1. Pulse Modulation Objectives
To explain sampling theorem and analyze sampling process
To study pulse amplitude modulation (PAM)
To illustrate time division multiplexing (TDM) method
To study pulse code modulation (PCM)
To describe quantization process
To determine quantization noise
To describe encoding process
To determine transmission bandwidth
To study differential pulse code modulation (DPCM)
To study delta modulation (DM)
To compare PCM and DPCM, PCM and DM systems

2. Pulse Modulation In amplitude modulation and angle modulation, some
parameter of sinusoidal carrier wave is varied continuously in
accordance with the message signal.
This is referred to as analog modulation.

3. Pulse Modulation
If the carrier consists of (discrete) pulse trains, some
parameter of the pulse train is varied in accordance with the
message signal, it is called pulse modulation.
After pulse modulation, the message signal becomes discrete.
One of the major concern:
How to recover the message signal from the discrete pulse trains?

4. Pulse Modulation
Sampling Theorem
A bandwidth limited signal having no frequency
components higher than fm Hz may be completely
recovered from its samples taken uniformly at a rate at
least 2fm samples per second, i.e. the sampling
frequency is fs = 2fm Hz.
How to prove the sampling theorem?
According to our knowledge, a pulse train is a periodic
function.
What is the spectrum of a pulse train?

5. Pulse Modulation Fourier transform of a periodic function
A periodic function f(t), of period Ts, or angular frequency s,
can be expressed in a Fourier series
where Fn is the Fourier coefficient defined by

6. Pulse Modulation Since 1  2()
from frequency shifting property, we have
then for a periodic function
its Fourier transform can be expressed as

7. Pulse Modulation
In particular, if the periodic function is a set of impulse trains,
its Fourier coefficient then becomes
thus
the Fourier transform of a set of impulse trains becomes

8. Pulse Modulation
For a message signal g(t) whose spectrum is band-
limited to fm Hz (or m = 2fm radian/s), in the
sampling process, g(t) is multiplied by a set of impulse
train,
the product becomes

9. Pulse Modulation
So the sampled signal is a product of message signal and a set
of impulse trains
According to the convolution theorem, the sampled signal
Spectrum is
where we used the properties
g(t)  (t) = g(t) and g(t)  (t – T) = g(t – T)
Proof

10. Pulse Modulation

11. Pulse Modulation

12. Pulse Modulation
The signal spectrum G() can be recovered by the
use of a low-pass filter if there is no overlap between
the successive cycles of Gs(). This requires
s  2m or fs  2fm
or the sampling interval
Ts  1 / 2fm
Up to this point, we have proved the sampling theorem.
*The minimum sampling rate 2fm is also called the Nyquist rate.

13. Pulse Modulation
What does sampling theorem tell us?
to convey the information contained in a band-limited signal it is necessary to send only a finite number of discrete samples.
a message signal that is band-limited to fm Hz is completely specified by its values at intervals spaced no greater than TS = 1/2fm seconds apart.
In pulse modulation, these discrete samples are used to vary a parameter of a pulse waveform.

14. Pulse Modulation
Discrete samples play the role of message signal.
Notice that these discrete samples are discrete in time but continuous in amplitude which means our pulse modulation is an analog pulse modulation.
The pulse amplitude takes an analog value.

15. Pulse Modulation Analog pulse modulation
Pulse amplitude modulation (PAM)
Pulse width modulation (PWM)
Pulse position modulation (PPM).

16. Pulse Modulation
Pulse amplitude modulation (PAM)
The amplitude of a train of constant-width pulses is varied in proportion to the sample values of the message signal.
In this scheme, the message is multiplied by a pulse train, which is similar to DSB-SC system.

17. Pulse Modulation
Pulse amplitude modulation (PAM)
The PAM signal spectrum is
A distortion is introduced because of the shape of the sampling pulse so that the spectral density G() has lost its original shape.
How to correct such a distortion?

18. Pulse Modulation
A method of signal recovery is to use a filter that has a transfer function
Equalization
The technique of correcting the frequency response of a system for a known distortion is called equalization.
Equalization is often used in correcting distortions which are known but over which one has little control.

19. Pulse Modulation

20. Pulse Modulation
The message is recoverable by lowpass filtering if fS > 2m, where m is the highest frequency component contained in the message.
If fS < 2m, spectral overlap occurs and lowpass filtering can recover only a distorted form of the message.
PAM signals can be multiplexed in the time domain.

21. Pulse Modulation Time division multiplexing (TDM)
The transmission of several sampled signals at a time-sharing basis is called time division multiplexing (TDM).
To recover the individual message at the receiver,
it is necessary to sample
in a synchronous manner
to that done at the
transmitter.
Can analog signal be used
in TDM?

22. Pulse Modulation
It is quite easy to see that the minimum bandwidth for TDM transmission is proportional to the product of the message signal bandwidth and the number of the multiplexed signals.
(This assumes that all signals have the same bandwidth.)

23. Pulse Modulation
Comparison between FDM and TDM
In TDM, multiple incoming signals are sliced into small time intervals, whereas in FDM the incoming signals are placed on different frequency ranges.
Therefore, in the time domain, all the signals overlap in FDM, whereas signals may overlap in the frequency domain in TDM.
This implies that an analog modulation system cannot use TDM unless sampling is performed (to change the system into a pulse modulation system.)

24. Pulse Modulation
Comparison between FDM and TDM
In TDM, all the channels require identical circuits, thus
providing an advantage in simplicity to TDM.
In FDM, different carriers are generated for different
channels. Also, different bandpass filters are required
because each channel occupies a different frequency band.
However, in TDM, sampling needs to be done at high
speeds and synchronization of timing between the
transmitter and the receiver must be achieved.
This is a disadvantage.

25. Pulse Modulation
PAM is still an analog pulse modulation. It is not
completely digital because the amplitudes of the
pulses takes analog value.
In analog pulse modulation, information is
transmitted in analog form, but the transmission
takes place at discrete times.
If the message signal is represented in a form that is
discrete in both time and amplitude, then we have
digital pulse modulation.
In digital pulse modulation, the signal transmission
is in digital form, as a sequence of coded pulses.

26. Pulse Modulation
In digital pulse modulation, PAM signals need to be
further digitized and then encoded for transmission.
This is achieved in a pulse code modulation (PCM)
system.
Binary PCM (where the pulses have only two
permissible values) is the most common.
We will confine our discussion to binary PCM system.
How to produce a PCM signal?

27. Pulse Modulation
A PCM signal is produced by an analog-to-digital
conversion process.

28. Pulse Modulation Pulse code modulation (PCM) system Quantization
Quantization is the process of transforming the sampled amplitude of a message signal into a discrete level taken from a finite set of possible amplitudes.

29. Pulse Modulation How to perform the quantization?
1. The amplitudes of signal m(t) lie in the range (- mp, mp), which is partitioned into L intervals, each of magnitude  = 2mp/L.
2. Each sample amplitude is approximated by the midpoint value of the interval in which the sample falls.

30. Pulse Modulation The amplitude range: (- mp, mp),
[mp is not necessarily the peak amplitude of m(t)]
Interval:  = 2mp/L
In quantization process, a sampling value is approximated by
the midpoint of the interval. This introduces an error q(t),
defined as the difference between the message signal m(t) and
the corresponding quantized sample mq(t),
q(t) = m(t) – mq(t)
This error is called the quantization noise.

31. Pulse Modulation
Quantization noise
Since q(t) is uniformly distributed over the interval (-/2, /2), i.e., the error has equal probability to lie in the range (-/2, /2), the probability density is then 1/, hence the mean square value of q(t) is given by

32. Pulse Modulation
Quantization noise power
Since we can assume that the time average is equal
to the statistical average, the quantization noise
power is then
the output signal-to-noise ratio is

33. Pulse Modulation
From above discussion, it is clear that quantization
results in a loss of information.
(Information can also be lost in PAM due to noise)
Such an information lost due to quantization may
be reduced by increasing the number of levels used, L.
e.g. 8 to 16 levels are sufficient for speech communication.

34. Example:
The digital audio compact optical disc (CD) system uses 16 bit
quantization and a sampling rate of kHz per channel. Assuming the audio signal has a peak to mean power ratio
of 13 dB, occupies the frequency band 0 to 20 kHz and that the recovery filter has an effective bandwidth, allowing for the finite cut-off rate of a practical filter of 22 kHz, estimate the signal to quantization noise ratio attainable.
Solution:
fs = 44.1 kHz, n = 16, thus we have L = 216 quantization levels.
So we have

35. Pulse Modulation
Encoding
After sampling and quantization, the analog message signal becomes discrete in values, but it is still not in the form best suited for transmission.
In order that the signal is best suited for transmission, i.e. more robust to noise and interference, an encoding process is required to translate the discrete samples to a more appropriate form, such as the binary digits.

36. Pulse Modulation
Encoding

37. Pulse Modulation
Encoding

39. Pulse Modulation
Transmission bandwidth
Digital signals use much more bandwidth than analog signals. This can be explained as follows:
The binary digits must be transmitted in the sampling interval originally allotted to one sample, the binary pulse widths are correspondingly narrower and therefore
occupy a larger bandwidth according to the inverse time–bandwidth relationship.
The transmission bandwidth increases proportionately to the number of binary pulses needed.

40. Pulse Modulation
Transmission bandwidth
For a binary PCM, a distinct group of binary digits (bits) is assigned to each of the L quantization levels.
As n binary digits can be arranged in 2n distinct patterns,
L  2n or n  log2L
Each quantized sample is thus encoded into n bits.

41. Pulse Modulation
Transmission bandwidth
According to sampling theorem, a signal m(t) band-limited to B Hz requires a minimum of 2B samples per second, a total of 2nB bits per second (bps) is required, that is, 2nB piece of information per second. Because a unit bandwidth (1 Hz) can transmit a maximum of two pieces of information (1 or 0) per second, a minimum channel bandwidth is given by
BT = nB Hz
This is the theoretical minimum transmission bandwidth required to transmit the PCM signal.

42. Bandwidth – Signal-to-noise ratio trade-off
Pulse Modulation
Bandwidth – Signal-to-noise ratio trade-off
Assuming that L = 2n, the output signal-to-noise ratio can be expressed as
where
since n = BT/B, we have
It shows that the signal-to-noise ratio increases exponentially with the transmission bandwidth BT.

43. Pulse Modulation
If n increases, then BT = nB also increases, this leads
to that the signal-to-noise ratio increases. In other
words, a larger channel bandwidth corresponds to a
higher signal-to-noise ratio.
If you want to improve the signal-to-noise ratio, you
have to use a larger channel bandwidth.
--- There is a trade-off between the SNR and the channel bandwidth.

44. Pulse Modulation
Demodulation of PCM signal
When the PCM signal is demodulated, the signal-to-
noise ratio obtained at the receiver should be identical
to that at the transmitter.
Any noise which may be added during transmission
can be eliminated because the binary signal can have
only two known values, 1 and 0.
If the value of the pulse is different from the set values,
we know that it is due to external noise and we can
readjust it to its original value.

45. Pulse Modulation Key advantage of PCM
In analog system, a message signal suffers the channel noise and
the signal distortion, which are cumulative.
Amplification is of little help because it enhances the signal and the
noise in the same proportion.
The analog signal can not cleaned periodically, and thus the
transmission is not reliable.
In PCM system, the new, clean signals can be completely
regenerated at repeater stations because all the information is
contained in the code.
The PCM signal can then be transmitted over long distance with
great reliability.

46. Pulse Modulation
Other advantages of PCM and digital communications
1. Allow us to use computer as a tool for communications. (Computers generate digital signals.)
2. Digital communications systems use a type of coding (error correction code) which can minimize noise and interference, thus producing high quality signals.
3. Digital systems can use both types of multiplexing (FDM and TDM) so that many different sources of information can be handled efficiently.

47. Pulse Modulation
4. Transmission media (such as optical fibres) that have wide bandwidths are available so we can cope with the large bandwidth requirements of digital systems.
5. Digital signal processing has become well established. Digital electronic circuits are now easy to design and to implement in integrated circuit (IC) form.

48. Disdvantages of digital communications:
Pulse Modulation
Disdvantages of digital communications:
Requires wider bandwidths than analog transmission
Requires synchronization between receiver and transmitter.

49. Pulse Modulation
Minimum Information Capacity (Bit Rate) of PCM Systems
The information capacity is defined as the number of bits that can be transmitted per second (bit rate). Since we are using the Nyquist rate for sampling, the minimum bit rate transmitted for a binary system is
That is, the minimum bit rate is equal to double the product of the signal bandwidth and the number of binary pulses.

50. Pulse Modulation Example: Plain-old-telephone system (POTS)
Voice bandwidth limited to: 3.4 kHz
Babdwidth including guard band: 4 kHz
Sampling frequency: 8 kHz
Sampling rate: 8,000 samples/s
Coding: binary
8 bits per sample
(L = 28 = 256 quantization levels)
Bit rate :

51. Pulse Modulation Speech waveform
Speech waveform for part of the word "compute"
cslu.cse.ogi.edu/tutordemos/ SpectrogramReading/waveform.html

52. Pulse Modulation
When audio or video signals are sampled, it is usually found that adjacent samples are close to the same value.
the difference signal is much less in amplitude than the actual sample
less number of quantization levels are needed.
 the number of bits per code is reduced
 resulting in a reduced bit-rate.
the bandwidth required in this case is less than the one required in PCM.
Differential pulse code modulation (DPCM)
to transmit PCM signals corresponding to the difference in adjacent sample values.

53. Pulse Modulation If m[k] is the kth sample, we transmit the difference
d[k] = m[k] – m[k-1].
At the receiver,
knowing d[k] and the previous sample
value m[k-1] gives m[k].

54. Pulse Modulation Comparison between PCM and DPCM
PCM system has a relatively large signal range, i.e. mp(PCM) > mp(DPCM)
if a constant SNR can be maintained, then from
n = log2L and BT = nB
we have L(PCM) > L(DPCM)  n(PCM) > n(DPCM)
 BT(PCM) > BT(DPCM)
To maintain the same S/N, DPCM system requires smaller channel bandwidth than PCM system.
The modulator and demodulator circuits for DPCM are more complicated than those in PCM.

55. Pulse Modulation Delta modulation (DM)
In DPCM, if the signal change is represented by just one bit, that
bit being used for the sign of the sample difference, then we have
delta modulation (DM).
In DM, since only 1 bit/sample is employed, it transmits information to only indicate whether the analog signal it encodes is to “go up” or “go down”.
DM has high noise as 1 bit/sample is used, thus fs is much higher
than Nyquist frequency 2fm.

56. Pulse Modulation
Delta modulation system

57. Pulse Modulation
Delta modulation (DM)

58. Pulse Modulation Delta modulation (DM) In DM, mq[k] = mq[k - 1] +dq[k]
hence mq[k - 1] = mq[k - 2] +dq[k - 1]
then mq[k] = mq[k - 2] +dq[k] + dq[k - 1] = …
= mq[0] +dq[k] + dq[k - 1] + … + dq[1]
proceeding iteratively in this manner, we have
and assuming zero initial condition, i.e. mq[0] = 0, yields
The receiver is just an accumulator (adder)! Simple circuit!

59. Pulse Modulation
Delta modulation (DM)
DM system have an advantage in that the electronic circuitry required for modulation at the transmitter and demodulation at the receiver is substantially simpler than that required for other PCM systems.

60. Pulse Modulation Comparison between PCM and DM PCM
Relatively complicated system
Good signal to quantization noise ratio DM
Simple system (simple encoder/decoder and does not require
synchronisation)
poor signal to quantization noise ratio
High sampling rate (4 times of the Nyquist rate)
Commonly used in the system with small capacity and low quality
requirement and military system
DPCM uses n binary digit to represent the signal difference.
DM uses only one binary digit to represent the signal difference.

61. Questions (Pulse Modulation)
How to sample a bandlimited analog signal to ensure distortion-free reconstruction?
What is Nyquist rate and Nyquist interval?
What is time division multiplexing (TDM)?
How to perform quantization in a PCM system?
What is quantization noise?
What is the difference between PAM and PCM? What type of signal PAM signal is? What type of signal PCM signal is?
What are the advantages of PCM?
What is non-uniform quantization?
To maintain the same S/N, which system requires smaller channel bandwidth, PCM or DPCM?
How to obtain delta modulation (DM) signal?

62. Exercise Problems (Pulse Modulation)
1. For a given signal f(t) = cos1t + cos21t,
(a) Draw the time waveform and the spectrum of the signal;
(b) Determine the minimum sampling frequency.
2. If the signal f(t) = 10cos20t cos200t, the sampling frequency is 450Hz,
(a) Determine the spectrum of the sampling signal;
(b) If an ideal low-pass filter is used to recover f(t) from the sampled signal, determine the bandwidth of the low-pass filter required;
(c) What is the Nyquist sampling rate for f(t)?

63. Exercise Problems (Pulse Modulation)
3. A TDM system consists of 24 transmission channels and a synchronization channel, a sampling rate of 8kHz is used. The bandwidth of the signal for each channel is below 3.3kHz. Determine the minimum channel bandwidth required to transmit TDM signal in the system.
4. Five signals are combined to be transmitted in a TDM system, the combined signals will pass through a low-pass filter. Three channels are used to transmit the signals of frequency range between 300 to 3300 Hz and the rest two channels transmit the signals of 50 Hz to 10 kHz range.
(a) What is the minimum sampling rate required?
(b) What is the minimum bandwidth of the low-pass filter required corresponding to the minimum sampling rate?

64. Exercise Problems (Pulse Modulation)
5. The information in an analog voltage waveform is to be transmitted over a PCM system with a 0.1% accuracy (full scale). The analog waveform has an absolute bandwidth of 100 Hz and an amplitude range of –10 to 10 V.
(a) Determine the minimum sampling rate needed;
(b) Determine the number of bits needed in each PCM word;
(c) Determine the minimum bit rate required in PCM signal;
(d) Determine the minimum channel bandwidth required for transmission of this PCM signal.

65. Exercise Problems (Pulse Modulation)
6. For a signal f(t) = 9 + Amcosmt, with Am  10, to be quantized into exactly 41 binary levels, with one level set at the smallest value of f(t).
(a) Determine the number of bits needed in each PCM word;
(b) What are the values of extreme quantized levels Vmax and Vmin if the quantized levels are centered to [f(t)max + f(t)min] / 2;
(c) If Am = 10 V, find the signal to quantized noise ratio.

66. Exercise Problems (Pulse Modulation)
7. A speech signal with frequency range between 50 to 3300 Hz, the sampling rate used is 8 kHz, the sampled signal is transmitted through a PAM or PCM system.
(a) Determine the minimum bandwidth required by PAM system;
(b) If a binary PCM system is used and the number of quantization level is 8, determine the transmission channel bandwidth;
(c) If the number of quantization level is now 128, recalculate the transmission channel bandwidth.