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Huiswerkopgaven 6a - Numerical problems

Q: Consider the binary data sequence 1 0 1 1 0 0 1 0. (d) For each code, comment on its DC component and its suitability for self-synchronisation.

(d) NRZ on–off keying Mean value is A/2 for equally likely bits – has a DC component. Long runs produce no transitions – poor self-synchronisation. Bipolar RZ (AMI) Zero mean always – no DC. Long runs of ‘0’s produce no transitions – moderate synchronisation; also provides single-error detection via AMI violation. Manchester Zero mean always – no DC. Every bit contains a mid-bit transition – excellent self-synchronisation.

8 important questions on Huiswerkopgaven 6a - Numerical problems

Consider the binary data sequence 1 0 1 1 0 0 1 0.
(a) Sketch the waveform using uni-polar NRZ (i.e. on–off keying) (1 → A, 0 → 0, held for the full bit period).

NRZ with on–off keying

Consider the binary data sequence 1 0 1 1 0 0 1 0.

b) Sketch the waveform using Bipolar RZ (AMI) signalling (0 → 0; successive 1s → +A, −A, +A, . . ., each pulse occupying the first half of the bit period)

(b) Bipolar RZ (AMI)

Consider the binary data sequence 1 0 1 1 0 0 1 0.
(c) Sketch the waveform using Manchester signalling (1 → low-to-high mid-bit transition; 0 → high-to-low mid-bit transition).

Consider the binary data sequence 1 0 1 1 0 0 1 0.
(d) For each code, comment on its DC component and its suitability for self-synchronisation.

(d) NRZ on–off keying Mean value is A/2 for equally likely bits – has a DC component. Long runs produce no transitions – poor self-synchronisation.
Bipolar RZ (AMI) Zero mean always – no DC. Long runs of ‘0’s produce no transitions – moderate synchronisation; also provides single-error detection via AMI violation.
Manchester Zero mean always – no DC. Every bit contains a mid-bit transition – excellent self-synchronisation.

(a) On the same axes, sketch the raised cosine spectrum HRC(f ) (for f ≥ 0) for roll-off factors α = 0, α = 0.5, and α = 1, labelling the key frequencies.
(b) For a symbol rate of Rs = 10 ksymbols/s, calculate the transmission bandwidth BT for each value of α.

Antwoord

A binary PAM system must transmit data at 6 kbits/s over a channel with a bandwidth of 4 kHz. Raised cosine pulse shaping is used.
(a) Determine the maximum roll-off factor α that can be used.
(b) How would your answer change if the data rate were increased to 7 kbits/s while the channel bandwidth remained 4 kHz? Is this achievable with binary PAM?

This would require α ≈ 0.14, which is positive and ≤ 1, so it is theoretically achievable. Note: a very small α is difficult to implement in practice due to filter complexity and sensitivity to timing errors

An 8-PAM (M = 8) system uses raised cosine pulse shaping.
(a) Determine the minimum transmission bandwidth (i.e. α = 0) required to transmit data at 318 kbits/s with zero ISI.
(b) Determine the transmission bandwidth when a roll-off factor of α = 0.25 is used.
(c) Compare these bandwidths with those required by a binary (M = 2) system carrying the same 318 kbits/s. Comment on the trade-off

c) For binary PAM with the same Rb = 318 kbps:
Rs = Rb = 318 ksymbols/s,
Bmin = 159 kHz,
BT (α = 0.25) = 198.75 kHz
8-PAM uses three times less bandwidth than binary PAM for the same bit rate. The trade-off is that 8-PAM requires a significantly higher SNR to achieve the same BER, because the eight amplitude levels are much more closely spaced for the same peak power.

A music signal of bandwidth 18 kHz is sampled at 44.1 kHz, quantised to 256 levels, and transmitted using M-ary PAM with raised cosine pulse shaping (α = 0.2). The available channel bandwidth is 24 kHz.
(a) Calculate the PCM bit rate.
(b) Show that binary PAM cannot support this bit rate within the available bandwidth for any α ∈ [0, 1].
(c) Determine the minimum value of M (as a power of 2) that allows transmission within the 24 kHz channel

Antwoord

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Telecommunicatie

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Huiswerkopgaven 6a - Numerical problems

8 important questions on Huiswerkopgaven 6a - Numerical problems

Consider the binary data sequence 1 0 1 1 0 0 1 0.(a) Sketch the waveform using uni-polar NRZ (i.e. on–off keying) (1 → A, 0 → 0, held for the full bit period).

Consider the binary data sequence 1 0 1 1 0 0 1 0.b) Sketch the waveform using Bipolar RZ (AMI) signalling (0 → 0; successive 1s → +A, −A, +A, . . ., each pulse occupying the first half of the bit period)

Consider the binary data sequence 1 0 1 1 0 0 1 0.(c) Sketch the waveform using Manchester signalling (1 → low-to-high mid-bit transition; 0 → high-to-low mid-bit transition).

Consider the binary data sequence 1 0 1 1 0 0 1 0.(d) For each code, comment on its DC component and its suitability for self-synchronisation.

(a) On the same axes, sketch the raised cosine spectrum HRC(f ) (for f ≥ 0) for roll-off factors α = 0, α = 0.5, and α = 1, labelling the key frequencies.(b) For a symbol rate of Rs = 10 ksymbols/s, calculate the transmission bandwidth BT for each value of α.

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Consider the binary data sequence 1 0 1 1 0 0 1 0.
(a) Sketch the waveform using uni-polar NRZ (i.e. on–off keying) (1 → A, 0 → 0, held for the full bit period).

Consider the binary data sequence 1 0 1 1 0 0 1 0.

b) Sketch the waveform using Bipolar RZ (AMI) signalling (0 → 0; successive 1s → +A, −A, +A, . . ., each pulse occupying the first half of the bit period)

Consider the binary data sequence 1 0 1 1 0 0 1 0.
(c) Sketch the waveform using Manchester signalling (1 → low-to-high mid-bit transition; 0 → high-to-low mid-bit transition).

Consider the binary data sequence 1 0 1 1 0 0 1 0.
(d) For each code, comment on its DC component and its suitability for self-synchronisation.

(a) On the same axes, sketch the raised cosine spectrum HRC(f ) (for f ≥ 0) for roll-off factors α = 0, α = 0.5, and α = 1, labelling the key frequencies.
(b) For a symbol rate of Rs = 10 ksymbols/s, calculate the transmission bandwidth BT for each value of α.