(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.
(CH 6) Pulse and Digital Modulation By Dr. Hikmat Al-Shamary & Dr. Tariq M. Salman
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The output of the sampler ππ (πππ )(π΄π·πΆ). Assuming that f(t) has Β±ππ peak voltage
level, (ADC full scale), the quantizer will divide the +ππ to βππ range into L equally
spaced intervals of size βπ (step size) then:
volt β¦. (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 β¦. (6-6)
βV =2ππ
πΏ
Nπ =ππ
2
3πΏ2
(CH 6) Pulse and Digital Modulation By Dr. Hikmat Al-Shamary & Dr. Tariq M. Salman
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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 Β±ππ 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 β¦ β¦ (6-7)
Note:
If n for a given value of L is not integer number,
Then n is computed using π = πππ‘(πππ2πΏ) + 1,
and L is corrected using πΏ = 2π
πΏ = 2π
π = πππ2πΏ
(CH 6) Pulse and Digital Modulation By Dr. Hikmat Al-Shamary & Dr. Tariq M. Salman
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The output SNR:
Volt2 β¦. (6-8)
Note:
ππ = ππ ; Sπ
Nπ=
Sπ
Nπ
β’ For tone modulation: π2(π‘)Μ Μ Μ Μ Μ Μ Μ =π΄2
2; ππ = π΄
F ... (6-9)
F ... (6-10)
Bandwidth Requirement of PCM
The information rate of PCM channel is πππ bits/sec, if message bandwidth is ππππ₯
and the sampling rate is ππ (β₯ 2ππππ₯) then πππ binary pulses must be transmitted per
second.
Assuming the PCM signal is a low-pass signal of bandwidth π΅πππΆπ , the required
minimum sampling rate is 2π΅πππΆπ. Thus:
2π΅πππΆπ = πππ
Hz ... (6-11)
Hz ... (6-12)
Sπ
Nπ=
3πΏ2π2(π‘)Μ Μ Μ Μ Μ Μ Μ
ππ2
Sπ
Nπ=
3πΏ2
2
(Sπ
Nπ)
ππ΅
= 1.76 + 20 πππ πΏ = 1.76 + 6.02 π
π΅πππΆπ =π
2ππ β₯ πππππ₯
π΅πππΆπππππππ’π = πππππ₯
(CH 6) Pulse and Digital Modulation By Dr. Hikmat Al-Shamary & Dr. Tariq M. Salman
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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) πΏ = 2π
Sπ
Nπ= 10000 = 40 ππ΅
Sπ
Nπ=
3πΏ2
2 (S.T)
β΄ πΏ = β2
3β 10000 = [81.6] = 82
π = log2 82 = [6.36] = 7
β΄ πΏ = 27 = 128
2) Minimum system bandwidth = πππππ₯=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
(CH 6) Pulse and Digital Modulation By Dr. Hikmat Al-Shamary & Dr. Tariq M. Salman
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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.
π(π‘) = π(π‘) β π(π‘) , where π(π‘) is a stair case approximation of π(π‘)
(CH 6) Pulse and Digital Modulation By Dr. Hikmat Al-Shamary & Dr. Tariq M. Salman
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The sampler with rate (ππ β« ππ¦ππ’ππ π‘ πππ‘π) produces pulse train π(π‘) where:
π(π‘) = βπ ππ[π(π‘)] = { βπ π(π‘) > 0
ββπ π(π‘) < 0
π(π‘) represents the derivative of π(π‘)
The demodulator will integrate π(π‘) to produce ππ (πππ ) smoothed by LPF with π΅π of
ππππ₯
Integrator
LPFRecovered
f(t)d(t)
f(t)
Slope overload problem:
Due to finite step size βπ of integrator and if the slope of π(π‘) is Larger thanποΏ½ΜοΏ½(π‘)
will not track π(π‘) in its value [(ποΏ½ΜοΏ½(π‘)) and π(π‘) will diverge from each other]. This will
produce distortion at Rx side when π(π‘) is used to construct ποΏ½ΜοΏ½(π‘).
To avoid slope overload, the step size must be kept such that:
F
... (6-13)
For single tone case π(π‘) = π΄ππππ πππ‘
|ππ(π‘)
ππ‘|
πππ₯= π΄πππ , therefore
F ... (6-14)
|ππ(π‘)
ππ‘|
πππ₯< βπ. ππ
βππππ =π΄πππ
ππ
(CH 6) Pulse and Digital Modulation By Dr. Hikmat Al-Shamary & Dr. Tariq M. Salman
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β’ 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 ππ=800 Hz, then we could
assume βππππ for speech to be:F
Edeqd ... (6-15)
where ππ 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 (-βV,βV)
Edeqd ... (6-16)
Where B is the preconstruction filter bandwidth
Output Signal to Noise Ratio:
Edeqd ... (6-17)
βππππ =2π(800)π΄π
ππ
ππ =π΅
ππ .(ΞV)2
3
ππ
ππ=
3ππ π2(π‘)Μ Μ Μ Μ Μ Μ Μ
(Ξπ)2π΅
(CH 6) Pulse and Digital Modulation By Dr. Hikmat Al-Shamary & Dr. Tariq M. Salman
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f(volt)
P(f)
-1 1
fp-fp
1/2
For single tone messageπ(π‘) = π΄ππππ πππ‘
Edeqd ... (6-18)
Ex 6-6:
A DM has sampling frequency of 64 kHz is used to encode speech signal of Β±1 volt:
1- Find minimum step size to avoid step overloading.
2- Find πππ π assuming speech has uniform probability density function (PDF) over
the interval [-1, 1] volt.
Solution:
1. For speech signal βππππ =2π(800)ππ
ππ
βππππ =2π(800)(1)
64000β 78 ππ
2. ππ
ππ=
3ππ π2(π‘)Μ Μ Μ Μ Μ Μ Μ Μ
(Ξπ)2π΅
π2(π‘)Μ Μ Μ Μ Μ Μ Μ = β« π2π(π)ππ1
β1
= 2 β«1
2π2ππ =
1
3
1
0
ππ
ππ=
(3)(64000)(1/3)
(0.078)2(3400)β 35 ππ΅
Note:
Compare this result of 35 dB with PCM at 64000 bps (ππ = 8 ππ»π§, π =
8 πππ‘π π πππππβ ) then πππ π β 48 ππ΅. i.e PCM is better than DM for the same bit rate.
ππ
ππ=
3ππ 2
8π2ππ2π΅
(CH 6) Pulse and Digital Modulation By Dr. Hikmat Al-Shamary & Dr. Tariq M. Salman
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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.
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