Analog Transmission Example 1

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Analog Transmission Example 1
What sampling rate is needed for a signal with a
bandwidth of 10,000 Hz (1000 to 11,000 Hz)?
Solution
The sampling rate must be twice the highest
frequency in the signal
Sampling rate 2 x (11,000) 22,000
samples/s
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Sampling Rate
According to Nyquist theorem sampling rate must
be at least 2 times the highest frequency in the
signal.
Reason
Perfect reconstruction is possible if the signal
is sampled at this rate.
Sample Rate 2 X Highest Frequency Used
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Analog Transmission Example 2
A signal is sampled. Each sample requires at
least 11 levels of precision (-5, -4, , -1, 0,
1, , 5) How many bits should be sent for each
sample?
Solution
We need 4 bits. A 3-bit value can represent 23
8 levels (000 to 111), which is not enough. A
4-bit value can carry up to 24 16 values.
Note A signal with L levels actually can carry
log2L bits per level
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Analog Transmission Example 3
We want to digitize the human voice range up
to 4000 Hz. What is the bit rate, assuming 8
bits per sample?
Solution
The human voice normally contains frequencies
from 0 to 4000 Hz. Sampling rate 4000 x 2
8000 samples/s Bit rate sampling rate x number
of bits per sample 8000 x 8 64,000 bps 64
Kbps
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Analog Transmission Example 4
An analog signal carries 4 bits in each signal
unit. If 1000 signal units are sent per second,
find the baud rate and the bit rate
Solution
Baud rate 1000 bauds per second (baud/s) Bit
rate 1000 x 4 4000 bps
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Analog Transmission Example 5
The bit rate of a signal is 3000 bits per second.
If each signal unit carries 6 bits, what is the
baud rate?
Solution
Baud rate 3000 / 6 500 baud/s
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Amplitude Shift Keying (ASK)

• Bandwidth is proportional to the signal rate
  (baud rate).

• ASK is normally implemented using only 2 levels.

• When each symbol is binary it carries just one
  bit, so baud and bit rate are equal.

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Analog Transmission Example 6
Find the minimum bandwidth for an ASK signal
transmitting at 2000 bps. The transmission mode
is half-duplex (i.e. in one direction at a time).
Solution
In ASK the baud rate and bit rate are the same.
The baud rate is therefore 2000 per second. An
ASK signal requires a minimum bandwidth equal to
its baud rate. Therefore, the minimum bandwidth
is 2000 Hz.
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Analog Transmission Example 7
Given a bandwidth of 5000 Hz for an ASK signal
with 2 signal levels, what are the baud rate and
bit rate?
Solution
In ASK the baud rate is the same as the
bandwidth, which means the baud rate is 5000 per
second. But because the baud rate and the bit
rate are also the same for ASK with 2 signal
levels, the bit rate is 5000 bps.
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Analog Transmission Example 7
Find the minimum bandwidth for an FSK signal
using 2 signal levels transmitting at 2000 bps.
Transmission is in half-duplex mode, and the
carriers are separated by 3000 Hz.
Solution
For FSK BW baud rate fc1 - fc0
BW bit rate fc1 - fc0 2000 3000 5000
Hz
fc1 higher frequency signal level fc0 lower
frequency signal level The difference is the
bandwidth (3000 Hz)
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Analog Transmission Example 8
Find the maximum bit rates for an FSK signal with
2 signal levels if the bandwidth of the medium is
12,000 Hz and the difference between the two
carriers is 2000 Hz. Transmission is in
full-duplex mode (i.e. simultaneous, two-way).
Solution
Because the transmission is full duplex, only
6000 Hz is allocated for each direction. BW
baud rate fc1 - fc0 Baud rate BW - (fc1 -
fc0 ) 6000 - 2000 4000 But because the baud
rate is the same as the bit rate, the bit rate is
4000 bps.
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Analog Transmission Example 9
Find the bandwidth for a 4-PSK signal
transmitting at 2000 bps. Transmission is in
half-duplex mode.
Solution
For 4-PSK the baud rate is one half of the bit
rate. Therefore the baud rate is 1000 per second.
A PSK signal requires a bandwidth equal to its
baud rate. Therefore the bandwidth is 1000 Hz.
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Analog Transmission Example 9 4PSK method
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Analog Transmission Example 10
A constellation diagram consists of eight equally
spaced points on a circle. If the bit rate is
4800 bps, what is the baud rate?
Solution
The constellation indicates 8-PSK with the points
45 degrees apart. Since 23 8, 3 bits are
transmitted with each signal unit. Therefore, the
baud rate is 4800 / 3
1600 baud per second
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Analog Transmission Example 10 8 PSK Method
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Analog Transmission Example 11
Compute the bit rate for a 1000-baud 16-QAM
signal.
Solution
A 16-QAM signal has 4 bits per signal unit since
log216 4. Thus,
(1000)(4) 4000 bps
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Multiplexing Example 1
Five channels, each with a 100-KHz bandwidth, are
to be multiplexed together. What is the minimum
bandwidth of the link if there is a need for a
guard band of 10 KHz between the channels to
prevent interference?
Solution
For five channels, we need at least four guard
bands. This means that the required bandwidth is
at least 5 x 100 4 x 10 540
KHz,
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Multiplexing Example 1
Five channels of 100 KHz
Guard Band of10 KHz
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Multiplexing Example 2
The Advanced Mobile Phone System (AMPS) uses two
bands. The first band, 824 to 849 MHz, is used
for sending and 869 to 894 MHz is used for
receiving. Each user has a bandwidth of 30 KHz in
each direction. The 3-KHz voice is modulated
using FM, creating 30 KHz of modulated signal.
How many people can use their cellular phones
simultaneously?
Solution
Each band is 25 MHz. If we divide 25 MHz into 30
KHz, we get 833.33. In reality, the band is
divided into 832 channels.
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Multiplexing Example 3
Four 1-Kbps connections are multiplexed together.
A unit is 1 bit. Find (1) the duration of 1 bit
before multiplexing, (2) the transmission rate of
the link, (3) the duration of a time slot, and
(4) the duration of a frame?
Solution

• The duration of 1 bit is 1/1 Kbps, or 0.001 s (1
  ms).
• 2. The rate of the link is 4 Kbps.
• 3. The duration of each time slot 1/4 ms or 250
  ms.
• 4. The duration of a frame 1 ms.

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Multiplexing Example 3 some points

• Bit duration is inverted bit rate 1/1kbps 1ms
• Each frame carries 4 time slots so the duration
  of each output time slot is 1/4th of the input
  time slot 1/4 ms or 250 ms.
• Each frame carries 4 time slots so duration is 4
  X 1/4 ms or 1ms . The duration of the frame is
  same as the duration of the input unit.

Frame
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Multiplexing Example 4
Four channels are multiplexed using TDM. If each
channel sends 100 bytes/s and we multiplex 1 byte
per channel, show the frame traveling on the
link, the size of the frame, the duration of a
frame, the frame rate, and the bit rate for the
link.
Solution
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Multiplexing Example 5
A multiplexer combines four 100-Kbps channels
using a time slot of 2 bits. Show the output with
four arbitrary inputs. What is the frame rate?
What is the frame duration? What is the bit rate?
What is the bit duration?
Solution
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Multiplexing Example 6
We have four sources, each creating 250 8-bit
characters per second. If the interleaved unit is
a character and 1 synchronizing bit is added to
each frame, find (1) the data rate of each
source, (2) the duration of each character in
each source, (3) the frame rate, (4) the duration
of each frame, (5) the number of bits in each
frame, and (6) the data rate of the link.
Solution
See next slide.
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Solution (continued)
1. The data rate of each source is 2000 bps 2
Kbps. 2. The duration of a character is 1/250 s,
or 4 ms. 3. The link needs to send 250 frames per
second. 4. The duration of each frame is 1/250 s,
or 4 ms. 5. Each frame is 4 x 8 1 33
bits. 6. The data rate of the link is 250 x 33,
or 8250 bps.
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Multiplexing Example 7
Two channels, one with a bit rate of 100 Kbps and
another with a bit rate of 200 Kbps, are to be
multiplexed. How this can be achieved? What is
the frame rate? What is the frame duration? What
is the bit rate of the link?
Solution
We can allocate one slot to the first channel and
two slots to the second channel. Each frame
carries 3 bits. The frame rate is 100,000 frames
per second because it carries 1 bit from the
first channel. The frame duration is 1/100,000 s,
or 10 ms. The bit rate is 100,000 frames/s x 3
bits/frame, or 300 Kbps.