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TUTORIAL 1 ANSWER

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EKT 231 COMMUNICATION SYSTEMS
TUTORIAL 1
CHAPTER 1
1. Define electronic communication system.
2. State the basic requirements of communication system.
3. Draw the simplified block diagram of an electronic communications system and
briefly explain the elements included in the system.
4. Define the term modulation.
5. Define the following terms: frequency, cycle, wavelength and period.
6. Differentiate between noise, interference and distortion.
7. What is the dB gain (or loss) for the following situation?
a. For a certain system, the output power is 100,000 times the input power.
b. When the power out of a certain system is 0.01 times the power in.
8. Convert a power gain of 64 dB into a ratio.
9. What is the overall gain of the system in Figure 1 below?
A1
A2
15 dB gain
10 dB gain
Figure 1
10. Convert a power level of 5 W to dBm.
11. Convert a power level of 500μV to dBμV.
12. For the system shown in Figure 2, if the input power is -12 dBm, what is the
power out?
Pin = -12 dBm
Pout = ?
+27 dB gain
Figure 2
13. For the system shown in Figure 3, the input voltage is +20dBμV. What is the
output voltage?
Vin=
+20dBμV
Vout=?
+80dB
voltage gain
Figure 3
14. Convert a power level of 23 dBm to an absolute power.
Ooo00000ooO
Answers:
1. Transmission, reception and processing of information between two or more
locations using electronic circuits.
2. The requirements are:
a. Rate of information transfer
b. Purity of signal received
c. Simplicity of the system
d. Reliability
3. Elements of communication system
 Information
 The communication system exists to convey a message.
 Message comes from information source
 Information forms - audio, video, text or data
 Transmitter:
 Processes input signal to produce a transmitted signal that suited the
characteristic of transmission channel.
 E.g. modulation, coding, mixing, translate
 Other functions performed - Amplification, filtering, antenna
 Message converted to into electrical signals by transducers
 E.g. speech waves are converted to voltage variation by a microphone
 Channel (transmission media):
 a medium that bridges the distance from source to destination.
Eg:Atmosphere (free space), coaxial cable, fiber optics, waveguide
 signals undergoes degradation from noise , interference and distortion
 Receiver:
 to recover the message signal contained in the received signal from the
output of the channel, and convert it to a form suitable for the output
transducer.
 E.g. mixing, demodulation, decoding
 Other functions performed: Amplification, filtering.
 Transducer converts the electrical signal at its input into a form desired by
the system used
4. Modulation is the process of changing one or more properties of the analog carrier
in proportion to the information signal.
5. Cycle - One complete occurrence of a repeating wave (periodic signal) such as
one positive and one negative alternation of a sine wave.
Frequency - the number of cycles of a signal that occur in one second.
Period - the time distance between two similar points on a periodic wave.
Wavelength - the distance traveled by an electromagnetic (radio) wave during one
period.
6. Noise
 Unwanted signals that coincide with the desired signals.
 Two type of noise: internal and external noise.
 Internal noise - caused by internal devices/components in the circuits.
 External noise - noise that is generated outside the circuit.
 E.g. atmospheric noise, solar noise, cosmic noise, man made noise.
Interference
 Contamination by extraneous signals from human sources.
 E.g. from other transmitters, power lines and machineries.
 Occurs most often in radio systems whose receiving antennas usually
intercept several signals at the same time
 One type of noise.
Distortion
 Signals or waves perturbation caused by imperfect response of the system
to the desired signal itself.
 May be corrected or reduced with the help of equalizers.
7. In this case,
a. Pout = 100,000 Pin
Therefore,
Gain (or loss) = 10log10(100,000) = 10log10(105) = 50 dB
Note that the log of 100,000 = 5. That is the same power one needs to
raise 10 to in order to get 100,000.
b. Pout = 0.01Pin
Therefore,
Gain (or loss) = 10log10(0.01) = 10log10(10-2) = -20 dB
Note that the answer is negative, indicating a system loss or an
attenuation of the input signal.
8. 64 dB = 10(64/10) = 10(6.4) = 2.5212 x 106
9. In this case, simply add the dB gains of each stage together:
Total dB Gain = gain of stage 1 + gain of stage 2 = 15 dB + 10 dB = 25 dB
10. Power in referenced dB = 10log10(5 W / 1mW) = 10log10(5000) = 36.99 dBm
11. Voltage in referenced dB = 20log10(500 μV / 1 μV) = 20log10(500) = 53.98 dBμV
12. Pin = -12 dBm, Power Gain = +27 dB
Simply add the values together:
Pout = -12 dBm + 27 dB = +15 dBm
13. Vin = +20 dBμV, Amplifier Voltage Gain = +80 dB
By simply adding the dB values together, we get
Vout = +20 dBμV + 80 dB = 100 dBμV
14. Solution:
23 dBm = 10log10(P / 0.001 W)
2.3 = log10(P / 0.001 W)
Take the antilog:
102.3 = (P / 0.001 W)
200 = (P / 0.001 W)
P = 0.2 W or 200 mW
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