What is a Signal ?

A signal is an entity that carries information.

A signal can be modeled as a function.

Examples :

Temperature is a function of time $t$ (Continuous)
hourly average temperature is a function of hour (Discrete)

Classification of a signal

Signal function $f(t)$

Domain defines the allowable values of the argument of a function ( $x$ in $y = f(x)$ )
Range defines the allowable values of a function ( $y$ in $y = f(x)$ )
A signal can be either discrete- or continuous-time
- Discrete-time function - The number of allowable values in the domain is countable
- Continuous-time function - The number of allowable values in the domain is not countable

$y[n]$ denotes a discrete-time signal $y$

n means particular time instance in the sequence

$y(t)$ denotes a continuous-time signal $y$

The argument can be any time-oriented variable

C.T.S. (Continous Time Signal)

C.T.S. is Not good, We cannot use a computer to handle it since there are infinite number of time instance.

Therefore we need to convert it into D.T.S. Through Sampling.

Sampling

Measuring the continuous-time signal to produce a new function whose number of domain values is countable.

D.T.S. (Discrete Time Signal)

Representation of a signal

Time-domain representation - Represented in the original form of the function.

Not only apply to functions of time
Can apply to a function of any variable as long as the function value changes according to an ordered sequence of time instants.

Any continuous periodic signal can be represented as a sum of scaled sinusoidal signals.

A Time Domain Representation Signal $x(t)$ is a combination of many sinusoidal signals (Frequency domain):

$a_0$ x Sinusoid $cos(\omega_0+\phi_0)$
$a_1$ x Sinusoid $cos(\omega_1+\phi_1)$
$a_2$ x Sinusoid $cos(\omega_2+\phi_2)$
$a_3$ x Sinusoid $cos(\omega_3+\phi_3)$
$a_4$ x Sinusoid $cos(\omega_4+\phi_4)$
$a_5$ x Sinusoid $cos(\omega_5+\phi_5)$
…

$a_i$ is the Scaling factors

$\phi_i$ is the phase angles

They are unique to a signal. We can use them to represent the signal.

DC coefficient - $a_0$

AC coefficient - $a_i$ where i > 0

Frequency-domain representation

Represented with the set of scaling factors + phase angles

Waveform vs Spectrum

Waveform - a plot of a function of time
Spectrum - a plot of a function of frequency

Each sinusoid is called a frequency component of the signal

Sinusoid is Identified by its frequency which specifies how many cycles it repeats in a second.

High frequency component

repeat quickly

Low frequency component

repeat slowly

DC component

Don’t change at all (frequency = 0 Hz, change 0 times/s)
Average value of the signal
is a constant term

AC component

Any components other than the DC one
AC components cannot share the same frequency

Bandwidth of a signal

The range of frequencies bounded by the DC component and the non-zero frequency component having the maximum frequency.

Bandwidth = $f_i - f_0$ = Highest frequency - Lowest frequency

Signal Representation

Continuous-time Signals (Analog Signals)

Also known as analog signals.

continuous signals =/= continuous-time signals

Domain variable
May not be continuous signals
Usually decomposed into a combination of basic functions for study.

Note: Vectical Line = Gap = Discontinuous

Basic functions

Impulse Function

It is a continuous-time signal that is not continuous

Unit Step Function

It is a continuous-time signal that is not continuous

Sign Function

It is a continuous-time signal that is not continuous

Ramp Function

Impulse Train Function

It is a continuous-time signal that is not continuous

Rectangle Function

It is a continuous-time signal that is not continuous

Triangle Function

Sinc Function

Dirichlet Function

Continuous Signals

The signal $x(t)$ is said to be discontinuous at the point $t_2$ if

$x(t_2^-)\neq x(t_2^+)$
The value of $x(t)$ jumps at point $t_2$

If the signal x(t) is continuous at all points $t$ , $x(t)$ is said to be a continuous signal.

Note this is different from continuous-time signal

Discrete-time Signals

Basically, A Continous Signal will become Discrete Signal after Sampling Process.

Note: Discrete-time Signals can never be a continuous function.

Sample $x(t)$ @ $t = nT_s$ to get $x[n] = x(nT_s)$

Note:

$T_s$ = Sampling Period

$\frac{1}{T_s} = f_s$ = Sampling Frequency

$t = nT_s = \frac{n}{f_s}$

Basic discrete functions

Basic discrete functions can be obtained by sampling basic functions.

Like continuous-time function, Discrete functions can also be represented as shifted, reflected, scaled versions of basic discrete functions or combinations of them.

Time-shifting

Given $x[n]$ and a positive integer $r$

$x[n-r]$ : $x[n]$ shift to the right by $r$ step
$x[n + r]$ : $x[n]$ shift to the left by $r$ step

Time-reflecting

Given $x[n]$

$x[-n]$ : a time-reflected version of $x[n]$ about $n = 0$ .

Time-scaling

Given $x[n]$

$x[kn]$ : a time-scaled version of $x[n]$ where k is generally an integer.
If $k>1$ , It is called Supsampling. All the values of $x[n]$ will not be retained.
If $k < 1$ , The extra values will be filled with $0$ s since the sample does not exist.

Even Function and Odd Function

Even function : Symmetrical (Mirrored)

Odd function : Not Mirrored

Note:

Sin component is a Odd function

Cos component is a Even function

Periodic Discrete-time Signals

A discrete-time signal x[n] is periodic if there exists a positive integer r such that

$x[n+r] = x[n]$

Fundamental period

smallest value of r

Examples:

A discrete cosine function which repeats every 2 samples: $x[n]=Acos(\pi n)$

$x[n] = Acos(\pi n) = Acos(\pi(n+2)) = x[n+2] = Acos(\pi(n+4)) = x[n+4]$

The Fundamental period is 2.

Note: A sampled periodic time-continuous signal may not be a periodic discrete signal.

After sampling, may not be still periodic**

Ex1

Determine whether the following signals are periodic or not. For periodic signals, determine also the fundamental period.

$x[n] = sin(n\pi/4)$
$x[n] = sin(n/4)$
$x[n] = cos(3n)$
$x[n] = cos(\pi n/10)$

Keypoints:

Determine the function cycle

If $x[n+r] = x[n]$ it is periodic, otherwise not periodic

$r$ must be positive integer value.

Q1: $x[n] = cos(\pi n/4)$

$x[n+r] = x[n]$ if $r$ is the period

where

$x[n+r] = sin(n\pi/4 + r\pi /4)$

$x[n] = sin(n\pi/4) = sin(n\pi/4 + 2\pi m)$ <- We used $2\pi$ because sin function cycle every $2\pi$

Therefore, $r\pi/4 = 2\pi m$

$r = 8m$ <- $m$ is any integer value

$r$ can be a interger -> Valid -> Periodic

To get the Minimizing $r$ value

By sub $m = 1$ ,

Fundamental period = $8 \times 1$ = 8

Q2: $x[n] = sin(n/4)$

$x[n+r] = sin(n/4 + r/4)$

$x[n] = sin(n/4) = sin(n/4 + 2\pi m)$ <- We used $2\pi$ because sin function cycle every $2\pi$

$r = 8\pi m$ <- $m$ is any integer value

$r$ cannot be a interger -> invalid -> Not Periodic

Q3: $x[n] = cos(3n)$

$x[n+r] = cos(3n + 3r)$

$x[n] = cos(3n) = cos (3n + 2\pi m)$ <- We used $2\pi$ because cos function cycle every $2\pi$

$r = \frac{2}{3}\pi m$ <- $m$ is any integer value

$r$ cannot be a interger -> invalid -> Not Periodic

Q4: $x[n] = cos(\pi n/10)$

$x[n+r] = cos(\pi n/10 + \pi r/10)$

$x[n] = cos(\pi n/10) = cos(\pi n/10 + 2\pi m)$ <- We used $2\pi$ because cos function cycle every $2\pi$

$r = 20m$ <- $m$ is any integer value

$r$ can be a interger -> Valid -> Periodic

To get the Minimizing $r$ value

By sub $m =1$ ,

Fundamental period = $20 \times 1$ = 20

Note: Since $r$ must be a positive integer, you might encounter situations like

$r = \frac{m}{3}$

In this case, You should sub $m = 3$ instead of $m = 1$ to get the positive integer $r$ value.

Discrete-time signal (DTS) $x[n] = cos(\pi n / 10)$ is obtained by sampling Continous-time signal (CTS) $x(t) = cos(5\pi t / 2)$ . Determine the sampling frequency and the sampling period of $x[n]$ . (By default, unit of t is second unless specified otherwise.)

Keypoints:

$x(t)$ -> $x[n] = x(nT_s)$

$n$ = Sample

$T_s$ = Sampling Period ( $T_s$ time for 1 sample)

$\frac{1}{T_s} = f_s$ = Sampling Frequency

$t = nT_s = \frac{n}{f_s}$

First you get the frequency from the continous-time signal (CTS):

$cos(5\pi t / 2) = cos(2\pi \times \frac{5}{4}t)$

Frequency = $\frac{5}{4}$ Hz

Period = 0.8 sec

Radian frequency = $2\pi f$ = $2.5\pi$

For discrete-time signal (DTS):

$x[n] = cos(\pi n / 10) = x(nT_s) = cos(5\pi \times \frac{nT_s}{2}])$

$\pi n/10 = 5\pi \times \frac{nT_s}{2}$

$T_s = \frac{1}{25}$

Sampling Period = 0.04 sec

Sampling Frequency = $\frac{1}{T_s} = f_s = 25$ Samples per second

Note: With $x[n] = cos(\pi n / 10)$ ,

now you know the fundamental period of $x[n] = 20$ samples.

If you want to represent it in unit seconds:

$t = nT_s$

$20$ samples $\times T_s$ = $\frac{4}{5}$ = 0.8 seconds

Ex 2

(a)

Sampling period = $\frac{1}{f} = 0.2$ sec

(b)

Yes. x(t) is continuous.

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Note : $x[n] = x(T_sn)$ .

Therefore $x[n] = x(0.2n) = 3 + 2sin(0.8n\pi) - cos(0.1n\pi)$

(d)

Look at $sin(0.8n\pi)$ first.

Let period of $sin(0.8\pi n)$ be $T_1$ samples <- It is basically $r$ in our previous questions

$sin(0.8\pi(n+T_1)) = sin(0.8\pi n + 2\pi m)$ <- We used $2\pi$ because sin function cycle every $2\pi$

$0.8T_1\pi = 2m\pi$

$T_1 = 2.5m$ <-- m is any integer value

Note : $T_1$ needs to be integer.

$T_1 = 2.5 \times 2 = 5$

Fundamental period of $sin(0.8n\pi)$ is 5 samples.

Now look after $cos(0.1n\pi)$

Let period of $cos(0.1n\pi)$ be $T_2$ samples <- It is basically $r$ in our previous questions

$cos(0.1\pi(n+T_2)) = cos(0.1\pi + 2\pi m)$ <- We used $2\pi$ because sin function cycle every $2\pi$

$0.1\pi T_2 = 2\pi m$

$T_2 = 20m$ <-- m is any integer value

$T_2 = 20 \times 1 = 20$

Fundamental period of $cos(0.1n\pi)$ is 20 samples.

$LCM(T_1,T_2) = LCM(5, 20) = 20$

Therefore the period of $x[n]$ is 20 samples.

(e)

The fundamental period in terms of second = sample times sampling period = $20 \times 0.2$ s $= 4$ sec

Digital Signals

A digital signal $x[n]$ is a discrete-time signal whose values belong to the finite set $\left\{ a_1,a_2,\cdots,a_N\right\}$ .

A digital signal can have only a finite number of different values

After Quantization,

Discrete-time signal -> Digital signal

Full Picture:

Continous sign —Sampling—> Discrete-time signal —Quantization—> Digital signal

This is what digital signal looks like.

Ex 3

(a)

The ans is 1. Dc is the constant term. it does not change with time.

(b)

The ans is 2. In Frequency domain analysis, we decompose a signal into sinosoids. Each contributes 1 ac component.

Note: Ac components cannot share the same frequency.

for example. if $x(t)=cos(2\pi t)+cos(2\pi t)$ , we consider 1 ac component only, because both cos function have the same frequency.

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The ans is 3.

(d)

The ans is $2 - 0 = 2Hz$ .

Note: Bandwidth = $f_i - f_0$ = Highest frequency - Lowest frequency

Find their highest frequency and lowest frequency.

$x(t) = 3 + 2sin(4\pi t) - cos(0.5\pi t)$

Frequency of $3$ : $f = 0$ (DC component)

Frequency of $2sin(4\pi t)$ : $2sin(4\pi t) = 2sin(2\pi \times 2t) \rightarrow f = 2$

Frequency of $cos(0.5\pi t)$ : $cos(0.5\pi t) = cos(2pi \times \frac{1}{4}t) \rightarrow f = \frac{1}{4}$

Highest = 2, Lowest = 0.

(e)

Note: Sin component is a Odd function, Cos component is a Even function

$x(t) = 3 + 2sin(4\pi t) - cos(0.5\pi t)$ contains both sin component and cos component.

Therefore $x(t)$ is neither even or odd function.

(f)

Note: we ignore DC term when finding fundamental period.

Period of $3$ : $T = 0$ (DC component)

Period of $2sin(4\pi t)$ : $0.5$

Period of $cos(0.5\pi t)$ : $4$

Fundamental period = 4 second (LCM of both periods)

(g)

$f = \frac{1}{T} = \frac{1}{4}$

Ex 4

Keypoint : Basic discrete functions.

Note They both have the height of 1 and stands at 0.

$\delta[n]$ Looks like this:

$u[n]$ Looks like this:

(a) $g[n]=5 \delta[n-2]+3 \delta[n+1]$

This is a easy one.

First, Look at $\delta[n-2]$ .

$\delta[n-2]$ is basically $\delta[n]$ but shifted to right by 2 units.

Now Look at $\delta[n+1]$ .

$\delta[n+1]$ is basically $\delta[n]$ but shifted to left by 1 unit.

Therefore, after multiply the height, this is how $5 \delta[n-2]+3 \delta[n+1]$ looks like:

(b) $g[n]=5 n \cdot(u[n-2]-u[n-5])+(8-n) \cdot(u[n-6]-u[n-8])$

First, Look at $(u[n-2]-u[n-5])$ .

$u[n-2]$ and $u[n-5]$ are basically $u[n]$ with right shift of 2 and 5 units respectively.

$(u[n-2]-u[n-5])$ means the $u[n]$ starts from 2 and stop at 4. ( $2 \leq n < 5$ )

Next, Look at $(u[n-6]-u[n-8])$ .

$u[n-6]$ and $u[n-8]$ are basically $u[n]$ with right shift of 6 and 8 units respectively.

$(u[n-6]-u[n-8])$ means the $u[n]$ starts from 6 and stop at 7. ( $6 \leq n < 8$ )

Multiply the height and join them together.

Since it is not easy to visualize like impulse functions, We can draw a table to plot it out.

n 1 2 3 4 5 6 7 8

5n 10 15 20

8-n 2 1

g[n] 0 10 15 20 0 2 1 0

you will get this graph.

n	1	2	3	4	5	6	7	8
5n		10	15	20
8-n						2	1
g[n]	0	10	15	20	0	2	1	0

Ex 5

This is a easy one. We break it into segments.

We can clearly see there is 2 operated $u[n]$ function there. the second one is fliped with x axis (times -1).

So we can interprete that the function look like this : (I denote the function with $x_1[n]$ and $x_2[n]$ )

$x_1[n] + (-1)x_2[n]$

First, Look at $x_1[n]$ .

It is a operated $u[n]$ function. $0 \leq n < 2$ thus it is $(u[n] - u[n-2])$ .

Now Look at $x_2[n]$ .

It is a operated $u[n]$ function. $3 \leq n < 6$ thus it is $(u[n - 3] - u[n-6])$ .

Therefore, $x_1[n] + (-1)x_2[n]$ = $(u[n] - u[n-2]) + (-1)(u[n - 3] - u[n-6])$

= $(u[n] - u[n-2]) - (u[n - 3] - u[n-6])$

Ex 6

Keypoint : Ramp Function.

Important thing to remember : in ramp function , $f(n) = n$ . You might say the height of ramp[n] = n.

In this question, we can break it into two segment.

First look at the first segment.

$-2 \leq n < 2$ , Thus it is $u[n+2] - u[n-2]$ .

Note: since it is a ramp function, remember to times $n$ .

So it is $n(u[n+2] - u[n-2])$ .

Second segment is only a impulse function with half height.

So it is $0.5 \delta[n-2]$ .

Therefore:

$x[n] = n(u[n+2] - u[n-2]) + 0.5 \delta[n-2]$