Intro Telecommunications (5ETA0)

Instruction 4 – Solutions
Problem 4
Solution problem 4
Max. PCM Signal-to-Noise ratio
Maximum value of average output S/N, if Pe negligible: Ch. 3.1_3 - 50

𝑆 𝑺
= 𝑀2 ⇒ in dB: = 𝟔. 𝟎𝟐 × 𝒏 [dB] ( M=2n )
𝑁 out 𝑵 out, dB

→ 6 dB S/N improvement per added PCM word bit

N.B. Valid only if distribution of signal is uniform over interval -V to +V,

i.e. rms value of signal is V /3
𝑆 𝑆
= 3𝑀2 ⇒ in dB: = 6.02 × 𝑛 + 4.77 [dB]
𝑁 pk out 𝑁 pk out, dB

𝑆
In general: = 6.02 × 𝑛 + 𝛼 [dB]
𝑁 out, dB

where 10 log 3 = 4.77, depending on the range used of the ADC
Output Signal-to-Noise Ratio*
Ch. 3.1_3 - 40
Average noise power: 𝑁 = 𝑛𝑘2 = 𝑒𝑏2 + 𝑒𝑞2
quantizing error eq , bit error due to channel noise eb
𝑉 2
1 2 𝑉
Average signal power (uniformly distributed signal): 𝑆 = var 𝑥𝑘 = 𝑥𝑘2 = න 𝑥𝑘 𝑑𝑥𝑘 =
2𝑉 −𝑉 3
where xk, max=V
where 𝑥𝑘,𝑚𝑎𝑥 = 𝑉

Signal-to-noise ratio: 𝑆 𝑀2
where Pe is the bit error rate = 𝑷𝒆 … 𝒃𝒊𝒕 𝒆𝒓𝒓𝒐𝒓 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚
𝑁 out 1 + 4𝑃𝑒 𝑀2 − 1
of the channel

Peak signal power: 𝑆pk = 𝑉 2

Max. signal-to-noise: 𝑆 3𝑀2

=
𝑁 pk out 1 + 4𝑃𝑒 𝑀2 − 1

* see section Ch. 7-7 for derivations

Appendix
Q function:

3.85x10-5
 Problem 5
A Video signal with a bandwidth of 5MHz is transmitted over a baseband channel. To
optimize noise rejection the analog receiver uses an ideal rectangular filter whose is identical
to the analog signal’s bandwidth. During transmission white Gaussian noise is added to the
video signal. The resulting analog signal to noise at the input to the receiver is 20dB.
In an attempt to improve the transmission quality the same video signal is now digitized using
PCM. The signal is sampled, quantized and coded into 400 levels.
In order to allow for the transmission of the PCM coded signal only the bandwidth of the
input filter is changed, while the transmit power remains the same!

a) What is the minimum bandwidth of the input filter of the PCM based system? With what
pulse shape is this accomplished?
b) Calculate the new SNR for the PCM channel and derive the bit error rate of the received
digital signal.
c) Calculate the signal to noise ratio at the output of the PCM receiver after the signal has
been reconstructed.
Solution problem 5
a) Minimum fs is 2 B = 2  5  10 6 = 10 MHz , see Ch.2.7.
400 levels so 9 bits per sample needed ( 28 = 256 2 9 = 512 )
Bit rate is 9 x 10 = 90 Mbit/s
Minimum bandwidth is 45 MHz with sinc pulses, see Ch.3.1.
b) Bandwidth becomes 9 times larger.
S
  = 100 (20 dB)
 N  r ,ana log
S 100
  = = 11.1
 N  r ,digital 9
 S 
( )
Pe = Q    = Q 11.1 = Q(3.33) = 5 10 −4
  N r 
 
(see Appendix)
S M2 400 2
c)   = = −4
= 498  27 dB
 N  out 1 + 4  Pe  ( M − 1) 1 + 4  5  10  (400 − 1)
2 2
 Bandwidth of PCM signal
Ch. 3.1_3 - 32
⚫ Bit rate R (using n bits per sample): 𝑅 = 𝑛𝑓𝑠 [bits/s]
with sample freq. fs  2B - M = 2𝑛

⚫ Dimensionality theorem*: 1 1
(so PCM requires a bandwidth at least n times 𝐵PCM ≥ 𝑅 = 𝑛𝑓𝑠 ≥ 𝑛𝐵
2 2
as large as that of the input analog signal )

⚫ BPCM depends on the line coding. (see next sheet, will be addressed in Week 2)
➢ Polar rectangular NRZ signal: 𝐵PCM = 𝑅 = 𝑛𝑓𝑠
(null bandwidth at f = 1/Tb = R) 1
➢ Unipolar NRZ with sinc pulses: 𝐵PCM = 𝑅 (see 2nd next sheet)
2
⚫ When too narrow bandwidth: pulses are smeared into
neighbouring bit slots, causing intersymbol interference, ISI
Impulse sampling (5/5)
Ch. 2.7 - 27
Choosing the sampling frequency:
𝑓𝑠 ≥ 2𝐵
⚫ spectra do not overlap
⚫ complete restoration

𝑓𝑠 < 2𝐵

⚫ aliasing or folding of spectra

⚫ produces distortion
⚫ prefiltering (anti-aliasing filter) to chop
components above fs/2 is better
Sinc pulse
Ch. 3.1_3 - 34

See sheet Ch. 2.7-7

𝑊(𝑓) = 1 if 𝑓 ≤ 𝑊
= 0 elsewhere
∞ 𝑊
2 𝑊 sin 2 𝜋𝑊𝑡
𝑤(𝑡) = න 𝑊(𝑓)𝑒 𝑗⋅2𝜋𝑓𝑡 d𝑓 = 2 න cos 2 𝜋𝑓𝑡d𝑓 = sin 2 𝜋𝑓𝑡 𝑓=0 = 2𝑊 = 2𝑊sinc 2𝑊𝑡
2𝜋𝑡 2𝜋𝑊𝑡
−∞ 0
1
Time interval between zeros : 𝑇=
2𝑊
1
Symbol rate : 𝑅 = = 2𝑊 → Spectral efficiency R/W = 2 Baud/Hz
𝑇
Appendix
PCM system
(S / N )out = M2
1 + 4 Pe ( M 2 − 1)
met Pe = Q ( (S / N )r )

Power spectrum of a digital signal

F( f )
2

Pulse shape f (t )  F ( f ) ; signal spectrum Ps ( f ) =  R(k )e j 2kfTs

Ts k = −

'Raised-cosine' spectrum
 1, f  f1
 
1   ( f − f1 ) 
H e ( f ) =  1 + cos   , f1  f  B
2
    2 f   
 0, f B

with
f
f 0 = f1 + f  = B − f  ; roll - off factor r =
f0
 Problem 6
A music signal with a spectrum till fx =16 kHz is transmitted using PCM over a baseband
channel with no bandwidth restrictions, but during the transfer white Gaussian noise is added.
The requirement is that at the output of the PCM system, the signal-to-noise ratio (S/N)O
should have a minimum value of 50 dB in case of an uniform distributed input signal of the
ADC.

a) How large, the number of bits per sample n of the PCM system and the bandwidth BT
at the input of the digital receiver at least should be in order to meet the system
requirement?
b) If under these conditions, the exact requirement is met, how large is the signal-to-
noise ratio (S/N)R at the input of the receiver?
c) Now the signal power is increased by 3 dB. How large is the signal-to-noise ratio at
the output of the receiver now, if n is unmodified?
d) What is the maximum achievable (S/N)O with another choice of n and BT?
Assume for part a) that Pe can be neglected!
Solution problem 6
a) n = 9 means maximum (S/N)out is 9 x 6.02 = 54.18 dB (see Ch.3.1)
(with n = 8 the maximum (S/N)out is 8 x 6.02 = 48.16 dB is not enough)
Sample rate is at least 2fx = 2 x 16 = 32 kHz
Bit rate is in that case 32 x 9 is 288 kbits/s
The minimum bandwidth is obtained with Sinc pulses and is BT =144 kHz (Ch.3.1)
S M2 218
b)   = = 10 =
5
 Pe = 1.55  10 −6
 N  O 1 + 4 Pe ( M − 1) 1 + 4  Pe  2
2 18

 S  S S

Pe = Q    = Q( z ) = 1.55  10 −6  z = 4.7 =      = 22.1  13.4 dB
 N   N R  N R
 R

(see Ch.3.1)
S
c) After increasing   with a factor of two:
 N R
S 1
  = 4.7 2 = 6.65  Pe = Q (6.65) = e −( 6.65) / 2 = 1.5  10 −11
2

 N R 6.65 2
S 218
  = −11
= 2.62  10 5  54.18 dB
 N  O 1 + 4  1.5  10  2
18

S S
d) Increase n until   no longer increasing.   decreases if the number of bits per
 N O  N R
sample n increases, because the bandwidth and thus the received noise power increases.
9
The optimum is at n = 12  z = 6.65  = 5.76  Pe = 4.35  10−9
12
(S/N)o = 71 dB.
Max. PCM Signal-to-Noise ratio
Maximum value of average output S/N, if Pe negligible: Ch. 3.1_3 - 50

𝑆 𝑺
= 𝑀2 ⇒ in dB: = 𝟔. 𝟎𝟐 × 𝒏 [dB] ( M=2n )
𝑁 out 𝑵 out, dB

→ 6 dB S/N improvement per added PCM word bit

N.B. Valid only if distribution of signal is uniform over interval -V to +V,

i.e. rms value of signal is V /3
𝑆 𝑆
= 3𝑀2 ⇒ in dB: = 6.02 × 𝑛 + 4.77 [dB]
𝑁 pk out 𝑁 pk out, dB

𝑆
In general: = 6.02 × 𝑛 + 𝛼 [dB]
𝑁 out, dB

where 10 log 3 = 4.77, depending on the range used of the ADC
Appendix
PCM system
(S / N )out = M2
1 + 4 Pe ( M 2 − 1)
met Pe = Q ( (S / N )r )

Power spectrum of a digital signal

F( f )
2

Pulse shape f (t )  F ( f ) ; signal spectrum Ps ( f ) =  R(k )e j 2kfTs

Ts k = −

'Raised-cosine' spectrum
 1, f  f1
 
1   ( f − f1 ) 
H e ( f ) =  1 + cos   , f1  f  B
 2   2 f  
 0, f B

with
f
f 0 = f1 + f  = B − f  ; roll - off factor r =
f0
 Bandwidth of PCM signal
Ch. 3.1_3 - 32
⚫ Bit rate R (using n bits per sample): 𝑅 = 𝑛𝑓𝑠 [bits/s]
with sample freq. fs  2B - M = 2𝑛

⚫ Dimensionality theorem*: 1 1
(so PCM requires a bandwidth at least n times 𝐵PCM ≥ 𝑅 = 𝑛𝑓𝑠 ≥ 𝑛𝐵
2 2
as large as that of the input analog signal )

⚫ BPCM depends on the line coding. (see next sheet, will be addressed in Week 2)
➢ Polar rectangular NRZ signal: 𝐵PCM = 𝑅 = 𝑛𝑓𝑠
(null bandwidth at f = 1/Tb = R) 1
➢ Unipolar NRZ with sinc pulses: 𝐵PCM = 𝑅 (see 2nd next sheet)
2
⚫ When too narrow bandwidth: pulses are smeared into
neighbouring bit slots, causing intersymbol interference, ISI
PCM transmission system
Ch. 3.1 - 22
Up until now… Coming up next…

Figure 3-7 Digital and analog communication systems, Leon Couch, 8th edition
 Illustration of waveforms in a PCM system
Ch. 3.1_3 - 24
a. Quantizer transfer characteristic

b. Flat-top sampling and

Quantization of an Analog Signal

c. Error signal
Peak value of error signal (± 1) is
one half the quantizer step size (=2)

d. PCM word in Gray code

7 7 5 5 3 3 3 3

Figure 3-8 Digital and analog communication systems, Leon Couch, 8th edition
Sinc pulse
Ch. 3.1_3 - 34

See sheet Ch. 2.7-7

1
Time interval between zeros : 𝑇=
2𝑊
1
Symbol rate : 𝑅 = = 2𝑊 → Spectral efficiency R/W = 2 Baud/Hz
𝑇
Important transforms
Ch. 2.7 - 7

Rectangular time pulse

=2Wsinc(2Wt)

Sinc time pulse

2 (bit/s) / Hz

Triangular time pulse

(is equivalent to the convolution of two
identical rectangular pulses)

Fig 2-6
 Spectral efficiencies of line codes
Ch. 3.1_3 - 36

*
* with L=2l levels
Unipolar NRZ with sinc pulses ½R 2

Table 3-6 Digital and analog communication systems, Leon Couch, 8th edition
PCM communication system
Quantized samples Ch. 3.1_3 - 39

S = M2
 
 N  out 1 + 4 Pe (M − 1)
2

 S 
Pe = Q   
  N in 
 
Output Signal-to-Noise Ratio*
Ch. 3.1_3 - 40
Average noise power: 𝑁 = 𝑛𝑘2 = 𝑒𝑏2 + 𝑒𝑞2
quantizing error eq , bit error due to channel noise eb
𝑉 2
1 𝑉
Average signal power (uniformly distributed signal): 𝑆 = var 𝑥𝑘 = 𝑥𝑘2 = න 𝑥𝑘2 𝑑𝑥𝑘 =
2𝑉 −𝑉 3
where xk, max=V
where 𝑥𝑘,𝑚𝑎𝑥 = 𝑉

Signal-to-noise ratio: 𝑆 𝑀2
where Pe is the bit error rate = 𝑷𝒆 … 𝒃𝒊𝒕 𝒆𝒓𝒓𝒐𝒓 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚
𝑁 out 1 + 4𝑃𝑒 𝑀2 − 1
of the channel

Peak signal power: 𝑆pk = 𝑉 2

Max. signal-to-noise: 𝑆 3𝑀2

=
𝑁 pk out 1 + 4𝑃𝑒 𝑀2 − 1

* see section Ch. 7-7 for derivations

The Q function
∞ Ch. 3.1_3 - 44
1 −𝑥 2 /2
𝑄(𝑧) = න 𝑒 𝑑𝑥
2𝜋
𝑧
1 𝑧
= erfc
2 2

Chernoff bound:
1 −𝑧 2 /2
𝑄(𝑧) ≤ 𝑒
𝑧 2𝜋
Q(6)=10-9
Q(7)=10-12 𝑆
𝑃𝑒 = 𝑄
𝑁 𝑖𝑛

Figure B-7 - Digital and analog communication systems, Leon Couch, 8th edition
Problem 7
An analog video signal with a maximum frequency of 2 MHz is sent over a transmission
line connection. The transmit power is 1 Watt. The transmission loss on the connection is
21 dB. The spectral density of the received noise is  / 2 = 2  10 −11 Watt / Hz. The
receiver is noise-free, and the receive filter is an ideal low-pass filter.
a) What is the signal-to-noise ratio at the input of the receiver?
The connection is now digitized, and the video signal is transmitted by means of a
(binary) PCM system. The quantization is uniform, the number of bits per sample is 5
and the transmission system is using ideal sinc pulses for transmission. The transmission
power remains the same (1 Watt), but the bandwidth of the receive filter is adapted, of
course.
b) What is the SNR at the input the PCM receiver?
c) What is at the output of the PCM receiver the signal to noise ratio, SNR, of the
recovered video signal?
You will now have the freedom to adjust the number of bits per sample (and the
transmission bandwidth) to achieve optimal transmission.
d) What is the number of bits per sample for a maximum signal-to-noise ratio at the
output of the PCM receiver and what is the signal-to-noise ratio?
The power density of the video signal is uniform over the whole band, however the
amplitude of the video signal is not uniformly distributed: the signal has a peak-to-
average ratio of 10 dB.
e) How can you use this to increase the quality of the transmission system even
further?
Solution problem 7
𝑁0a) N =  / 2  2B = 8  10 −5 Watt → -40.97 dBW
A loss of 21dB is a factor of 10-2.1 = 7.94 x 10-3.
S = PT x 7.94 x 10-3 = 1 x 7.94 x 10-3 = 7.94 x 10-3 W
Or S = 0 - 21 = -21 dBW → 7.94 x 10-3 W
S/N analog = 7.94 x 10-3/(8 x 10-5) = 99.25 → 19.97 dB

b) BT = nB = 5 x 2 = 10 MHz (see Ch. 3.1)

S/N' = 7.94 x 10-3/(2 x 10-11x 2 x 107) =19.85

( ) ( )
c) Pe = Q S / N ' = Q 19.85 = Q(4.46) = 4.4 10−6 (see Appendix)
M = 2 5 = 32
S M2 32 2
  = = = 1018 → 30.0 dB
 N  out 1 + 4 Pe  (M − 1) 1 + 4  4.4  10  (32 − 1)
2 −6 2
Solution problem 7
d) for n=5:10
n
SNRa=7.94e-3/(2e-11*n*2e6)
Z=(SNRa)^0.5
Pe=exp(-SNRa/2)/(Z*(2*pi)^0.5)
SNRout=2^(2*n)/(1+4*Pe*(2^(2*n)-1))
SNRoutdB=10*log10(SNRout)
end
n= 5 n= 7 n= 9
SNRa = 39.7000 SNRa = 28.3571 SNRa = 22.0556
Z = 6.3008 Z = 5.3251 Z = 4.6963
Pe = 1.5162e-10 Pe = 5.2108e-08 Pe = 1.3799e-06
SNRout = 1.0240e+03 SNRout = 1.6328e+04 SNRout = 1.0713e+05
SNRoutdB = 30.1030 SNRoutdB = 42.1294 SNRoutdB = 50.2992

n= 6 n= 8 n = 10
SNRa = 33.0833 SNRa = 24.8125 SNRa = 19.8500
Z = 5.7518 Z = 4.9812 Z = 4.4553
Pe = 4.5410e-09 Pe = 3.2780e-07 Pe = 4.3818e-06
SNRout = 4.0957e+03 SNRout = 6.0350e+04 SNRout = 5.4110e+04
SNRoutdB = 36.1233 SNRoutdB = 47.8068 SNRoutdB = 47.3327

e) If the amplitudes of the inputs signal to the quantizer are not uniformly
distributed you can choose to use a compander (non-linear quantizer) so that
the mapping of input analog values is not mapped linearly to output values.
 Bandwidth of PCM signal
Ch. 3.1_3 - 32
⚫ Bit rate R (using n bits per sample): 𝑅 = 𝑛𝑓𝑠 [bits/s]
with sample freq. fs  2B - M = 2𝑛

⚫ Dimensionality theorem*: 1 1
(so PCM requires a bandwidth at least n times 𝐵PCM ≥ 𝑅 = 𝑛𝑓𝑠 ≥ 𝑛𝐵
2 2
as large as that of the input analog signal )

⚫ BPCM depends on the line coding. (see next sheet, will be addressed in Week 2)
➢ Polar rectangular NRZ signal: 𝐵PCM = 𝑅 = 𝑛𝑓𝑠
(null bandwidth at f = 1/Tb = R) 1
➢ Unipolar NRZ with sinc pulses: 𝐵PCM = 𝑅 (see 2nd next sheet)
2
⚫ When too narrow bandwidth: pulses are smeared into
neighbouring bit slots, causing intersymbol interference, ISI
Appendix
PCM system
(S / N )out =
M2
1 + 4 Pe ( M 2 − 1)
met Pe = Q ( (S / N )r )

Power spectrum of a digital signal

F( f )
2

Pulse shape f (t )  F ( f ) ; signal spectrum Ps ( f ) =  R(k )e j 2kfTs

Ts k = −

'Raised-cosine' spectrum
 1, f  f1
 
1   ( f − f1 ) 
H e ( f ) =  1 + cos   , f1  f  B
2
    2 f   
 0, f B

with
f
f 0 = f1 + f  = B − f  ; roll - off factor r =
f0
The Q function
∞ Ch. 3.1_3 - 44
1 −𝑥 2 /2
𝑄(𝑧) = න 𝑒 𝑑𝑥
2𝜋
𝑧
1 𝑧
= erfc
2 2

Chernoff bound:
1 −𝑧 2 /2
𝑄(𝑧) ≤ 𝑒
𝑧 2𝜋
Q(6)=10-9
Q(7)=10-12 𝑆
𝑃𝑒 = 𝑄
𝑁 𝑖𝑛

Figure B-7 - Digital and analog communication systems, Leon Couch, 8th edition
Output SNR of 8-bit PCM systems with and without companding
Ch. 3.1_3 - 58

With companding, output SNR

relatively insensitive to input level

Figure 3 - 10
Problem 8
In a digital transmission system an analog music signal is sampled at a frequency of 16 kHz,
the samples are binary coded and transmitted as a PCM signal. The number of levels of the
A/D converter M = 64. The input signal of the ADC is uniform distributed.
a) What is the maximum achievable signal to noise ratio of the regenerated signal at the
output of the receiver?
The transmission channel has an effective bandwidth of 75 kHz, and the received power is 8
mW. By the transmission channel and the receiver, white Gaussian noise with a power
spectral density of 𝑁/20 = 1.4 x 10-8 W/Hz is added.
b) Calculate successively:
The signal to noise ratio at the input of the receiver,
The bit error rate of the detector,
The resulting signal to noise ratio at the output of the detector.
Now, the number of levels of the A/D converter is decreased to 32 without adapting the
signal power and the bandwidth of the transmission system.
c) What is the resulting signal to noise ratio at the output of the receiver?
Now, the number of levels of the A/D converter is increased to 128, without adjusting the
transmission parameters.
d) What is the effect on the signal quality in your opinion? Explain your answer.
Assume for part a) that Pe can be neglected!
Solution problem 8
S
a) The absolute maximum   = 3M 2 = 3  64 2 = 12288  40.89 dB (see Ch.3.1)
 N out,max
In case of a uniform distributed input signal the maximum is
S
  = M 2 = 64 2 = 4096  36.1 dB
 N  out,max

b) S in = 8 mW
N in = 𝑁/02  2 B = 1.4  10 −8  2  75  10 3 = 2.1 mW
S 8  10 −3
  = −3
= 3.81
 N  in 2.1  10
 S 
( )
Pe = Q    = Q 3.81 = Q(1.95) = 3  10 −2 (see Ch.3.1 or Appendix)
 N 
 in 

S M2 4096
  = = = 8.32 (see Ch.3.1 or Appendix)
  out
N 1 + 4 P e (
 M 2
− )
1 1 + 4  3  10 −2
(4096 − 1)

S M2 32 2
c)   = = = 8.27
  out
N 1 + 4 Pe  M (
2
− 1 1)+ 4  3  10 −2
32 (
2
− 1 )
d) 128 levels so number of bits n = 7.
Due to this increase of the number of bits, also the required bandwidth increases. This
required bandwidth would be too large to be used for a transmission channel of 75 kHz.
By increasing the number of bits per sample the output S/N will not increase anyway,
due to the relatively high Pe.
Max. PCM Signal-to-Noise ratio
Maximum value of average output S/N, if Pe negligible: Ch. 3.1_3 - 50

𝑆 𝑺
= 𝑀2 ⇒ in dB: = 𝟔. 𝟎𝟐 × 𝒏 [dB] ( M=2n )
𝑁 out 𝑵 out, dB

→ 6 dB S/N improvement per added PCM word bit

N.B. Valid only if distribution of signal is uniform over interval -V to +V,

i.e. rms value of signal is V /3
𝑆 𝑆
= 3𝑀2 ⇒ in dB: = 6.02 × 𝑛 + 4.77 [dB]
𝑁 pk out 𝑁 pk out, dB

𝑆
In general: = 6.02 × 𝑛 + 𝛼 [dB]
𝑁 out, dB

where 10 log 3 = 4.77, depending on the range used of the ADC
Appendix
PCM system
(S / N )out =
M2
1 + 4 Pe ( M 2 − 1)
met Pe = Q ( (S / N )r )

Power spectrum of a digital signal

F( f )
2

Pulse shape f (t )  F ( f ) ; signal spectrum Ps ( f ) =  R(k )e j 2kfTs

Ts k = −

'Raised-cosine' spectrum
 1, f  f1
 
1   ( f − f1 ) 
H e ( f ) =  1 + cos   , f1  f  B
2
    2 f   
 0, f B

with
f
f 0 = f1 + f  = B − f  ; roll - off factor r =
f0
Appendix
Q function:

3x10-2

1.95