Noise in PCM

• There are 2 major sources of noise in a PCM

system: (i) Transmission noise outside the
transmitter (generally assumed to be thermal noise
which is white & Gaussian) & (ii) Quantization
noise in the transmitter.
• During pulse transmission if the peak noise
amplitude is greater than half the pulse size, then a
bit error occurs, i.e., a pulse 0 is received as 1 or
vice versa, giving rise to the probability of error.
• In the PCM transmitter, a quantized value of the
sample is encoded instead of its actual value. Hence
an error occurs known as the quantization
 S
error, qe  m(t )  mk  ...(i) , since the difference
2
between the actual value & the quantized value of
the sample is random with S as the step size.
• Then the mean square quantization error, or
quantization noise is : N  S 2 ...(ii )
q
12
I2 S2
• The sampled quantization error N sq  2 ...(a)
Ts 12
where, I is the impulse strength, i.e. area of an
impulse.
• But, the mean square value of the quantized o/p
(S0) = to the mean square value of the quantized
samples ( mk2 ).
i.e.,

 2 2
1  S   3S   5 S 
2
 (2 M  1) S  
2

S0  mk 2
         ...    
M  2   2   2   2  
S2 2 2 2 S 2
4 M 3
 1  3  5  ...  (2 M  1) 2   . ; l arg e M
4M 4M 3
S 2M 2
 ....(iii )
3
S0
 o / p signal power to quantization noise, 
Nq
S 2 M 2 12 S0
 M 2  22 N
. 2  4 M 2 ....(iv) i.e.,
Nq
...(b)
3 S
Where, N is no. of bits used to represent a sample.
• To find the NF, we must find i/p SNR, Si/Ni.

• Let the mean square value of input noise, Ni= n
2

• Assuming ‘0’ by 0-volt level & ‘1’ by A-volt level

with an equal probability for 0 & 1, then the
average signal power, A2 K 2 2
Si   n
( say )
2 2
• Where K is a constant having large value to
reduce probability of error.
Si K 2 n2 K 2
Now, i / p signal to noise ratio    ...(v)
Ni 2 n
2
2
Si / N i K2
 Noisefigure( F )   ...(vi)
S 0 / N 0 8M 2
• The effect of Thermal noise:- The effect of additive
thermal noise is to cause the matched filter detector
to make an occasional error in determining whether a
binary ‘1’ or ‘0’ was transmitted.
• If the thermal noise is white & Gaussian, the
probability of such an error depends on the ratio
Eb/η, where Eb is the bit energy & η/2 is the 2-sided
PSD of the noise.
• The error probability, Pe also depends on the type of
modulation employed, i.e., direct transmission, PSK,
FSK, etc. Typically, PCM systems operate with error
probabilities which are small enough so that we may
ignore.
• Let us assume that a N-bit code word is used to
identify a quantization level in PCM & assuming that
an error may occur with equal likelihood in any bit
of the word, then for N ≥ 2, the o/p power due to the
thermal noise error is:-
fM M 2 S 2 Pe 2 2 N S 2 Pe
N th   Gth ( f )df   ...(c)
 fM 3 3
• O/p SNR in PCM:- Thus, the o/p SNR, including
both quantization & thermal noise is:-
S0 S0 M 2 S 2 / 12
 
N0 N q  N th ( S 2 / 12)  ( Pe M 2 S 2 / 3)
M2 22 N
  ...( d )
1  4 Pe M 2
1  4 Pe 2 2 N
• In PSK (or for direct transmission),
1 Eb 2 z

t 2
Pe  erfc ...(e) ( erf z  e dt & erfc  1  erf z )
2   0

• Assuming N bits occupies the entire interval between

samples, then a bit has a duration Ts/N. If the
received signal power is Si, then the Eb is:-
Ts 1
Eb  Si  Si ...( f )
N 2 fM N

• Then, Eq.(d) will be:-

S0 22 N
 ...( g )
N0 1  2 2 N 1 erfc (1 / 2 N )( Si / f M )
• Typically, for Si/ηfM >> 1 & N = 8, o/p SNR = S0/N0
= 48 dB.
• PCM systems exhibit a threshold. The threshold
point is arbitrarily defined as the Si/ηfM value at
which S0/N0 has fallen 1 dB from the value
corresponding to a large Si/ηfM.
• Experimentally, the onset of threshold in PCM is
marked by an abrupt increase in a crackling noise
analogous to the clicking noise heard below
threshold in analog FM systems.
• Ex-01: Consider a PCM system where maximum
value of the analog sample is to be represented
within 0.05% accuracy, find the maximum no. of
binary digits required ?
• Ans:- As the desired accuracy is within 0.05%,
hence the minimum no. of quantization levels M
will be = 100/0.05= 2000.
Thus, M = 2N ≥ 2000 or N = 11 (Ans).
• Ex-02: A PCM system has a uniform quantizer
followed by a n-bit encoder. Show that mean
square signal to quantization noise ratio is
approximately given by (1.8+6n) dB for a
sinusoidal input ?
• Ans:- Assume that the sinusoidal i/p is varying
between ±A.
The no. of quantization levels = M = 2n
 Then, step size, S = 2A/M & the signal power, Ps
= A2/2.
The quantization noise, Nq = S2/12 = A2/3M2.
Thus, SNR  A2 / 2  3 M 2  3 2 2 n
A 2 / 3M 2 2 2
 3 2n 
 SNR ( dB )  10 log 1 0  2 
2 
3
 10 log1 0    10 log 1 0 ( 2 2 n )
2
 1.76  20n(0.301)
 1.76  6.02n  (1.8  6n) dB ( Ans.)
• NB:- Go through sufficient numericals from
various text books ?
 Noise in Delta Modulation
• Quantization error in DM is Δ(t) = m(t) - mq(t) ≤
±S & the maximum quantization error is ± S.
• If it is assumed that the error Δ(t) takes on all
values between –S & +S with equal likelihood,
then the probability density of Δ(t) is:-
1
p ( )  ;  S     S ...(vii) Then the normalized
2S
power of Δ(t) is:- S S 
2
S 2
[(t )]2   2 p()d   d  ...(viii)
• Hence the S  S 2S 3
output quantization noise power in baseband
frequency range 0 to fm is:-
 2  The 2  sided PSD of (t ) is : 
S fm
Nq  ....(a) G ( f )  S / 3  S ; f  f  f ...(ix )

2 2

3 fb 2f 6fb b
b b

where fb is the bit rate in the DM system.

• O/p signal Power:- Let m(t) = Asinωmt, with ωm=
2πfm , with fm as the upper limit of the baseband
frequency range. Then, the o/p signal power, S0 :-
2 2 2
A S f
S0  m 2 (t )   b
...(b)
2 2m 2

(since, the maximum slope of m(t) is ωm A & the

maximum average slope of DM should be S/Ts =
Sfb, to avoid the slope overload error, i.e., ωm A =
Sfb Sf b A2 S 2 f b2
 A   ...(b)
m 2 2 2
m
• Thus o/p signal to quantization noise ratio in the
worst case is :-
3 3
S0 3  fb  3  fb 
 2      ....(c)
N q 8  f m  80  f m 
3
3  fb 
& the actual value is :    ....(d )
64  f m 
(3/64 corresponds to 1 dB higher than 3/80 values.)
• HT-01: Find the signal-to-quantization noise ratio
of a DM for a sinusoidal signal with a bit rate of
64kbps & i/p signal bandwidth of 4 kHz ?
• Ans:- 21.86 dB & 22.83 dB.
 Thermal noise in DM
• When the thermal noise is present (as shown in
Fig.1), the matched filter in the receiver occasionally
make an error in determining the polarity of the
transmitted waveform.
• The PSD of the thermal noise at the i/p to the
baseband filter is:- 4S 2 Pe
Gth ( f )  ;   0 ...(1)
• Generally, Gth(f) Ts 2

is ∞ at ω →0, hence we should not include f = 0 &

take low-frequency cut-off of baseband filter at f1.
• Thus, the thermal noise o/p will be:-
Fig.-1:-. DM system including the thermal noise
 S 2 Pe   f1 df f m df 
N th  2    
 Ts   fm f
2 f1 f 2

2S Pe2
 1 1  2
2S Pe f b
 2     ( f1  f m ) ...(2)
 Ts  f1 f m   f12

• Thus, the thermal noise o/p in DM depends on the

low-frequency cut-off, f1 than the high-frequency
limit of the baseband range.
• O/p SNR in DM including thermal noise:-
S0 S0 ( S 2 / 8 2 )( f b / f m ) 2
 
N0 N q  N th ( S 2 f m / 3 f b )  (2 S 2 Pe f b /  2 f1 )
0.0375( f b / f m ) 3
 ...(3)
1  0.6 Pe ( f b2 / f m f1 )
• If transmission is direct or by means of PSK,
1 Es 2 z

t 2
Pe  erfc ...(4) ( erf z  e dt & erfc  1  erf z )
2   0

where Es is the signal energy in a bit, relates to the

received signal power Si as follows:- Es = SiTb = Si Ts
= Si /f b …(5)
• Thus,
S0 0.0375( f b / f m ) 3
 ...(6)
N0 1  [0.3( f b2 / f m f1 )]erfc Si / f b

• Comparison between PCM & DM:- DM needs a

simple ckt compared to PCM, but for a good quality
transmission, the BW needed by DM is more than
 PCM. Hence, the use of DM is recommended for the
following 2 situations:-
(i) When the BW conservation is desirable at the cost
of the quality of transmission & (ii) when a simple
circuitry is required & the allowable/available BW is
large.
• When the probability of a bit error is very small, the
PCM system has a higher o/p SNR than the DM
system.
• But, ADM performance is better than PCM. As a result,
many Satellite Business Systems (SBSs) employ ADM
instead of companded PCM to accommodate twice as
many voice channels in a given frequency band.