Digital Communications KEEE346_02 Extra Note Lecture Note 12 2011. 11. 14. Prof. Young-Chai Ko

Digital Communications KEEE346_02

Extra Note

Lecture Note 12

2011. 11. 14.

Prof. Young-Chai Ko

Signal Design for Band-Limited Channels

So far, we considered the transmission of digital information through an additive Gaussian noise channel. In effect, no bandwidth constraint was imposed on the signal design and the communication system design.

Now we consider the problem of signal design when the channel is band-limited to some specified bandwidth of Hz.

Under this condition, the channel may be modeled as a linear filter having an equivalent low-pass frequency response that is zero for .

W

C(f ) |f| > W

Band-Limited Channels + C(f )

C(f ) 2W 2W

0 f

f_c f_c

r(t)

n(t) s(t) = <[v(t)e^j2⇡fct]

W W

C_l(f )

Signal Expression for Band-Limited Channels

Transmit signal

where is the equivalent low-pass signal given for the linear modulation such as PAM, PSK, and QAM as

Received signal in equivalent low-pass representation

Letting , we can write s(t) = <[v(t)e^j2⇡fct]

v(t)

v(t) =

X1 n=0

I_ng(t nT )

r_l(t) =

Z ₁

1

v(⌧ )c_l(t ⌧ ) d⌧ + z(t)

=

Z ₁

1

X1 n=0

I_ng(⌧ nT )c_l(t ⌧ ) d⌧ + z(t)

=

X1 n=0

I_n

Z ₁

1

g(⌧ nT )c_l(t ⌧ ) d⌧ + z(t)

h(t) =

Z ₁

1

g(⌧ )c(t ⌧ ) d⌧ r_l(t) =

X1 n=0

I_nh(t nT ) + z(t) AWGN

Let us suppose that the received signal is passed first through a filter and then sampled at a rate samples/s.

We shall show later on that the optimum filter from the point of view of signal detection is one matched to the received pulse.

That is, the frequency response of the receiving filter is . We denote the output of the receiving filter as

where is the pulse representing the response of the receiving filter to the input pulse and is the response of the receiving filter to the noise .

If is sampled at times , we have

or equivalently 1/T

H^⇤(f )

y(t) =

X1 n=0

I_nx(t nT ) + ⌫(t)

x(t)

h(t) ⌫(t) z(t)

y(t) t = kT, k = 0, 1, ...

y(kT ) ⌘ y^k =

X1 n=0

I_nx(kT nT ) + ⌫(kT )

y_k =

X1 n=0

I_nx_{k n} + ⌫_k, k = 0, 1, ...

The sampled values can be expressed as

We regard as an arbitrary scale factor, which we arbitrarily set equal to unity for convenience. Then,

y_k = x₀ 0

@I_k + 1 x₀

X1 n=0,n6=k

I_nx_{k n} 1

A + ⌫_k, k = 0, 1, ...

x₀

y_k = I_k +

X1 n=0,n6=k

I_nx_{k n} + ⌫_k, k = 0, 1, ...

k

desired information symbol rate

at the th sampling instant Inter-symbol Interference (ISI)

k

additive Gaussian noise variables at the th sampling instant

Assume that the band-limited channel has ideal frequency-response characteristics, i.e.,

Then the pulse has a spectral characteristic , where

We are interested in determining the spectral properties of the pulse and hence, the transmitted pulse , that results in no intersymbol interference.

Since

the condition for no intersymbol interference is

Design of Band-Limited Signals for No Intersymbol Interference - The Nyquist Criterion -

C_l(f ) = 1 for |f|  W

x(t) X(f ) = |G(f)|²

x(t) =

Z W W

X(f )e^{j2⇡f t} df

x(t) g(t)

y_k = I_k +

X1 n=0,n6=k

I_nx_{n k} + ⌫_k

x(t = kT ) ⌘ x^k =

⇢ 1 (k = 0) 0 (k 6= 0)

Below we derive the necessary and sufficient condition on in order for to satisfy the no intersymbol interference condition.

This condition is known as the Nyquist pulse-shaping criterion or Nyquist condition for zero ISI.

Theroem (Nyquist)

X(f ) x(t)

x(t) x(nT ) =

⇢ 1 (n = 0) 0 (n 6= 0) X(f )

The necessary and sufficient condition for to satisfy

is that its Fourier transform satisfy X1

m= 1

X(f + m/T ) = T

Proof. In general, is the inverse Fourier transform of . Hence,

At the sampling instants , the relation becomes

Let us break up the integral above into integrals covering the finite range of . Thus, we obtain

where we have defined as

x(t) X(f )

x(t) =

Z ₁

1

X(f )e^{j2⇡f t} df t = nT

x(nT ) =

Z ₁

1

X(f )e^{j2⇡f nT} df

1/T

B(f )

B(f ) =

X1 m= 1

X(f + m/T ) x(nT ) =

X1 m= 1

Z (2m+1)/2T (2m 1)/2T

X(f )e^{j2⇡f nT} df

=

X1 m= 1

Z 1/2T 1/2T

X(f + m/T )e^{j2⇡f nT} df

=

Z 1/2T 1/2T

" ₁ X

m= 1

X(f + m/T )

#

e^{j2⇡f nT} df

=

Z 1/2T 1/2T

B(f )e^{j2⇡f nT} df (1)

Obviously is a periodic function with period , and, therefore, it can be expanded in terms of its Fourier series coefficients as

where

Comparing equations (2) and (1), we obtain

Therefore, the necessary condition for no-ISI to be satisfied is that

which, when substituted into , yields , or equivalently,

This concludes the proof of theorem.

B(f ) 1/T

{bⁿ} B(f ) =

X1 n= 1

b_ne^{j2⇡nf T}

b_n = T

Z 1/2T 1/2T

B(f )e ^{j2⇡nf T} df (2)

b_n = T x( nT )

b_n =

⇢ T (n 6= 0) 0 (n = 0)

B(f ) =

X1 n= 1

b_ne^{j2⇡nf T} B(f ) = T X1

m= 1

X(f + m/T ) = T

Now suppose that the channel has a bandwidth of .

Then for and consequently, for . We distinguish three cases.

(1) When , or, equivalently, , since consists of nonoverlapping replicas of , separated by .

There is no choice for to ensure in this case and there is no way that we can design a system with no ISI.

(2) When , or, equivalently, , (the Nyquist rate), the replications of separated by .

W

C_l(f ) ⌘ 0 |f| > W X(f ) = 0 |f| > W

T < 1/2W 1/T > 2W ^{B(f ) =}

X1 n= 1

X(f + n/T )

X(f ) 1/T

X(f ) B(f ) ⌘ T

0 W

W ¹

T

1

T + W 1

T W

1

T W 1

T + W 1

T

X1 n= 1

X(f + n/T )

1/T = 2W X(f )

T = 1/2W

1/T ₁

X

n= 1

X(f + n/T )

W = 1 2T

1 T 1

T

... ... ... ...

... ...

It is clear that in this case there exists only one that results in , namely,

•

This means that the smallest value of for which transmission with zero ISI is possible is , and for this value, has to be a sinc function.

The difficulty with this choice of is that it is noncausal, and, therefore, non- realizable.

To make it realizable, usually a delayed version of it, i.e., is used and is chosen such that for , we have .

Of course, with this choice of , the sampling time must be shifted to . A second difficulty with this pulse shape is that its rate of convergence to zero is

low. The tails of decays as , consequently, a small mistiming error in sampling the output of the matched filter at the demodulator results in an infinite series of ISI components. Such a series is not absolutely summable because of the rate of

decay of the pulse, and, hence, the sum of the resulting ISI does not converge.

X(f ) B(f ) = T

X(f ) =

⇢ T (|f| < W ) 0 (otherwise)

x(t) = sin(⇡t/T )

⇡t/T ⌘ sinc

✓⇡t T

◆

T

T = 1/2W x(t)

x(t)

sinc[⇡(t t₀)/T ]

t₀ t < 0 sinc[⇡(t t₀)/T ] ⇡ 0

x(t) mT + t₀

x(t) 1/t

1/t

(3) When , consists of overlapping replications of separated by .

In this case, there exists numerous choices for such that

A particular pulse spectrum, for the case, that has desirable spectral properties and has been widely used in practice in the raised cosine spectrum.

where is called the roll-off factor and takes values in the range .

T > 1/2W B(f ) X(f )

1/T

X(f ) B(f ) ⌘ T

X1 n= 1

X(f + n/T )

. . . . . .

T > 1/2W

X_rc(f ) = 8>

>>

<

>>

>:

T ⇣

0  |f|  ¹_2T ⌘

T 2

n1 + cos h

⇡t ⇣

|f| ¹_2T ⌘io ⇣

1

2T  |f|  ¹⁺_2T ⌘

0 ⇣

|f| > ¹⁺_2T ⌘

0   1

Raised cosine pulse

Note that is normalized so that .

For , the pulse reduces to and the symbol rate For , the symbol rate is .

In general, the tails of decays as for Consequently, a mistiming error in sampling leads to a series of ISI components that converges to a finite value.

0 T 2T 3T 4T

−0.2 0 0.2 0.4 0.6 0.8

−T −0.5T 0 0.5T T

0 0.2 0.4 0.6 0.8 1

= 1

= 0

= 0.5 = 1

= 0.5

= 0

x(t) X(f )

x(t) = sin(⇡t/T )

⇡t/T

cos(⇡ t/T )

1 4 ²t²/T² = sinc(⇡t/T ) cos(⇡ t/T ) 1 4 ²t²/T²

x(t) x(0) = 1

= 0 x(t) = sinc(⇡t/T ) 1/T = 2W.

= 1 1/T = W

x(t) 1/t³ > 0.

Because of the smooth characteristics of the raised cosine spectrum, it is possible to design practical filters for the transmitter and the receiver that approximate the overall desired frequency response.

In the special case where the channel is ideal, i.e., , we have

where and are the frequency responses of the two filters. In this case, if the receiver filter is matched to the transmitter filter, we have

Ideally,

and , where is some nominal delay that is required to ensure physical realizability of the filter.

Thus the overall raised cosine spectral characteristic is split evenly between the transmitting filter and the receiving filter.

Note also that an additional delay is necessary to ensure the physical realizability of the receiving filter.

C_l(f ) = 1, |f|  W X_rc(f ) = G_T (f )G_R(f )

G_T(f ) G_R(f )

X_rc(f ) = G_T(f )G_R(f ) = |G^T(f )|²

G_T(f ) = p

|X^rc(f )|e ^{j2⇡f t0} G_R(f ) = G^⇤_T (f ) t₀

So far, we studied signal design criteria for the modulation filter at the transmitter and the demodulation filter at the receiver when the channel is ideal.

Now we perform the the signal design under the condition that the channel distorts the transmitted signal.

We assume that the channel frequency-response is known for and is zero for .

The filter and may be selected to minimize the error probability at the detector.

The additive channel noise is assumed to be Gaussian with power spectral density

For the signal components at the output of the demodulator, we must satisfy the condition where is the desired frequency response of the cascade of the modulator,

Signal Design for Channels with Distortion

C_l(f ) |f|  W

|f| > W

G_T (f ) G_R(f )

nn(f ).

Modulation filter

G_T (f )

Channel

C_l(f ) ⁺

Demodulation filter

G_R(f ) ^Detector

Input data

Gaussian noise

G_T (f )C_l(f )G_R(f ) = X_d(f )e ^{j2⇡f t0}, |f|  W X_d(f )

The desired frequency response may be selected to yield zero ISI by selecting

where is the raised cosine spectrum with an arbitrary roll-off factor.

The noise at the output of the demodulation filter may be expressed as

where is the input to the filter. Since is zero-mean Gaussian, is zero- mean Gaussian, with a power spectral density

For simplicity, we consider binary PAM transmission. Then, the sampled output of the matched filter is

where is normalized to unity, , and represents the noise term, which is zero mean Gaussian with variance

X_d(f )

X_d(f ) = X_rc(f ) X_rc(f )

⌫(t) =

Z ₁

1

n(t ⌧ )g_R(⌧ ) d⌧

n(t) n(t) ⌫(t)

⌫⌫(f ) = _nn(f )|G^R(f )|²

y_m = x_oI_m + ⌫_m = I_m + ⌫_m

x₀ I_m = ±d ⌫_m

v2 =

Z ₁

1

nn(f )|G^R(f )|² df

Consequently, the probability of error is

The probability of error is minimized by maximizing the ratio or equivalently, minimizing the noise-to-signal ratio .

Let us consider two possible solutions for the case in which the additive Gaussian noise is white, so that .

First, suppose that we pre-compensate for the total channel distortion at the transmitter, so that the filter at the receiver is matched to the received signal.

In this case, the transmitter and receiver filters have the magnitude characteristics

The phase characteristic of the channel frequency response may also be compensated at the transmitter filter.

P₂ = 1 p2⇡

Z ₁

d/ _⌫²

e ^y2/2 dy = Q

s d²

⌫2

!

⌫2/d²

d²/ _⌫²

nn(f ) = N₀/2

C(f )

|G^T(f )| =

pX_rc(f )

|C^l(f )| , |f|  W

|G^T(f )| = p

X_rc(f ), |f|  W

(1)

For these filter characteristics, the average transmitted power is

and, hence,

The noise at the output of the receiver filter is and, hence, the SNR at the detector is

As an alternative, suppose we split the channel compensation equally between the transmitter and receiver filter, i.e.,

P_av = E(I_m² ) T

Z ₁

1

g_T² (t) dt = d² T

Z W

W |G^T (f )|² df = d² T

Z W W

X_rc(f )

|C^l(f )|²

d² = P_avT

"Z W W

X_rc(f )

|C(f)|² df

# 1

⌫2 = N₀/2,

d²

⌫2

= 2P_avT N₀

"Z W W

X_rc(f )

|C(f)|² df

# 1

|G^T(f )| =

pX_rc(f )

|C^l(f )|^1/2 , |f|  W

|G^T(f )| =

pX_rc(f )

|C^l(f )|^1/2 , |f|  W

(2)

(3)

(4)

In this case, the average transmitting power is

Hence, the SNR at the detector is

We observe from (3) and (5) that when we express the SNR in terms of the average transmitter power , there is a loss incurred due to channel distortion.

In the case of the filters given by (1), the loss is

and in the case of the filters given by (4), the loss is

We observe that when , the channel is ideal and , so that no loss incurred.

P_av = d² T

Z W W

X_rc(f )

|C^l(f )|

d²

⌫2

= 2P_avT N₀

"Z W W

X_rc(f )

|C(f)| df

# 2

d²/ _⌫² P_av

10 log

Z W W

X_rc(f )

|C^l(f )|² df

(5)

10 log

"Z W W

X_rc(f )

|C^l(f )|

#2

df C_l(f ) = 1

Z W W

X_rc(f ) df = 1

On the other hand, when there is amplitude distortion, for some range of frequencies in the band , and there is a loss in SNR given by (3) and (5). ^|Cl

(f )| < 1

|f|  W

Let us determine the transmitting and receiving filters given by (4) for a binary

communication system that transmits data at a rate of 4800 bits/s over a channel with frequency (magnitude) response

where . The additive noise is zero-mean white Gaussian with spectral density W/Hz.

Since , we use a signal pulse with a raised cosine spectrum and Thus,

Then,

and otherwise.

Example

|C(f)| = 1

p1 + (f /W )² , |f|  W

W = 4800 Hz N₀/2 = 10 ¹⁵

W = 1/T = 4800 = 1

X_rc(f ) = T

2 [1 + cos(⇡T|f|)] = T cos²

✓ ⇡|f|

9600

◆

|G^T(f )| = |G^R(f )| =

"

1 +

✓ f 4800

◆2#1/4

cos

✓ ⇡|f|

9600

◆

, |f|  4800

|G^T (f )| = |G^R(f )| = 0,

−4800 0 4800 0

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

|G^T (f )|

f