Transcript CS412 Computer Networks - Winona State University

CS412 Introduction to
Computer Networking &
Telecommunication
Theoretical Basis of
Data Communication
Chi-Cheng Lin, Winona State University
Topics

Analog/Digital Signals

Time and Frequency Domains

Bandwidth and Channel Capacity

Data Communication Measurements
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Signals
Information must be transformed into
electromagnetic signals to be
transmitted
 Signal forms

Analog or digital
Periodic or aperiodic
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Analog/Digital Signals

Analog signal
Continuous waveform
Can have a infinite number of values in a
range

Digital signal
Discrete
Can have only a limited number of values
E.g., 0 or 1
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Figure 3.1
Comparison of analog and digital signals
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Periodic/Aperiodic Signals

Periodical signal
Contains continuously repeated pattern
Period (T): amount of time needed for the
pattern to complete

Aperiodical signal
Contains no repetitive signals
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Analog Signals

Simple analog signal
Sine wave
3 characteristics
1. Peak amplitude (A)
2. Frequency (f)
3. Phase ()

Composite analog signal
Composed of multiple sine waves
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Figure 3.2
A sine wave
s(t )  A sin(2ft   )
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Figure 3.3
Amplitude
s(t): instantaneous amplitude
t
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Characteristics of Analog Signal
Peak amplitude: highest intensity
 Frequency (f)

Number of cycles/rate of change per second
Measured in Hertz (Hz), KHz, MHz, GHz, …
Period (T): amount of time it takes to
complete one cycle
f = 1/T

Phase: position of the waveform relative
to time 0
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Figure 3.4
Period and frequency
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Table 3.1 Units of periods and frequencies
Unit
Seconds (s)
Equivalent
1s
Unit
hertz (Hz)
Equivalent
1 Hz
Milliseconds (ms)
10–3 s
kilohertz (KHz)
103 Hz
Microseconds (ms)
10–6 s
megahertz (MHz)
106 Hz
Nanoseconds (ns)
10–9 s
gigahertz (GHz)
109 Hz
Picoseconds (ps)
10–12 s
terahertz (THz)
1012 Hz
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Figure 3.5
Relationships between different phases
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Figure 3.6
Sine wave examples
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Figure 3.6
Sine wave examples (continued)
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Figure 3.6
Sine wave examples (continued)
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Characteristics of Analog Signal

Changes in the three characteristics
provides the basis for
telecommunication
Used by modems (later …)
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Time Vs. Frequency Domain
The sine waves shown previously are
plotted in its time domain.
 An analog signal is best represented in
the frequency domain.

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Figure 3.7
Time and frequency domains
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Composite Signals
A composite signal can be decomposed
into component sine waves - harmonics
 The decomposition is performed by

Fourier Analysis
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Figure 4-13
WCB/McGraw-Hill
Signal with DC Component
 The McGraw-Hill Companies, Inc., 1998
Figure 3.8-3.10 Square wave and the first three harmonics
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Figure 3.11 Frequency spectrum comparison
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Frequency Spectrum and Bandwidth

Frequency spectrum
Collection of all component frequencies it
contains

Bandwidth
Width of frequency spectrum
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Figure 3.13
Bandwidth
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Example 3
If a periodic signal is decomposed into five sine waves
with frequencies of 100, 300, 500, 700, and 900 Hz,
what is the bandwidth? Draw the spectrum, assuming all
components have a maximum amplitude of 10 V.
Solution
B = fh - fl = 900 - 100 = 800 Hz
The spectrum has only five spikes, at 100, 300, 500, 700,
and 900 (see Figure 13.4 )
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Figure 3.14
Example 3
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Example 4
A signal has a bandwidth of 20 Hz. The highest frequency
is 60 Hz. What is the lowest frequency? Draw the
spectrum if the signal contains all integral frequencies of
the same amplitude.
Solution
B = fh - fl
20 = 60 - fl
fl = 60 - 20 = 40 Hz
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Figure 3.15
Example 4
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Example 5
A signal has a spectrum with frequencies between 1000
and 2000 Hz (bandwidth of 1000 Hz). A medium can pass
frequencies from 3000 to 4000 Hz (a bandwidth of 1000
Hz). Can this signal faithfully pass through this medium?
Solution
The answer is definitely no. Although the signal can have
the same bandwidth (1000 Hz), the range does not
overlap. The medium can only pass the frequencies
between 3000 and 4000 Hz; the signal is totally lost.
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Digital Signals
0s and 1s
 Bit interval and bit rate

Bit interval: time required to send 1 bit
Bit rate: #bit intervals in one second
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Example 6
A digital signal has a bit rate of 2000 bps. What is the
duration of each bit (bit interval)
Solution
The bit interval is the inverse of the bit rate.
Bit interval = 1/ 2000 s = 0.000500 s
= 0.000500 x 106 ms = 500 ms
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Digital Signal - Decomposition

A digital signal can be decomposed into
an infinite number of simple sine waves
(harmonics), each with a different
amplitude, frequency, and phase
A digital signal is a composite signal
with an infinite bandwidth.

Significant spectrum
Components required to reconstruct the
digital signal
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Figure 4-20
Harmonics of a Digital Signal
WCB/McGraw-Hill
 The McGraw-Hill Companies, Inc., 1998
Bandwidth-Limited Signals

(a) A binary signal and its root-meansquare Fourier amplitudes.
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Bandwidth-Limited Signals (2)

(b) – (e) Successive approximations
to the original signal.
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Figure 4-21
WCB/McGraw-Hill
Exact and Significant Spectrums
 The McGraw-Hill Companies, Inc., 1998
Channel Capacity

Channel capacity
Max. bit rate a transmission medium can
transfer

Nyquist theorem
C = 2H log2V
where C: channel capacity (bit per second)
H: bandwidth (Hz)
V: signal levels (2 for binary)
C is proportional to H
 Significant bandwidth puts a limit on
channel capacity
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Figure 3.18
Digital versus analog
To transmit 6bps, we need a bandwidth = 3 - 0 = 3Hz
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Channel Capacity
Nyquist theorem is for noiseless (errorfree) channels.
 Shannon Capacity

C = H log2(1 + S/N)
where C: (noisy) channel capacity (bps)
H: bandwidth (Hz)
S/N: signal-to-noise ratio
dB = 10 log10 S/N

In practice, we have to apply both for
determining the channel capacity.
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Example 7
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
Bit Rate = 2  3000  log2 2 = 6000 bps
Example 8
Consider the same noiseless channel, transmitting a signal
with four signal levels (for each level, we send two bits).
The maximum bit rate can be calculated as:
Bit Rate = 2 x 3000 x log2 4 = 12,000 bps
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Example 9
Consider an extremely noisy channel in which the value
of the signal-to-noise ratio is almost zero. In other words,
the noise is so strong that the signal is faint. For this
channel the capacity is calculated as
C = B log2 (1 + S/N) = B log2 (1 + 0)
= B log2 (1) = B  0 = 0
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Example 10
We can calculate the theoretical highest bit rate of a
regular telephone line. A telephone line normally has a
bandwidth of 3000 Hz (300 Hz to 3300 Hz). The signalto-noise ratio is usually 35dB, i.e., 3162. For this channel
the capacity is calculated as
C = B log2 (1 + S/N) = 3000 log2 (1 + 3162)
= 3000 log2 (3163)
C = 3000  11.62 = 34,860 bps
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Example 11
We have a channel with a 1 MHz bandwidth. The S/N for
this channel is 63; what is the appropriate bit rate and
signal level?
Solution
First, we use the Shannon formula to find our upper
limit.
C = B log2 (1 + S/N) = 106 log2 (1 + 63) = 106 log2 (64) = 6 Mbps
Then we use the Nyquist formula to find the
number of signal levels.
4 Mbps = 2  1 MHz  log2 L  L = 4
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Data Communication Measurements

Throughput
How fast data can pass through an entity

Propagation speed
Depends on medium and signal frequency

Propagation time (propagation delay)
Time required for one bit to travel from one
point to another

Wavelength
Propagation speed = wavelength X frequency
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Figure 3.25
Throughput
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Figure 3.26
Propagation time
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Figure 3.27
Wavelength
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