ETM2046 Analog and Digital Communications
Tutorial 5: Baseband Data Transmissions and
Digital Modulation Techniques (Solutions)
1. p(t) = sinc(t/Tb), where Tb is the bit duration of the input binary data
Sequence of sinc pulses with the tails crossing zero level at all sampling instants
Tb
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ETM2046 Analog and Digital Communications
2. Bit rate Rb = 56 kbits/s, number of bits per symbol k = 3, 8-level signalling
Transmission bandwidth
= 0.25 BT = 11.67 kHz
= 0.50 BT = 14 kHz
= 0.75 BT = 16.4 kHz
= 1.00 BT = 18.7 kHz
3. No. of signalling levels M = 4 = 2k
No. of bits per symbol k = 2
Channel bandwidth = transmission bandwidth BT = 6 kHz
= 0.5 (50% roll-off)
(a) , as RS = Rb / k
(b) Only ¾ symbols carry information out of 8,000 symbols in one second
Info symbol rate = ¾ x 8,000 = 6,000 info symbols/s
Info bit rate = info symbol rate x k = 6,000 x 2 = 12,000 info bits/s
(c) Assume sampling is at min. rate of (fs)min
(fs)min = info bit rate/no. of info bits per sample
= (12 kbits/s)/(6 bits/sample)
= 2,000 Hz
Source bandwidth fM = (fs)min / 2 = 1,000 Hz
No. of quantisation levels = 26 = 64
No. of quantisation levels reduced by a factor of 4, gives new no. of levels = 26/4 = 16
New (fs)min = info bit rate/no. of info bits per sample
= 12,000/4
= 3,000 Hz
New source bandwidth fM = new (fs)min / 2 = 1,500 Hz
Conclusion: no. of quantisation levels is decreased for the increase of source bandwidth
4. Bit rate Rb = 4,800 bits/s, roll-off factor = 0.5, and B = (Rb / 2)(1 + )
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ETM2046 Analog and Digital Communications
(a) BASK
BT = 2B = Rb(1 + ) = 4,800(1 + 0.5) = 7,200 Hz
(b) BFSK; Frequency deviation f = 3,600 Hz
BT = 2f + 2B
= 2f + Rb(1 + )
= 2 x 3,600 + 7,200
= 14,400 Hz
(c) 16-state QAM; Number of bits per symbol k = log216 = 4
BT = 2B / k
= Rb(1 + ) / 4
= 7,200 / 4
= 1,800 Hz
5. Binary FSK transmission; Rb = 1,200 bps and BT = | 500 – 2,900 Hz | = 2,400 Hz
f1 = fc f = 1,200 Hz --------(1)
f2 = fc + f = 2,200 Hz --------(2)
Sub. (2) into (1),
2,200 f f = 1,200
-2f = 1,200 – 2,200
f = 500 Hz
BT = 2f + 2B
2,400 = 2 x 500 + 2B
Baseband bandwidth B = 700 Hz
B = (Rb / 2)(1 + )
700 = (1,200 / 2)(1 + )
= 0.17
6. 16 QAM; Rb = 9,600 bps, BT = | 300 – 3,000 Hz | = 2,700 Hz, = 0.125, k = 4, and
B = (Rb / 2)(1 + )
BT = 2B / k
BT = Rb(1 + ) / k
2,700 = Rb(1 + 0.125) / 4
Rb = 9,600 bps (shown)
