Solution of problem 16

1. 1) Entropy H (S)

Entropy bitrate:

Fixed-length code C₁ (length L₁): we have five symbols to encode, hence: L₁ = 3 bits.

Efficiency:

Bitrate of code C₁:

1. 2) Huffman code C₂.

Average length of the codewords:

Efficiency of code C₂

Bitrate of code C₂:

1. 3) Sequence S B of bits obtained at the output of the Huffman coding.

More bits are observed at zero than at one and, in addition, we have sequences of three consecutive zeros from time to time.

1. 4) Information for baseband signal encoding.
   1. a) Bipolar CODBip encoder of RZ type (see the graph in Table 2.4). The problems encountered in reception are related to the difficulties of getting a correct clock recovery in some cases, here three consecutive zeros, because the encoder produces no pulse for duration 3T.
   2. b) CODHDB coder of HDB-2 type: look at the graph in Table 2.4.

      Sequences of three consecutive zeros are replaced by sequences of type “ 0 0 V” or “B 0V “ and thus, we can have a maximum duration of 2T without impulse.

   3. c) The power spectral density of the RZ bipolar code is given by:

The zeros of I_RZ (f) occur every 1/T_b, so its bandwidth is 1/T_b. It is substantially the same for the HDB-n code (here HDB-2).

Furthermore, in the vicinity of the frequency f = 0, the power is zero, this code can therefore be used for long-distance cable transmission. However, the presence of long sequences of zeros is detrimental to the clock recovery, therefore we use the code HDB – n.

1. 5) Partial response code (PRC).
   1. a) Yes, it is necessary to use a pre-encoder so that on reception, the decoding is instantaneous (without recursion) and therefore, if a decoding error occurs, it does not propagate recursively.
   2. b) Pre-encoder output (symbol ) as a function of its input b_k

      

      
   3. c) Relationship between the encoder output (symbol c_k) and its input a_k:
      
   4. d) Graphical representation of the signal portion associated with SBB transmitted by this whole partial response coder: look at the graph in Table 2.4.
   5. e) The power spectral density of the partial response code concerned is given by (see Volume 1, Chapter 5):
      

The zeros of take place every 1/2T_b, its bandwidth is then approximately 1/2T_b

Thus, there is a reduction of the bandwidth of the transmitted signal by a factor of 2, compared to question 4, because of the introduction of the correlation. In addition, one has no continuous component.

1. 6) We have:
   

   hence, an instantaneous decoding (no recursion). Thus:

   1. – if no decision error on ĉ_k, then, no decision error on
   2. – if decision error on ĉ_k, then decision error on only (not on subsequent ones).

Solutionof problem 17

Let us make the following form of the signal s (t) (with θ < T/2 on Figure 2.3).

Such a signal can be represented by:

With the deterministic function (pulse):

and:

since t₀ is uniform on time interval [0, T[, then s (t) is a second-order stationary random signal.

Moreover, the autocorrelation function R_s (τ) is an even function: R_s (–τ) = R_s (τ) and thus the calculation can be done with suitability with τ ≥ 0 or τ ≤ 0.

The autocorrelation function R_s (τ) is written:

hence:

Moreover, using the theorem of compound probability, R_s(τ) is also written:

To calculate R_s(τ), it is sufficient to calculate Pr{V at t} and Pr{V at t–τ/V at t}

A. First case, where 0 < θ ≤ T/2

There are three situations to consider.

1. – First situation: 0 ≤ τ ≤ θ

The only possibility is that t and t – τ belong to the same first part of time slice (hatched region).

Let:

and:

or, since a_n = 1

that is to say:

Or:

with random t₀, of a posteriori law uniform over a measurement interval θ therefore with probability density (ddp(t₀)) equal to 1/θ

Hence:

So finally:

NOTE.– In general, if a priori t belongs to a time interval, of measurement T (t₀ is random, of uniform law on time interval [0, T[), we know a posteriori that t belongs to a first part of the time interval, of measure θ. It is therefore this posterior uniform law that is used in time conditional probabilities.

1. – Second situation: θ < τ ≤ T

The only possibility is that s(t) and s(t – τ) come from two first parts of adjacent time slots, hence independence between the variables a considered.

In general, we will always have:

and:

turns into:

Or:

Which depends on the value of τ with respect to θ and T

Two situations can occur (see graph in Figure 2.5).

We know that if a real random variable follows a uniform law over a given interval [c, d] , then its probability density is equal to 1/Mes [c, d] and that:

and if Mes[c, d] = 0, then the probability density will be zero because we are faced here with a continuous random variable

if: θ < τ ≤ T, then:

So if θ < τ ≤ T – θ, the measure of the interval is zero, thus q = 0

Otherwise, if T – θ < τ ≤ T, then the measure of the interval is given by:

τ + θ – T, and:

So if:

and if:

1. – Third situation: τ > T

In summary, for 0 < θ ≤ T/2, if:

For values τ > T + θ, we see that the intervals intervening in the conditional probability relating to instants (t – τ) and t are different and therefore n – k < n. This implies that conditional probability Pr{a_n–k/a_n}_k≠0 = Pr{a_n–k}, and that the probability relating (t – τ) conditionally to t will evolve periodically, from T to T strictly as we have just described it. More precisely, we have:

1. – for: θ + kT ≤ τ < (k + 1) T – θ, then q = 0
2. – for: (k + 1) T – θ ≤ τ < (k + 1) T, then:
   
3. – for:(k+1) T leq τ < (k+1) T+ θ , then
   
4. – for: (k + 1) T + θ ≤ τ < (k + 2) T – θ, then q = 0

It is therefore a periodic function of period T

Note that the autocorrelation function breaks down into a sum of two functions:

with R₁ (τ) , a non-periodic function, and R₂ (τ), a periodic function of period T

Calculation of the power spectral density Γ_s(f)

Calculation of the power spectral density Γ₁(f)

R₁ (τ) is an even symmetric function, hence:

but:

and:

Integration by part, with:

u = τ and v = cos(2πfτ)dτ

hence:

which is a classic result. Indeed, recall that if:

then:

Γ₁ (f) has a continuous spectrum.

Calculation of Γ₂ (f)

The basic form of R₂ (τ) is identical to that of R₁ (τ) and in addition, it is periodic with period T, then:

It is a discrete spectrum, the discrete spectral components being spaced 1/T apart from each other.

Special case where θ = T/2: binary RZ code.

In this case, we then have:

and:

The continuous component is such that:

Γ₂(0) = [continuous component]² × δ(f)

One has:

Thus, the continuous component is then equal to V/4

The function sin(kπ/2) is non zero for odd integer k, so by setting k = 2n + 1 the expression of Γ₂ (f) is then written:

This is also written:

Γ₂ (f) has discrete components at odd frequencies and in particular at the frequency f = 1/T which makes it possible to recover the clock.

B. Second case, the one where T/2 < θ ≤ T

There are also three situations to consider.

1. – First situation: 0 < τ ≤ θ

Calculation of p and q. There are two cases:

1. a) τ + θ ≤ T → t and t – τ∈ the same first part of time slice:
   
2. b) τ + θ >T, there are two possibilities:

First possibility: t and t – τ do not belong to the same first part of the time slice, but to the first parts of adjacent time slices:

Or:

Second possibility: t and t – τ belong to the same first part of the time slice, hence:

NOTE.– Both hypotheses a) and b) exclude each other.

Thus, for:

Note that in the case a), the expression of R_s (τ) is also valid throughout the time interval τ ∈ [0, θ]

1. – Second situation: θ ≤ τ < T

The only possibility is that s (t) = V and s (t – τ) = V come from the first two parts of adjacent time slices.

Let’s set:

and:

hence:

1. – Third situation: T ≤ τ < T + θ.

Calculation of p and q. There are two cases:

1. a) T < τ ≤ 2T – θ → t and t – τ ∈ to the first two parts of adjacent time slices:
   

   Following the same approach as in the previous situation, we obtain:

   
2. b) 2T – θ < τ < T + θ , there are two possibilities:

NOTE.– Both hypotheses a) and b) exclude each other.

Thus, for T ≤ τ < T + θ:

Note that in case a), the expression of R_s (τ) is also valid throughout the time interval [T ≤ τ ≤ T + θ]

In summary, for T/2 < θ ≤ T:

1. – for 0 < τ ≤ θ
   
2. – for θ ≤ τ < T:
   
3. – for T ≤ τ ≤ T+ θ:

As in the case 0 < θ ≤ T/2, the probability relating to (t – τ) conditionally at t will evolve periodically from T to T, strictly as we have just described it above.

In the same way as for 0 < θ ≤ T/2, the autocorrelation function breaks down into a sum of two functions:

with R₁(τ) , a non-periodic function, and R₂(τ), a periodic function, of period T. In addition, we note that the expression of R₁(τ) and R₂(τ) are the same as previously (case 0 < θ ≤ T/2).

Calculation of the power spectral density, case where: T/2 < θ ≤ T

The power spectral density of the signal s (t) can be broken down into the sum of two functions:

with:

This is a continuous spectrum.

And:

This is a discrete spectrum.

Special case: θ = T (NRZ code)

In this case, we obtain:

So:

Thus, finally we get:

Notice that for θ = T (NRZ code), the signal does not have a discrete spectrum component at 1/T (Γ₂ (f) is zero for the other values of k ≠ 0). Therefore, clock recovery is not easy with NRZ code.

NOTE.– One could easily generalize the problem to the situation where the symbols of information to be transmitted are not equiprobable: Pr {a = 0} = λ and: Pr{a = 1} =1 – λ

Solutionof problem 18

The signal s (t) can be represented by:

where:

is a deterministic function, and:

The autocorrelation R_s(τ) is written:

Using the theorem of compound probabilities, R_s(τ) is written as:

Calculation of the simple probabilities Pr{V at t} and Pr{–V at t}

since there is independence between t₀ and the symbols of information, we have:

Likewise:

and:

Besides:

And yet:

as we have two mutually exclusive hypotheses, hence:

thus:

and:

Let’s now calculate the other probabilities, representing four mutually exclusive hypotheses. Note that for reasons of symmetry, we have:

and:

1. – First case: 0 ≤ τ ≤ T/2
   1. 1) Let:

The only possibility is that t and t – τ belong to the same first half of the time slice:

We have:

therefore:

1. 2) Let:

So, we have: q₂ = 0

So, we have also:

and:

1. – Second case: T/2 < τ ≤ T

The only possibility is that t and t – τ∈ first halves of adjacent time slices.

Since two successive bits to 1 are issued in alternating polarity, then:

yet:

(due to the independence between binary information symbols).

And on the other hand:

hence:

and, by symmetry, we have:

1. – Third case: T < τ ≤ T + T/2

The only possibility is that t and t – τ belong to the first two halves of adjacent time slices.

Since two successive bits at “1” are issued in alternating polarity, then as explained previously:

hence:

We have:

and, on the other hand:

hence:

and, by symmetry, we have:

1. – Fourth case: 3T/2 < τ ≤ 2T.

The only possibility is that (t – τ) and t belong to two first time slices separated by a time slot of intermediate duration T:

Moreover, since we have the conditional event: {V at (t – τ)/V at t} , we must also have in the intermediate time slot a negative impulse, and therefore a symbol a_n–1 = –1

Therefore:

Because of the statistical independence between the pairs of instants considered and the values of the symbols considered, we have:

The first conditional probability on the instants gives:

The second conditional probability on the symbols gives (because of the independence between the symbols):

and due to the independence between the symbols b:

hence:

and, by symmetry, we have: q₃ = q₁ and therefore:

As before, the only possibility is that t – τ and t each belong to two first time slices separated by a time slice of intermediate duration T but, unlike in the previous case, in the intermediate time slice, it must not have transmitted a pulse, therefore a symbol: a_n–1 = 0 . The conditional probability for the instants remains the same. It is the same for conditional probabilities on symbols a. As a result, we have:

and therefore:

And by symmetry, we also have q₂ = q₄ and therefore:

Consequently, we finally get:

Actually, we show that for:

Indeed, let’s take for example the case: 2T ≤ τ < 5T/2.

From the study of the previous case, we see that:

Likewise:

So in the general case where: kT < τ ≤ (k + 1/2)T and k ≥ 2, we have: R_s(τ) = 0

Indeed, (for clarity, see the case: 3T/2 < τ ≤ 2T), we have:

That is:

Because of the independence between t₀ and the information symbols a, we therefore have:

or:

with:

and:

This last conditional probability implies implicitly the fact that between n – k and n, we have an odd number of bits b_n–k′ at 1 (0 < k′ < k)

Thus:

Because of the independence between the symbols, then one obtains successively:

because it is obvious that on a given bit length k, the probability of having an even number of bits at 1 is identical to having an odd number. Indeed, for each given configuration of k – 1 bits, there are two configurations of the additional bit (whatever its position in the k bits): one with 0, the other with 1.

So we have: .

By symmetry, we also have:

In addition, we also have:

i.e.: q₄ = q₁₁ × q₄₂

with:

For the same reasons as previously with the calculation of q₁₃, it is easy to show that:

and with the independence of the information symbols:

By symmetry, we also have:

because:

Thus, we obtain:

In summary:

for:

The autocorrelation function R_s (τ) of the bipolar RZ code represented in the Figure 2.19 a shows that it can be decomposed into the sum of two functions R₁ (τ) and R₂ (τ) represented in the Figure 2.19b

The power spectral density Γ_s (f) of the bipolar code is therefore given by:

with:

–Γ₁ (f) , power spectral density of the signal whose autocorrelation function is R₁ (τ) of triangular shape. It is given by:

–Γ₂ (f) , power spectral density of the signal whose autocorrelation function is R₂ (τ) of trapezoidal shape. And if g (τ) has a trapezoidal form (see Figure 2.20)

It is obtained by convolution between two even rectangular functions that do not have the same support in general. Hence:

Let’s apply this to R₂ (τ) , with:

Thus, we have successively:

Hence:

Finally, it gives:

Properties of Γ_s (f)

– no continuous component;

– no energy at frequency f = 1/T, however a double alternation rectification of the bipolar RZ code gives a RZ code which has a discrete component at frequency f = 1/T in its power spectral density and therefore, a rather easy clock recovery;

– more than 90% of the energy is located in the physical frequency band [0, 1/T]

Solution of problem 19

1. 1) It makes a spectrum shaping. This is performed by the introduction of a certain correlation. The effect is a reduction of the frequency bandwidth of the signal transmitted.
2. 2) A precoder is associated with the encoder to ensure that the decoding of the symbols b_n is instantaneous. The precoder is not necessary in reception because the decoding is instantaneous.
3. 3) since we have:

then:

1. 4) Time sequence.

1. 5) From the previous relation, we have:
   
2. 6) We have:

The chronogram of the signal s_e (t) is shown in Table 2.5 above (answer to question 4).

In this sequence, we have a continuous component, since there are five +V pulses and three –V pulses. This continuous component has the value:

Value 2 in the denominator comes from the fact that the signal x (t) is of type RZ

On a statistical average: E { s_e (t) } = 0 . (i.e. no continuous component)

1. 7) since {b_n} are equiprobable, then:
   
2. 8) The power spectral density Γ_se (f) is given by:

hence:

since:

Furthermore:

hence:

finally, we have:

This code is well suited to long distance cable transmissions because:

• – it has no continuous component, and no power spectral density at very low frequencies;
• – from a practical point of view, most of its power is distributed in the frequency band [0, 1/2T_b].

1. 9) High density pulse code: HDB-3 (s_HDB-3(t) signal).

With V: polarity alternation violation bit (bit of violation); B: stuffing bit.

Solution of problem 20

Chronograms of the different signals:

1. 1) Look at chronograms of the different signals.

The power spectral density Γ₁ (f) of the RZ code is:

This code is not interesting for transmission over long distance cable because:

• – it has a continuous component equal to V/4 and a high power spectral density at low frequencies;
• – the spectral occupancy of the code is practically 2/T_b, twice that of the NRZ code.

However, the presence of a discrete component at the frequency 1/T_b in the power spectral density facilitates the recovery of the clock rate in reception.

1. 2) See the graph of the temporal representations of the different signals in Table 2.7. The power spectral density Γ₂ (f) of the bipolar code RZ is:

Benefits of this code:

• – it has no continuous component, and no power spectral density at very low frequencies;
• – its spectral occupancy is only 1/T_b.

Disadvantage of this code: it does not produce pulses to encode a sequence of consecutive 0, therefore the receiver may lose synchronization.

1. 3) See the graph of the temporal representations of the different signals in Table 2.7. The HDB code has the same advantages as the bipolar code RZ, but in addition we always have pulses even if a long sequence of 0 is presented. So, it has a good match for the transmission of binary information over long distance cable.
2. 4) We have:

The precoding makes it possible to perform in reception (after transmission) an instantaneous decoding.

1. 5) We have:
   
2. 6) and 7) See the graph of the temporal representations of the different signals in Table 2.7.
3. 8) The power spectral density Γ₄ (f) is:

Since:

Furthermore:

hence:

This code is well suited for long distance cable transmission because:

• – it has no continuous component, and no power spectral density at very low frequencies;
• – its spectral occupancy is only 1/2T_b.

1. 9) We have:

1. 10) If error on ĉ_n, we have:
   

   Thus, decision error on b_n if e_n is odd (e_n = 1) and no decision error on b_n if e_n is even (e_n = 0).

2. 11) We have:

Independence:

For the reasoning that follows, see the graph of the temporal representations of the different signals:

Solution of problem 21

1. 1) The power spectral density is:

The noise power b₀ (t) is given by:

The power of the noise b₁ (t) is given by:

1. 2) The signal transmitted s_e (t) is:

The signal received s_r (t) is:

with:

The signal y (t) is the response of the channel to the basic pulse (rectangular shape) x (t) , of duration θ, in the noiseless case.

The equalized signal (corrected) s_c (t) is:

with:

The noise b₁ (t) is the result of noise b₀ (t) filtering by the equalizer whose impulse response is g_c (t).

At sampling times t_k = kT, the signal s_c (t) is written:

The term a_kp (0) represents the useful response of the system (channel + equalization) to the transmission of the symbol a_k associated with the time interval kT.

The term is the intersymbol interference. It is a disturbing signal depending on all of the symbols transmitted {a_n}, except for the symbol a_k which is related to the time interval considered.

The term b₁ (kT) is the noise at the time of decision.

1. 3) The messages of only the form m_l = [a_k–1, a_k+1] interferes with symbol a_k thus:

I_ml (kT) depends on p [(k – n) T] , here on p (± T) . We must first calculate p (t) from P (f).

By definition, we have: p (t) = F^–1 {P (f)} , hence:

This gives:

and:

ml = [ak–1,+ ak+1 ]			pml
-1	-1	V/3
-1	1	0
1	-1	0
1	1	–V/3

1. 4) The two expressions of conditional probabilities of error are:
   
2. 5) To calculate these conditional probabilities for each message m_l, it is necessary to express the integration domains as a function of σ_b1. As:

Furthermore:

hence the following Table 2.9.

1. 6) The average probability of error P_e,1 is given by:
   
2. 7) As there is no intersymbol interference, this means that the first Nyquist frequency criterion is verified (α = 0) . So, from the previous results we get:

The two simplified expressions of conditional probabilities of error are now:

The two simplified expressions of conditional probabilities of error are now:

As we have the same signal-to-noise ratio as before, it means that:

And we can keep (as the approximation remains rather good) the value of the optimal threshold μ₀ as a function of the noise power :

Thus, in this case, the previous Table 2.9 is replaced by Table 2.10:

Finally, we get:

Thus, in this case (with the same signal-to-noise ratio), the probability of transmission error without intersymbol interference is approximately 10 times lower than it was in the presence of intersymbol interference.

Solution of problem 22

1. 1) The calculation of energy bandwidth Δf_b of noise b₀ (t) is made from the expression of the power :

On one hand:

On the other hand, we have to calculate from the expression of Γ_b0 (f) , which is the result of filtering Γ₀ by the first-order RC low pass filter:

Calculation of the transfer function of a 1 st order low pass R – C filter:

For:

and:

then:

Recall:

Finally, we get:

1. 2) Power spectral density Γ_b1 (f) and power of noise b₁ (t).

We have:

and:

Since the support of G_c is included in [– Δf_b, Δf_b] then:

hence:

1. 3) The equalized signal (corrected) s_c (t) is:

with:

The noise b₁ (t) is the result of filtering b₀ (t) by the equalizer filter whose impulse response is g_c (t)

At the sampling times t_k = kT, the signal s_c (t) is written:

The term a_kp (0) represents the response of the system (channel + equalization) to the transmission of the symbol a_k associated with the time interval kT.

The term is the intersymbol interference.

It is a disturbing signal depending on all the transmitted symbols {a_n } , except for the symbol a_k which is related to the time interval considered.

The term b₁ (kT) is the noise at the output of the equalizer at the decision instant.

1. 4) Only messages in the form m_l = [a_k–1 , a_k+1] interfere with symbol a_k, so:

I_ml (kT) depends on p [(k – n) T] , here on p (± T) . We have to calculate p (t) from its Fourier transform P (f) which is defined (see Figure 2.33).

We have: p (t) = F^–1 {P (f)} . This gives:

Finally, we get:

So:

This must be true for α non null integer.

1. 5) Conditional probabilities of error:

Relation between V and σ_b1. We have

Since μ₀ = 0 (equiprobable symbols) and I_ml (kT) =0, then:

Knowing that:

and:

Finally we then get:

Solution of problem 23

1. 1) The transmitted signal s_e(t) is:

The received signal s_r(t) is:

with: y(t) = x(t)⊗h(t).

The signal y(t) is the output of the channel when its input is the basic impulse (rectangular shape) x(t) of period T and without noise.

The equalized (corrected) signal s_c(t) is:

with: p(t) = y(t)⊗g_c(t) , that is: p(t) = x(t)⊗h(t)⊗g_c(t).

The noise b₁(t) is the result of filtering the noise b₀(t) by the equalizer whose impulse response is g_c(t).

At the sampling times t_k = kT, the signal s_c(t) is written:

The term a_kp (0) represents the useful response of the system (channel + equalization) to the transmission of the symbol a_k associated with the time interval kT.

The term is the intersymbol interference.

It is a disturbing signal which depends on all the symbol {a_n} transmitted, except for the symbol a_k which is related to the time interval considered.

The term b₁(kT) is the noise at the decision instant.

1. 2) The intersymbol interference I_ml(kT) is given by:

It depends on p [(k – n) T] , so we have to calculate p(t) from P(f)

By definition, we have: p(t) = F^–1 {P(f)}, hence:

From these values, we can conclude that:

1. 3) Possible values of I_ml(kT).

In the case where the messages only of the form m_l = [a_k–1 , a_k+1] interfere with the symbol a_k, we have (with π ≅ 3):

Thus, there are seven distinct values of I_ml (kT).

1. 4) Neglecting the noise, after equalization the signal s_c(kT) is written:

The decision thresholds are given by:

It can easily be seen that to change the decision class, it is sufficient that

• – an erroneous decision on the transmitted symbol a_k = 3 is made for all the values of I_ml(kT) < – V, that is in 6/16 of cases;
• – similarly, an erroneous decision on the transmitted symbol a_k = –3 is made for all the values of I_ml(kT) > V, which is also in 6/16 cases;
• – An erroneous decision on the symbols transmitted a_k = ±1 is made for all the values of |I_ml(kT) | > V, that is in 12/16 of the cases.

In view of these results, even without noise, the probability of error is extremely high.

1. 5) Null value of the intersymbol interference:

   Under these new conditions, we have:

   
2. 6) Calculation of the 16 following conditional probabilities:

Let us first express V as a function of and calculate the new values of the decision thresholds μ_–1 , μ₀, μ₁:

At the output of the equalizer, the signal is then written:

and the threshold values of the decision blocks are:

The calculation of the conditional error probabilities is based on the knowledge of the noise intervals given by the course formulas (see the relations in Chapter 6 of Volume 1). For each value of the symbol a_k transmitted, we have four possible decisions (three erroneous, and a correct one) on the estimated value ȃ_k of the symbol a_k.

Recall that for the intermediate values of the symbol a_k transmitted, the conditional error probability is given by:

and for the extreme values of the transmitted symbol a_k, the two conditional error probabilities are:

The conditional probability matrix is then obtained like this:

• – for the symbol transmitted a_k = –1 → i = –1 → P_e–1, and the four decisions in reception are:
  
  
• – for the symbol transmitted a_k = 1 → i = 0 → P_e1, and the four decisions in reception are:
  
• – for the symbol transmitted a_k = 3 → i = 1 → P_e3, and the four decisions in reception are:
  
• – for the symbol transmitted a_k = –3 → i = –2 → P_e-3, and the four decisions in reception are:

Furthermore, we have:

Hence see the matrix of conditional decision probabilities in Table 2.14 below.

Solution of problem 24

1. 1) Major reasons:
   • – integration of services, therefore lower costs;
   • – performance in terms of error / distortion not cumulable, because the regeneration of signals can be exactly performed.
2. 2) The frequency bandwidth of the standard telephone channel is:

The maximum possible symbol rate (according to Nyquist criterion) is:

The maximum possible bitrate is:

If coding on two levels, then:

NOTE.– Usually, one uses a M-ary coding system with M ⪢ 2 where the number M is a function of the signal-to-noise ratio at the input of the decision block which ensures a probability of a wrong decision lower than a given level of admissible errors.

1. 3) Rule of transformation of a binary symbol into a ternary symbol (bipolar code).

   The bipolar code is a three-level code such as:

   
2. 4) Generation of the sequence {a_n} (bipolar code).

1. 5) The interests in using a bipolar code in baseband transmission are:
   • – no continuous component;
   • – the spectrum of the transmitted signal vanishes at all the multiples of 1/T_b and limitation of the spectral occupation.
2. 6) Bipolar code limitation:

If we have a long series of bits at zero, then there are no pulses issued over a period that can be significant. This causes the loss of clock synchronization at the receiver side. To overcome this drawback, the bipolar code with a high pulse density is used.

If on a period of 4T_b, there are no pulses, we have to use the HDB– 3 code.

1. 7) Block diagram of the RZ bipolar encoder and decoder.

1. 8) The ultimate goal of equalization is to cancel the influence of the transmission channel in order to have an ISI as small as possible (and even null).

The Nyquist frequency criterion states that the equalization must ensure that we have:

1. 9) The noise power b₁(t) is:

The power spectral density b₁(t) is given by:

The noise power b₀(t) is:

hence:

and:

1. 10) Number and value(s) of the decision threshold(s) of the bipolar code: two decision thresholds, denoted: μ⁺ = – μ^- = p (0)/2
2. 11) Expression of the signal p(t) obtained by the inverse Fourier transform of the amplitude spectrum P(f).

1. 12) Expression of the received and corrected signal:

with:

hence:

1. 13) For each possible value of a_k, determination of possible interfering messages and their conditional probabilities: p_(ml/i) = Pr{ m_l/a_k = i }.

The summary of all these values is given in Table 2.26.

1. 14) Expression of the probability p_ml:

with:

and we actually have:

1. 15) Possible values of intersymbol interference and associated probabilities.

And we have:

1. 16) Expression of the probability of error on the symbols a:
   
2. 17) Expression of each of the four conditional probabilities of possible error.

The corrected and sampled signal is:

The decision thresholds are such that:

The noise b₁(kT) is written:

If:

Then, we have:

• – Case of transmission a_k = 0.

If we decide:

If we decide:

• – Case of transmission a_k = –1.

  If we decide:

  
• – Case of transmission a_k = 1.

If we decide:

1. 18) For each message m_l, calculation of intervals of noise dynamics and values of the conditional probabilities of error, are given in Table 2.29.

The noise amplitude intervals should be expressed as a function of σ_b1. We have:

1. 19) Calculation of the probability of error P_e,a :
   
2. 20) We keep only the most unfavorable case of intersymbol interference according to the table giving the values of the conditional probabilities for each message:

hence:

1. 21) Expression and value of the probability of error P_e,b:

therefore: P_e,b = P_e1,a.

This result was predictable with the simplification of the statement because:

But these errors on a_k do not introduce errors on b_k, hence: P_e,b = P_e1,a.

1. 22) Probability of error P_e2,a in the absence of ISI. We have:

hence:

We have, from the normalized Gaussian law table:

Solution of problem 25

1. 1) Transfer function P(z) of the precoder and equation giving as a function of b_k:
   
2. 2) Equation of the coder giving c_k as a function of a_k:
   
3. 3) Equation allowing the estimation of the emitted symbols from the received symbols c_k : ĉ_k.

With precoding, the transcoder provides:

hence:

Thus, we get a direct estimation of the emitted sequence {b_k} from the received sequence {ĉ_k}.

Without precoding:

This leads to a propagation of decision errors. Indeed, if b_k–1 is badly decoded, it will also be the case for b_k.

1. 4) Block diagram of the precoder, transcoder and duobinary coder.

1. 5) Chronograms of duobinary coding and decoding.

1. 6) Expression of the signal s_c(kT + t₀):
   
2. 7) Duobinary partial response transmission and reception system: expression of s_c(kT + t₀).

   With p(T + t₀) = p(t₀) = V, the intersymbol interference is now given by:

   
3. 8) New expression of s_c(kT + t₀):
   
4. 9) Conditional probabilities of error:

From the result obtained in response 8, we have:

The decision thresholds are located in the middle of two adjacent levels obtained without noise:

Let’s express the decision thresholds according to σ_b1:

We have three possible decision values of symbol ĉ_k one without errors and two with errors.

Transmission of c_k = 2, we decide on reception:

Transmission of c_k = 0, we decide on reception:

Transmission of c_k =–2, we decide on reception:

1. 10) Calculation of the total probability of error:

The total probability of error is then given by:

The binary symbols b_k are independent and identically distributed on the alphabet {0,1}, hence:

The probabilities of transmitting the symbols c_k are respectively:

Hence, the total probability of error:

Calculation of integrals:

Solution of problem 26

1. 1) Transfer function of the encoding filter:
   
2. 2) Transfer function of the precoder:
   
3. 3) Equation defining the encoder:
   
4. 4) Equation giving the estimate of b_k.

With precoding, the transcoder gives: hence:

So a direct estimate of the sequence {b_k} issued from the sequence {c_k} received.

Without precoding:

This leads to a propagation of decision errors. Indeed, if b_k-3 is badly decoded, it will be also the same case for b_k.

1. 5) Block diagram of the combined precoder, transcoder and encoder:

1. 6) Temporal sequences:

1. 7) Expression of the signal s_c(kT):
   
2. 8) In this case, the intersymbol interference is:
   
3. 9) In this particular case, we have:
   
4. 10) The different conditional probabilities of errors are as follows:

1. a) Calculation of conditional probabilities of errors. According to the result obtained in question 9, we have:

The decision thresholds [c, d [ are such that:

Let’s express the decision thresholds according to σ_b1:

1. b) Application of the decision rules. We have three possible decision values on c_k, one without errors and two with errors.
2. c) Transmission of symbol c_k = 2, we decide on reception:
   
3. d) Transmission of symbol c_k=0, we decide on reception:
   
4. e) Transmission of symbol c_k=–2, we decide on reception:
   

Calculation of integrals:

1. 11) The total probability of error is then:

The symbols b_k are independent and identically distributed on the alphabet {0,1}, hence:

The probabilities of emission of the symbols c_k are:

Hence finally: