initial voltage:
+$$V(t) = {1\over C} \int\limits_0^t I(\tau) d\tau + V_0$$
+
+##### Homogeneity
+$$S[aI(t)] = {1\over C}\int\limits_0^t aI(\tau) d\tau + V_0$$
+$$aV(t) = {a\over C}\int\limits_0^t I(\tau) d\tau + aV_0$$
+
+
+#### Impulse Response
+$$i(t) = \delta (t), h(t) = V(t) = ?$$
+$$h(t) = {1 \over C} \int\limits_0^t \delta (\tau) d\tau = 
+\begin{cases} 
+{1\over C} & t \ge 0 \\
+0 & t < 0
+\end{cases}$$
+$$h(t) = {1 \over C} u(t)$$
+
+### Averager
+$$y(t) = S[x(t)] = {1 \over T} \int\limits_{t-T}^t x(\tau) d\tau$$
+
+#### Impulse response
+$$h(t) = {1 \over T} \int\limits_{t-T}^t \delta(\tau) d\tau$$
+
+3 possibilities for impulse response:
+1. $0 < T - t$ means that the impulse is to the left of the limits of integration, and $h(t)$ is 0.
+1. $T - t \le 0 \le t$ means that the impulse is within the bounds of integration, and $h(t)$ takes on a non-zero value.
+1. $t > 0$ means that the impulse is to the right of the bounds of integration, nad $h(t)$ is 0.
+
+For the second case, $T - t \le 0 \le t$, gives the response:
+$$h(t) = \begin{cases}
+{1 \over T} & 0 \le t \le T \\
+0 & \text{elsewhere}
+\end{cases} = {1\over T} \left[u(t) - u(t - T)\right]$$
+
+Since $h(t)$ is 0 for $t < 0$, it is a causal signal.
+
+## Convolution
+$$y(t) = x(t)*h(t) = \int x(\tau) h(t - \tau)d\tau$$
+
+##### Commutative:
+$$x(t) * h(t) = \int h(\tau) x(t-\tau)d\tau = h(t) * x(t)$$
+
+#### Case 1:
+Two finite support wich have the same region of support (domain).
+$$x(t) = h(t)$$
+
+##### Step 1:
+Decide to manipulate $x$ or $h$. 
+$$h(t - \tau) = h(-\tau + t)$$
+Since we are looking at $\tau$ as the variable, there is a reflection and an advance by $t$.
+
+As $h(t-\tau)$ slides along the real number line, first there will be no overlap between $x(\tau)$ and $h(t - \tau)$. They multiply to 0, so $y(t) = 0$. Then, there will be some overlap, so $y(t)$ becomes:
+$$y(t) = \int\limits_0^t d\tau = t$$
+$h(t-\tau)$ continues to slide and soon there is full overlap, so $y(t) = 1$.
+![](Images/OverlapConvolution)
+

5th-Semester-Fall-2023/Signals-and-Systems/Notes/Chapter3.md

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+# Chapter 3 - The Laplace Transform
+
+$f(t)$ is causal, $f(t) = 0$ for $t < 0$.
+
+$$F(s) = L[f(t)] = \int\limits_0^\infty f(t) e^{-st}dt$$
+
+$s$ is a complex variable of the form:
+$$s = \sigma + j\omega$$
+
+### The Region of Convergence
+The ROC is the set of all values of $s$ for which the one-sided Laplace transform exists.
+
+If the causal $f(t)$ has finite support in the temporal region:
+$$0 \le t_1 \le t_2 < \infty$$
+$$F(s) = L[f(t)] = \int\limits_{t_1}^{t_2} f(t)e^{-st}dt$$
+
+If $f(t)$ is causal, the ROC includes $s = \infty$.
+
+If $f(t)$ has infinite support, $F(s)$ can be written as the ratio of two functions $N(s)$ and $D(s)$.
+$$F(s) = {N(s) \over D(s)}$$
+
+### Example (Finite Support)
+$$F(s) = {s \over s^2 + 2s + 2}$$
+Zero at $s=0$, poles at $s = -1 \pm j$.
+
+ROC does not contain any poles, and is not influenced by zeros.
+
+### Example (Infinite Support)
+$$F(s) = {s+1 \over (s+2)(s+5)}$$
+
+ROC is $\Re\{s\} > -2$
+
+### Laplace Transform of Impulse
+$$L[\delta(t)] = \int\limits_0^\infty \delta(t)e^{-st}dt = 1$$
+$$L[\delta(t - \alpha)] = e^{-\alpha s}$$
+
+### Example
+Find the Laplace transform of a causal version of the complex exponential
+$$L[e^{ja t}u(t)] = \int\limits_0^\infty e^{ja t}e^{-st}dt$$
+$$F(s) = \lim_{v \to \infty} \int\limits_0^v e^{(ja-s)t}dt$$
+$$F(s) = -{1\over s-ja} \lim_{v\to\infty}[e^{(ja-s)v}-1]={1\over s- ja}$$
+
+Since $a$ is generally a complex number, $s = \sigma + j\omega$ and $a = \alpha + j\omega$.
+
+## The Inverse Laplace Transform
+
+$$F(s) = {3s + 5 \over (s+1)(s+2)} = {A \over s+1} + {B \over s+2}$$
+
+Region of convergence:
+$$\Re\{s\} > -1$$
+
+Solve for $A$ and $B$:
+$$3s + 5 = A(s+2) + B(s+1)$$
+Eliminate $B$ by setting $s$ to -1:
+$$3(-1) + 5 = A(-1 + 2) + B(-1 + 1)$$
+$$2 = A$$
+Eliminate $A$ by setting $s$ to -2:
+$$3(-2) + 5 = A(-2 + 2) + B(-2 + 1)$$
+$$-1  = -B \therefore B=1$$
+
+Plug back in for $F(s)$:
+$$F(s) = {2 \over s+1} + {1 \over s+2}$$
+$$\therefore f(t)=\left[2e^{-t} + e^{-2t}\right]u(t)$$
+**Note:** multiplying by $u(t)$ is required to make $f(t)$ causal.
+
+### Example of the Delay Property
+$$F(s) = {4e^{-10s} \over s(s+2)^2}$$
+Take out the $e^{-10s}$ term and evaluate.
+$$G(s) = {4 \over s(s+2)^2} = {A \over s} + {B \over s+2} + {C \over (s+2)^2}$$
+$$4 = A(s+2)^2 + Bs(s+2) + Cs$$
+Setting $s=0$ eliminates $B$ and $C$
+$$4 = A(0+2)^2 + B(0)(0+2) + C(0)$$
+$$4 = 4A \therefore A = 1$$
+Setting $s=-2$ eliminates $A$ and $B$:
+$$4 = A(-2 + 2)^2 + B(-2)(-2+2) + C(-2)$$
+$$4 = -2C \therefore C=-2$$
+Set an easy $s$ value to solve for $B$, $s=1$:
+$$4 = 1(1+2)^2 + B(1)(1+2) + -2(1)$$
+$$4 = 9 + 3B-2 \therefore B=-1$$
+Plug back in:
+$$G(s) = {4 \over s(s+2)^2} = {1 \over s} - {1 \over s+2} - {2 \over (s+2)^2}$$
+Solve for $g(t)$
+$$g(t) = \left[1 -e^{-2t} -2te^{-2t}\right]u(t)$$
+By taking out the delay term, $e^{-10t}$, we offset $g(t)$ with respect to $f(t)$. To solve for $f(t)$, we must take this offset into account. The delay is by $10$, so for each $t$ in $g(t)$, there is a $(t-10)$ in $f(t)$. Plug in and solve:
+$$f(t) = \left[1-e^{-2(t-10)}-2(t-10)e^{-2(t-10)}\right] u(t-10)$$
+
+## Differential Equations using Laplace Transforms
+$${df \over dt} + 6f(t) = u(t), f(0^-) = 1$$
+$$s F(s) - f(0^-) + 6F(s) = {1 \over s}$$
+$$(s + 6) F(s) = {1\over s} + 1$$
+$$F(s) = {1 + s \over s(s + 6)}$$
+$$F(s) = {A \over s} + {B \over s + 6}$$
+$$1 + s = (s +6)A + sB$$
+$$A = {1\over6}, B = {5\over6}$$
+$$F(s) = {1 \over 6s} + {5 \over 6(s + 6)}$$
+$$f(t) = \left[{1 \over 6} + {5 \over 6} e^{-6t}\right]u(t)$$

5th-Semester-Fall-2023/Signals-and-Systems/Notes/Chapter4.md

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+# Complex Exponential Fourier Series
+
+Inner product is just Ravi's fancy math way of saying dot product for functions, and orthogonal is his fancy math way of saying perpendicular.
+
+#### Definition: Orthonormal
+$V_1$ and $V_2$ are orthonormal if:
+1. $V_1 \cdot V_2 = 0$
+2. $\|V_1\| = \|V_2\| = 1$
+
+If $\psi_l$ and $\psi_k$ are orthonormal on $[a,b]$:
+$$\int\limits_a^b \psi_k(t) \psi_l^*(t)dt = \begin{cases} 1 & k=l \\ 0 & k \ne l \end{cases}$$
+
+### Example
+Consider the interval $[0,1]$:
+$$\text{Let } \psi_1(t) = 1; t \in [0,1]$$
+$$\text{Let } \psi_2(t) = \begin{cases} \end{cases}$$
+
+### Fourier Series in Exponential Form
+$$x(t) = \sum_{k=-\infty}^\infty X_k e^{jk\omega_0 t}$$
+
+$$T_0 = 1$$
+$$\omega_0 = 2\pi$$
+
+$$x(t) = 1 + \sum_{k=-\infty, k\ne0}^\infty {2 \sin\left({ \pi k \over 2}\right) \over \pi k}e^{2\pi k t}$$
+
+$$x(t) = 1 + \sum_{k=-\infty, k\ne0}^\infty {2 \sin\left({\pi k \over 2}\right) \over \pi k} e^{jk2\pi t}$$
+
+$$x(t - 0.1) = 1 + \sum_{k=-\infty, k\ne0}^\infty {2 \sin\left({\pi k \over 2}\right) \over \pi k} e^{jk2\pi (t - 0.1)}$$
+
+### Fourier Series in Trigonometric Form
+$$x(t) = c_0 + 2\sum_{k=1}^\infty \left[c_k \cos(k \omega_0 t) + d_k \sin(k \omega_0 t)\right]$$
+
+### Odd Signal Example
+$A = 1$, $T = 2$, $T_0 = 2$, $\omega_0 = \pi$
+
+$$x(t) = t$$
+$$x(t) = \sum_k X_k e^{j\pi k t}$$
+$$X_k = {1\over2} \int\limits_{-1}^1 t e^{-j \pi k t} dt$$
+This gives an imaginary $X_k$.
+$$X_0 = {1\over2} \int\limits_{-1}^1 t dt = 0$$
+$$x(t) = c_0 + 2\sum_{k=1}^\infty c_k \cos(\pi k t) + d_k \sin(\pi k t)$$
+Since it's odd:
+$$c_k = 0$$
+$$d_k = {1\over2} \int\limits_{-1}^1 t \sin(\pi k t) dt$$
+$$x_t = 2 \sum_{k=1}^\infty {(-1)^{k+1} \over \pi k} \sin(\pi k t)$$

5th-Semester-Fall-2023/Signals-and-Systems/Notes/Chapter5.md

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+# Chapter 5
+
+The Fourier Transform:
+$$X(\omega) = F[x(t)] = \int\limits_{-\infty}^\infty x(t) e^{-j\omega t} dt$$
+
+The Inverse Fourier Transform:
+$$x(t) = F^{-1}[X(\omega)] = {1 \over 2\pi} \int\limits_{-\infty}^{\infty} X(\omega)e^{j\omega t} d\omega$$
+
+The Fourier transform exists if:
+1. $x(t)$ is abolutely integrable
+2. $x(t)$ has only a finite number of discontinuities and a finite number of minima and maxima in any finite interval.
+
+If $x(t)$ is even, $X(\omega)$ is real.
+
+If $x(t)$ is odd, $X(\omega)$ is imaginary.
+
+Otherwise, $X(t)$ has a real and imaginary part.
+
+### Example
+Consider $x(t) = \delta(t)$:
+$$X(\omega) = \int\limits_{-\infty}^\infty \delta(t)e^{-j\omega t} dt = 1$$
+
+### Example
+Consider $x(t) = \delta(t - \alpha)$:
+$$X(\omega) = \int\limits_{-\infty}^\infty \delta(t - \alpha)e^{-j\omega t}dt = e^{-j\omega \alpha}$$
+
+### Example
+Consider $x(t) = u(t+T) - u(t-T)$:
+
+This is a pulse whose value is 1 on the interval $[-T,T]$.
+$$X(t) = \int\limits_{-T}^T e^{-j\omega t}dt = -{1 \over j \omega} \left[e^{-j\omega T} - e^{-j\omega (-T)}\right] = {2 \over \omega}\left[{e^{j\omega T} - e^{-j\omega T} \over 2j}\right] = 2T\operatorname{sinc}(\omega T)$$
+
+### Example
+Consider $x(t) = e^{-|t|}$
+$$X(\omega) = \int\limits_
+
+## Method 2 - Laplace Transfor
+$$x(t) \to X(s) \to X(\omega)$$
+$$s = j\omega$$
+
+Using a Laplace transform requires that $x(t)$ is a causal signal and the region of convergence of $X(s)$ includes the imaginary axis.
+
+### Example - Finite Support Signals
+The region of convergence is the entire s-plane (must check $s=0$). It definitely includes $s=j\omeaga$.
+
+$$X(s) = {0.5 \over s^2} - {0.5e^{-2s} \over s^2} - {e^{-2s} \over s}$$
+$$X(j\omega) = {0.5 \over j\omega^2} - {0.5e^{-2j\omega} \over j\omega^2} - {e^{-2j\omega} \over j\omega}$$