SIGNALS and SYSTEMS

CT vs DT

CT(continuous-time)

For a continuous-time (CT) signal, the independent variable is always enclosed by a parenthesis $(N)$
Example: $x(t),y(t),z(t),A(x,y)$

DT(discrete-time)

For a discrete-time (DT) signal, the independent variable is always enclosed by a brackets $[N]$
Example: $x[n],y[n],z[n],A[m,n]$

Signal Energy and Power

$$p(t)=v(t)i(t)=\frac{1}{R}{v}^2(t)$$ $$\triangleq \text{is “by definition“symbol}$$

Energy

$$E={\int }_{t_1}^{t_2}p(t)dt={\int }_{t_1}^{t_2}\frac{1}{R}{v}^2(t)dt$$

average power

$$P=\frac{E}{t_2-t_1}={\int }_{t_1}^{t_2}p(t)dt={\int }_{t_1}^{t_2}\frac{1}{R}{v}^2(t)dt$$

ex1

$$\begin{gathered} x(t)=\{\begin{array}{ll}x& 0<x<3\\ 6-x& 3<x<6\end{array} \\ E={\int }_0^{6}x(t{)}^{2}dt={\int }_0^3(t{)}^{2}dt+{\int }_3^6(6-t{)}^{2}dt \\ P=\frac{E}{6-0}=\frac{1}{6}({\int }_0^3(t{)}^{2}dt+{\int }_3^6(6-t{)}^{2}dt) \end{gathered}$$

ex2

$$\begin{gathered} x[t]=\{\begin{array}{ll}x& 0<x<3\\ 6-x& 3<x<6\end{array} \\ E=\sum _{t=0}^{3}x(t{)}^2=\sum _{t=0}^3(t{)}^2+\sum _{t=3}^6(6-t{)}^2 \\ P=\frac{E}{6-0}=\frac{1}{6}(\sum _{t=0}^3(t{)}^2+\sum _{t=3}^6(6-t{)}^2) \end{gathered}$$

odd vs even

$$x(t)=E_v\{x(t)\}+O_d\{x(t)\}$$

even

$$E_v\{x(t)\}=\frac{x(t)+x(-n)}{2}$$

odd

$$O_d\{x(t)\}=\frac{x(t)-x(-n)}{2}$$

some prove

$$\begin{array}{rl}& \\ {\int }_{-\infty }^{\infty }{E}_v\{x(t)\}\times O_d\{x(t)\}dt& ={\int }_{-\infty }^{\infty }\frac{x(t)+x(-t)}{2}\times \frac{x(t)-x(-t)}{2}dt\\ & ={\int }_{-\infty }^{\infty }\frac{x(t{)}^2-x(-t{)}^2}{4}dt\\ & ={\int }_{-\infty }^{\infty }\frac{x(t{)}^2}{4}dt-{\int }_{-\infty }^{\infty }\frac{x(-t{)}^2}{4}dt\\ & =0\end{array}$$

unit step function and unit impulse function

basic system properties

memory and memoryless

memoryless systems

only the current signal
$$y[n]=(2x[n]-x[n{]}^2{)}^2$$

memory systems

only the current signal
$$\begin{array}{rlr}& & \\ & y[n]=\sum _{k=-\infty }^{n}x[k]& \text{(accumulator)}\\ & y(t)={\int }_{-\infty }^{t}x(t)dt& \text{(integral)}\\ & y[n]=x[n]& \text{(delay)}\end{array}$$

invertibility

function is invertable $$\begin{array}{rlrl}& & & \\ & y(t)=2x(t)& ,& y(t{)}^{-1}=\frac{1}{2}x(t)\\ & y[n]=\sum _{-\infty }^{n}x[n]& ,& y[t{]}^{-1}=x[n]-x[n-1]\end{array}$$

causality

only the current and past signal are relate then it is causal system

causal systems

$$\begin{gathered} y(t)=x(t-1) \\ y[n]=x[n]-x[n-1] \end{gathered}$$

non-causal systems

$$\begin{gathered} y(t)=x(t+1) \\ y[n]=x[n]-x[n+1] \end{gathered}$$

stability (BIBO stable )

can find BIBO(bounded-input and bounded-output) in another word the function is diverage or not. $$|x(t)|\le \infty \text{ and }|y(t)|\le \infty \text{ for all t}$$

BIBO stable

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

BIBO unstable

$$y(t)=1.01x[n-1]$$

time invariance

the function shift input will only shift and dont have any effect

example

$$\begin{array}{rlrlr}& & & & \\ & ,& & y(t)=tx(t)& \\ \text{define }{y}_1(x)\text{ let }x(t)=x(t+t_0)& ,& & y_1(t)=tx(t+t_0)& \\ \text{if }{y}_1(t)==y(t+t_0)\text{then it is invariance}& ,& & \because (t+t_0)x(t+t_0)\ne tx(t+t_0)& \therefore \text{not invariance}\end{array}$$

linearity

if $ay(x(t))+by(x(t))==y(ax(t)+bx(t))$ then is linearty

test

complex plane

$$\begin{gathered} j=\sqrt{-1} \\ \begin{array}{rl}z& =r\times e^{j\theta }\\ & =r\times (\sin (\theta )+j\cos (\theta ))\\ & =r\times \sin (\theta )+jr\cos (\theta )\\ & =\alpha +j\omega \end{array} \end{gathered}$$

exponential signal & sinusoidal signal

$$x(t)=C{e}^{\alpha t}$$

periods

CT

$$\begin{gathered} x(t)=r{e}^{j{w}_{0}t} \\ \text{fundamental period }{T}_0=\frac{2\pi }{w_0} \end{gathered}$$

example

$$\begin{gathered} x(t)=e^{j2t}+e^{j3t} \\ \text{fundamental period of }{e}^{j2t}=\pi \\ \text{fundamental period of }{e}^{j3t}=\frac{2\pi }{3} \\ \text{LCD(Least Common Denominator) of }\pi \text{ and }\frac{2\pi }{3}\text{ is }2\pi \\ \text{fundamental period of }{e}^{j2t}+e^{j3t}=2\pi \end{gathered}$$

DT

fundamental period $T_0$ is integer that $x[n]=x[n+i\%T_0]$ for all integer $i$

$T_0$ have to be integer.
not every “sinusoidal signal” have $T_0$

$$\begin{gathered} x[n]=r{e}^{j{w}_{0}t}=r{e}^{j\frac{m}{N}n} \\ \text{fundamental period }{T}_0=N \\ \text{fundamental frequency }=\frac{2\pi }{N} \end{gathered}$$

example

$$\begin{gathered} x[n]=e^{j(\frac{2\pi }{3})n}+e^{j(\frac{3\pi }{4})n} \\ \text{fundamental period of }{e}^{j(\frac{2\pi }{3})n}=3 \\ \text{fundamental period of }{e}^{j(\frac{3\pi }{4})n}=8\because (2\pi =\frac{24\pi }{4}) \\ \text{LCD(Least Common Denominator) of }3\text{ and }8\text{ is }24 \\ \text{fundamental period of }{e}^{j(\frac{2\pi }{3})n}+e^{j(\frac{3\pi }{4})n}=24 \end{gathered}$$

convolution

CT

$$(f\ast g)(t)={\int }_{-\infty }^{\infty }f(\tau )g(t-\tau )d\tau$$

DT

$$(f\ast g)[n]=\sum _{k=-\infty }^{\infty }f[k]g[n-k]$$

$$\begin{array}{rl}x[n]=& \{\begin{array}{ll}0.5^n& ,0\le n\\ 0& ,else\end{array}\\ h[n]=& \{\begin{array}{ll}0.5^n& ,0\le n<3\\ 0& ,else\end{array}\end{array}$$

$$\begin{array}{rl}(x\ast h)[0]& =x[0]\times h[0]\\ (x\ast h)[1]& =x[0]\times h[1]+x[1]\times h[0]\\ (x\ast h)[2]& =x[0]\times h[2]+x[1]\times h[1]+x[2]\times h[0]\end{array}$$

commutative

$$\begin{array}{rl}x(t)\ast h(t)& =h(t)\ast x(t)\\ {\int }_{-\infty }^{\infty }x(\tau )h(t-\tau )d\tau & ={\int }_{-\infty }^{\infty }h(\tau )x(t-\tau )d\tau \end{array}$$

distributive

$$\begin{array}{rl}x(t)\ast (h_1(t)+h_2(t))& =x(t)\ast h_1(t)+x(t)\ast h_2(t)\end{array}$$

associative

$$\begin{array}{rl}x(t)\ast (h_1(t)\ast h_2(t))& =(x(t)\ast h_1(t))\ast h_2(t)\end{array}$$

LTI(Linear Time-Invariant)

Linear

$$\begin{array}{rl}y(t)& =x(t)\ast h(t)\\ & ={\int }_{-\infty }^{\infty }x(t-\tau )h(\tau )d\tau \\ & ={\int }_{-\infty }^{\infty }x(\tau )h(t-\tau )d\tau \end{array}$$

Time-invariant

$$\begin{array}{rl}y(t)& =x(t)\ast h(t)\\ y(t-k)& =x(t-k)\ast h(t)\end{array}$$

LTI systems and convolution

stability for LTI Systems

$$\begin{gathered} \text{if }|x(t)|\text{ is bounded, then }|y(t)|\text{ is also bounded.} \\ \text{Sufficient condition for a continuous-time LTI system to be stable} \end{gathered}$$

Unit Step Response of an LTI System

CT

$$\begin{gathered} \begin{array}{rl}y[n]& =h[n]\ast u[t]\\ & ={\int }_{-\infty }^{\infty }u[n-\tau ]h[\tau ]d\tau \\ & ={\int }_{-\infty }^{n}h[\tau ]d\tau \end{array} \\ h[n]=y[n]-y[n-1] \end{gathered}$$

DT

$$\begin{gathered} \begin{array}{rl}y(t)& =h(t)\ast u(t)\\ & =\sum _{-\infty }^{\infty }u(t-\tau )h(\tau )d\tau \\ & =\sum _{-\infty }^{t}h(\tau )d\tau \end{array} \\ h(t)=y^{\prime }(t) \end{gathered}$$

eigen function and eigen value of LTI systems

The response of an LTI system to a eigen function is the same eigen function with only a change in amplitude(eigen value)

$$\begin{gathered} x(t)\to H(t)x(t) \\ x(t)\text{ is eigenfunction } \\ H(t)\text{ is eigen value} \end{gathered}$$

The Response of LTI Systems to Complex Exponential Signals

$$\begin{gathered} x(t)=e^{st} \\ \begin{array}{rl}y(t)& =h(t)\ast x(t)\\ & ={\int }_{-\infty }^{\infty }h(\tau )x(t-\tau )dt\\ & ={\int }_{-\infty }^{\infty }h(\tau )e^{s(t-\tau )}dt\\ & =e^{st}{\int }_{-\infty }^{\infty }h(\tau )e^{-s\tau }dt\end{array} \\ H(t)={\int }_{-\infty }^{\infty }h(\tau )e^{-s\tau }dt \end{gathered}$$

example

$$\begin{gathered} x(t)=e^{j2t} \\ y(t)=e^{j2(t-3)}=e^{j6}{e}^{j2t} \\ H(t)={\int }_{-\infty }^{\infty }\delta (\tau -3)e^{-s\tau }dt=e^{-3s} \end{gathered}$$

system

delay system

$$\begin{gathered} x(t)\ast \delta (t-t_0)=x(t-t_0) \\ x[t]\ast \delta [t-t_0]=x[t-t_0] \end{gathered}$$

difference system

$$x[n]\ast \delta [t-t_0]=x[t-t_0]$$

accumulation system

$$y[n]=\sum _{m-\infty }^{n}x[m]$$

example

$${\int }_{T_0}{e}^{j{w}_{0}t}dt$$ $${\int }_{T_0}{e}^{j{w}_{0}t}dt$$