2. data and signals

Chapter 2
Data and Signals

M. Mozammel Hoque Chowdhury, Dept. of CSE, JU

1

DATA
Data can be Analog or Digital.
The term Analog data refers to information that is
continuous;
Digital data refers to information that has discrete
states.
Analog data take continuous values. Digital data
take discrete values.

2

Signals
To be transmitted, data must be transformed
to electromagnetic signals.
A signal is defined as any physical quantity that varies
with time, space or any other independent variables.
By a signal we mean any variable that carries some
information.
Continous signal:

x(t)

Discrete signal (sequence):

x[n]
3

Signals…
Signals can be analog or digital.
Analog signals can have an infinite number of values
in a range.
Digital signals can have only a limited number of
values.

4

PERIODIC & NON-PERIDIC SIGNALS


Periodic Signal


A continuous-time periodic signal is
unchanged/repeated by a time shift
of T0. T0 is called the fundamental
period.

x( t ) = x( t + T0 )



A sine wave: x(t) = A sin(2Πft+Φ)

A discrete-time periodic signal is
unchanged by a time shift of N0.
N0 is the fundamental period.

x[ n ] = x[ n + N 0 ]



Non-Periodic Signal


Not periodic

In data communications, we commonly use periodic
analog signals and non-periodic digital signals.

5

Properties of a Signal








Peak Amplitude (A ) – the highest value/intensity of
the signal.
Period (T ) – the amount of time (in sec) needed by a
signal to complete one cycle.
Frequency (f ) – number of cycles (periods) per sec.
f=1/T
Phase (Φ) – position of a signal relative to time zero.
Phase describes the position of the waveform relative to
time zero.



Wavelength(λ ) – the distance traveled by a wave
(signal) in one cycle.
λ = 1/f

6

Properties of a Signal…
Frequency is the rate of change with respect to time.
Change in a short span of time means high frequency.
Change over a long span of time means low frequency.
If a signal does not change at all, its frequency is zero.
If a signal changes instantaneously, its frequency is infinite.

Example
The power we use at home has a frequency of 60 Hz. The period
of this sine wave can be determined as follows:

7

Fig. Two signals with the same phase and frequency, but different amplitudes

8

Figure Three sine waves with the same amplitude and frequency,
but different phases

9

Figure Wavelength and period

10

The time-domain and frequency-domain plots of a sine
wave

A complete sine wave in the time domain can be represented by
one single spike in the frequency domain.

11

The time domain and frequency domain of three sine
waves
The frequency domain is more compact and useful when we are
dealing with more than one sine wave.
Following Figure shows three sine waves, each with different
amplitude and frequency. All can be represented by three spikes in
the frequency domain.

A single-frequency sine wave is not useful in data communications; we need
to send a composite signal, a signal made of many simple sine waves.

12

Composite Periodic Signal

According to Fourier analysis, any composite signal
is a combination of simple sine waves with different
frequencies, amplitudes, and phases.

Fig A composite periodic signal

13

Decomposition of a composite periodic signal in the time
and frequency domains

14

The time and frequency domains of a non-periodic signal

The bandwidth of a composite signal is the difference
between the highest and the lowest frequencies
contained in that signal.
15

Figure The bandwidth of periodic and nonperiodic composite signals

16

Example
If a periodic signal is decomposed into five sine waves with frequencies of 100,
300, 500, 700, and 900 Hz, what is its bandwidth? Draw the spectrum,
assuming all components have a maximum amplitude of 10 V.
Solution
Let fh be the highest frequency, fl the lowest frequency, and B the bandwidth.
Then

The spectrum has only five spikes, at 100, 300, 500, 700, and 900 Hz

17

Figure The bandwidth for Example 3.10

18

TRANSMISSION IMPAIRMENT
Signals travel through transmission media, which are not
perfect. The imperfection causes signal impairment. This
means that the signal at the beginning of the medium is not
the same as the signal at the end of the medium. What is sent
is not what is received. Three causes of impairment are
attenuation, distortion, and noise.

Topics discussed in this section:
Attenuation
Distortion
Noise
19

Example
Suppose a signal travels through a transmission medium
and its power is reduced to one-half. This means that P 2
is (1/2)P1. In this case, the attenuation (loss of power)
can be calculated as

A loss of 3 dB (–3 dB) is equivalent to losing one-half
the power.
22

Example
A signal travels through an amplifier, and its power is
increased 10 times. This means that P2 = 10P1 . In this
case, the amplification (gain of power) can be calculated
as

23

Example
The power of a signal is 10 mW and the power of the
noise is 1 μW; what are the values of SNR and SNRdB ?
Solution
The values of SNR and SNRdB can be calculated as
follows:

26

Example
The values of SNR and SNRdB for a noiseless channel
are

We can never achieve this ratio in real life; it is an ideal.

27

Figure Two cases of SNR: a high SNR and a low SNR

28

DATA RATE LIMITS
A very important consideration in data communications is
how fast we can send data, in bits per second, over a
channel. Data rate depends on three factors:
1. The bandwidth available
2. The level of the signals we use
3. The quality of the channel (the level of noise)

29

Nyquist theorem for Sampling

The Nyquist rate is the minimum sampling rate required to avoid aliasing, equals to twice
the highest frequency contained within the signal.

where is B the highest frequency of the signal.

To avoid aliasing, the sampling rate must exceed the Nyquist rate:

30

Noiseless Channel and Nyquist theorem

Bit rate = 2X BX log2L
Where B is the bandwidth, L is the number of signal levels used to
represent data.

31

Example
Consider a noiseless channel with a bandwidth of 3000
Hz transmitting a signal with two signal levels. The
maximum bit rate can be calculated as:

32

Example
Consider the same noiseless channel transmitting a signal
with four signal levels (for each level, we send 2 bits). The
maximum bit rate can be calculated as:

33

Example
We need to send 265 kbps over a noiseless channel with
a bandwidth of 20 kHz. How many signal levels do we
need?
Solution
We can use the Nyquist formula as shown:

Since this result is not a power of 2, we need to either
increase the number of levels or reduce the bit rate. If we
have 128 levels, the bit rate is 280 kbps. If we have 64
levels, the bit rate is 240 kbps.
34

Noisy Channel and Shannon Capacity
Consider an extremely noisy channel in which the value of the signalto-noise ratio is almost zero.

For noisy channel the capacity C (highest data rate) is
calculated as:

This means that the capacity of this channel is zero regardless of
the bandwidth. In other words, we cannot receive any data
through this channel.
35

Example
We can calculate the theoretical highest bit rate of a
regular telephone line. A telephone line normally has a
bandwidth of 3000. The signal-to-noise ratio is usually
3162. For this channel the capacity is calculated as

This means that the highest bit rate for a telephone line
is 34.860 kbps. If we want to send data faster than this,
we can either increase the bandwidth of the line or
improve the signal-to-noise ratio.
36

Example
We have a channel with a 1-MHz bandwidth. The SNR
for this channel is 63. What are the appropriate bit rate
and signal level?
Solution
First, we use the Shannon formula to find the upper
limit.

37

Note

The Shannon capacity gives us the
upper limit; the Nyquist formula tells us
how many signal levels we need.

38

2. data and signals

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