Contemporary communication systems 1st edition mesiya solutions manual

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Chapter 10 10.1

Binary data is transmitted over an AWGN channel with a power spectral density N o / 2 = 10−9 W/Hz. Determine the signal amplitude A required to achieve a BER = 10 −6 , when the data rate is (a) 1 Mb/s, (b) 10 Mb/s, and (c) 100 Mb/s. Evaluate for unipolar NRZ and RZ line coding schemes. What is the channel first- null bandwidth required in each case?

Solution: For unipolar NRZ signaling,

⎛ Eb ⎞ BER = Q ⎜⎜ ⎟⎟ using a MF receiver ⎝ No ⎠ N o / 2 = 10−9 ⇒ N o = 2 ×10−9 Using Table 6.1, we get x = 4.7535 for BER = 10-6.

∴

Eb 2 = ( 4.7535) = 22.6 ⇒ Eb = 4.519 ×10−8 No

Now Eb =

A 2Tb 2

(a) R = 1 Mbps, ∴Tb = 10−6 4.519 ×10 −8 × 2 A = ⇒ A = 0.3 V 10−6 2

(b) R = 10 Mbps, ∴Tb = 10−7 A2 =

4.519 ×10−8 × 2 ⇒ A = 0.95 V 10−7

(c) R = 100 Mbps, ∴Tb = 1×10−8 4.519 ×10−8 × 2 A = ⇒ A=3V 10−8 2

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⎛ Eb ⎞ A2Tb and BER = Q ⎜⎜ ⎟⎟ using a N 4 o ⎝ ⎠ MF receiver. From Table 6.1, we get x = 4.7535 for BER = 10-6.

For unipolar RZ line coding scheme Eb =

∴

Eb 2 = ( 4.7535) = 22.6 ⇒ Eb = 4.519 ×10−8 No

Now (d) R = 1 Mbps, ∴Tb = 10−6 A2 =

4.519 × 10−8 × 4 ⇒ A = 0.425 V 10−6

(e) R = 10 Mbps, ∴Tb = 10−7 A2 =

4.519 ×10−8 × 4 ⇒ A = 1.34 V 10−7

(f) R = 100 Mbps, ∴Tb = 1×10−8 A2 =

4.519 ×10−8 × 4 ⇒ A = 4.25 V 10−8

The following table summarizes the results.

Bit Rate (Mbps)

Unipolar NRZ

Unipolar RZ

Amplitude (V) BW (MHz)

Amplitude (V) BW (MHz)

1

0.3

1

0.425

2

10

0.95

10

1.34

20

100

3.0

100

4.25

200

,

10.2 Consider a binary signaling system using sinusoidal pulses

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⎛ πt ⎞ s1 (t ) = A sin ⎜ ⎟ , 0 ≤ t ≤ Tb ⎝ Tb ⎠ s 2 (t ) = 0

A correlation detector is used for the detection of transmitted symbols at the output of an AWGN channel with noise spectral density N o / 2 = 2.8 × 10−11 W/Hz. a. Determine an expression for the average probability of bit error assuming equiprobable binary data.

Solution: The BER for a correlation or MF detector is given by ⎛ Eb ⎞ BERMF = Q ⎜ ⎜ N ⎟⎟ o ⎠ ⎝

where Eb = Average energy/bit =

1 1 E E1 + 0 = 1 2 2 2

⎛ πt ⎞ ⎛ 2π t ⎞ ⎤ A2Tb A2 Tb ⎡ E1 = ∫ A sin ⎜ ⎟ dt = ⎢1 − cos ⎜ ⎟ ⎥ dt = 0 2 ∫0 ⎢⎣ 2 ⎝ Tb ⎠ ⎝ Tb ⎠ ⎥⎦ Tb

2

2

N o / 2 = 2.8 ×10−11 ⇒ N o = 5.6 × 10−11 Substituting ⎛ A2T ⎞ b ⎟ BER = Q ⎜ ⎜ 4 No ⎟ ⎝ ⎠

b. Evaluate the bit error rate for A = 50 mV and the bit rate of 1 Mb/s.

Solution: ⎛ ⎜ BER = Q ⎜ ⎜ ⎝

( 50 ×10 ) ×10 ⎞⎟ = Q ( 3.34 ) = 4.2 ×10 −3

2

4 × 5.6 × 10

−6

−11

⎟ ⎟ ⎠

−4

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c. How much does the received pulse amplitude A have to increase for the same BER performance when the bit rate is doubled?

Solution: To assure the same BER performance, the argument of Q function in the BER expression must remain the same when the bit rate is doubled. Therefore,

A2 × 0.5 ×10−6 11.16 × 4 × 5.6 ×10−5 2 = 3.34 ⇒ A = = 5 ×10−3 ⇒ A = 70.7 mV 4 × 5.6 ×10−11 0.5 10.3

A binary signaling system uses pulses s1 (t ) and s2 (t ), 0 ≤ t ≤ Tb to transmit binary data occurring with probabilities p and 1 − p , respectively . A matched filter detector is used for the detection of transmitted symbols at the output of an AWGN channel with noise spectral density N o / 2 W/Hz. a. Show that the optimum threshold value Vopt is given by

Vopt =

No s +s p log e + o1 o 2 2 1− p 2

where so1 and so 2 are given by (10.41) and (10.42), respectively.

Solution: ⎛ s −V ⎞ ⎛V − s ⎞ BER = pQ ⎜ o1 T ⎟ + (1 − p )Q ⎜ T o 2 ⎟ ⎝ σo ⎠ ⎝ σo ⎠

d ( BER) p 1 1− p 1 − (V − s ) 2 /2σ 2 − (V − s )2 /2σ o2 = e opt o1 o − e opt o 2 =0 2 dVT V =V 2 2πσ o2 2 2 πσ o T opt 2 2 ⎛ p ⎞ (V − s ) ⎛ 1 − p ⎞ (Vopt − so 2 ) log e ⎜ ⎟ − opt 2 o1 = log e ⎜ − ⎟ 2σ o 2σ o2 ⎝2⎠ ⎝ 2 ⎠ ⎛ p⎞ ⎛ 1− p ⎞ 2 2σ o2 log e ⎜ ⎟ + 2Vopt so1 − so21 = 2σ o2 log e ⎜ ⎟ + 2Vopt so 2 − so 2 ⎝2⎠ ⎝ 2 ⎠

⎛ p ⎞ 2 2 2σ o2 log e ⎜ ⎟ + 2Vopt so1 − so1 = 2Vopt so 2 − so 2 ⎝ 1− p ⎠ Now solving for Vopt , we obtain

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(

)

so22 − so21 ⎛ p ⎞ Vopt = + log ( so 2 − so1 ) e ⎜⎝ 1 − p ⎟⎠ 2 ( so 2 − so1 )

σ o2

=

σ o2

⎛ p ⎞ so1 + so 2 + log e ⎜ 2 ( so 2 − so1 ) ⎝ 1 − p ⎟⎠

(*)

For MF detection

d 2 = so1 − so 2 = so 2 − so1

σ o2 =

N (s − s ) 2 No ∞ N H opt ( f ) df = o d 2 = o o 2 o1 ∫ 2 −∞ 2 2

Substituting into (*) yields Vopt =

No s +s p log e + o1 o 2 2 1− p 2

b. What is optimum threshold value when the binary signals are equiprobable?

Solution: When the binary signals are equiprobable, p = 1 − p = Vopt =

so1 + so 2 2

1 . Substituting yields 2

c. Show that the average probability of error for the system is given by ⎛ so1 − so 2 N o ⎛ No s −s ⎞ p ⎞ p log e log e − + o1 o 2 ⎟ ⎜ ⎟ ⎜ 2 2 1− p ⎟ 2 1− p 2 ⎟ BER = pQ ⎜ + (1 − p )Q ⎜ ⎜ ⎟ ⎜ ⎟ N o ( so 2 − so1 ) N o ( so 2 − so1 ) ⎜ ⎟ ⎜ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠

Solution: Substituting Vopt into the BER expression yields

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No so1 + so 2 ⎞ s +s p p ⎛ ⎛ No ⎞ + o1 o 2 − so 2 ⎟ log e ⎜ so1 − 2 log e 1 − p − ⎟ ⎜ 2 2 1− p 2 ⎟ + (1 − p )Q ⎜ ⎟ BER = pQ ⎜ σo σo ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ so1 − so 2 N o ⎛ No s −s ⎞ p ⎞ p − + o1 o 2 ⎟ log e log e ⎜ ⎟ ⎜ 2 2 1− p ⎟ 2 1− p 2 ⎟ = pQ ⎜ + (1 − p )Q ⎜ ⎜ ⎟ ⎜ ⎟ N o ( so 2 − so1 ) N o ( so 2 − so1 ) ⎜ ⎟ ⎜ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠

10.4 The binary signaling system in Problem 10.3 uses antipodal pulses s1 (t ) = AΠ ( t / Tb ) = − s2 (t ) . a. Show that the optimum threshold value Vopt is given by Vopt =

No p log e 2 1− p

Solution: For antipodal signaling, so1 = − so 2 . Substituting

σ o2

⎛ p ⎞ so1 + so 2 log e ⎜ + 2 ( so 2 − so1 ) ⎝ 1 − p ⎟⎠ N p = o log e 2 1− p

Vopt =

b. Calculate the value of optimum threshold Vopt for following a priori bit probabilities (i) p = 0.5,1 − p = 0.5 , (ii) p = 0.3,1 − p = 0.7 , p = 0.8,1 − p = 0.2 .

Solution:

( p,1 − p ) ( 0.5, 0.5) ( 0.3,0.7 ) ( 0.8, 0.2 )

Vopt 0 −0.423N o 0.693 N o

c. Show that the average probability of error for the system is given by

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No p ⎞ p ⎛ ⎛ No ⎞ ⎜ 2 Eb − 2 log e 1 − p ⎟ ⎜ 2 log e 1 − p + 2 Eb ⎟ ⎟ + (1 − p )Q ⎜ ⎟ BER = pQ ⎜ 2 N o Eb 2 N o Eb ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

Solution: By combining (10.43), (10.44), and (10.54), we obtain the following the following relationship for antipodal signaling . ∞

so1 − so 2 = d 2 = ∫ s1 (t ) − s2 (t ) dt = 4Eb .

2

−∞

Substituting No p ⎞ p ⎛ ⎛ No ⎞ ⎜ 2 Eb − 2 log e 1 − p ⎟ ⎜ 2 log e 1 − p + 2 Eb ⎟ ⎟ + (1 − p )Q ⎜ ⎟ BER = pQ ⎜ 2 N o Eb 2 N o Eb ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

When the binary signals are equiprobable

BER =

10.5

⎛ 2 Eb ⎞ 1 ⎛ 2 Eb ⎞ 1 ⎛ 2 Eb ⎞ Q⎜ + Q⎜ = Q⎜ ⎟ ⎟ ⎜ N ⎟⎟ 2 ⎜⎝ N o ⎟⎠ 2 ⎜⎝ N o ⎟⎠ o ⎠ ⎝

The binary signaling system in Problem 10.3 uses unipolar NRZ pulses s1 (t ) = AΠ ( t / Tb ) and s2 (t ) = 0 . a. Show that the optimum threshold value Vopt is given by

Vopt =

⎛ p ⎞ No log e ⎜ ⎟ + Eb 2 ⎝ 1− p ⎠

Solution:

Vopt =

σ o2

⎛ p ⎞ so1 + so 2 + log e ⎜ 2 ( so 2 − so1 ) ⎝ 1 − p ⎟⎠

For unipolar signaling

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d 2 = so 2 − so1 = 2 Eb No 2 d = N o Eb 2 so1 + so 2 = Eb 2 A2Tb is average energy/bit. Substituting where Eb = 2 ⎛ p ⎞ N E Vopt = o b log e ⎜ ⎟ + Eb 2 Eb ⎝ 1− p ⎠

σ o2 =

=

⎛ p ⎞ No log e ⎜ ⎟ + Eb 2 ⎝ 1− p ⎠

b. Calculate the value of optimum threshold Vopt for following a priori bit probabilities (i) p = 0.5,1 − p = 0.5 , (ii) p = 0.3,1 − p = 0.7 , p = 0.8,1 − p = 0.2 .

Solution:

( p,1 − p ) ( 0.5, 0.5) ( 0.3,0.7 ) ( 0.8, 0.2 )

Vopt Eb Eb − 0.423 N o Eb + 0.693N o

c. Write an expression for the average probability of error at the detector output.

Solution: s −s ⎞ p ⎞ p ⎛ so1 − so 2 N o ⎛ No − + o1 o 2 ⎟ log e log e ⎜ ⎟ ⎜ 2 2 1− p 2 1− p 2 ⎟ + (1 − p )Q ⎜ ⎟ BER = pQ ⎜ σo σo ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ No p ⎞ p ⎛ ⎛ No ⎞ ⎜ Eb − 2 log e 1 − p ⎟ ⎜ 2 log e 1 − p + Eb ⎟ ⎟ + (1 − p )Q ⎜ ⎟ = pQ ⎜ N o Eb N o Eb ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

When the binary signals are equiprobable

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BER =

10.6

⎛ Eb ⎞ 1 ⎛ Eb ⎞ 1 ⎛ Eb ⎞ Q⎜ + Q⎜ = Q⎜ ⎟ ⎟ ⎜ N ⎟⎟ 2 ⎜⎝ N o ⎟⎠ 2 ⎜⎝ N o ⎟⎠ o ⎠ ⎝

Consider the binary digital communication system using antipodal signaling. The received signal corresponding to binary “1” is given by r (t ) = s1 (t ) + n(t ) where s1 (t ) is shown in Figure P10.1 and n(t ) is AWGN with power spectral density N o / 2 Watts/Hz.

hopt (t )

(a)

s1 (t )

(b)

A

A

0

Tb / 2

Tb

t

−A

0

t

Tb

Tb / 2

−A

so1 (t )

(c)

(d)

A2Tb

φ1 (t ) − A Tb 0 −

Tb / 2

Tb

3Tb / 2

2Tb

A Tb

t

A2Tb 2

a. Sketch the impulse response of the filter matched to s1 (t ) .

Solution: The impulse response of the filter matched to s1 (t ) is sketched in Figure (b). b. Sketch the output of the filter matched to the input s1 (t ) .

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Solution:

The output of the filter matched to s1 (t ) is sketched in Figure (c). c. Draw the signal constellation. What is the d min for the signal set?

Solution:

The signal constellation is sketched in Figure (d).

d min = 2 A Tb d. Write an expression for BER in terms of A and No. Solution:

The BER for antipodal signaling using a matched filter detector is given by ⎛ 2 Eb ⎞ BER = Q ⎜ ⎜ N ⎟⎟ o ⎠ ⎝

In the present case, Eb =

1 2 1 A Tb + A2Tb = A2Tb . Substituting 2 2

⎛ 2 A2T ⎞ b ⎟ BER = ⎜ ⎜ ⎟ N o ⎝ ⎠

10.7

A binary digital communication system uses orthogonal pulses shown in Figure P10.2. The channel noise is white Gaussian with a power spectral density N o / 2 W/Hz. a. Draw block diagram of the ML detector. Sketch impulse responses of matched filters.

Solution:

In the matched-filter implementation of the ML detector, the impulse responses of matched filters are given by

h1 (t ) = φ1 (Tb − t ) h2 (t ) = φ2 (Tb − t ) where

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φ1 ( t ) =

s1 (t ) A Tb

φ2 ( t ) =

s2 (t ) A Tb

The block diagram of ML detector is shown in Figure (a). The impulse responses of matched filters are displayed in Figure (b).

T

r1 = ∫ r (t )φ1 (t )dt

(a)

0

h1 (t ) = φ1 (Tb − t )

Threshold Comparator

t = Tb

r (t ) = s1 (t ) or s2 (t )

+

+

+ −

n(t ) h2 (t ) = φ2 (Tb − t )

ro

ro

t = Tb T

r2 = ∫ r (t )φ2 (t )dt 0

(b)

h1 (t )

h2 (t )

1/ Tb

1/ Tb

0

−1/ Tb

Tb 4

Tb 2

Tb

t

0

Tb / 2

t

Tb

−1/ Tb

b. Plot the output of the filter matched to s1 (t ) when the input is s1 (t ) . Repeat when the input is instead s2 (t ) .

Solution:

The outputs of the matched filter with impulse response h1 (t ) to input pulses s1 (t ) and s2 (t ) are illustrated in Figure.

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h1 (t ) * s1 (t ) / A Tb

1 0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2

0

0

-0.2

-0.2

-0.4

-0.4

-0.6

-0.6

-0.8

-0.8

-1

0

0.2

0.4

0.6

0.8

h1 (t ) * s2 (t ) / A Tb

1

1 1.2 time (t/Tb)

1.4

1.6

1.8

2

-1

0

0.2

0.4

0.6

0.8

1 1.2 time (t/Tb)

1.4

1.6

1.8

1.6

1.8

2

c. Plot the output of the filter matched to s2 (t ) when the input is s2 (t ) . Repeat when the input is instead s1 (t ) . What do you conclude from (b) and (c)?

Solution:

The outputs of the matched filter with impulse response h2 (t ) to input pulses s2 (t ) and s1 (t ) are illustrated in Figure.

h2 (t ) * s2 (t ) / A Tb

h2 (t ) * s1 (t ) / A Tb

1

1

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2

0

0

-0.2

-0.2

-0.4

-0.4

-0.6

-0.6

-0.8

-0.8

-1

0

0.2

0.4

0.6

0.8

1 1.2 time (t/Tb)

1.4

1.6

1.8

2

-1

0

0.2

0.4

0.6

0.8

1 1.2 time (t/Tb)

1.4

2

We observe that response of the filter matched to one signal pulse (say, s1 (t ) ) is zero at the sampling instant Tb when the input is other signal pulse (say s2 (t ) ).

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d. Draw the signal constellation. What is the d min for the signal set?

Solution:

φ2 (t )

(

s 2 = 0, Eb

)

d min s1 =

( E , 0)

φ1 (t )

b

The d min for the signal set is

d min = 2 Eb Substituting Eb = A2Tb , d min can be expressed as

d min = A 2Tb e. Write an expression for BER in terms of A and No. Solution:

Since the waveforms s1 (t ) and s2 (t ) are orthogonal, the BER is given by

⎛ A2T ⎞ ⎛ Eb ⎞ b BER = Q ⎜ = Q ⎜ ⎟ ⎜ N ⎟⎟ ⎜ ⎟ N o o ⎝ ⎠ ⎝ ⎠ 10.8 Consider the signal set displayed in Figure P10.3. a. Apply the Gram-Schmidt procedure to determine an orthonormal basis for the signal set.

Solution:

We take s1 (t ) as the first basis function.

T

T

E1 = ∫ s (t )dt = A ∫ dt = 2 1

0

2

T /2

A2T 2

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Figure P10.3

s1 (t )

s2 (t )

A

A

0

T

T 2

0

t

t

T 2

s3 (t )

T

s4 (t )

A

A

0 T 2

T

−A

t

T

0 T 2

t

−A

The unit energy function φ1 (t ) is obtained by dividing s1 (t ) by E1 . That is, ⎧ 2 s1 (t ) , 0≤t ≤T ⎪ φ1 (t ) = ⎨ T A ⎪0, otherwise ⎩

Although, it is obvious that s2 (t ) is orthogonal to s1 (t ) , we will proceed to compute the function θ 2 (t ) using (10.88).

θ 2 (t ) = s2 (t ) − c21φ1 (t ) where

c21 = ( s2 (t ) • φ1 (t ) ) = ∫ s2 (t )φ1 (t )dt = ∫ 0dt = 0 T

T

0

0

This proves that s2 (t ) is orthogonal to s1 (t ) . Therefore,

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θ 2 (t ) = s2 (t ) T

T /2

0

0

2 2 ∫ s2 (t )dt = A ∫ dt = A

Since θ2 = E2 =

T , the second orthonormal 2

function is obtained as

φ2 (t ) =

⎧ 2 s2 (t ) , 0≤t ≤T ⎪ =⎨ T A E2 ⎪ otherwise ⎩0,

θ 2 (t )

To obtain the next orthonormal function, the function θ 3 (t ) is computed as

θ3 (t ) = s3 (t ) − c31φ1 (t ) − c32φ2 (t ) where T

c31 = ( s3 (t ) • φ1 (t ) ) = ∫ s3 (t )φ1 (t )dt = 0

T

c32 = ( s3 (t ) • φ2 (t ) ) = ∫ s3 (t )φ2 (t )dt = − 0

T

1 2 2 AT T A2 dt = =A ∫ A T T /2 T 2 2 T /2

1 2 2 AT T A2 dt = − = −A ∫ A T 0 T 2 2

Substituting the values of c31 and c32 , we obtain

θ3 (t ) = s3 (t ) − A

T T φ1 (t ) + A φ2 (t ) = s3 (t ) − s1 (t ) + s2 (t ) = 0 2 2

This implies that s3 (t ) is a linear combination of φ1 (t ) and φ2 (t ) . That is, s3 (t ) = A

T T φ1 (t ) − A φ2 (t ) 2 2

Similarly, it is easy to conclude that s4 (t ) is a linear combination of φ1 (t ) and φ2 (t ) . That is, s4 (t ) = − A

T T φ1 (t ) + A φ2 (t ) 2 2

Figure (a) displays the orthonormal basis of the signal set in Figure P10.3. b. Plot the signal constellation corresponding to the signal set.

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Solution:

The signal constellation for the signal set in Figure P10.3 is displayed in Figure (b).

(a)

φ1 (t )

φ2 (t )

2/T

2/T

0

T /2

0

t

T

T /2

φ2 (t )

T

t

2A T / 2

(b)

s4

s2

A T /2

s1 −A T / 2

2A T / 2

A T /2

−A T / 2

φ1 (t )

s3

10.9 Consider the signal set si (t ) =

2 Es cos ( 2π f c t + ψ i ) , 0 ≤ t ≤ T T

i = 1,......,8

where 5π 3π 7π ⎫ ⎧ π π 3π , , ,π , , , ⎬ 4 2 4 ⎭ ⎩ 4 2 4

ψ i ∈ ⎨0,

a. What is the dimensionality of signal space?

Solution:

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b. Determine the basis vectors for the signal set.

Solution:

By choosing the basis functions

φ1 (t ) =

2 cos ( 2π f c t ) T

φ2 (t ) =

2 sin ( 2π f c t ) , T

s i = ( si1 Es , si 2 Es )

si1 = cos(ψ i ) si 2 = sin(ψ i ) c. Draw the constellation diagram of the signal set.

s 2 = (0, Es )

φ2 (t )

s 3 = (− Es / 2, Es / 2)

s1 = ( Es / 2, Es / 2)

s 4 = (− Es , 0)

φ1 (t ) s 0 = ( Es , 0)

s 5 = (− Es / 2, − Es / 2)

s 7 = ( Es / 2, − Es / 2)

s 6 = (0, − Es )

10.10 Consider the 2 basis functions shown in Figure P10.4 (a). Sketch the waveforms corresponding to the points in the constellation shown in Figures P10.4 (b) and (c).

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Figure P10.4

φ2 (t )

φ1 (t ) 1/ T

1/ T 0 T 4

3T 4

T

0

t

t

T 2

T

−1/ T

−1/ T

φ2 (t )

(a)

φ2 (t )

A − 2

s2 •

(c)

• s1

A

A − 2

A

s2

33A A

φ1 (t )

s1

A

−A

φ1 (t )

A

−A

s3

(b)

s4

−3A

Solution:

For constellation (b), the waveforms corresponding to signal points s1 and s 2 are obtained as s1 = ( A, A) ⇒ s1 (t ) = s11φ1 (t ) + s12φ2 (t ) = Aφ1 (t ) + Aφ2 (t ) s 2 = (− A / 2, − A / 2) ⇒ s2 (t ) = s21φ1 (t ) + s22φ2 (t ) = −

A A φ1 (t ) − φ2 (t ) 2 2

The waveforms are displayed in Figure.

s1 (t )

s2 (t )

2A / T A/ T 3T / 4

0

T /4

−2 A / T

t T

0

T

T /4

t

3T / 4

−A / T 18

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For constellation (c), the waveforms corresponding to signal points s1 , s 2 , s 3 and s 4 are obtained as s1 = ( A, A) ⇒ s1 (t ) = s11φ1 (t ) + s12φ2 (t ) = Aφ1 (t ) + Aφ2 (t ) s 2 = (− A,3 A) ⇒ s2 (t ) = s21φ1 (t ) + s22φ2 (t ) = − Aφ1 (t ) + 3 Aφ2 (t ) s 3 = (− A, − A) ⇒ s2 (t ) = s31φ1 (t ) + s32φ2 (t ) = − Aφ1 (t ) − Aφ2 (t ) s 4 = ( A, −3 A) ⇒ s4 (t ) = s41φ1 (t ) + s42φ2 (t ) = Aφ1 (t ) − 3 Aφ2 (t )

The waveforms are displayed in Figure.

s1 (t )

s3 (t ) 2A / T

2A / T

0

3T / 4

0 T /4

3T / 4 T

T

−2 A / T

−2 A / T

s2 (t )

s4 (t )

4A / T

4A / T

2A / T 0

−2 A / T −4 A / T

T /4

T T /2

2A / T

t

0

−2 A / T

T /2

t

T

−4 A / T

10.11 Using the basis functions in Figure P10.5(a), sketch the waveforms corresponding the points of the constellation shown in Figure P10.5(b).

Solution:

The signal waveforms corresponding to the constellation in Figure P10.5(b) are shown in Figure.

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φ2 (t ) 2A

s3

s2

s1

s0

φ1 (t )

2A

s5

s4

s7

s6

T s0 (t ) 2 A 0

T s1 (t ) 1 A 0

s 1(t)

2

s 0(t)

4

0

0.5

1

T s2 (t ) -1 A -2

T s3 (t ) -2 A

0.5

1

0

0.5

1

T s4 (t ) -2 A -4

T s5 (t ) -1 A

0

0.5

1

0

0.5 time (t/T)

1

s 5(t)

0

s 4(t)

0

-2 0

0.5

1 4

2 1

s 7(t)

s 6(t)

1

-4 0

T s6 (t ) A

0.5

s 3(t)

0

s 2(t)

0

0

T s7 (t ) 2 A 0

0 -1 0

0.5 time (t/T)

1

10.12 Consider the signal constellation shown in Figure P10.5(b). a. Derive the exact average probability of symbol error for the constellation.

Solution:

For the constellation in Figure 10.5(b), d min = 2 A . The probability of symbol error can be exactly computed by noting that the conditional probability of a correct decision falls into one of two categories:

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(i) Four corner points ( s 0 , s 3 , s 4 and s 7 ) P {correct decision s 0 sent} = P { n1 > − d min / 2 ∩ n2 > − d min / 2} = P { n1 > − d min / 2} P { n2 > − d min / 2} ⎛ ⎛ d ⎞⎞⎛ ⎛ d ⎞⎞ = ⎜ 1 − Q ⎜ min ⎟ ⎟ ⎜ 1 − Q ⎜ min ⎟ ⎟ ⎜ 2N ⎟ ⎟ ⎜ 2N ⎟ ⎟ ⎜ ⎜ o ⎠⎠⎝ o ⎠⎠ ⎝ ⎝ ⎝ 2

⎛ d ⎞ ⎡ ⎛ d ⎞⎤ ⎛ d ⎞ = 1 − 2Q ⎜ min ⎟ + ⎢Q ⎜ min ⎟ ⎥ ≈ 1 − 2Q ⎜ min ⎟ ⎜ 2 N ⎟ ⎢ ⎜ 2 N ⎟⎥ ⎜ 2N ⎟ o ⎠ o ⎠⎦ o ⎠ ⎝ ⎝ ⎣ ⎝

⎛ d ⎞ P {error s 0 sent} = 2Q ⎜ min ⎟ ⎜ 2N ⎟ o ⎠ ⎝

= P {error s 3 sent} = P {error s 4 sent} = P {error s 7 sent}

(ii) Four inner points ( s1 , s 2 , s 5 and s 6 ) P {correct decision s1 sent} = P {− d min / 2 < n1 < d min / 2 ∩ n2 > − d min / 2} = P {− d min / 2 < n1 < d min / 2} P {n2 > − d min / 2} ⎛ ⎛ d ⎞⎞⎛ ⎛ d ⎞⎞ = ⎜ 1 − 2Q ⎜ min ⎟ ⎟ ⎜ 1 − Q ⎜ min ⎟ ⎟ ⎜ 2N ⎟ ⎟ ⎜ ⎜ 2N ⎟ ⎟ ⎜ o ⎠⎠⎝ o ⎠⎠ ⎝ ⎝ ⎝ 2

⎡ ⎛ d ⎛ d ⎞ ⎞⎤ ⎛ d ⎞ = 1 − 3Q ⎜ min ⎟ + 2 ⎢Q ⎜ min ⎟ ⎥ ≈ 1 − 3Q ⎜ min ⎟ ⎜ 2N ⎟ ⎜ 2N ⎟ ⎢⎣ ⎜⎝ 2 N o ⎟⎠ ⎥⎦ o ⎠ o ⎠ ⎝ ⎝

⎛ d ⎞ P {error s1 sent} = 3Q ⎜ min ⎟ ⎜ 2N ⎟ o ⎠ ⎝

= P {error s 2 sent} = P {error s 5 sent} = P {error s 6 sent}

We now obtain the following expression for the average probability of symbol error Pe = =

⎛ d ⎞⎞ 4 ⎛ ⎛ d ⎞⎞ 1 M 4⎛ P {error s i sent} = ⎜ 2Q ⎜ min ⎟ ⎟ + ⎜ 3Q ⎜ min ⎟ ⎟ ∑ ⎜ 2N ⎟ ⎟ 8 ⎜ ⎜ 2N ⎟ ⎟ M i =1 8 ⎜⎝ o ⎠⎠ o ⎠⎠ ⎝ ⎝ ⎝ 5 ⎛ d min ⎞ 5 ⎛ 2 A2 ⎞ Q⎜ ⎟ ⎟ = Q⎜ 2 ⎜⎝ 2 N o ⎟⎠ 2 ⎜⎝ N o ⎟⎠

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b. Compute the average probability of symbol error using the nearest neighbor estimate. Comment.

Solution:

The number of minimum distance pairs for the constellation is M. Substituting K = M = 8 into (10.156) yields ⎛ 2 A2 ⎞ ⎞ ⎛ d ⎞ 8 ⎛ d Pe ≈ 2 Q ⎜ min ⎟ = 2Q ⎜ min ⎟ = 2Q ⎜ ⎟ ⎜ 2N ⎟ ⎜ No ⎟ 8 ⎜⎝ 2 N o ⎟⎠ o ⎝ ⎠ ⎝ ⎠ The average probability of symbol error estimate using the nearest neighbor approximation is very accurate for this constellation.

10.13 Consider the signal constellations shown in Figure P10.6. a. Determine the average energy of each signal constellation.

Solution:

The average energy of a signal constellation Es is

Es =

1 M si 2 ∑ M i =1

Constellation (a) Es =

1 [ 2 Eb + 2Eb + 2Eb + 2 Eb ] = 2 Eb 4

Constellation (b) Es =

1 [0 + 4 Eb + 4 Eb + 8Eb ] = 4 Eb 4

Constellation (c) Es =

1 [ 2 Eb + 2Eb + 2Eb + 2 Eb ] = 2 Eb 4

b. Sketch decision regions in each case.

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Solution:

φ2(t)

φ2(t) (− Eb , Eb )

( Eb , Eb )

s2

s1

D2

D1

D3

D2

D1

D3

D4

s3 (0, 0)

s4

(− Eb , − Eb )

s1

φ1(t)

D4

s3

(2 Eb , 2 Eb )

(0, 2 Eb ) s2

( Eb , − Eb )

φ1(t)

(2 Eb , 0)

(b)

φ2(t)

(a)

s4

D2

s2 (0, 2 Eb )

D3 s3

D1

(− 2 Eb , 0)

( 2 Eb , 0)

s4

D4

(0, − 2 Eb )

(c)

c. Write the average probability of symbol error for each constellation using the nearest neighbor bound.

Solution:

For each constellation, the number of minimum distance pairs is 4. Also d min = 2 Eb in every case. Substituting K = 4 into (10.156) yields ⎞ ⎛ 2 Eb ⎞ ⎛ 2 Eb ⎞ 4 ⎛ d Pe ≈ 2 Q ⎜ min ⎟ = 2Q ⎜ ⎟ = 2Q ⎜⎜ ⎟⎟ ⎜ 2N ⎟ 4 ⎜⎝ 2 N o ⎟⎠ o ⎠ ⎝ No ⎠ ⎝

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d. From this example, what conclusions can you make about the constellation rotation and translation on the error performance of the signal set.

Solution:

Although average energy of the signal set in (b) is twice that in (a) and (c), d min remains the same. Consequently, twice the energy is required in (c) for the same error performance. Since the translation and rotation of a signal constellation does not change distances between signal pairs, the average probability of symbol error remains the same. The signal energy efficient constellations, however, are centerd about the centroid (origin in the present example) of the signal set.

10.14 Consider the 8-PAM signal with constellation in Figure P10.7. a. Draw the block diagram of the correlation detector.

Solution:

r (t ) = s (t ) + n (t ) ×

∫

ro

T 0

t =T

3-bit ADC

Bit sequence

φ1 (t ) ADC: Mid-rise Analog-to Digital converter

b. Sketch the transfer characteristic of the threshold comparator.

Solution:

The transfer characteristic of the threshold comparator is displayed in Figure. c. Determine the estimated symbol sequence and corresponding bit sequence for the correlator output samples {+0.12, −0.201, +0.71, −1.55, −0.6,1.25} .

Solution:

The estimated symbol sequence is {+0.25, −0.25, +0.75, −1.75, −0.75,1.25} .

The corresponding bit sequence is {000,100, 001,110,101, 011} .

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y[n] 010

1.75 011 1.25

001

0.75

0.25 −2.0

−1.5

−1.0

−0.5 100

101

111

000

ro[n] 0 0.5 −0.25

1.0

1.5

2.0

−0.75

−1.25

−1.75

110

10.15 Consider the 8-point signal constellation shown in Figure P10.8. a. Determine the symbol rate if the desired bit rate is 45 Mbps.

Solution:

Since each symbol conveys 3 bits of information, the resultant symbol rate is D=

45 ×106 = 15 × 106 symbols/sec 3

b. Calculate the average energy of the signal set.

Solution: 1 44 2 11 2 Es = ⎡⎣ 4 × ( 2 A2 ) + 4 × ( 9 A2 ) ⎤⎦ = A = A 8 8 2

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c. What is the d min for the signal set?

Solution: d min = 2 A =

8 Es 11

d. Use the nearest neighbor bound to estimate the probability of bit error. Can you improve the estimate by including additional terms?

Solution:

Pe ≈

2 K ⎛ d min ⎞ Q⎜ ⎟ M ⎜⎝ 2 N o ⎟⎠

Here K = 4, M =8. Therefore,

Pe ≈

⎛ 4 Es ⎞ 8 Es ⎞ 2× 4 ⎛ Q ⎜⎜ ⎟⎟ = Q ⎜⎜ ⎟⎟ 8 ⎝ 11× 2 N o ⎠ ⎝ 11N o ⎠

Probability of bit error =

Pe 1 ⎛ 4 Es ⎞ 1 ⎛ 4 × 3Eb ⎞ ≈ Q⎜ ⎟ = Q⎜ ⎟ k 3 ⎜⎝ 11N o ⎟⎠ 3 ⎜⎝ 11N o ⎟⎠

If we include 8 signal pairs at distance 5, we get a better estimate

1 ⎛ 4 × 3Eb ⎞ 2 ⎛ 5 × 3Eb ⎞ Probability of bit error ≈ Q ⎜⎜ ⎟ + Q⎜ ⎟ 3 ⎝ 11N o ⎟⎠ 3 ⎜⎝ 11N o ⎟⎠ 10.16 Consider the 8-point signal constellation shown in Figure P10.9. a. Assign the bits to each point in the signal constellation using Gray coding.

Solution:

See Figure

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φ2 (t ) D5

D4 101

•

•

111 3A

011

A

D1

D0

•

•

001

φ1 (t ) D2

D3

•

D6

110

•

−3A

000

•

−3A

010

•

−A

100

−A

D7 A

3A

b. Calculate the average energy of the signal set.

Solution: 1 Es = ⎡⎣ 4 × ( 2 A2 ) + 4 × (18 A2 ) ⎤⎦ = 10 A2 8

c. Sketch ML decision regions Di for the signal set.

Solution:

See Figure d. Use the nearest neighbor bound to estimate the probability of bit error.

Solution:

The number of minimum distance pairs for the constellation is 4. Substituting K = 4 into (10.156) and applying (10.163) yields 2 × 4 ⎛ d min ⎞ 1 ⎛ 2 A ⎞ 1 ⎛ 2 A2 ⎞ BER ≈ Q⎜ ⎟ ⎟ = Q⎜ ⎟ = Q⎜ 3 × 8 ⎜⎝ 2 N o ⎟⎠ 3 ⎜⎝ 2 N o ⎟⎠ 3 ⎜⎝ N o ⎟⎠

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