2P6 Information Engineering

Question 1

How do you find the energy of a signal?

Answer 1

Parsaval’s theorem: integral(|x(t)|^2) over all real numbers

Question 2

How do you find the power of a signal?

Answer 2

P = lim T->inf (1/T integral(|x(t)|^2)) from -T/2 to +T/2

Question 3

Define bandwidth

Answer 3

The range of frequencies over which its spectrum is non-zero. |X(f)| where X(f) is the fourier transform of x(t)

Question 4

What is a passband signal?

Answer 4

A signal where its spectral content is centered around +- fc,

Usually fc+W to fc-W. so the bandwidth is 2W.

Question 5

What is rect(t/T)?

Answer 5

Rectangular pulse, which =1 for -T/2 <t<T/2

Question 6

What are some alternative definitions of bandwith which are more realistic?

Answer 6

90% bandwidth, the range of frequencies which contain 90% of the energy of the spectrum

3- dB bandwith, range of frequencies that contain 50% o the energy of teh spectrum.

null->null bandwidth, the width of the main lobe

Question 7

How can you model a channel?

Answer 7

y(t) = h(t)*x(t) + n(t)

Question 8

How is noise in a channel modelled?

Answer 8

As a Gaussian random process, at each time t, n(t) is a Gaussian random variable.

Question 9

For normal AM modulation what is the moduluation index limited to?

Answer 9

mA = max(x(t))/a0

mA <1 because otherwise there will be phase reversal.

Question 10

What is a reciever for a normal AM receiver?

Answer 10

diode into a capacitor and resistor in parallel.

Positive half cycle, capacitor chargers up rapidly, when input falls below, diode is reverse biased so slowly dischargers through Rl

Question 11

What is the spectrum of an ordinary AM sigal?

Answer 11

delta function at carrier frequency, and the information bandwidth shifted to symetrically around the carrier frequency.

Question 12

What is the power of an ordinary AM signal withoriginal power Px, and offset a0?

Answer 12

a0^2/2 + Px/2

Question 13

What is DSB-SC?

Answer 13

Double Side Band Suppressed Carrier,

get rid of a0, to get rid of delta function at fc,

s(t) = x(t) cos(2pi fc t)

Question 14

How does a DSB-SC reciever work?

Answer 14

modulator + lowpass filter

s(t) * cos(2pifct)

Question 15

What is SSB-SC?

Answer 15

Only need to specify X(f) for f>0

Question 16

What is the power of of SSB-SC and DSB-SC?

Answer 16

DSB-SC -> Px/2

SSB-SC -> Px/4

Question 17

What is the expression for an FM modulated signal?

Answer 17

s(t) = Ac cos [ 2pi fc t + 2pi kf integral(x(u)du)]

because x determines instantaenous frequency, so needs to be integrated.

Question 18

How do you obtain the original signal from an FM signal?

Answer 18

differentiate giving passband signal with AM, then can use envelope detection

differentiator + envelope detector.

Question 19

Why is FM more robust than AM?

Answer 19

message is hidden infrqeucny rather than amplitude, which is less affected by noise. But this leads to increased bandwidth.

Question 20

What is the moduluation index of an FM signal?

Answer 20

Δf/fx, where Δf is the max deviation of teh carrier frequency f(t) from fx.

Question 21

What is the spectrum of an FM wave?

Answer 21

Bessel function multipled by delta functions either side of fc and -fc.

Question 22

What is the effective bandwidth of an FM signal?

Answer 22

Bfm = 2Δf + 2fx

Question 23

What’s the difference between sampling and Quantisation?

Answer 23

Sampling discretises the time axis.

Quantisation discretises the signal amplitude axis

Question 24

What is the fourier transform of the sampled signal?

Answer 24

Xs(f) = 1/T * Sum (X(f-nfs))

Question 25

What is uniform quantisation?

Answer 25

Finite set of levels, even step between each level, assign bits to each level.

Question 26

Loss rate of sampling and quantisation?

Answer 26

Sampling is a lossless procedure as long as the sampling rate is higher than the Nyquist rate. Quatisation is always lossy.

Question 27

Define quantisation noise?

Answer 27

Q(x(nT)) is the quantisation of x(nT) then eQ = x(nT) - Q(x(nT)) if step size is Δ then eq(z) ε [-Δ/2 , +Δ/2]

Question 28

What is noise power for a uniformlly distributed noise with a step size of Δ?

Answer 28

Δ^2/12 RMS noise = Δ/root(12)

Question 29

What is signal to noise ratio?

Answer 29

signal power/noise power = (RMS)^2/(RMS)^2 = (V^2/2) / (Δ^2/12) = 3 x 2^(2n-1)

Question 30

What is the rate of a digitised uniform quantiser?

Answer 30

n2W (bits/sec) bandwidth W, n-bit uniform quantiser

Question 31

How can we reduce the bit-rate of signals by using quantisation?

Answer 31

non-uniform step size, more steps at smaller values, less steps at larger (rarer) values

Question 32

What is a constellation?

Answer 32

The set of value the bits are mapped to

Question 33

In a constellation with M symbol, each symbol represents how many bits?

Answer 33

log2(M)

Question 34

What is the orthonomal shifts property of digital pulses?

Answer 34

integral(p(t-kT)p(t-mT)dt) is only equal to 1 when k=m

Question 35

What shapes satisfy the orthonormal shift property?

Answer 35

sinc and rect

Question 36

What is a common pulse shape?

Answer 36

root raised cosine spectrum, quick decay but larger bandwidth

Question 37

What is a matched filter?

Answer 37

y(t) is passed through filter with impulse response p(-t) the filter output is then xb(t) * h(t) = sum Xk integral(p(τ-kT)p(τ-t)dτ) which is sampled only at t=mT so you only get Xm out

Question 38

What happens when there is noise at the reciever of a matched filter?

Answer 38

you get r(mT) = Xm + Nm where Nm = integral(n(τ)p(τ-mT)dτ) decoder usually picks te most likely one.

Question 39

How does does probability of detection error depend on snr?

Answer 39

Pe = e^(-snr/2)

Question 40

What is the difference between digital baseband and digital passband communications?

Answer 40

Passband -> the baseband signal is moduluated by a high frequency carrier. x(t) = sum(Xk p(t-kT)) cost(2pifc t)

Question 41

What is quadrature amplitude modulation?

Answer 41

make the constellations complex valued x(t) = Re[xb(t)e ^(j2pifct)] the cosine component carriers the real part of the signal and the sine component carriers the imaginary part of the signal

Question 42

How is QAM demodulated?

Answer 42

By using two seperate de modulators. One multiples by a cos and the other by a -sin, they then end up seperately with the real and imaginary parts of the original signal. and the constellation can be extracted.

Question 43

Where are the decision regions for QAM?

Answer 43

Bisextors inbetween the points.

Question 44

What happens when you increase the number of constellation poins with QAM?

Answer 44

the transmission rate increases, however, the power increases

Question 45

What is FSK?

Answer 45

Frequency shift keying, shifting the frequecny for each symbol of constellation.

Question 46

What are repetition codes?

Answer 46

Repeating each information bit odd number of times, and then picking the majority one. Slow rate 1/n, not hugely improved probability of correct.

Question 47

What is a block coding?

Answer 47

Passing a block of bits larger than actual number of bits, rate>0.5, but keep redundancy up. first bits are the original, the other are dependent

Question 48

What is Hamming codes?

Answer 48

N=7 K=4, first 4 bits are information, last three are redundant bits. c5 = s1+s2+s3 c6 = s2 + s3 + s4 c7 = s3 + s4 + s1 where + is moduluo 2 addition. rate is 4/7. Can only correct one flipped bit.

Question 49

What is TDMA,FDMA,CDMA?

Answer 49

TDMA - Time division multiple access FDMA - frequency division multiple access CDMA - code division multiple access

Question 50

How to demodulate FDMA?

Answer 50

modulate with different carrier frequencies depending on which signal you want to extract.

Question 51

How do you demodulate CDMA?

Answer 51

Integrate with the code that you want to extract, due to orthogonality of the codes, you will only end up with the one you are looking for.

Question 52

How does BPSK?

Answer 52

Binary phase shift keying, to alternate 1s and 0s, phase is flipped 180o, carrier is usually a sine wave.