Broadband Communication Systems

521316S

2

Exercise 1

• Determine parameters for OFDM system operating under the following conditions:

• bit rate Rb 20 Mbps

• tolerable rms delay spread 200 ns

• System bandwidth B 15 MHz

• Loss due to cyclic prefix max. 1 dB

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Example Design for OFDM (1)

• Let’s choose Tcp=800 ns*), to allow for timing errors<~0.6 s

• In order to have loss due to cyclic prefix less than 1 dB:

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• Let’s choose T = 4.8 s => SNRloss=0.8 dB

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Example Design for OFDM (2)

• Then subcarrier spacing becomes Df=1/TS= 250 kHz

• Number of subcarriers becomes:

– N=B/Df = 15 MHz/250 kHz = 60

• IFFT/FFT of size 64 should be chosen

• To achieve 20 Mbps, each transmitted OFDM symbol (including cp) must carry 96 bits of information (96/4.8s = 20Mbps)

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Example Design for OFDM (3)

• 16-QAM with rate 1/2 coding gives

4*1/2*N/4.8s= 20 Mbps => N=48 subcarriers for data

• QPSK with rate 3/4 coding gives

2*3/4*N/4.8s= 20 Mbps => N=64 subcarriers for data

• However, 64 subcarriers would mean bandwidth of 64*250kHz=16MHz

• The first option is selected to maintain under 15MHz bandwidth (48*250kHz=12MHz).

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Example Design for OFDM (4)

• The receiver operates by using samples. Hence an integer number of samples must be collected both from FFT interval and OFDM symbol interval:

64-FFT => sampling freq. 64*250kHz=16MHz, BUT

16MHz*0.8s=12.8 samples for cp

• The parameters need to be readjusted to meet this requirement.

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Example Design for OFDM (5)

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Example Design for OFDM (6)

• The bandwidth constraint needs to be rechecked since the carrier spacing is slightly modified: Df=253.90625kHz => (16-QAM needs 48 carriers

for data) B=12.1875MHz.

• Data rate requirement is achieved since OFDM symbol interval is now a bit shorter.

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Example Design for OFDM (7)

• How the system parameters would be changed if the channel delay spread is 1s and the number of subcarriers is kept the same?

• Let’s assume that 0.6 s is enough to cover timing errors etc. => Tcp=1.6s

• Let’s choose T = 8.4 s => SNRloss= 0.9dB

• Then subcarrier spacing becomes Df=1/TS = 147 kHz

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Example Design for OFDM (8)

• 48 subcarriers results in 48* 147 kHz = 7 MHz system bandwidth.

• 16-QAM with rate 1/2 coding gives

4*1/2*48/8.4s= 11.4 Mbps

• QPSK with rate 5/6 coding gives 2*5/6*48/8.4s= 9.5 Mbps

• The larger delay spread causes lowering the data rate and

lowering the system bandwidth if the number of subcarriers is kept constant.

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Example Design for OFDM (9) • How the data rate of 20Mbps can be maintained with system

bandwidth of 15 Mbps for delayspread of max. 1 s?

• Tcp=1.6s and TS = 6.8 s

• The subcarrier spacing is Df=1/TS ~ 147 kHz

• Number of subcarriers is N=B/Df = 15 MHz/147 kHz = 102

• IFFT/FFT of size 128 should be chosen.

• 16-QAM with rate 1/2 coding gives 4*1/2*102/8.4s=24.3Mbps

• QPSK with rate 5/6 coding gives 2*5/6*102/8.4s= 20.2 Mbps

When higher delayspreads must be tolerated, OFDM symbol length must be increased to avoid large performance loss due to cyclic prefix.

This in turn results in narrower subcarrier spacing => synchronisation problems and larger number of subcarriers => increased complexity due to larger FFT

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Discussion on the Choice of Parameters

• The guard interval often isn't negligible compared to the OFDM data symbol length (often, it's 1/4th of the useful symbol size). Why not use a very long OFDM data symbol after a guard interval in order to decrease the redundancy (i.e. to minimise the loss due to cyclic prefic) ?

– Subcarrier spacing is inverse of the OFDM symbol length

– Subcarriers would be more closely spaced to keep bandwith constant

=> tighter frequency and phase synchronisation requirements

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Discussion on the Choice of Parameters (2)

• If we define an OFDM system for a quasi-AWGN-channel context (i.e. channel impulse response is short) - so, the data throughput can be increased by choosing a short guard interval.

– Long enough cyclic prefix relaxes timing requirements

– TX and RX filters also cause extra delay, i.e., lengthening of channel impulse response

Longer cyclic prefix makes system implementation easier

• CP >> channel delay spread

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Exercise 2 • Consider a multicarrier (MC) system, which has a total

passband bandwidth of 1 MHz. Assume that the channel delay spread Tm = 20 µs. How many subchannels are needed, so that each subchannel can be expected to be approximately flat fading?

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Exercise 2 solution • The coherence bandwidth of the channel Bc can be

calculated from the channel delay spread Tm,

• To ensure that each subchannel has flat fading, the bandwidth is limited to BN = 0,1 Bc << Bc. Thus the number of subchannels is

• Keeping in mind that N must be a power of 2 for the sake of FFT and IFFT, it must be rounded upwards to the closest suitable number. Thus the minimum value for N is 256.

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Exercise 3 • Consider a multicarrier FDM system (without and with

overlapping) operating in a channel with coherence bandwidth Bc = 10 kHz.

a) Find a subchannel symbol time TN = 1/BN = 10Tm, assuming channel delay spread Tm = 1/Bc. This should insure flat fading on the subchannels.

b) Assume the system has N = 128 subchannels. If raised cosine pulses with β = 1.5 are used, and the required additional bandwidth due to time limiting to insure minimal power outside the signal bandwidth is ε = 0.1, what is the total bandwidth of the system?

c) Find the total required bandwidth of the system using overlapping carriers separated by 1/TN, and compare with your answer in part b).

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Exercise 3 solution a) • The subchannel symbol duration is

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Exercise 3 solution b) • If we assume raised cosine pulse, we get a symbol time

for each subchannel, where β is the rolloff factor of the pulse shape.

• In a realistic implementation, the subchannels occupy a larger bandwidth than under ideal raised cosine pulse shaping due the time limiting of these pulse shapes. Let ε/TN denote the additional bandwidth required due to time-limiting of these pulse shapes. Hence the subchannels must be separated by

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Exercise 3 solution b) • The entire occupied bandwidth for non-overlapping

subchannels is then given by

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Exercise 3 solution c) • This time β and ε do not affect the total system bandwidth

due to the subchannels overlap except in the first and last subchannels. The total system bandwidth with overlapping subchannels is given by

• where the approximation holds for N large.

• Thus we get (without approximation)

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Exercise 3 solution c) • The total bandwidth using overlapping carries is less than

half of the non-overlapping bandwidth.

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Exercise 4 • Consider a high-speed data signal with bandwidth 0.5 MHz

and a data rate of 0.5 Mbps. The signal is transmitted over a wireless channel with a delay spread of 10 µs.

a) If multicarrier modulation with non-overlapping subchannels is used to mitigate the effects of ISI, approximately how many subcarriers are needed? What is the data rate and symbol time on each subcarrier? (We do not need to eliminate the ISI completely, so BN = Bc is enough for ISI mitigation)

b) Assume that the average received SNR (γs) on the nth subcarrier is 1000/n (linear units) and that each subcarrier experiences flat Rayleigh fading (so ISI is completely eliminated). Suppose BPSK modulation is used for each subcarrier. If a repetition code is used across all subcarriers (i.e. if a copy of each bit is sent over each subcarrier), then what is the BER after majority decoding? What is the data rate of the system?

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Exercise 4 solution a) • Aim: diversity gain obtained from subcarriers. The

condition imposed for ISI mitigation in this time [compare to BN = 0,1xBc << Bc ->TN = 10Tm] is TN

= Tm = 10 μs

• The bandwidth of a subchannel BN = 1/TN = 100 kHz. Thereby the number of subchannels is

• and each subcarrier must carry

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Exercise 4 solution b) • The bit error probability of BPSK in a Rayleigh

fading channel [1],(6.58) is

• where

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Exercise 4 solution b) • Thus, for the 5 subchannels, it is possible to

calculate the following BERs

• Majority decoding means that the data is detected on each subcarrier and that the decision is done by selecting the symbol, which is the most common amongst the detected symbols. Since N = 5, three symbols is enough to form the majority.

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Exercise 4 solution b) • BER after decoding can be calculated from equation

• where the first term is probability that there is an error in every 5 subcarriers, the second term is probability that there is an error in 4 subcarriers (one is correct), and the third term is probability that there is an error in 3 subcarriers (two are correct).

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Exercise 4 solution b) • There is noticeable gain compared to a case where

n = 1. This is known as diversity gain.

• The total data rate of the system is the same as the data rate of any of the subcarriers (since they all have the same bits transmitted over them), i.e., R = 0,1 Mbps.

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Exercise 5 • Consider a channel with impulse response

h(t) = α0δ(t) + α1δ(t − T1) + α2δ(t − T2).

• Assume that T1 = 10 µs and T2 = 20 µs. You want to design a multicarrier system for the channel, with subchannel bandwidth BN = Bc/2. If raised cosine pulses with β = 1 are used, and the subcarriers are separated by the minimum bandwidth necessary to remain orthogonal, then what is the total bandwidth occupied by a multicarrier system with 8 subcarriers? Assuming a constant SNR on each subchannel of 20 dB, find the maximum constellation size for MQAM modulation that can be sent over each subchannel with a target BER of 10−3, assuming M is restricted to be a power of 2. Find also the corresponding total data rate of the system.

• Use as the upper bound for BER of an M-ary QAM, where γ denotes the SNR.

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Exercise 5 solution • Channel impulse response is h(t) = α0δ(t) + α1δ(t − T1) +

α2δ(t − T2).

• T1 = 10 µs and T2 = 20 µs.

• Subchannel bandwidth BN = Bc/2.

• Raised cosine pulses with β = 1 are used.

• Subcarriers are separated by the minimum bandwidth necessary to remain orthogonal.

• Constant SNR on each subchannel of 20 dB = 100.

• Target BER of 10−3.

• MQAM modulation, where M is restricted to be a power of 2.

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Exercise 5 solution • What is the total occupied bandwidth, if the

multicarrier system has 8 subcarriers?

• The delay spread of the channel Tm = T2 = 20 µs. Thus the coherence bandwidth of the channel

• and the subcarrier bandwidth (this time)

• Note that this condition is not sufficient to avoid ISI (the proper condition would be BN << Bc

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Exercise 5 solution • The minimum frequency separation which allows the

subchannels to remain orthogonal over symbol interval [0,TN] is 1/TN. For a raised cosine pulse with β = 1, the total occupied bandwidth with 8 subcarriers is ([1] (12.4), where ε = 0)

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Exercise 5 solution • What is the maximum constellation size for MQAM

modulation that can be sent over each subchannel with a target BER?

• The BER of an M-ary QAM constellation is upper bounded by

• Inserting the SNR (γ = 100) and the target BER (10–3), the appropriate modulation order can be calculated as

• The closest power of 2 is 16, and therefore the maximum constellation size for MQAM modulation is M = 16.

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Exercise 5 solution • What is corresponding total data rate of the system?

• The associated data rate is

• Note that in this instance, no CP was defined, which means that 1/T = 1/TN = BN.