Communication Systems

Question 1

Why do we need communication systems?

Answer 1

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

Question 2

Draw the box diagram for a comms system

Answer 2

Question 3

What is modulation?

Answer 3

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

Question 4

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

Answer 4

Amplitude

Frequency

Phase

Question 5

Why do we use modulation?

Answer 5

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

Question 6

What is noise?

Answer 6

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

Question 7

What is the equation for thermal noise power?

Answer 7

Question 8

What is NR and NF for an amplifier?

What are their equations?

What values would an ideal amp have?

Answer 8

NR- Noise ratio

NF- Noise figure

NR = 1, NF = 0

Question 9

What is the Quality factor of components?

What about for a bandpass filter?

Answer 9

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.

Question 10

What is bandwidth?

Answer 10

Bandwidth is the range of frequencies occupied by a signal.

Question 11

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

Answer 11

x_c(t) = A(t)cos(2πf_ct + ø(t))

We typically set the phase ø(t) to 0

Question 12

What is DSB-SC?

Answer 12

Double-Sideband Modulation with suppressed carrier

Question 13

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

Answer 13

Question 14

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

Answer 14

Question 15

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

Also comment on the bandwidth

Answer 15

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

The bandwidth of the modulated signal is 2W

Question 16

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

Answer 16

f_c > W

where W is BW of message signal

Question 17

Describe the process of demodulation for DSB-SC.

Answer 17

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

We multiply the modulated signal by a scaled copy of the carrier (2cos(2πf_ct)) 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.

Question 18

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

Answer 18

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.

Question 19

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

Answer 19

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.

Question 20

What is the equation for DSB-AM?

x_DSB-AM( t ) = ?

Answer 20

x_DSB-AM( t ) = (A_c + m( t ))cos( 2πf_ct )

Question 21

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

Answer 21

Envelope Detection

Question 22

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

Answer 22

(A_c + m( t )) > 0

Question 23

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

Example:

f_c = 1.4MHz, m(t) = 20Hz→10kHz

Answer 23

Upper sideband:

f_c + a → f_c + b

Lower sideband:

-f_c - b → -f_c - a

Example:

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

Question 24

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

Answer 24

A_c ≥ m_p

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

Question 25

What is the AM modulation index **µ**? What are the constraints on µ for envelope detection?

Answer 25

µ = m_p / A_c 0 ≤ µ ≤ 1 for envelope detction

Question 26

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

Answer 26

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.

Question 27

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

Answer 27

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

Question 28

What is the formula for Power efficiency η ?

Answer 28

_Useful Power_ Total Power η = µ² / (2 + µ²) x 100% η_max = 33% where µ = 1

Question 29

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

Answer 29

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.

Question 30

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

Answer 30

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

Question 31

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

Answer 31

Increase the time constant RC

Question 32

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

Answer 32

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

Question 33

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

Answer 33

RC \> 1 / ω_c and RC \< 1 / 2πW (where W is the highest frequency of m( t ) )

Question 34

What is SSB-AM?

Answer 34

**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.

Question 35

What are the two types of **angle modulation**?

Answer 35

**FM** - Frequency modulation **PM** - Phase modulation

Question 36

What do we vary in PM? What is the equation for this parameter? Therefore what is the equation for a PM signal?

Answer 36

The phase θ(t) θ(t) = ω_ct + θ₀ + k_pm( t ) (k_p is constant) x_PM( t ) = A_ccos(ω_ct + k_pm( t ))

Question 37

What do we vary in FM? What is the equation for this parameter? Therefore what is the equation for a FM signal?

Answer 37

The instantaneous frequency ω_i ω_i(t) = ω_c + k_fm( t ) (k_f is constant) x_FM( t ) = A_ccos(ω_ct + k_f ∫m(α)dα)

Question 38

What can be said of the power of **angle modulated** wave, regardless of the value of k_p or k_f?

Answer 38

A_c² / 2 As the amplitude remains constant in angle modulation

Question 39

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

Answer 39

2B | (Narrow-band FM)

Question 40

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

Answer 40

2B( β+1 ) | (Wide-band FM)

Question 41

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

Answer 41

B_FM = 2(∆f + B) using β = ∆f / B B_FM = 2B( β+1 )

Question 42

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

Answer 42

_Advantages:_ High-noise immunity Constant envelope good for non-linear amplifiers _Disadvantages:_ Higher bandwidth occupancy Higher implementation complexity

Question 43

When might we use FM?

Answer 43

High-fidelity music/audio broadcasting

Question 44

What happens with increasing values of the deviation ratio β?

Answer 44

By increasing β, m_f increases. This improves the signal-to-noise ratio But needs more bandwidth.