Optimum Receiver for Binary Data Transmission

P5.1 First consider Z _Tb

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an orthonormal set.

What if the minus sign is put elsewhere, i.e.,

· cos θ − sin θ Are {φ₁(t), φ₂(t)} still orthonormal? If so does the matrix represent just a rotation?

(b) Need to find θ such thatR_Tb

Of course θ is determined by the signal set. You must know the signal set in order to rotate the original basis set as desired: ˆφ₂(t) is perpendicular to the line joining s₁(t) and s₂(t).

The above result of θ can also be obtained geometrically (see the signal space diagram on Fig. 5.1). Use the following fact from geometry: Given the equation of a straight line y = mx + b, then the perpendicular to the line has a slope of −_m¹.

The slope of the line joining s₁(t) and s₂(t) is ^s_s²²₂₁^−s_−s¹²₁₁. Therefore the slope of ˆφ₁(t) is tan θ = −^s_s²¹₂₂^−s_−s¹¹₁₂, or θ = tan^{−1 s}_s²¹₁₂^−s_−s¹¹₂₂.

Since we choose θ such that

Tb

Z

0

φˆ₁(t) [s₁(t) − s₂(t)] dt = 0, (5.7)

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Figure 5.1: Determining the rotation angle θ.

It follows that

Tb T < µ. Of course the argument of Q(x) is now negative and one would use the relationship Q(−x) = 1 − Q(x).

Stated in English, the area is a Q function whose argument is the distance from the mean to the threshold divided by the RMS value (standard deviation).

Technology Adoption Process

"The Chasm"

Laggards Late

Majority Early

Majority Early

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Figure 5.2: Writing the shaded area in terms of the Q function.

Of course one must pay attention on which side of the mean the threshold lies and write the expression accordingly. For instance in terms of the Q function what is the shaded area in Fig. 5.2?

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P5.4 (a) Since s_i(t) = s_i1φ₁(t) + s_i2φ₂(t), i = 1, 2, one has:

E_i = Z _Tb

0

s²_i(t)dt = Z _Tb

0

[s_i1φ₁(t) + s_i2φ₂(t)]²dt

= Z _Tb

0

[s²_i1φ²₁(t) + 2s_i1s_i2φ₁(t)φ₂(t) + s²_i2φ²₂(t)]dt

= s²_i1 Z _Tb

0

φ²₁(t)dt

| {z }

=1

+2s_i1s_i2 Z _Tb

0

φ₁(t)φ₂(t)dt

| {z }

=0

+s²_i2 Z _Tb

0

φ²₂(t)dt

| {z }

=1

= s²_i1+ s²_i2 (b) Now let

φ₁(t) = (q2

Tbcos(2πf_ct), 0 ≤ t ≤ T_b

0, otherwise (5.9)

and

φ₂(t) = (q2

Tbsin(2πf_ct), 0 ≤ t ≤ T_b

0, otherwise (5.10)

Compute Z _Tb

0

φ₁(t)φ₂(t) = 2 T_b

Z _Tb

0

cos(2πf_ct) sin(2πf_ct)dt

= 1

T_b Z _Tb

0

sin(4πf_ct)dt = − 1

4T_bπf_ccos(4πf_ct)

¯¯

¯¯

Tb

0

= − 1

4T_bπf_c[cos(4πf_cT_b) − 1] (5.11) Thus, φ₁(t) and φ₂(t) are orthogonal if cos(4πf_cT_b) = 1 ⇒ 4πf_cT_b = 2kπ ⇒ f_c= _2T^k

b, where k is a positive integer (k = 1, 2, . . .). There must be an integer number of half cycles of the cos and sin in the interval [0, T_b] for them to be orthogonal over the interval of T_b seconds. No other frequency suffices.

Finally, (f_c)_min= _2T¹

b when k = 1. It is graphically illustrated in Fig. 5.3.

0 T_b t

sin

cos

Figure 5.3: Orthogonality of sin and cos with a minimum frequency.

P5.5 Consider the following two signals s₁(t) and s₂(t) (this is the same signal set considered in Example 5.5):

s₁(t) = V cos(2πf_ct)

s₂(t) = V cos(2πf_ct + θ), 0 ≤ t ≤ T_b, f_c= k

2T_b, k integer (5.12)

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(a) The energies of two signals can be computed as:

E₁ =

cos(4πf_ct)dt

¸

where we have used the fact that cos(4πf_ct+β) is a periodic function with a fundamental period of T_b/k (regardless of the value of the phase β) and an integration over multiple periods of this function equals to zero.

The two signals s₁(t) and s₂(t) have unit energy if E₁ = E₂ = ^V2₂^Tb = 1. Therefore, V =

q2 Tb.

Remark: In communications we invariably take f_c to be an integer multiple of _T¹

b typ-ically, sometimes _2T¹

b so the results for E₁ and E₂ hold (and those in P5.4). Typically f_c is very large, on the order of 10⁶ (MHz) or 10⁹ (GHz) so that even if there aren’t an integer number of cycles (or half cycles) in the time interval, for engineering purposes E₁ and E₂ still are closely approximated by V²T_b/2.

(b) Since E₁= E₂= E, the correlation coefficient of the two signals is

ρ = 1

Two signals are orthogonal when the correlation coefficient ρ = 0, which is equivalent to cos θ = 0. So, θ = k^π₂, k = ±1, ±2, ±3, . . . results in −φ₂(t) of P5.4b. The signal space of the signal set is shown in Fig. 5.12 of the text and it is reproduced in Fig. 5.5.

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Figure 5.5: Signal space representation of Problem 5.5.

P5.6 (a)

cos(2π(2f_c)t)dt

| {z }

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Plot of ρ versus the normalized frequency ∆f T_b is shown in Fig. 5.6.

0 ∆fT^b

ρ

1 1

or (minimum frequency separation

2 2

for orthogonality)

b

b

fT f

∆ = ∆ = T

minimum point maximum distance point between the 2 points

ρ ^⇒

ρmin

Figure 5.6: Plotting of ρ(∆f ).

(b) One can differentiate ρ with respect to the (normalized) variable ∆f T_b to find where the minimum occurs. From the plot of ρ we expect it to be between 0.5 and 0.75. Formally let x ≡ 2π∆f T_b. Then ρ(x) = ^{sin x}_x and ^dρ(x)_dx = ^{cos x}_x −^{sin x}_x2 = 0 (for maximum/minimum points) or x = tan x. Solve this numerically.

Another approach (still numerical) is to set up a table of values of ^{sin x}_x between x = 2π × 0.65 and 2π × 0.75 in increments of say 0.01 and pick out value of x where the minimum occurs.

The simplest approach is to trust the answer and check it by seeing if x= tan x at x = 2π×^? 0.715 (within acceptable engineering numerical accuracy), i.e., 2π×0.715= tan(2π × 0.715).^? For ∆f T_b = 0.715, ρ = −0.2172. Therefore d =p

2E(1.2172).

When ρ = 0, d =√

2E. The distance has increased by a factor of 1.1033.

A more relevant way to “quantify” this increase is to state that for ρ = 0 the energy E needs to be increased by 1.2172 to achieve the same distance between the 2 signals as in the ρ = −0.2172 case. This is a 10 log₁₀(1.2172) = 0.85 dB increase in energy (& hence power).

Nonetheless for other reasons, namely synchronization, ∆f is chosen so that ρ = 0.

Remarks:

(i) When the frequency separation ∆f is chosen to be _2T¹

b the two signals are said to be “coherently” orthogonal, when ∆f is chosen to be k 1

T_b (k ≥ 2) they are called

“noncoherently” orthogonal. Chapter 7 discusses this further.

(ii) To obtain the signal space representation is not that straightforward, except for ρ = 0.

All that can be said of the top of one’s head is that the signal space is 2–dimensional, and that the 2 signals lie at a distance of √

E (because of how we adjust V₁, V₂) from the origin. One basis function can be chosen to be, say φ₁(t) = ^s√1(t)

E . The other basis function is a function of ∆f .

P5.7 Using the Gram-Schmidt procedure, construct an orthonormal basis for the space of quadratic polynomials {a₂t²+ a₁t + a₀; a₀, a₁, a₂ ∈ R} over the interval −1 ≤ t ≤ 1.

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It is simple to see that the equivalent problem is to find an orthonormal basis for three signals s₁(t) = 1, s₂(t) = t and s₃(t) = t² over the interval −1 ≤ t ≤ 1.

The first orthonormal basis function is simply φ₁(t) = ^√1

2 since the energy of s₁(t) = 1 over the interval −1 ≤ t ≤ 1 is 2.

Next observe that s₂(t) = t is an odd function, while φ₁(t) is an even function. It follows that s₂(t) is already orthogonal to φ₁(t). Thus, we set: The orthogonal signal is given by

φ⁰₃(t) = s₃(t) − 2 3√

2φ₁(t) − 0 · φ₂(t) = t²−1 3 Finally, the third orthonormal basis function is

φ₃(t) =

In summary, (r

1

is an orthonormal basis set for the quadratic functions over [−1, 1]. The set is plotted in Fig.

5.7. One, of course, does not achieve very much by approximating a quadratic polynomial by another quadratic polynomial. Of more practical interest is the situation where we are given

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−1 −0.5 0 0.5 1

−1.5

−1

−0.5 0 0.5 1 1.5 2

t

Figure 5.7: Orthonormal basis set for quadratic polynomials.

a more general time function, say s(t) of finite energy, and wish to approximate it by a quadratic polynomial over the interval of [−1, 1]. Then the set of {φ₁(t), φ₂(t), φ₃(t)} found above can be used to approximate s(t) where the coefficients in the approximation s(t) ≈ ˆ

s(t) = s₁φ₁(t) + s₂φ₂(t) + s₃φ₃(t) are found by:

s₁= Z ₁

−1

s(t)φ₁(t)dt, s₂= Z ₁

−1

s(t)φ₂(t)dt, s₃= Z ₁

−1

s(t)φ₃(t)dt.

Furthermore the mean-squared error [s(t) − ˆs(t)]² is minimum (see P5.9).

Remark: The 3 polynomials {φ₁(t), φ₂(t), φ₃(t)} are normalized versions of the first three members of what are referred to as Legendre polynomials. These are orthogonal polynomials over [−1, 1] that are solutions of Legendre’s differential equations:

(1 − t²)d²p_n(t)

dt² − 2tdp_n(t)

dt + n(n + 1)p_n(t) = 0.

The nth member is given by p_n(t) = ₂n¹n! dⁿ

dtⁿ(t²− 1)ⁿ (Rodrique’s formula) and Z _∞

−∞

p_n(t)p_m(t)dt =

½ ₂

2n+1, n = m 0, n 6= m .

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& Shwedyk

P5.8 Orthogonal : Z _T

0 Now let us find the correlation coefficient:

ρ_mn = signals are antipodal with energy ˆE = E/2.

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Therefore the signals {ˆs_m(t)}^M_m=1 are linearly independent since any one of them is a linear combination of the remaining M − 1 of them. They therefore lie in an (M − 1)-dimensional signal space at distance

q

M −1M from the origin. Further, since they are equally correlated the angle between any 2 vectors from the origin to the signal points is the same for each pair.

Putting this all together we have for:

M = 3: A 2-dimensional signal space. Signal points are at distance q2

3E from origin, laying at a angle spacing of 120^◦, i.e., on vertices of an equilateral triangle.

M = 4: Lying in a 3-dimensional space, distance of q3

4E, on the vertices of a regular tetra-hedron.

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energy of the estimation error is:

E_e =

Obviously E_e is a function of s_k and therefore s_k can be chosen to minimize E_e. To find the optimal s_k, differentiate E_e with respect to each s_k and set the result to zero:

dE_e (b) SubstitutingR_+∞

−∞ s(t)φ_k(t)dt = s_k into (5.20), the minimum mean square approximation error is given by:

(E_e)_min = E_s− 2

Basically the energy of the approximation error is the signal energy minus the energy in the approximation signal.

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(c) Determine the new set of coefficients s^R_ij, i, j ∈ {1, 2} in the representation:

· s₁(t)

• One method is a straightforward calculation: s^R_ij = Z ₁

• A second method is based on s_ij and θ directly:

· s₁(t)

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(d) Geometrical picture for both the basis function sets (original and rotated) and for the two signal points is shown in Fig. 5.10. Note: As seen from the geometrical picture,

2^R( )t

Figure 5.10: Geometrical representation of the basis function sets (original and rotated) and the two signals.

s^R₁₁ = −s^R₂₂ and s^R₁₂ = s^R₂₁. This is seen in (c) above and is true for any θ. But do not jump to conclusions that this is true all the time. It comes about because of the locations of s₁(t), s₂(t) in the signal space.

(e) Determine the distance d between the two signals s₁(t) and s₂(t) in two ways:

(i) Algebraically: (ii) Geometrically: From the signal space plot of (d) one has d =√

2 + 2 = 2 (Note that both signal have energy E₁= E₂= 2).

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(f) Thought φ₁(t) and φ₂(t) can represent the two given signals, they are by no means a complete basis set because they cannot represent an arbitrary, finite-energy signal defined on the time interval [0, 1]. As a start to complete the basis, we wish to plot the next two possible orthonormal functions φ₃(t) and φ₄(t). Obviously there are may possible choices for φ₃(t) and φ₄(t). One simple choice is shown below.

t

P5.11 (a) The first orthonormal basis function is chosen, somewhat arbitrarily, as φ₁(t) = ^s√1_E^(t)

1 = The third orthogonal signal is given by

φ⁰₃(t) = s₃(t) −√

2φ₁(t) − 0 · φ₂(t) =

½ −1, 2 ≤ t ≤ 3 0, elsewhere Since the energy of φ⁰₃(t) equal 1, one sets φ₃(t) = φ⁰₃(t).

Finally, the fourth orthonormal basis function is obtained by fist projecting s₄(t) onto φ₁(t), φ₂(t), φ₃(t): The orthogonal signal is given by

φ⁰₄(t) = s₄(t) +√

2φ₁(t) − 0φ₂(t) − φ₃(t) = 0

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In conclusion, we need only 3 orthonormal basis functions to represent the signal set.

They are plotted in Fig. 5.12.

φ₁(t) =

Figure 5.12: Three basis functions.

In terms of the above basis functions, the four signals are:



(b) First the 3 functions are certainly orthogonal (they are what one calles “time orthogonal”

since they do not overlap in time) and obviously each of them has unit energy. The question is do they represent the 4 time functions exactly. Well,

s₁(t) = v₁(t) + v₂(t) + 0 · v₃(t), s₂(t) = v₁(t) − v₂(t) + 0 · v₃(t) s₃(t) = v₁(t) + v₂(t) − v₃(t), s₄(t) = −v₁(t) − v₂(t) − v₃(t) (c) Geometrical representation of the four signals is plotted in Fig. 5.13.

P5.12 The signal set is given as:

s₁(t) =

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1( ) v t

2( ) v t

3( ) v t )

1(t s

3( ) s t

2( ) s t

4( ) s t

1 1

−1

−1

0

−1

Figure 5.13: Geometrical representation of the 4 signals by {v₁(t), v₂(t), v₃(t)}.

Tb 2

Tb

0 t

sin

cos

Figure 5.14

(a) There is one cycle of the cos and one cycle of the sin in the interval [0, T_b]. Graphically:

Obviously the area of the product is zero, regardless of the amplitudes of the sine and cos. Therefore s₁(t) and s₂(t) are orthogonal.

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The energies of the two signals are:

E₁ =

(b) The optimum decision rule for the minimum distance receiver is:

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(c) From the signal space diagram, the squared distance between the two signals is:

d²₂₁ = E₁+ E₂=

r3A²T_b

2 + A²T_b

2 = 2A²T_b = 2T_b (when A = 1 volt) It follows that the bit error rate bit is:

P [error] = Q

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(d) To implement the optimum receiver that uses only matched filter, one needs to rotate the basis functions as shown in the signal space diagram to obtain ˆφ₁(t) and ˆφ₂(t): ˆφ₁(t) is perpendicular to the line joining s₁(t) and s₂(t). Since

tan(θ) = It follows that

φˆ₂(t) = − sin θφ₁(t) + cos θφ₂(t)

The block diagram of the optimum receiver is shown below:

Received signal The impulse response of the filter is found as:

h(t) = ˆφ₂(T_b− t) =

The threshold can be found from the signal space diagram to be:

T = −·p

For a given (fixed) energy per bit E_b, antipodal signalling achieves the smallest error probability amongst all binary signalling schemes. This is because the distance between the two signals in antipodal signalling is the largest (again, for a fixed E_b). Thus, the two signals can be modified as:

ˆ

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−V

t )

1(t s

V

0

Tb

t

)

2(t s

0 T_b

V

2 Tb

2 Tb

2

b

Vt T

2 2

b

Vt V

−T +

2

1 3

V Tb

E = E2=V T² b

Figure 5.18

0 t

Tb

0 T_b t

2 Tb

2 Tb 1( )t

φ φ2( )t

0 t

Tb

2 Tb

1 Tb

1 Tb

− 3

Tb 1( ) ( )2

s t s t V2

V2

− total area

=0

Figure 5.19

P5.13 (a) To show s₁(t) is orthogonal to s₂(t), one needs to show that R_Tb

0 s₁(t)s₂(t)dt = 0.

The above is quite obvious by plotting s₁(t)s₂(t).

Since s₁(t) and s₂(t) are orthogonal, choose:

φ₁(t) = s₁(t)

√E₁ =

√3 V√

T_bs₁(t); φ₂(t) = s₂(t)

√E₂ = 1 V√

T_bs₂(t) Plots of φ₁(t) and φ₂(t) are shown above.

(b) Plots of signal space diagram and the optimum decision regions are shown below:

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The optimum decision rule is that of the minimum distance rule:

(r₁− s₂₁)²+ (r₂− s₂₂)²

(c) The squared distance between the two signals:

d²₂₁ = E₁+ E₂ = V²T_b

3 + V²T_b= 4V²T_b 3 = 4T_b

3 (when V = 1 volt) The error probability of the minimum distance receiver is:

P [error] = Q

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To realize the optimum receiver that uses only one matched filter, one needs to obtain φˆ₂(t) and the threshold T . The block diagram of the optimum receiver:

( )t

where the threshold can be found from the signal space diagram to be:

T = d₂₁

Note: One can also make use of the result obtained in Question 2 to come up with the following implementation:

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(e) Need to modify s₂(t) so that the Euclidean distance d₂₁ is maximized. Since the energy of s₂(t) cannot be changed ⇒ move s₂(t) to the point (−√

P5.14 (a) The two signals are (time) orthogonal with equal energy E = ^A₂^2T (joules). The orthog-onal basis is φ₁(t) = ^s√1(t) (b)+(c) See above plots.

(d) P [bit error] = Q

(e) Rotate the orthogonal basis by 45⁰ x and look at projection along φ^R₂(t) = normalizing factor^{s2(t)−s1(t)} . (f) We want s₂(t) as far away from s₁(t) as possible. Therefore make it antipodal, i.e.,

s₂(t) = −s₁(t).

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)

1(t φ )

2(t φ

E

E

( )

1 0_T

s t ↔

( )

2 1_T

s t ↔ 1_D

r2

r1

0

0_D r2

Figure 5.26

t=Tb

( )^t

r

( )

0

d

Tb

• t

∫

( )

0

d

Tb

• t

∫

t=Tb

Comparator

-+ Threshold = 0

2− 1

r r r1

r2

)

1(t φ

)

2(t φ

Figure 5.27

0 t

T 1 T

1

− T

2 T

2^R( )t φ

Figure 5.28

t=Tb

( )

0

d

Tb

• t

∫

²

rR

2^R( )t φ

( )^t

r

Figure 5.29

P5.15 The noise, n(t) = n, is simply a DC level but the amplitude of the level is random and is

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Figure 5.30: A binary communication system with additive noise.

Gaussian distributed with a probability density function given by f_n(n) = 1 The received signal in the first bit interval can be written as:

r(t) = s_i(t) + n(t) = s_i(t) + n, 1 ≤ t ≤ T_b (5.27) (a) The autocorrelation function of the noise n(t) is

R_n(τ ) = E{n(t)n(t + τ )} = E{n × n} = E{n²} = N₀,

Since R_n(τ ) = N₀ for every τ , it follows that any two noise samples, no matter how far they are, are correlated. In fact, because the noise is modeled as a random DC level, knowing the value of the noise at one time instant gives a full knowledge about the noise at any other time instants.

The power spectral density of the noise is

S_n(f ) = F{R_n(τ )} = F{N₀} = N₀δ(f )

Note that S_n(f ) 6= 0 only when f = 0, or S_n(f ) = 0 for f 6= 0. This is of course due to the noise being modeled as a random DC level!

(b) Consider the following signal set and the receiver (as usual, s₁(t) is used for the trans-mission of bit “0” and s₂(t) is for the transmission of bit “1”). In essence, the system

∑

uses antipodal signalling. The decision variable is r₁ = _T¹ R_T

0 r(t)dt. More specifically,

½ r₁= V + n₁ if “0” was transmitted r₁= −V + n₁ if “1” was transmitted

Solutions Page 5–26

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Nguyen & Shwedyk A First Course in Digital Communications

where the noise n₁ is Gaussian, zero-mean and has a variance of N₀/T . Therefore the error probability of the receiver is:

P [error] = Q

(c) Consider the signal set plotted in Fig. 5.32.

∑

Figure 5.32: A signal set.

To achieve a probability of zero, the receiver needs to remove the noise completely. There are many possibilities. Perhaps the simplest is to take 2 samples of r(t), at r(t₁) and r(t₂), where 0 < t₁ < T /2 and T /2 < t₂ < T , i.e., they are in the first half and second half of the bit interval, respectively. The decision rule is as follows:

If sample r(t₁) equals r(t₂), then decide s₁(t).

If sample r(t₁) is not equal to r(t₂), then decide s₂(t) (note that r(t₁) > r(t₂)).

When s₁(t) is transmitted, r(t) = V + n, 0 ≤ t ≤ T ⇒ r(t₁) = r(t₂).

When s₂(t) is transmitted, then r(t) =

½ V + n, 0 ≤ t ≤ T /2

−V + n, T /2 ≤ t ≤ T ⇒ r(t₁) 6= r(t₂).

Therefore it is possible to determine whether s₁(t) or s₂(t) was transmitted perfectly.

Another receiver implementation is as follows:

∑

The noise process in this problem is an extreme example of colored noise. Using the signal space ideas developed in this chapter one can see that the noise lies along one

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axis only, namely that given by φ₁(t) = ^s√1(t)_E . It does not have any component along φ₂(t) = ^s√2(t)_E. This is illustrated in Fig. 5.34. This allows you to determine which signal is sent exactly. In this context, another receiver implementation is shown in Fig. 5.34.

T t=

2

2

0 1

0 0

D D

r r

> ⇒

= ⇒ ( )t

r ( )

0

d

T

∫

t )

1(t φ )

2(t φ E

E

( )

s t1

0

( )

s t2

Noise projection is on ( ) onlyφ1 t φ2( )t

Figure 5.34: Noise projection and another receiver implementation.

P5.16 (Matched Filter )

) ( ) (t H f

h ⇔

( ) ( )

s t +w t sout( )t +wout( )t

Figure 5.35

(a) The response of the filter to the input signal component can be written as s_out(t) = F⁻¹{S(f )H(f )} =

Z _∞

−∞

S(f )H(f )e^{j2πf t}df

⇒ s_out(t₀) = Z _∞

−∞

S(f )H(f )e^{j2πf t0}df ⇒ [s_out(t₀)]²=

·Z _∞

−∞

S(f )H(f )e^{j2πf t0}df

¸₂

(5.28) (b) Since the input noise is white with power spectral density (PSD) N₀/2, it follows that the PSD of the output noise is N2 |H(f )|^{0 2}. The total power of the output noise P_wout is thus given by,

P_wout = Z _∞

−∞

N₀

2 |H(f )|²df = N₀ 2

Z _∞

−∞

|H(f )|²df (5.29) From (5.28) and (5.29) the SNR_out can be written as,

SNR_out= [s_out(t₀)]² P_wout =

hR_∞

−∞S(f )H(f )e^{j2πf t0}df i₂

N0

2 Z _∞

−∞

|H(f )|²df

(5.30)

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(c) NowR_∞

−∞[a(t) + λb(t)]²dt ≥ 0 (always). Considered as a quadratic in λ:

αλ²+ βλ + γ ≥ 0 (5.31)

with coefficients α, β and γ given as:

α =

The above inequality requires that the quadratic αλ²+βλ+γ cannot have any real roots, otherwise it was cross the real axis for some value of λ (actually at two points) and would be less than 0. From the formula for the roots this means that ∆ = β²− 4αγ < 0, which, upon substituting for α, β, γ is

·Z _∞

which is known as Cauchy-Schwartz inequality. The equality holds when a(t) = Kb(t), where K is an arbitrary (real) scaling factor, i.e., the shapes of a(t) and b(t) are the same.

(d) Using Parseval’s relationship, Z _∞ where equality holds when A(f ) = KB(f ), K an arbitrary (real) scaling constant.

(e) Now let A(f ) = H(f ) and B^∗(f ) = S(f )e^{j2πf t0}, then (5.36) becomes: Combining (5.30) and (5.37) gives:

SNR_out ≤ (f) The maximum value of SNR_out is achieved by choosing A(f ) = KB(f ), i.e.,

H(f ) = KS^∗(f )e^{−j2πf t0}. Finally, the corresponding impulse response of the filter is

h(t) = Ks(t₀− t) (5.39)

where K is any real number. Note that to arrive at the above relation, the following two properties of the Fourier transform have been used (x(t) is a real function here):

1. Time shift: x(t − t₀) ⇔ X(f )e^{−j2πf t0}.

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0

t

1( ) s t K

3 T

T 2T 3 0

t

2( ) s t

−K

3 T

T 2T 3

Figure 5.36: Signal set matched to the filter’s impulse response.

2. Time reversal: x(−t) ⇔ X^∗(f ).

In our situation we take t₀ to be T_b, the bit duration. Therefore h(t) = Ks(T_b− t).

P5.17 (a) The signal set would be ±Kh(T_b− t) where K is a scaling factor that determines the transmitted energy (or average power). They plot as in Fig. 5.36. This makes h(t) a matched filter to the signal set.

(b) Assume K = 1. The transmitted energy of either signal is E =

Z _Tb

0

s²₁(t)dt = Z _Tb/3

0

µ 3 T_bt

¶₂ dt +

Z _Tb

Tb/3

·

− 3 2T_bt +3

2

¸₂

dt = T_b

3 (joules) P [bit error] = Q

Ãr2E N₀

!

= Q

Ãr2T_b 3N₀

!

= Q

µr 2

3r_b× 10⁻⁹

¶

≤ 10⁻³

r_b≤ 2 × 10⁹

3[Q⁻¹(10⁻³)]² = 6.98 × 10⁷ bits/second = 69.8 Mbps P5.18 (Antipodal signalling)

) (t s

Tb

0 A

/ 3 t

Tb 2T_b/ 3 )

(t y 2A T2 _b 3

2 _b 3 A T

Figure 5.37

(a) The impulse response of the filter matched to s(t) is

h(t) = s(T_b− t) = s(t) (5.40)

where to arrive at the last equality we recognize that s(t) is even about t = T_b/2.

Therefore, the impulse response h(t) looks exactly the same as s(t).

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(b) The output of the matched filter when s(t) is applied at the input is y(t) = s(t) ∗ h(t) =

The output y(t) is sketched in Fig. 5.38. Observe that the output peaks at the sampling time t = T_b = 3. This is of course not a coincidence but due to the property of the matched filter, namely maximizing the signal-to-noise ratio at the sampling instant (see P5.16).

Figure 5.38: Output of the matched filter: only signal s(t) is present at the input.

(c) The output of the matched filter when −s(t) is applied at the input is the negative version of the waveform in Fig. 5.38. This is simply because [−s(t)] ∗ h(t) = −[s(t) ∗ h(t)].

(d) To compute the SNR, the variance (i.e., average power) of the noise at the output of the filter at t = T_b needs to be found. This can be accomplished as follows.

Approach 1: At the output of the matched filter and for t = T_b the noise is n = The variance of the noise is

σ_n² = E

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Approach 2: The PSD of the noise at the output of the matched filter is N₀

2 |H(f )|² =

The signal-to-noise ratio (SNR) at the output of the matched filter at the sampling instant is given by

SNR = [±y(T_b)]²

σ²_n = (2A²T_b/3)²

N₀A^{2 T}₃^b = 4A²T_b

3N₀ (5.45)

(e) For antipodal signalling, the distance between the two signals is d₂₁= 2√ E = 2

q2A²Tb

3 . Therefore the probability of bit error is

P [error] = Q

Finally, it follows from (5.46) and (5.45) that P [error] = Q³√

SNR

´

(5.47) Equation (5.47) holds for general antipodal signalling, i.e., regardless of what s(t) is used. The relationship in (5.47) also clearly shows that for antipodal signalling over an AWGN channel maximizing the SNR at the output of a receiving filter is the same as minimizing the probability of error!

(f) Plot of P [error] as a function of SNR is shown in Fig. 5.39. The minimum SNR to achieve a probability of error of 10⁻⁶ is

SNR = [Q⁻¹(10⁻⁶)]² = 22.595 = 13.54 (dB) (5.48) P5.19 (Bandwidth) Define W to be the bandwidth of the signal s(t) if ²% of the total energy of s(t)

is contained inside the band [−W, W ]:

Z _W

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Figure 5.39: Plot of P [error] as a function of SNR.

(ii) Half-sine: s_(ii)(t) = Using the sifting property of the impulse function we have

S_(ii)(f ) =

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Doing all the necessary algebra, one obtains:

S_(iii)(f ) = Substituting the above into (5.49) and changing the variable λ = f T_b one has the fol-lowing equations to solve for W T_b:

(i) Rectangular pulse:

2

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(iii) Raised cosine:

2 Z _W

0

2T_b 3π²

· sin(πf T_b) (f T_b)[1 − (f T_b)²]

¸₂

df = 4 3π²

Z _{W T}

b

0

sin²(πλ)

λ²(1 − λ²)²dλ = ²

100 (5.55) (b) The above equations can only be solved numerically. Using the numerical integration routine in MATLAB (quad or quad8 in MATLAB 5.3 and quadl in MATLAB 6.0), the following table is obtained.

Table 5.1: Values of W T_b

²% Rectangular Half-Sine Raised Cosine 90.0% 0.8487 0.7769 0.9501 95.0% 2.0740 0.9116 1.1146 99.0% 10.2860 1.1820 1.4093 99.9% 31.1677 2.7355 1.7290

Note that the bandwidth is normalized in terms of the signalling rate r_b= 1/T_b(bits/second).

Asymptotically, as f becomes very large, the energy spectral densities behave as follows:

Rectangular: 1/f²; Half-sine: 1/f⁴ and Raised cosine: 1/f⁶.

Thus the raised cosine decays the fastest. This is related to the number of times a signal can be differentiated before impulses appear. In the above, impulses appear in the rectangular pulse after the first derivative, in the half-sine after the second derivative and in the raised cosine after the third derivative. Thus one would feel that the raised cosine is the smoothest, then the half-sine and finally the rectangular pulse. This agrees with the plot of all three signals in Figure 5.40.

In the frequency domain, this translates to the smoother signal occupies the lesser amount of frequency axis (though this may also depends on the value of ² used for bandwidth definition. Note from Table 5.1 that though the raised cosine’s spectral den-sity decays (asymptotically) the fastest, it does not start to occupying “less frequency axis” until the energy bandwidth criterion is 99.9%, a value that is probably not of engi-neering interest. A criterion of 95% or 99% would be more realistic values for engiengi-neering design and analysis.

Figures 5.41 and 5.42 plot the energy spectral densities of all three signals in linear and logarithmic scales, respectively. It can be seen that the faster the density decays with f , the wider the main lobe. This is reasonable because all three signals have the same energy (of 1 joule). Thus the energy of the respective signals has to wide up someplace on the frequency axis. If not at the higher frequency then the main lobe will do.

P5.20 (a) When the two bits are equally likely (uniform source), the error performance of a digital binary communications system only depends on the Euclidean distance between the two signals. This distance is 2√

E for antipodal signalling, where√

E is the energy of s(t).

It follows that, for the two communication systems to have the same error performance, the energies of the two waveforms in Figure 5.43 have to be the same. That is:

It follows that, for the two communication systems to have the same error performance, the energies of the two waveforms in Figure 5.43 have to be the same. That is:

In document A First Course in Digital Communication Solution Manual by Nguyen HH (Page 101-150)