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Unit 2 Asignment 2

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Unit 2 Assignment 1 Blank Answer Sheet
Name: Date: Electronics ET2530
Chapter 2 (pp. 111-113)

2. A 1500 kHz carrier and a 2 kHz intelligence signal are combined in a non-linear device. List all the frequency components produced.
- 1498, 1500, and 1520KHz

3.If a carrier is amplitude modulated what causes the sideband frequencies?
- The non-linear mixing of the carrier and intelligence frequencies. 


4.What determines the bandwidth of emission for an AM transmission?
- It is twice the frequency of the highest audio frequency transmitted.
The upper sideband is fc +FM where fc is the carrier frequency and FM is the modulation (audio) frequency .The lower sideband is fc – FM. The total bandwidth is 2*FM

5.Explain the difference between a sideband and a side frequency.
- A sideband carries the modulation or "information". in old fashioned AM and FM transmission schemes, a carrier is broadcast in addition to the sideband(s). Only the sideband is needed to communicate.
A spurious frequency that happens when a carrier is transmitted has no added "information" and just adds to the transmission bandwidth. Sometimes these occur by intermixing within the transmitter. a common example is CB splatter from overdrive of RF amps. 9.Draw a diagram (or indicate which text book Figure # shows this) of a carrier wave envelope when modulated by 50% by a sinusoidal wave. Indicate on the diagram the dimension from which the percentage of modulation is determined.
- Figure 2-8 on page 77 textbooks

10.What are the possible results of over-modulation?
- If this occurs, the modulated carrier will go to more than double its unmodulated value but will fall to zero.

13.Determine Ec and Em from problem 12 (Ec = 65 Vpk, Em = 35 Vpk)
-

15. A 100 Volt carrier is modulated by a 1 KHz sine wave. Determine the side frequency amplitude when m = 0.75 (37.5V)
- SideFrequencyAmplitude = (ModulationIndex * CarrierAmplitude)/2 x = (0.75 * 100)/2 = 37.5 Volt

19.Why is a high percentage of modulation desirable?
- A high percentage of modulation is important so that the power is the sidebands.

20. During 100% modulation, what percentage of the average output power is in the sidebands? (33.33%)
- Pt = Pc (1+m2/2)

35. Draw a block diagram (or indicate which text book Figure # shows this) of an AM transmitter.
Chapter 3 (pp. 159-160)
- Figure 3-1 pg. 118

1. Draw a diagram (or indicate which text book Figure # shows this)of a tuned radio-frequency (TRF) radio receiver.

2. Explain the following: Sensitivity of a receiver, selectivity of a receiver. Why are these important characteristics? In what units are these usually expressed?
- Sensitivity: the minimum input RF signal to a receiver required to produce a specified audio signal at the output.
Selectivity: the extent to which a receiver can differentiate between the desire signal and

3.Explain why a receiver can be overly selective.
- A receiver that is overly selective can result in alack of fidelity because part of the intelligence is not included (filtered out)

6.Describe the advantages and disadvantages of a diode detector. * Advantages * Handle high power * Acceptable distortion levels * Efficient * Provide a usable dc voltage for the AGC circuits * Disadvantage * Reduce Q and selectivity * No amplification

10. Provide the advantages of a synchronous detector compared to a diode detector. Explain its principle of operation. * Synchronous detector advantages: * Low distortion – well under 1 percent * Fast modulation waveforms * High – fidelity applications * Ability to provide gain instead of attenuation * The principle of operation involves mixing in a nonlinear fashion just as in AM generation.

11. Draw a block diagram (or indicate which text book Figure # shows) this of a superheterodyne AM receiver. Assume an incident signal and explain briefly what happens at each stage.
- Figure 3-6 pg. 127 textbooks
-

15.Explain how variable tuned circuits in a superheterodyne receiver are adjusted with a single control.
-

Chapter 4 (pp. 199-200) 1. An AM transmission of 1000 W is fully modulated. Calculate the power transmitted as a SSB signal (167 W)
-
Use: m = 1 PSBS = 1/3 Pt

5. List and explain the advantages of SSB over conventional AM transmissions. Are there any disadvantages?
- The SSB system has a noise advantage over AM due to the bandwidth reduction (one half)). Taking into account the selective fading improvement, noise reduction, and power savings, SSB offers about 10- to 12-dB advantage over AM.

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