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Signals and Communications 2 > Communication Systems > Flashcards

Communication Systems Flashcards

1
Q

Why do we need communication systems?

A

To transmit information from a source to a destination.

This is important for humanitarian reasons as it alows global communication and makes human rights violations very difficult.

Also commercially very important; the internet and mobile phones both have comms at their heart

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2
Q

Draw the box diagram for a comms system

A
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3
Q

What is modulation?

A

Impressing the information signal onto a carrier wave so that the information signal is then transmitted at teh carrier frequency (usually of higher frequency)

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4
Q

What are the three characteristics of the carrier wave that we alter with the information signal?

A

Amplitude

Frequency

Phase

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5
Q

Why do we use modulation?

A

To be able to use more of the available bandwidth

To allow multiple transmitters without interference

To shift to higher frequencies to enable efficient power radiation with antennas of reasonable size

To take advantage of good propagation characteristics at the carrier frequency

To improve the signal-to-noise ratio

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6
Q

What is noise?

A

An unwanted signal perturbation that is always present in any communication system

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7
Q

What is the equation for thermal noise power?

A
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8
Q

What is NR and NF for an amplifier?

What are their equations?

What values would an ideal amp have?

A

NR- Noise ratio

NF- Noise figure

NR = 1, NF = 0

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9
Q

What is the Quality factor of components?

What about for a bandpass filter?

A

Q of a component is the ratio of the energy stored to the energy lost.

Q of a tuned circuit or bandpass filter is the ratio of the filter centre frequency to its bandwidth.

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10
Q

What is bandwidth?

A

Bandwidth is the range of frequencies occupied by a signal.

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11
Q

For a sinusoidal carrier, how do we typically represent the modulated carrier xc(t) ?

A

xc(t) = A(t)cos(2πfct + ø(t))

We typically set the phase ø(t) to 0

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12
Q

What is DSB-SC?

A

Double-Sideband Modulation with suppressed carrier

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13
Q

What is the output of a DSB-SC modulator with message signal m(t)

A
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14
Q

Give the equation for the frequency spectrum of a DSB-SC

A
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15
Q

Describe what happens to the spectrum of the message signal after DSB-SC.

Also comment on the bandwidth

A

M(f) is shifted to the left and right by fc

The bandwidth of the modulated signal is 2W

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16
Q

What condition on fc must be met to avoid distortion with DSB-SC?

A

fc > W

where W is BW of message signal

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17
Q

Describe the process of demodulation for DSB-SC.

A

This is the process of recovering the message signal m(t) from the modulated signal xDSB-SC(t).

We multiply the modulated signal by a scaled copy of the carrier (2cos(2πfct)) and then low-pass filter.

The desired component that is removed with the filter is then therefore a scaled version of the original message signal.

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18
Q

What are the problems with demodulating a DSB-SC signal?

A

We need to generate a copy of the carrier at the receiver that is in frequency and phase coherence with the transmitter carrier.

This complicates the receiver design, so increases costs and size of receivers.

19
Q

What is the main difference between DSB-SC and Amplitude Modulation (DSB-AM)?

A

DSB-AM is an approach where we also trasmit the carrier along with the modulated signal.

Thus, unlike DSB-SC, we don’t have to generate a carrier at the receiver too.

Thus the transmitter is much more complex (expensive/large) but allows for simple (cheap/small) receivers.

20
Q

What is the equation for DSB-AM?

xDSB-AM( t ) = ?

A

xDSB-AM( t ) = (Ac + m( t ))cos( 2πfct )

21
Q

What technique do we use to recover the message signal in AM?

A

Envelope Detection

22
Q

What condition must be satisifed to ensure that envelope detection of an AM signal contains all of the information of m( t ).

A

(Ac + m( t )) > 0

23
Q

Given the frequency of the carrier signal (fc) and the message signal range (a →b) how do we calculate the range of frequencies generated by the upper/lower sidebands?

Example:

fc = 1.4MHz, m(t) = 20Hz→10kHz

A

Upper sideband:

fc + a → fc + b

Lower sideband:

-fc - b → -fc - a

Example:

usb: 1,400,020Hz → 1,410,000Hz
lsb: 1,390,000Hz → 1,399,980Hz

24
Q

If we define the peak amplitude of m(t) as mp, what is the condtion for viable envelope detection?

A

Ac ≥ mp

The mimimum carrier amplitude required for the viability of envelope detction is mp.

25
Q

What is the AM modulation index µ?

What are the constraints on µ for envelope detection?

A

µ = mp / Ac

0 ≤ µ ≤ 1 for envelope detction

26
Q

If the condition for envelope detection is not met, is the signal useless?

A

No, we can still demodulate using coherent detction but this defeats the main advantage of AM over DSB-SC and is considerably more complex and expensive.

27
Q

What is useful power and wasted power in DSB-AM?

A

The carrier power is a waste as it does not carry any meaningful information.

The sidebands power is useful power as it contains the message signal

28
Q

What is the formula for Power efficiency η ?

A

Useful Power

Total Power

η = µ2 / (2 + µ2) x 100%

ηmax = 33% where µ = 1

29
Q

Describe the method of envelope detection using an RC circuit with diode

A

The diode clips the signal so only the positive half remains.

During each rising positive cycle the capacitor charges up to the peak voltage of the input.

The voltage then decays until the next positive cycle.

30
Q

What is the problem with non-coherent detection methods such as envelope detection?

A

The capacitor discharge between positive cycles creates an undesirable ripple signal of frequency ωc in the output.

31
Q

How do we reduce the ripple signal in non-coherent detection?

A

Increase the time constant RC

32
Q

What happens if, to reduce ripple, the time contant RC is raised too high in envelope detection?

A

Diagonal clipping

It is impossible for the capacitor output to discharge fast enough to follow the signal envelope

33
Q

What conditions on RC have to be met for good envelope detection

A

RC > 1 / ωc

and

RC < 1 / 2πW

(where W is the highest frequency of m( t ) )

34
Q

What is SSB-AM?

A

Single side-band Amplitude Modulation

Only one sideband is transmitted so uses half the bandwidth.

Usually the carrier is not transmitted so can be called SSB-SC and demodulated coherently.

35
Q

What are the two types of angle modulation?

A

FM - Frequency modulation

PM - Phase modulation

36
Q

What do we vary in PM?

What is the equation for this parameter?

Therefore what is the equation for a PM signal?

A

The phase θ(t)

θ(t) = ωct + θ0 + kpm( t ) (kp is constant)

xPM( t ) = Accos(ωct + kpm( t ))

37
Q

What do we vary in FM?

What is the equation for this parameter?

Therefore what is the equation for a FM signal?

A

The instantaneous frequency ωi

ωi(t) = ωc + kfm( t ) (kf is constant)

xFM( t ) = Accos(ωct + kf ∫m(α)dα)

38
Q

What can be said of the power of angle modulated wave, regardless of the value of kp or kf?

A

Ac2 / 2

As the amplitude remains constant in angle modulation

39
Q

If a message signal has bandwidth B then what is the bandwidth of the NBFM of the signal?

A

2B

(Narrow-band FM)

40
Q

If a message signal has bandwidth B then what is the bandwidth of the WBFM of the signal?

You can use the deviation ratio β = ∆f / B

A

2B( β+1 )

(Wide-band FM)

41
Q

What is Carson’s rule for effective bandwidth in FM?

A

BFM = 2(∆f + B)

using β = ∆f / B

BFM = 2B( β+1 )

42
Q

What are the advantages and dsadvantages of using FM vs. using AM?

A

Advantages:

High-noise immunity

Constant envelope good for non-linear amplifiers

Disadvantages:

Higher bandwidth occupancy

Higher implementation complexity

43
Q

When might we use FM?

A

High-fidelity music/audio broadcasting

44
Q

What happens with increasing values of the deviation ratio β?

A

By increasing β, mf increases.

This improves the signal-to-noise ratio

But needs more bandwidth.

Signals and Communications 2 (3 decks)

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