Digital Transmission of Analogue Signals

(CH 6) Pulse and Digital Modulation By Dr. Hikmat Al-Shamary & Dr. Tariq M. Salman

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LPF Sampler Trans. Decoder

DAC

Quantizer Encoder LPFf(t)

fsADC

Channel

Digital Communication System:

Digital communication has several advantages over analog communication:

1- Digital communication has high immunity to channel noise and channel

distortion.

2- Regenerative repeaters along the transmission path can detect and retransmit a

new, clean signal.

3- Digital hardware implementation is flex able (it may use microprocessors, digital

switching and LSI-ICs)

4- Digital signals can be added to yield low error and high fidelity as well as privacy.

5- It is easier to multiplex digital signals.

6- Exchange of SNR and BW can be done more effectively.

Digital Transmission of Analogue Signals:

1- Pulse Code Modulation (PCM):

This is widely used in digital transmissions. Its block diagram is as shown below:

ADC: Analogue to digital converter.

DAC: Digital to Analogue converter.


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The output of the sampler 𝑓𝑠(𝑘𝑇𝑠)(𝐴𝐷𝐶). Assuming that f(t) has ±𝑓𝑝 peak voltage

level, (ADC full scale), the quantizer will divide the +𝑓𝑝 to −𝑓𝑝 range into L equally

spaced intervals of size ∆𝑉 (step size) then:

volt …. (6-5)

Quantizing Noise:

Since the quantization process introduces some fluctuations about the true value,

these fluctuations can be regarded as noise. As the number of quantization levels L

increases, the quantization noise decreases.

Volt2 …. (6-6)

∆V =2𝑓𝑝

𝐿

N𝑞 =𝑓𝑝

2

3𝐿2


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Encoding:

ADC will then encode the quantized values according to a certain binary code. The

uniform PCM with equal step size mostly uses the signed binary code of n bits.

t

0100

0011

0010

0001

+

+

+

+

-

-

-

-

V

fS(kTS)0101

1100

1011

1010

1001

0000

+

+

V

n bits

Sign bit

Magnitude bits

{ - 1

+ 0

For n=4, then the ±𝑓𝑝 values will be encoded as shown above, this is called transfer

characteristic of the PCM encoder. The relation between number of quantizing levels and number

of bits of encoder is:

Or … … (6-7)

Note:

If n for a given value of L is not integer number,

Then n is computed using 𝑛 = 𝑖𝑛𝑡(𝑙𝑜𝑔2𝐿) + 1,

and L is corrected using 𝐿 = 2𝑛

𝐿 = 2𝑛

𝑁 = 𝑙𝑜𝑔2𝐿


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The output SNR:

Volt2 …. (6-8)

Note:

𝑁𝑜 = 𝑁𝑞 ; S𝑜

N𝑜=

S𝑜

N𝑞

• For tone modulation: 𝑓2(𝑡)̅̅ ̅̅ ̅̅ ̅ =𝐴2

2; 𝑓𝑝 = 𝐴

F ... (6-9)

F ... (6-10)

Bandwidth Requirement of PCM

The information rate of PCM channel is 𝑛𝑓𝑠 bits/sec, if message bandwidth is 𝑓𝑚𝑎𝑥

and the sampling rate is 𝑓𝑠(≥ 2𝑓𝑚𝑎𝑥) then 𝑛𝑓𝑠 binary pulses must be transmitted per

second.

Assuming the PCM signal is a low-pass signal of bandwidth 𝐵𝑊𝑃𝐶𝑀 , the required

minimum sampling rate is 2𝐵𝑊𝑃𝐶𝑀. Thus:

2𝐵𝑊𝑃𝐶𝑀 = 𝑛𝑓𝑠

Hz ... (6-11)

Hz ... (6-12)

S𝑜

N𝑞=

3𝐿2𝑓2(𝑡)̅̅ ̅̅ ̅̅ ̅

𝑓𝑝2

S𝑜

N𝑞=

3𝐿2

2

(S𝑜

N𝑞)

𝑑𝐵

= 1.76 + 20 𝑙𝑜𝑔 𝐿 = 1.76 + 6.02 𝑛

𝐵𝑊𝑃𝐶𝑀 =𝑛

2𝑓𝑠 ≥ 𝑛𝑓𝑚𝑎𝑥

𝐵𝑊𝑃𝐶𝑀𝑚𝑖𝑛𝑖𝑚𝑢𝑚 = 𝑛𝑓𝑚𝑎𝑥


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Ex 6-5:

In a binary PCM system, the output signal-to-quantization ratio is to be hold to a

minimum of 40 dB. If the message is a single tone with fm=4 kHz. Determine:

1- The number of required levels, and the corresponding output signal-to-quantizing noise ratio.

2- Minimum required system bandwidth.

Solution:

1) 𝐿 = 2𝑛

S𝑜

N𝑞= 10000 = 40 𝑑𝐵

S𝑜

N𝑞=

3𝐿2

2 (S.T)

∴ 𝐿 = √2

3∗ 10000 = [81.6] = 82

𝑛 = log2 82 = [6.36] = 7

∴ 𝐿 = 27 = 128

2) Minimum system bandwidth = 𝑛𝑓𝑚𝑎𝑥=7*4 kHz=28 kHz

H.W:

Consider a single tone signal of frequency 3300 Hz. A PCM is generated with a sampling

rate of 8000 sample/sec. the required output signal-to-quantizing noise ratio is 30 dB.

1) What the minimum number of uniform quantizing levels needed?. And what the

minimum number of bits per sample needed?

2) Calculate minimum system bandwidth required.

Ans: (a) 26.5 (b) 20 kHz


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-

+f(t)

Sampler fs

e(t) V

- V

d(t)

Integrator

Comparator

f(t)

2- Delta Modulation:

It is a sampling way to convert analog signal into digital with reduced bandwidth

t

Ts

f(t)&f(t)

V

d(t)

e(t)

t

t

f(t)

f(t)

Slope overload

Larger error

Its produces information about the difference between successive samples.

𝑒(𝑡) = 𝑓(𝑡) − 𝑓(𝑡) , where 𝑓(𝑡) is a stair case approximation of 𝑓(𝑡)


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The sampler with rate (𝑓𝑠 ≫ 𝑁𝑦𝑞𝑢𝑖𝑠𝑡 𝑟𝑎𝑡𝑒) produces pulse train 𝑑(𝑡) where:

𝑑(𝑡) = ∆𝑠𝑔𝑛[𝑒(𝑡)] = { ∆𝑉 𝑒(𝑡) > 0

−∆𝑉 𝑒(𝑡) < 0

𝑑(𝑡) represents the derivative of 𝑓(𝑡)

The demodulator will integrate 𝑑(𝑡) to produce 𝑓𝑠(𝑘𝑇𝑠) smoothed by LPF with 𝐵𝑊 of

𝑓𝑚𝑎𝑥

Integrator

LPFRecovered

f(t)d(t)

f(t)

Slope overload problem:

Due to finite step size ∆𝑉 of integrator and if the slope of 𝑓(𝑡) is Larger than𝑓�̃�(𝑡)

will not track 𝑓(𝑡) in its value [(𝑓�̃�(𝑡)) and 𝑓(𝑡) will diverge from each other]. This will

produce distortion at Rx side when 𝑑(𝑡) is used to construct 𝑓�̃�(𝑡).

To avoid slope overload, the step size must be kept such that:

F

... (6-13)

For single tone case 𝑓(𝑡) = 𝐴𝑚𝑐𝑜𝑠𝜔𝑚𝑡

|𝑑𝑓(𝑡)

𝑑𝑡|

𝑚𝑎𝑥= 𝐴𝑚𝜔𝑚 , therefore

F ... (6-14)

|𝑑𝑓(𝑡)

𝑑𝑡|

𝑚𝑎𝑥< ∆𝑉. 𝑓𝑠

∆𝑉𝑚𝑖𝑛 =𝐴𝑚𝜔𝑚

𝑓𝑠


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• For speech signal, the typical frequency analysis show that about 70% of total

energy lies between 600 and 1000 Hz indicating that peak energy is located that

almost at frequency of 800 Hz called response frequency 𝑓𝑟=800 Hz, then we could

assume ∆𝑉𝑚𝑖𝑛 for speech to be:F

Edeqd ... (6-15)

where 𝑓𝑝 in the maximum amplitude of the speech signal.

f (Hz)

x(f)

70% of the

total area

Quantizing Error:

Assuming quantizing error is equally likely in the interval (-∆V,∆V)

Edeqd ... (6-16)

Where B is the preconstruction filter bandwidth

Output Signal to Noise Ratio:

Edeqd ... (6-17)

∆𝑉𝑚𝑖𝑛 =2𝜋(800)𝐴𝑚

𝑓𝑠

𝑁𝑞 =𝐵

𝑓𝑠.(ΔV)2

3

𝑆𝑜

𝑁𝑞=

3𝑓𝑠𝑓2(𝑡)̅̅ ̅̅ ̅̅ ̅

(Δ𝑉)2𝐵


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f(volt)

P(f)

-1 1

fp-fp

1/2

For single tone message𝑓(𝑡) = 𝐴𝑚𝑐𝑜𝑠𝜔𝑚𝑡

Edeqd ... (6-18)

Ex 6-6:

A DM has sampling frequency of 64 kHz is used to encode speech signal of ±1 volt:

1- Find minimum step size to avoid step overloading.

2- Find 𝑆𝑁𝑅𝑞 assuming speech has uniform probability density function (PDF) over

the interval [-1, 1] volt.

Solution:

1. For speech signal ∆𝑉𝑚𝑖𝑛 =2𝜋(800)𝑓𝑝

𝑓𝑠

∆𝑉𝑚𝑖𝑛 =2𝜋(800)(1)

64000≅ 78 𝑚𝑉

2. 𝑆𝑜

𝑁𝑞=

3𝑓𝑠𝑓2(𝑡)̅̅ ̅̅ ̅̅ ̅̅

(Δ𝑉)2𝐵

𝑓2(𝑡)̅̅ ̅̅ ̅̅ ̅ = ∫ 𝑓2𝑝(𝑓)𝑑𝑓1

−1

= 2 ∫1

2𝑓2𝑑𝑓 =

1

3

1

0

𝑆𝑜

𝑁𝑞=

(3)(64000)(1/3)

(0.078)2(3400)≅ 35 𝑑𝐵

Note:

Compare this result of 35 dB with PCM at 64000 bps (𝑓𝑠 = 8 𝑘𝐻𝑧, 𝑛 =

8 𝑏𝑖𝑡𝑠 𝑠𝑎𝑚𝑝𝑙𝑒⁄ ) then 𝑆𝑁𝑅𝑞 ≅ 48 𝑑𝐵. i.e PCM is better than DM for the same bit rate.

𝑆𝑜

𝑁𝑞=

3𝑓𝑠2

8𝜋2𝑓𝑚2𝐵


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H.W:

A DM system is designed to operate at 3 times the Nyquist rate for the signal with a 3

kHz bandwidth. The quantization step size is 250 mV. Determine:

a) Maximum amplitude of a 1 kHz input sinusoid for which the delta modulator does not

show slop over load.

b) The post filter output signal-to-quantizing noise ratio for the signal in part a.

Digital Transmission of Analogue Signals

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