Signal and System Analysis [1] - Fundamentals

Mathematics Stuffs (Basic Concepts)

Calculations in rectangular form

Calculations in rectangular form

$$z = a+jb$$

$$w = c + jd$$

Addition/subtraction:

• $$z+w = (a+c) + j(b+d)$$

• $$z-w = (a-c) + j(b-d)$$

Multiplication:

• $$zw = (a+jb)(c+jd)=(ac-bd) + j (ad+bc)$$

Note $j^2= -1$.

Division:

• $$\frac{z}{w}=\frac{a=jb}{c+jd}=\frac{(a+jb)(c-jd)}{(c+jd)(c-jd)}=\frac{ac+bd+j(bc-ad)}{c^2+d^2}$$

Polar Form

Note p (magnitude) can only be postive.

materials support

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Khan’s Polar Form Review

examples

Conjugate

Euler’s Relation

$e^{j\theta} = cos\theta + jsin\theta$

$e^{-j\theta} = cos\theta - jsin\theta$

$cos\theta = \cfrac{e^{j\theta}+e^{-j\theta}}{2}$

$sin\theta = \cfrac{e^{j\theta}-e^{-j\theta}}{2j}$

Plotting $e^{j\pi}$

when we calculate Euler’s Formula for $x = \pi$ we get:

note here $i$ is acutally $j$.

$$e^{j\pi} = -1$$

$$e^{j2\pi} = 1$$

$$e^{j\pi/2} = j$$

$$e^{j\frac{3\pi}{2}} = -j$$

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Fun Game

Now we know that :

Exponential functions	equals to
$e^{j\pi}$	$-1$
$e^{j\frac{\pi}{2}}$	$j$
$e^{-j\frac{\pi}{2}}$	$-j$
$e^{j\frac{3\pi}{2}}$	$-j$
$e^{j\pi}+e^{-j\pi}$	$(-1)+(-1) = -2$
$e^{j\frac{\pi}{2}}+(e^{j\frac{\pi}{2}})^*$	$j + (-j) = 0$
$e^\pi + e^{-\pi}$	23.184 (note there are no $j$… just normal calulcation.)

Determine phase shift in a signal function (examples)

Steps:

1. Draw the x(t) = 1and y(t) = 1 graph
2. Sub $\tau$ = the number question required
3. merge into the x(t) function and generate a new graph
4. use the new graph to see whether the number match with $v(t)$
5. The overlapping part is the $y(\tau)$

Sinusoids

A sinusoid is a signal that has its magnitude changes in time according to a sine function sin($\theta$)

It is the basis of all signals.

sin($\theta$) repeats itself for every 2$\pi$ change in $\theta$

Signal

$\Alpha sin(2\pi ft + \phi)$

• $\Alpha$ - Amplitude

• $f$ - Frequency

• $\phi$ - Phase Shift

$\Alpha cos(2\pi ft)$

• $\Alpha$ - Amplitude

• $f$ - Frequency

Note:

Frequency $f = \frac{1}{T}$ where T = period.

Radian Frequency $\omega = 2\pi f$

Cos Signal and Sin Signal

Tan Signal

More about Graphs

Conversion of Signals

Note there is a 90 degree difference between sin and cos.

$cos(\phi) = sin(\phi + \cfrac{\pi}{2})$

$sin(\phi) = cos( \cfrac{\pi}{2} - \phi )$

Transformations of Continuous-Time Signals

The rules same as graph transformation.

More About Transformations of Continuous-Time Signals

Lets say there is a continuous-time signal $x(t)$.

Function	Transform
$x(t)+ 2$	Translation by 2 units upwards
$x(t)- 2$	Translation by 2 units downwards
$x(t+2)$	Translation by 2 units to Left Hand Side
$x(t-2)$	Translation by 2 units to Right Hand Side
$-x(t)$	Reflection over x-axis
$x(-t)$	Reflection over y-axis
$2x(t)$	Vertial stretch by a factor of 2 (y axis become twice)
$\frac{1}{2}x(t)$	Vertial compression by a factor of $\frac{1}{2}$ (y axis become half)
$x(\frac{1}{2}t)$	Horizontal stretch by a factor of $\frac{1}{2}$ (x axis become twice)
$x(2t)$	Horizontal compression by a factor of 2 (x axis become half)

Note:

$x(2t-2)$ = $x(2(t-1))$

• $w(t)$ = $x(2t)$

• $y(t)$ = $w(t-1)$

Therefore shrinking by a factor of 2, then translate to right by 1 unit.

How to determine the value of a,b and c so that $y(t)=ax(bt+c)$ by looking at the graph:

$y(t)=ax(bt+c)$

Note:

a = length of y axis

b = length of x axis

c = translation $y(t) = w(t-"actual\space translation\space you\space see\space in\space graph")$

Period of combined signals

The Period equals to the LCM of the period of the signals.

Radian

• Standard unit of angular measure

• Equal to the length of the arc of a unit circle

• $\frac{2\pi}{360}$

Some Special Trig function to remember:

$$sin(\pi) = 0$$

$$tan^{-1}(0) = 0$$

$$tan^{-1}(\infty) = \pi/2$$

Signal Representation (examples)

Partial Fraction

It is a trick for linear functions and quadratic functions.

Basically, find A,B,C,D… (the variables) to get the job done.

https://www.youtube.com/watch?v=oq8YZz7EqpU

Distinct Linear Factors

example :

You have this function $\frac{1}{x^2-5x+6}$

since $x^2-5x+6 = (x-3)(x-2)$

$\frac{1}{x^2-5x+6} = \frac{A}{x-3} + \frac{B}{x-2}$

Multiply the whole thing with this $x^2-5x+6$, and you will get :

$1 = (x-2)A + (x-3)B$

Now, find A and B. Make 1 side in the right become 0, and then another side.

let $x=2$ , $1 = (2-2)A + (2-3)B$

Therefore $B = -1$.

Let $x=3$ , $1 = (3-2)A + (3-3)B$

Therefore $A = 1$.

So the decomposition is :

$\frac{1}{x^2-5x+6} = \frac{1}{x-3} + \frac{-1}{x-2}$

Distinct Linear Factors with imaginary

Example :

You have this function $\frac{1}{j+1} \cdot \frac{1}{j+2}$

$\frac{1}{(j+1)(j+2)} = \frac{A}{j+1} \cdot \frac{B}{j+2}$

Multiply the whole thing with this $(j+1)(j+2)$, you will get :

$1 = (j+2)A + (j+1)B$

note the form $a+bj$.

$1 + 0j = (j+2)A + (j+1)B$

$1 + 0j = 2A+B + Aj+Bj$

Looking at the imaginary part, $A+B = 0$, $A = -B$

Looking at the real part, $1 = 2A + B$

Therefore $2A - A = 1$

$A = 1, B = -1$