T324 activities block1

T324 Activities of Block 1 Page 1 of 10

T324 Activities of Block 1

Activity 4, B1, p26
If each UHF TV channel occupies the same bandwidth in the 470-854 MHz frequency range, this
frequency range is shared out equally into 48 channels, what is the bandwidth available to each
channel? Which frequencies are occupied by channel 21? Which frequencies are occupied by channel
68?

Answer:

ℎ= 2− 1

854 − 470 = 384

The bandwidth available for each channel

384
=8
48

The bandwidth available for channel 21

As this channel is the lowest frequency, so

The lowest frequency (470 MHz) up to (8 MHz) plus

470 + 8 = 478

So the frequency will be 470 – 478 MHz

The bandwidth available for channel 68

As this channel is the highest frequency, so

The lowest frequency (854 MHz) up to (8 MHz) minus

854 – 8 = 846

So the frequency will be 846-854 MHz

Activity 6, B1, p32
If an antenna in free space receives 16 µW of power at distance of 2 km from an isotropic transmitter,
how much will it receive at 4 km? How much at 8 km?

Answer:

At distance (4 km)

Disclaimer: this paper(s) is prepared by a AOU student of KSA, (Yaseen Alhashim). Moreover, I am not responsible of any mistake (if
any). So, kindly consider it as un-official paper & refer to the book for more information. You might notify me for any mistake by e-
mail: yalhashim@gmail.com , http://alhashem.blogSpot.com | May, 2010


1 2
=
2 1

16 4
=
2 2
16 4
=
2 1

4 2 = 16

16
2=
4

2=4µ

At distance (8 km)

1 2
=
2 1

16 8
=
2 2
16 16
=
2 1

16 2 = 16

16
2=
16

2=1µ

Activity 7
A 1.8 GHz radio wave propagates 16 km through the atmosphere, and then through two brick walls,
each of 100 mm thickness. Rain leads to an atmospheric loss of 1.5 dB/km, and brick attenuates at 40
dB/m at this frequency. Calculate the total power loss in dB over this path that is due to attenuation by
the atmosphere and be the wall. (Ignore the inverse square law).

Answer
Given information:
(1) At atmosphere; 16 km
(2) Atmosphere loss; 1.5 dB/km
(3) 2 brick wall; 100 mm (100 x 10-3 m)
(4) Brick attenuation; 40 dB/m

Requests:
(1) Power loss in dB by (a) atmosphere, (b) by the wall =?
(2) Total power loss (a+b) in dB = ?



NOTE: to find the total loss of a path in dB, you simply multiply the attenuation coefficient in dB/km by the
path length in km or path in dB/m in path length in m

(a) 16 1.5 = 24 Refers to 2 brick wall
(b) 2 100 10 40 =8
To convert (mm) to (m)
So, the total power loss = 24 + 8 = 32 dB

Activity 11, B1, p50
Estimate the physical length of a physical length of a λ/4 rod to be used for a 100 MHz FM radio station.
How long would it be for a 2.4 GHz Wi-Fi link? (Hint: in each case, assue a free space wavelength).

Answer:

In a free space wavelength c = 3 x 108

3 10
λ= = =3
100 10

λ 3
. : . = = 0.75
4 4

For 2.4 GHz Wi-Fi

3 10
λ= = = 0.125 = 0.125 10 = 125
2.4 10

λ 125
.:. = = 31.25
4 4

Activity 3, B1, p69
What is the rate of transmission if the signaling rate is 4500 baud, and each symbol represents 4 bits?

Answer

4 x 4500 = 18000 bit/s

or 18 kbit/s



Activity 4, B1, p69
A carrier is modulated at signaling rate of 6000 baud. It is used with a modulation system known as 16-
QAM, in which there are 16 symbols.

a) How many bits does each symbol represent?
b) What is the data rate of this system in bits per second (or kilobits per second)?
Answer:

Given information:
(1) 6000 baud (bit/s)
(2) 16 symbols

Requests:
(1) bit rate = [ log10 number of symbols ] / 0.301
(2) data rate = symbols per second x bit per symbol (bit
rate)
a)
16
=
0.301
=4

b)

So,

= 6000 4
= 24000 / 24 /


Suppose that a version of MSK operates at 10 kbaud and one symbol uses a segment of a 10 MHz
sinusoid. If the other symbol uses a segment of a higher frequency, what frequency should that higher
frequency be?

Answer:
The frequency spacing is half the signaling rate, so is 5 kHz (half of 10 kbaud), which is 0.005 MHz. Hence the
other symbol should be a segment of a sinusoid with frequency 10 MHz + 0.005 MHz = 10.005 MHz.

The frequency spacing is half the signaling rate
10 kbaud = 10 kHz
So, the half is 5 kHz

As we have the segments in MHz, so we will convert the 5 kHz to MHz
5000 Hz x 10-6 = 0.005 MHz

.:. 10 MHz + 0.005 MHz = 10.005 MHz



A transmission at 2 GHz has a bandwidth of 4 MHz. Which of the receiver responses is the best suited
for a receiver of this transmission mentioned figures (a, b and c)?

Figure a: f1=1.999, f2=2.001 GHz,
Figure b: f1=1.998, f2=2.002 GHz,
Figure c: f1=1.6, f2=2.4 GHz

Answer:
We are requested to find the bandwidth of each figure. The bandwidth which is close to given bandwidth in
the question (4 MHz) is the best suited for receiver.
The

a) 2 − 1 = 2.001 − 1.999 = 0.002 = 0.002 10 = 2
b) 2 − 1 = 2.002 − 1.998 = 0.002 = 0.004 10 = 4
c) 2 − 1 = 2.4 − 1.6 = 0.8 = 0.8 10 = 800

So, the beast one suited to transmission with bandwidth of 4 MHz is (b)

(a) In a 64-QAM system, how many bits are there per symbol?
(b) If the symbol rate is 10000 baud, what is the data rate?

Answer:
(a)

Number of symbol = 2
2 = 64
.:. there is 6 bits per symbol

You might find it by [ log10 64 ] / 0.301 = 6

(b) data rate = symbol per second * bit per symbol
= 10000 x 6
= 60000 bit/s

What is the spectral efficiency if (f2 = 701.02 MHz, f1 = 701 MHz) and data rate is 800 kbit/s?

Answer:
Spectral efficiency =
= 2− 1 = 701.02 − 701 = 0.02 = 0.02 10 = 20

/
= = 40 / / = 40 bit/s/Hz



Calculate the signal-to-noise ratio for devices A, B and C, and hence stat which of A, B and C has the best
signal-to-noise ratio.

• Bandwidth of A device = 10 MHz,
• Bandwidth of B device = 20 MHz,
• Bandwidth of C device = 30 MHz,
• Signal = 150 pW/Hs,
• Noise = 50 pW/Hz

Answer:

Single-to-noise ratio =

Signal power = ℎ

Noise power = ℎ

For (A) device:

Signal power = 150 x 10 = 1500 W = 1500 x 10-6 = 1.5 x 10-3 µW = 1.5 mW

Noise power = 50 x 10 = 500 W = 500 x 10-6 = 5 x 10-4 µW

.
Single-to-noise ratio = .
= 30 = 30 10 = 3

For (B) device:

Signal power = 150 x 20 = 3000 W = 3000 x 10-6 = 3 x 10-3 µW = 3 mW

Noise power = 50 x 20 = 1000 W = 1000 x 10-6 = 1 x 10-3 µW = 1 mW

Single-to-noise ratio = =3

For (C) device:

For device C, the passband is 30 MHz wide, but the signal spectrum occupies only 20 MHz of this. Hence the
signal power is exactly as for device B, 3 mW

Signal power = 150 x 20 = 3000 W = 3000 x 10-6 = 3 x 10-3 µW = 3 mW

Noise power = 50 x 30 = 1500 W = 1500 x 10-6 = 1.5 x 10-3 µW = 1.5 mW

Single-to-noise ratio = =2
.

Devices A and B equally have the best signal-to-noise ration of 3:1



Activity 24
If S/N has a value 10, and W is 5 MHz, what is the theoretical maximum data rate?

= 2 1+

=5 2 1 + 10

=5 2 11

11
=5
2

= 5 3.45

= 17.29 / = 17.29 10 /

If 16-QAM is used, what is the signaling rate for each subcarrier? Use the approximate data rate per
subcarrier of 110 bit/s.

Answer:

Each signaling represents 4 bits (24 = 16).

The signaling rate = 110/4 = 27.5 baud (bit/s)

A signal has a flat power spectrum extending from 3 GHz to 8 GHz. The power density is -20 dBm/MHz.
what is the total power in the signal?

Answer

Note: 0 dBm is equal to 1 mW, and each additional 10 dBm represents an
increase of 10 times the power. Subtracting is similar. So, 1 W is 30 dBm,
and 1 µW is -30 dBm.
• dBm = log10 (mW)*10
• mW =10^(dBm/10)

-20 dBm = 10(-20/10) = 10(-2) = 0.01 mW (milliwatt)

The bandwidth = 8-3 = 5 GHz = 5000 MHz

The total power = 0.01 x 5000 = 50 mW



Given that the slot time is 20 µs and range of multipliers is 0 to 31, calculate the range of values for
backoff time.

Answer

Minimum time is 0, and maximum time is give by the product of the contention window and the slot time

= . = . = .

Calculate the transmission time for a frame consisting of 2364 byte over a wireless link operating at a
data rate of 2 Mbit/s. Repeat the calculation for a frame consisting of 14 bytes.

Answer:

=

=

Note that it is important to compatible units, so we will convert the byte to bit.

2364 8
=
2 10

= 9.384 10 = 9.384

For 14 byte

14 8
=
2 10

= 5.6 10 = 5.6 10 10 = 56 µ

Calculate the propagation time for signals over a wireless link of 20 m. you should assume that the
speed of propagation is 3 x 108 m/s.

=

=



20
=
3 10

= 6.667 10 = 6.667 10 10 = 66.67

Given a data rate of 2 Mbit/s, calculate how many bits are transmitted within 66.67 ns. Repeat the
calculation for a data rate of 34 Mbit/s.

Answer:

=

= 2 10 66.67 10 = 0.1333

For data rate 34 Mbit/s

= 34 10 66.67 10 = 2.267

For data rate 54 Mbit/s

= 54 10 66.67 10 = 3.6

Calculate the minimum amount of time that is wasted if a 2364 byte frame is involved in a collision. You
should assume that the data rate is 2 Mbit/s and the period of time a station waits for an
acknowledgment is 100 µs. repeat the calculation for a 14 byte.

Answer:

Note that it is important to compatible units, so we will convert the byte to bit.

=

2364 8
=
2 10

= 9.384 10 = 9.384

Transmission time for 2364 byte frame at 2 Mbit/s is 9.384 ms

The minimum wasted time is

9384 + 100 = 9484 µ



Let P=0111 0100 0110 1000 and K=0011 0001 0011 0100. Calculate C= P + K and confirm P = C + K

Answer:

C=P+K

0 1 1 1 0 1 0 0 0 1 1 0 1 0 0 0
0 0 1 1 0 0 0 1 0 0 1 1 0 1 0 0
0 1 0 0 0 1 0 1 0 1 0 1 1 1 0 0

P=C+K

0 1 0 0 0 1 0 1 0 1 0 1 1 1 0 0
0 0 1 1 0 0 0 1 0 0 1 1 0 1 0 0
0 1 1 1 0 1 0 0 0 1 1 0 1 0 0 0

The initialization vector is 24 bits long. Assuming that the initialization vector is incremented by 1 for
each frame transmitted by an access point operating at 11 Mbit/s and frames are, on average, 2000
bytes long, estimate how long it will take for an initialization vector value to be repeated.

Answer:

With 24 bits, the number of different initialization vector value is

2 = 16777216 ℎ ℎ ℎ

The maximum rate of transmitting frame is

11 10
= = 687.5 /
2000 8

The time taken to repeat a value

16777216 24403
= 24403 = = 6.77 ℎ
687.5 60 60


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