1 outline analog and digital data analog and digital signals amplitude modulation (am) frequency...
Post on 05-Jan-2016
247 Views
Preview:
TRANSCRIPT
1
Outline
Analog and Digital Data
Analog and Digital Signals
Amplitude Modulation (AM)
Frequency Modulation (FM)
2
What is a Signal ?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.
3WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998
Periodic Signals
4WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998
Aperiodic Signals
5
Analog and Digital Signals
In data communication, we commonly use periodic analog signals and
aperiodic digital signals.
6
Periodic signal: Sine Wave – Period(T) & Frequency(F)
T = 1 / F
Peak Amplitude
7
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.
Frequency
8
Frequencies
Unit Equivalent Unit Equivalent
Seconds (s) 1 s hertz (Hz) 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
9
Example 1Example 1
Express a period of 100 ms in microseconds, and express the corresponding frequency in kilohertz.
SolutionSolution
From Table 3.1 we find the equivalent of 1 ms.We make the following substitutions:100 ms = 100 10-3 s = 100 10-3 10 s = 105 s
Now we use the inverse relationship to find the frequency, changing hertz to kilohertz100 ms = 100 10-3 s = 10-1 s f = 1/10-1 Hz = 10 10-3 KHz = 10-2 KHz
10
Phase
Phase describes the position of the waveform relative to time zero.
11
Example 2Example 2
A sine wave is offset one-sixth of a cycle with respect to time zero. What is its phase in degrees and radians?
SolutionSolution
We know that one complete cycle is 360 degrees.
Therefore, 1/6 cycle is
(1/6) 360 = 60 degrees = 60 x 2 /360 rad = /3 rad or
1.046 rad
12
ftA 2sin
where
A = Amplitude
f = frequency
= phase
Time-Domain Signal Representation
13
14
An analog signal is best represented in the frequency domain.
15
Time-Frequency Domain
16
Time-Frequency Domain
17
Time-Frequency Domain
18WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998
Examples
19
Fourier Decomposition for Periodic Signals
k
kkk tfAts 2sin
...2sin2sin2sin 333222111 tfAtfAtfA
20
1
2sin4
k
tkfk
A
...52sin5
432sin
3
42sin
4 tf
Atf
Aft
Ats
Example(1)
21
Example(2)
22
Example(3)
23
Frequency Content of a Square Wave
24
Figure 4-19
WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998
Harmonics of a Digital Signal
25
Transmission Medium Imperfection
The bandwidth is a property of a medium: The bandwidth is a property of a medium: It is the difference between the highest and It is the difference between the highest and the lowest frequencies that the medium can the lowest frequencies that the medium can
satisfactorily pass.satisfactorily pass.
26
Frequency Response of a Medium
Signal frequency content Frequency response of the medium
27
Example 3Example 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.
SolutionSolution
B = fh fl = 900 100 = 800 HzThe spectrum has only five spikes, at 100, 300, 500, 700, and 900
28
29
Example 4Example 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.
SolutionSolution
B = fB = fhh f fll
20 = 60 20 = 60 ffll
ffll = 60 = 60 20 = 40 Hz20 = 40 Hz
30
31
Example 5Example 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?
SolutionSolution
The answer is definitely no. Although the signal can have The answer is definitely no. Although the signal can have the same bandwidth (1000 Hz), the range does not the same bandwidth (1000 Hz), the range does not overlap. The medium can only pass the frequencies overlap. The medium can only pass the frequencies between 3000 and 4000 Hz; the signal is totally lost.between 3000 and 4000 Hz; the signal is totally lost.
32
Digital Signals
33
Bit rate (R) & period of a bit (T)
1 1T or R
R T
The capacity of a data communication system can be expressed in terms of the number of data bits sent per second in time. – bits per second or (bits/sec)
This is also refers to the data rate (R) of the system – speed of transmission.
If the period of a data bit is T then R is the inverse of T.
The period of data bit is called bit interval.
34
Example 6Example 6
A digital signal has a bit rate of 2000 bps. What is the duration of each bit (bit interval)
SolutionSolution
The bit interval is the inverse of the bit rate.
Bit interval = 1/ 2000 s = 0.000500 s = 0.000500 x 106 s = 500 s
35
36
Transmission Impairment
Attenuation
Distortion
Noise
37
Attenuation
1
2log10P
PdB
38
Imagine a signal travels through a transmission medium and its power is reduced to half. This means that P2 = 1/2 P1. In this case, the attenuation (loss of power) can be calculated as
SolutionSolution
10 log10 log1010 (P2/P1) = 10 log (P2/P1) = 10 log1010 (0.5P1/P1) = 10 log (0.5P1/P1) = 10 log1010 (0.5) (0.5)
= 10(–0.3) = –3 dB = 10(–0.3) = –3 dB
Example 12Example 12
39
Example 13Example 13
Imagine a signal travels through an amplifier and its power is increased ten times. This means that P2 = 10 ¥ P1. In this case, the amplification (gain of power) can be calculated as
10 log10 log1010 (P2/P1) = 10 log (P2/P1) = 10 log1010 (10P1/P1) (10P1/P1)
= 10 log= 10 log1010 (10) = 10 (1) = 10 dB (10) = 10 (1) = 10 dB
40
Example 14Example 14
One reason that engineers use the decibel to measure the changes in the strength of a signal is that decibel numbers can be added (or subtracted) when we are talking about several points instead of just two (cascading). In Figure 3.22 a signal travels a long distance from point 1 to point 4. The signal is attenuated by the time it reaches point 2. Between points 2 and 3, the signal is amplified. Again, between points 3 and 4, the signal is attenuated. We can find the resultant decibel for the signal just by adding the decibel measurements between each set of points.
41
dB = –3 + 7 – 3 = +1
42
Distortion
43
Noise
44
Analog Modulation
45
Amplitude Modulation
46
Amplitude Modulation
47
Amplitude Modulation
The total bandwidth required for AM can be determined from the bandwidth
of the audio signal: rule-of thumb: BWt = 2 x BWm.
48
AM band allocation
49
Example 13Example 13
We have an audio signal with a bandwidth of 4 KHz. What is the bandwidth needed if we modulate the signal using AM?
SolutionSolution
An AM signal requires twice the bandwidth of the original signal:
BW = 2 x 4 KHz = 8 KHz
50
Frequency Modulation
51
Frequency Modulation
52
Frequency Modulation
The total bandwidth required for FM can be determined from the bandwidth
of the audio signal: rule-of thumb: BWt = 10 x BWm.
53
Frequency Modulation
The bandwidth of a stereo audio signal is usually 15 KHz. Therefore, an FM station needs at least a bandwidth of
150 KHz. The FCC requires the minimum bandwidth to be at least 200
KHz (0.2 MHz).
54
FM Band Allocation
55
Example 14Example 14
We have an audio signal with a bandwidth of 4 MHz. What is the bandwidth needed if we modulate the signal using FM?
SolutionSolution
An FM signal requires 10 times the bandwidth of the original signal:
BW = 10 x 4 MHz = 40 MHz
56
Tutorial and Examples1- Describe the three characteristics of a sine wave.
a. amplitude,b. frequency/period,c. phase
57
Tutorial and Examples2- Describe a sine signal using the unit circle.
xy sin where x is in radians.
58
Tutorial and Examples2- Describe a cosine signal using the unit circle.
xy cos where x is in radians.
59
Tutorial and Examples3-What are the amplitude, phase, and frequency characteristics of the cosine signal that is shown below.
31002cos10
t
Amplitude= 10Frequency=100 HzPhase= Π/3 radians=60 degrees
60
Tutorial and Examples4-What is the Fourier transform of the signal that is shown below.
cAtx
dtetxfX tfj
2
fAdteAfX ctfj
c
2 Dirac-Delta function
Dirac-Delta function is very useful as an approximation for tall narrow spike functions.
61
Properties of Dirac-Delta functions:
0,0
0,
xfor
xforx1-
2-
1dxx
0
As red lines approaches to zero, the green line approaches to infinity.
62
cAtx
t
x(t)
Ac
0
fAfX c
63
Tutorial and Examples5-What is the Fourier transform of the cosine signal shown below?
tfAtx cc 2cos
tfjtfe cctfj c 2sin2cos2
tfjtfe cctfj c 2sin2cos2
2
2cos22 tfjtfj
c
cc eetf
Euler’s Equations
64
xjxz sincos
xjxdx
dzcossin
xjxjxjxjdx
dzsincoscossin2
zjdx
dz
dxjdzz
1
dxjdzz
1
xjz ln xjxez xj sincos
To prove Euler’s equations
65
dtetfAdtetxfX tfjcc
tfj
22 2cos
dteee
AfX tfjtfjtfj
c
cc
2
22
2
dteA
dteA
fX tffjctffjc cc 22
22
cff cff
cc
cc ff
Aff
AfX
22
Hence, the Fourier transform of : tfAtx cc 2cos
top related