HW #3

1. (a) Since s0(t) and s1(t) are antipodal signals, γ = 0

(b) Pb, 0 = Pb, 1 = Pb = Q(

)0(2

)()( 0100

∧

∧∧

−

nR

TsTs ), Let, T0 = T

)(0 Ts∧

= - )(1 Ts∧

, Pb = Q(

)0(

)( 00

∧

∧

nR

Ts )

nR∧

(0) = 20N || h || 2 =

20N

2

|||| 2g , Since || g || 2 = ∫

2

0

T

t2 dt + ∫T

T

2

( T – t) 2 dt = 12

3T

Hence, nR∧

(0) = 48

30TN

)(0 ts∧

= ( s0 * h )(t) = 2

A[ pT(t) * g(t) ] cos( 2πfct + θ -

∧

θ - π )

= - 2

A[ pT(t) * g(t) ] cos( 2πfct + θ -

∧

θ )

( pT * g )(T) = ∫ 20T

t dt + ∫T

T

2

( T – t) dt = 4

2T

)(0 Ts∧

= - 2

A4

2T cos( 2πfct + θ -

∧

θ ) = - 8

2AT cos( 2πfcT + θ -

∧

θ )

Hence, Pb = Q( - 0

2

4

3

N

TA cos( 2πfcT + θ -

∧

θ ) )

2. s0(t) = A[ pT( t + T ) + pT( t ) ] cos( 2πfct + θ + α )

s1(t) = A[ pT( t + T ) - pT( t ) ] cos( 2πfct + θ + α )

In Figure 2.16 in the notes,

At point (1)

∫−T

T A[ pT( t + T ) + pT( t ) ] cos( 2πfct + θ + α ) [ pT( t + T ) + pT( t ) ] cos( 2πfct ) dt

= ∫−T

T A2

1[ cos( 4πfct + θ + α ) + cos(θ + α ) ]dt

= AT cos(θ + α )

At point (2)

∫−T

T A[ pT( t + T ) + pT( t ) ] cos( 2πfct + θ + α ) [ pT( t + T ) + pT( t ) ] sin( 2πfct ) dt

= ∫−T

T A2

1[ sin( 4πfct + θ + α ) + sin( -θ - α ) ]dt

= - AT sin(θ + α )

Similarly, at point (3) and at point (4 ), they have 0.

Var(n1) = E[n12 ]

= E[ ∫−T

Tn(t)[ pT( t + T ) + pT( t ) ] cos( 2πfct )dt ∫−

T

Tn(s) [ pT( s + T ) +

pT( s ) ] cos( 2πfcs ) ds ]

= ∫−T

TE[n(t)n(s)]cos( 2πfct ) cos( 2πfcs ) ds dt

= 20N ∫−

T

Tcos2( 2πfct ) dt =

20TN

Similarly,

E[n22 ] = E[n3

2 ] = E[n42 ] =

20TN = σ2

R1 = 24

23 nn +

1Rf ( r ) =

2σr

exp( - 2

2

2σr

), r ≥ 0

0, r < 0

R0 = 22

21 ))sin(())cos(( nATnAT ++−+++ αθαθ

Let, 1n′ = n1 cos(θ + α ) – n2sin(θ + α )

2n′ = n1 sin(θ + α ) + n2cos(θ + α )

R0 = 22

21 )()( nnAT ′+′+

0Rf ( r ) =

2σr

exp( - 2

222

2σTAr +

)I0( 2σATr

) , r ≥ 0

0, r < 0

Pb, 0 = Pr { R0 < R1 }

= ∫∞

0 ∫∞

0r 0Rf

1Rf ( r0, r1 ) dr0 dr1

= ∫∞

0 0Rf ( r0 ) dr0 ∫

∞

0r 1Rf ( r1 ) dr1

Since, ∫∞

0r 1Rf ( r1 ) dr1 = ∫

∞

0r21

σ

rexp( -

2

2

2σr

) dr1

= exp( - 2

20

2σ

r )

Pb, 0 = ∫∞

0 0Rf ( r0 ) exp( -

2

20

2σ

r ) dr0

= ∫∞

0 2σr

exp( - 2

222

2σTAr +

)I0( 2σATr

) exp( - 2

2

2σ

r ) dr

= ∫∞

0 2σr

exp( - 2

222 )

2(

σ

TAr +

)I0( 2σATr

) dr

Let, σ~ = 2

σ

= ∫∞

0 2~2σr

exp( - 2

222

~2

)2

(

σ

TAr +

)I0( 2~2σATr

) dr

= 2

1exp( -

2

22

4σTA

) ∫∞

0 2~σr

exp( - 2

222

~2

)4

(

σ

TAr +

)I0( 2~2σATr

) dr

Since, ∫∞

0 2~σr

exp( - 2

222

~2

)4

(

σ

TAr +

)I0( 2~2σATr

) dr = 1

Pb, 0 = 2

1exp( -

2

22

4σTA

) and Eb = 2

22TA

Pb, 0 = 2

1exp( -

0N

Eb )

In the same way, Pb, 1 = 2

1exp( -

0N

Eb )

Hence, Pb = 2

1exp( -

0N

Eb )