Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 9 pdf

As it is observed the µ-law compressor is insensitive to the dynamic range of the input signal for E[ ˘ X2] > 1... Each sample is quantized using 16 bits so the total number of bits per

or since N = 2 ν

Dtotal= x

2 max

3· 4 ν (1 + 4p b(4ν − 1)) = x2max

3N2(1 + 4p b (N2− 1))

4)

SNR = E[X

2]

Dtotal

2]3N2

x2 max(1 + 4pb(N2− 1))

If we let ˘X = x X

max, then E[X x2 2]

max = E[ ˘ X2] = ˘X2 Hence,

2X˘2

1 + 4p b (N2− 1) =

3· 4 ν X˘2

1 + 4p b(4ν − 1)

Problem 6.57

1)

g(x) = log(1 + µ

|x|

xmax)

log(1 + µ) sgn(x)

Differentiating the previous using natural logarithms, we obtain

g  (x) = 1

ln(1 + µ)

µ/xmax (1 + µ |x|

xmax)sgn

2(x)

Since, for the µ-law compander ymax= g(xmax) = 1, we obtain

D ≈ ymax2

3× 4 ν

∞

−∞

f X (x) [g  (x)]2dx

2 max[ln(1 + µ)]2

3× 4 ν µ2

∞

−∞



1 + µ2 |x|2

x2 max

+ 2µ |x|

xmax



f X (x)dx

2 max[ln(1 + µ)]2

3× 4 ν µ2



1 + µ2E[ ˘ X2] + 2µE[ | ˘ X |]

2 max[ln(1 + µ)]2

3× N2µ2



1 + µ2E[ ˘ X2] + 2µE[ | ˘ X |] where N2 = 4ν and ˘X = X/xmax

2)

2]

D

2]

x2 max

µ23· N2

[ln(1 + µ)]2(µ2E[ ˘ X2] + 2µE[ | ˘ X |] + 1)

2N2E[ ˘ X2]

[ln(1 + µ)]2(µ2E[ ˘ X2] + 2µE[ | ˘ X |] + 1)

3) Since SQNRunif = 3· N2E[ ˘ X2], we have

2

[ln(1 + µ)]2(µ2E[ ˘ X2] + 2µE[ | ˘ X |] + 1)

= SQNRunifG(µ, ˘ X)

where we identify

2

[ln(1 + µ)]2(µ2E[ ˘ X2] + 2µE[ | ˘ X |] + 1)

3) The truncated Gaussian distribution has a PDF given by

f Y (y) = √ K

2πσx e

− x2

2σ2x

where the constant K is such that

K

4σ x

−4σ x

1

√ 2πσx e

− x2

2σ2x dx = 1 = ⇒ K = 1

1− 2Q(4) = 1.0001

Hence,

E[ | ˘ X |] = √ K

2πσx

4σ x

−4σ x

|x|

4σ x e

− x2

2σ2x dx

4√ 2πσ2

x

4σ x

0

xe − x2

2σ2x dx

2√ 2πσ2

x



−σ2

x e − x2

2σ2x

4σ x

0

2√ 2π(1− e −2 ) = 0.1725

In the next figure we plot 10 log10SQNRunif and 10 log10SQNRmu −law vs 10 log10E[ ˘ X2] when the latter varies from −100 to 100 db As it is observed the µ-law compressor is insensitive to the dynamic range of the input signal for E[ ˘ X2] > 1.

-50 0 50 100 150 200

mu-law uniform

E[X^2] db

Problem 6.58

The optimal compressor has the form

g(x) = ymax



2

x

−∞ [f X (v)]1dv

∞

−∞ [fX (v)]

1

3dv −





where ymax= g(xmax) = g(1).

∞

−∞ [fX (v)]

1

dv =

1

−1 [fX (v)]

1

dv =

0

−1 (v + 1)

1

dv +

1

0

(−v + 1)1dv

= 2

1

0

x13dx = 3

2

If x ≤ 0, then

x

−∞ [f X (v)]

1

3dv =

x

−1 (v + 1)

1

3dv =

x+1

0

z13dz = 3

4z

4

3 

x+1

0

4(x + 1)

4 3

If x > 0, then

x

−∞ [f X (v)]

1

dv =

0

−1 (v + 1)

1

dv +

x

0

(−v + 1)1dv = 3

4 +

1

1−x z

1

dz

4 +

3 4

1− (1 − x)4

Hence,

g(x) =







g(1)



(x + 1)4 − 1 −1 ≤ x < 0 g(1)



1− (1 − x)43



0≤ x ≤ 1 The next figure depicts g(x) for g(1) = 1 Since the resulting distortion is (see Equation 6.6.17)

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

x

12× 4 ν

∞

−infty [f X (x)]

1

3dx

3

12× 4 ν

3 2

3

we have

SQNR = E[X

2]

32

9 × 4 ν E[X2] = 32

9 × 4 ν ·1

6 =

16

274

ν

Problem 6.59

The sampling rate is fs= 44100 meaning that we take 44100 samples per second Each sample is quantized using 16 bits so the total number of bits per second is 44100× 16 For a music piece of

duration 50 min = 3000 sec the resulting number of bits per channel (left and right) is

44100× 16 × 3000 = 2.1168 × 109

and the overall number of bits is

2.1168 × 109× 2 = 4.2336 × 109

Chapter 7

Problem 7.1

The amplitudes Am take the values

A m = (2m − 1 − M) d

2, m = 1, M Hence, the average energy is

Eav = 1

M

M



m=1

s2m= d

2

4M Eg

M



m=1 (2m − 1 − M)2

2

4M Eg M m=1 [4m2+ (M + 1)2− 4m(M + 1)]

2

4M Eg



4

M



m=1

m2+ M (M + 1)2− 4(M + 1)

M



m=1 m



2

4M Eg 4M (M + 1)(2M + 1)

2− 4(M + 1) M (M + 1)

2

2− 1

3

d2

4 Eg

Problem 7.2

The correlation coefficient between the mth and the nth signal points is

γmn= sm · sn

|sm||sn|

where sm = (s m1 , s m2 , , s mN ) and s mj =±%E s

N Two adjacent signal points differ in only one

coordinate, for which smk and snk have opposite signs Hence,

sm · sn =

N



j=1

s mj s nj=

j =k

s mj s nj + s mk s nk

= (N − 1) Es

N − Es

N =

N − 2

N Es

Furthermore,|sm| = |sn| = (Es)1 so that

γmn= N − 2

N

The Euclidean distance between the two adjacent signal points is

d =

%

|sm − sn|2 =

1



±2%Es /N

2 =

1

4Es

N = 2

1

Es N

Problem 7.3

a) To show that the waveforms ψ n(t), n = 1, , 3 are orthogonal we have to prove that

∞

−∞ ψm(t)ψn(t)dt = 0, m = n

Clearly,

c12 =

∞

−∞ ψ1(t)ψ2(t)dt =

4

0

ψ1(t)ψ2(t)dt

=

2

0

ψ1(t)ψ2(t)dt +

4

2

ψ1(t)ψ2(t)dt

4

2

0

dt −1

4

4

2

dt = 1

4× 2 − 1

4× (4 − 2)

= 0 Similarly,

c13 =

∞

−∞ ψ1(t)ψ3(t)dt =

4

0

ψ1(t)ψ3(t)dt

4

1

0

dt −1

4

2

1

dt −1

4

3

2

dt +1

4

4

3

dt

= 0 and

c23 =

∞

−∞ ψ2(t)ψ3(t)dt =

4

0

ψ2(t)ψ3(t)dt

4

1

0

dt −1

4

2

1

dt +1

4

3

2

dt −1

4

4

3

dt

= 0

Thus, the signals ψn(t) are orthogonal.

b) We first determine the weighting coefficients

x n=

∞

−∞ x(t)ψ n (t)dt, n = 1, 2, 3

x1 =

4

0

x(t)ψ1(t)dt = −1

2

1

0

dt + 1

2

2

1

dt −1

2

3

2

dt +1

2

4

3

dt = 0

x2 =

4

0

x(t)ψ2(t)dt = 1

2

4

0

x(t)dt = 0

x3 =

4

0

x(t)ψ3(t)dt = −1

2

1

0

dt −1

2

2

1

dt +1

2

3

2

dt +1

2

4

3

dt = 0

As it is observed, x(t) is orthogonal to the signal waveforms ψn(t), n = 1, 2, 3 and thus it can not

represented as a linear combination of these functions

Problem 7.4

a) The expansion coefficients{cn}, that minimize the mean square error, satisfy

cn=

∞

−∞ x(t)ψn(t)dt =

4

0

sinπt

4 ψn(t)dt

c1 =

4

0

sinπt

4 ψ1(t)dt =

1 2

2

0

sinπt

4 dt −1

2

4

2

sinπt

4 dt

= −2

π cos

πt

4



2

0

+ 2

πcos

πt

4



4

2

= −2

π(0− 1) + 2

π(−1 − 0) = 0

Similarly,

c2 =

4

0

sinπt

4 ψ2(t)dt =

1 2

4

0

sinπt

4 dt

= −2

π cos

πt

4



4

0

=−2

π(−1 − 1) = 4

π

and

c3 =

4

0

sinπt

4 ψ3(t)dt

2

1

0

sinπt

4 dt −1

2

2

1

sinπt

4 dt +

1 2

3

2

sinπt

4 dt −1

2

4

3

sinπt

4 dt

= 0

Note that c1, c2 can be found by inspection since sinπt4 is even with respect to the x = 2 axis and

ψ1(t), ψ3(t) are odd with respect to the same axis.

b) The residual mean square error Emin can be found from

Emin =

∞

−∞ |x(t)|2dt −3

i=1

|ci|2

Thus,

Emin =

4

0

sinπt 4

2

dt − 4 π

2

= 1 2

4

0

1− cos πt

2

dt −16

π2

= 2− 1

πsin

πt

2



4

0

− 16

π2 = 2−16

π2

Problem 7.5

a) As an orthonormal set of basis functions we consider the set

ψ1(t) =

1 0≤ t < 1

1 1≤ t < 2

0 o.w

ψ3(t) =

1 2≤ t < 3

1 3≤ t < 4

0 o.w

In matrix notation, the four waveforms can be represented as







s1(t)

s2(t)

s3(t)

s4(t)





=







2 −1 −1 −1













ψ1(t)

ψ2(t)

ψ3(t)

ψ4(t)







Note that the rank of the transformation matrix is 4 and therefore, the dimensionality of the waveforms is 4

b) The representation vectors are

s1 = 

2 −1 −1 −1 

s2 =



−2 1 1 0 

s3 =



1 −1 1 −1 

s4 =



1 −2 −2 2 

c) The distance between the first and the second vector is

d 1,2=

%

|s1− s2|2 =

(

 4 −2 −2 −1 2

=√

25 Similarly we find that

d 1,3 =

%

|s1− s3|2 =

(

 1 0 −2 0 2

=√

5

d 1,4 =

%

|s1− s4|2 =

(

 1 1 1 −3 2

=√

12

d 2,3 =

%

|s2− s3|2 =

(

 −3 2 0 1 2

=√

14

d 2,4 =

%

|s2− s4|2 =

(

 −3 3 3 −2 2

=√

31

d 3,4 =

%

|s3− s4|2 =

(

 0 1 3 −3 2

=√

19

Thus, the minimum distance between any pair of vectors is dmin =√

5

Problem 7.6

As a set of orthonormal functions we consider the waveforms

ψ1(t) =

1 0≤ t < 1

0 o.w ψ2(t) =

1 1≤ t < 2

0 o.w ψ3(t) =

1 2≤ t < 3

0 o.w The vector representation of the signals is

s1 =



2 2 2



s2 =



2 0 0



s3 = 

0 −2 −2 

s4 =



2 2 0



Note that s3(t) = s2(t) − s1(t) and that the dimensionality of the waveforms is 3.

Problem 7.7

The energy of the signal waveform s 

m (t) is

E  =
∞

−∞

s 

m (t) 2

dt =

∞

−∞





sm(t) − 1

M

M



k=1 sk(t)







2

dt

=

∞

−∞ s

2

m (t)dt + 1

M2

M



k=1

M



l=1

∞

−∞ sk(t)sl(t)dt

− 1 M

M



k=1

∞

−∞ s m (t)s k (t)dt − 1

M

M



l=1

∞

−∞ s m (t)s l (t)dt

= E + 1

M2

M



k=1

M



l=1

Eδkl − 2

M E

= E + 1

M E − 2

M E = M − 1

M

E

The correlation coefficient is given by

γ mn =

∞

−∞ s  m (t)s 

n (t)dt

∞

−∞ |s 

m (t) |2dt

 1

2 ∞

−∞ |s 

n (t) |2dt

 1 2

E 

∞

−∞



sm(t) − 1

M

M



k=1

s k (t)

 

sn(t) − 1

M

M



l=1

s l (t)



dt

E 


∞

−∞ sm(t)sn(t)dt +

1

M2

M



k=1

M



l=1

∞

−∞ sk(t)sl(t)dt



−1

E 



1

M

M



k=1

∞

−∞ sn(t)sk(t)dt +

1

M

M



l=1

∞

−∞ sm(t)sl(t)dt



=

1

M2M E − 1

M E − 1

M E

M −1

1

M − 1

Problem 7.8

Assuming that E[n2(t)] = σ n2, we obtain

E[n1n2] = E


T

0

s1(t)n(t)dt

 
T

0

s2(v)n(v)dv



=

T

0

T

0

s1(t)s2(v)E[n(t)n(v)]dtdv

= σ2n

T

0

s1(t)s2(t)dt

= 0

where the last equality follows from the orthogonality of the signal waveforms s1(t) and s2(t).

Problem 7.9

a) The received signal may be expressed as

r(t) =

n(t) if s0(t) was transmitted

A + n(t) if s1(t) was transmitted

Assuming that s(t) has unit energy, then the sampled outputs of the crosscorrelators are

r = sm + n, m = 0, 1 where s0 = 0, s1 = A √

T and the noise term n is a zero-mean Gaussian random variable with

variance

σ2n = E



1

√ T

T

0

n(t)dt √1

T

T

0

n(τ )dτ



T

T

0

T

0

E [n(t)n(τ )] dtdτ

= N0

2T

T

0

T

0

δ(t − τ)dtdτ = N0

2 The probability density function for the sampled output is

f (r |s0) = √1

πN0e

− r2 N0

f (r |s1) = √1

πN0

e − (r−A √ T )2 N0

Since the signals are equally probable, the optimal detector decides in favor of s0 if

PM(r, s0) = f (r |s0) > f (r |s1) = PM(r, s1)

otherwise it decides in favor of s1 The decision rule may be expressed as

PM(r, s0)

PM(r, s1) = e

(r−A √ T )2−r2 N0 = e − (2r−A √ T )A √

T N0

s0

>

<

s1

1

or equivalently

r

s1

>

<

s0

1

2A

√ T

The optimum threshold is 12A √

T

b) The average probability of error is

P (e) = 1

2P (e |s0) +1

2P (e |s1)

2

∞

1A √ T

f (r |s0)dr +1

2

1

2A

√ T

−∞ f (r |s1)dr

2

∞

1A √ T

1

√

πN0

e − r2 N0 dr +1

2

1A √ T

−∞

1

√

πN0

e − (r−A √ T )2

2

∞

1 2

%

2

N0 A

√ T

1

√ 2π e

− x2

2 dx +1

2

−1 %

2

N0 A

√ T

−∞

1

√ 2π e

− x2

2 dx



1 2

1

2

N0

A √ T



= Q

√

SNR



SNR =

1

2A2T

N0

Thus, the on-off signaling requires a factor of two more energy to achieve the same probability of error as the antipodal signaling

Problem 7.10

Since the rate of transmission is R = 105 bits/sec, the bit interval T b is 10−5 sec The probability

of error in a binary PAM system is

P (e) = Q

1

2Eb

N0



where the bit energy is Eb = A2Tb With P (e) = P2 = 10−6, we obtain

1

2Eb

N0 = 4.75 = ⇒ Eb = 4.75

2N0

2 = 0.112813 Thus

A2T b = 0.112813 = ⇒ A =&0.112813 × 105= 106.21

Problem 7.11

a) For a binary PAM system for which the two signals have unequal probability, the optimum

detector is

r

s1

>

<

s2

N0

4√

Ebln

1− p

p = η

The average probability of error is

P (e) = P (e |s1)P (s1) + P (e |s2)P (s2)

= pP (e |s1) + (1− p)P (e|s2)

= p

η

−∞ f (r |s1)dr + (1 − p)
∞

η

f (r |s1)dr

= p

η

−∞

1

√

πN0e

− (r− √ N0 Eb)2 dr + (1 − p)
∞

η

1

√

πN0e

− (r+ √ N0 Eb)2 dr

= p √1

2π

η1

−∞ e

− x2

2 dx + (1 − p) √1

2π

∞

η2

e − x2

2 dx

where

η1 =−

1

2Eb

N0 + η

1

2

N0 η2=

1

2Eb

N0 + η

1

2

N0

Thus,

P (e) = pQ

1

2Eb

N0 − η

1

2

N0



+ (1− p)Q

1

2Eb

N0

+ η

1

2

N0



b) If p = 0.3 and E b

N0 = 10, then

P (e) = 0.3Q[4.3774] + 0.7Q[4.5668] = 0.3 × 6.01 × 10 −6 + 0.7 × 2.48 × 10 −6

= 3.539 × 10 −6

If the symbols are equiprobable, then

P (e) = Q[

1

2Eb

N0

] = Q[ √

20] = 3.88 × 10 −6

Problem 7.12

a) The optimum threshold is given by

η = N0

4√

Eb ln

1− p

N0

4√

Eb ln 2

b) The average probability of error is (η = N0

4√

E bln 2)

P (e) = p(a m =−1)
∞

η

1

√

πN0e

−(r+ √ E b) 2/N0

dr +p(am = 1)

η

−∞

1

√

πN0

e −(r− √ E b) 2/N0

dr

3Q



η + √ Eb

&

N0/2



+ 1

3Q

√

Eb − η

&

N0/2



3Q

&

2N0/ Ebln 2

1

2Eb

N0



+1

3Q

1

2Eb

N0 −

&

2N0/ Ebln 2 4



Problem 7.13

a) The maximum likelihood criterion selects the maximum of f (r |sm ) over the M possible trans-mitted signals When M = 2, the ML criterion takes the form

f (r|s1)

f (r |s2)

s1

>

<

s2

1

or, since

f (r |s1) = √1

πN0

e −(r− √ E b) 2/N0

f (r |s2) = √1

πN0e

−(r+ √ E b) 2/N0 the optimum maximum-likelihood decision rule is

r

s1

>

<

s2

0

b) The average probability of error is given by

P (e) = p

∞

0

1

√

πN0e

−(r+ √ E b) 2/N0

dr + (1 − p)
0

−∞

1

√

πN0e

−(r− √ E b) 2/N0

dr

= p

∞

√

2E b /N0

1

√ 2π e

− x2

2 dx + (1 − p)

− √

2E b /N0

−∞

1

√ 2π e

− x2

2 dx

= pQ

1

2Eb

N0



+ (1− p)Q

1

2Eb

N0



1

2Eb

N0



Problem 7.14

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

h(t) = s(T − t) = s(3 − t) = s(t) where we have used the fact that s(t) is even with respect to the t = T2 = 32 axis

b) The output of the matched filter is

y(t) = s(t) s(t) =

t

0

s(τ )s(t − τ)dτ

=























0 t < 0

A2t 0≤ t < 1

A2(2− t) 1 ≤ t < 2 2A2(t − 2) 2 ≤ t < 3 2A2(4− t) 3 ≤ t < 4

A2(t − 4) 4 ≤ t < 5

A2(6− t) 5 ≤ t < 6

A scetch of y(t) is depicted in the next figure

@

@ @

A A A A AA

@

@ @

2A2

A2

c) At the output of the matched filter and for t = T = 3 the noise is

nT =

T

0

n(τ )h(T − τ)dτ

=

T

0

n(τ )s(T − (T − τ))dτ =
T

0

n(τ )s(τ )dτ

The variance of the noise is

σ n2T = E


T

0

T

0

n(τ )n(v)s(τ )s(v)dτ dv



=

T

0

T

0

s(τ )s(v)E[n(τ )n(v)]dτ dv

= N0 2

T

0

T

0

s(τ )s(v)δ(τ − v)dτdv

= N0 2

T

0

s2(τ )dτ = N0A2

d) For antipodal equiprobable signals the probability of error is

P (e) = Q S

N

o



where

S

N

o is the output SNR from the matched filter Since

S N

o

= y

2(T ) E[n2T] =

4A4

N0A2

we obtain

P (e) = Q





1

4A2

N0





Problem 7.15

a) Taking the inverse Fourier transform of H(f ), we obtain

h(t) = F −1 [H(f )] = F −1 1

j2πf



− F −1



e −j2πfT j2πf



= sgn(t) − sgn(t − T ) = 2Π



t − T

2

T



b) The signal waveform, to which h(t) is matched, is

s(t) = h(T − t) = 2Π



T − t − T

2

T



= 2Π

T

2 − t T



= h(t)

where we have used the symmetry of Π t − T

2

T

with respect to the t = T2 axis

Problem 7.16

If g T (t) = sinc(t), then its matched waveform is h(t) = sinc( −t) = sinc(t) Since, (see Problem

2.17)

sinc(t) sinc(t) = sinc(t)

the output of the matched filter is the same sinc pulse If

g T (t) = sinc(2

T (t − T

2)) then the matched waveform is

h(t) = gT (T − t) = sinc(2

T( T

2 − t)) = gT (t)

where the last equality follows from the fact that gT (t) is even with respect to the t = T2 axis The output of the matched filter is

y(t) = F −1 [g T (t) g T (t)]

= F −1



T2

4 Π(

T

2f )e

−j2πfT



2sinc(

2

T (t − T )) = T

2g T (t − T

2) Thus the output of the matched filter is the same sinc function, scaled by T2 and centered at t = T

Problem 7.17

1) The output of the integrator is

y(t) =

t

0

r(τ )dτ =

t

0

[s i (τ ) + n(τ )]dτ

=

t

0

s i (τ )dτ +

t

0

n(τ )dτ

At time t = T we have

y(T ) =

T

0

si (τ )dτ +

T

0

n(τ )dτ = ±

1

Eb

T T +

T

0

n(τ )dτ The signal energy at the output of the integrator at t = T is

Es=



±

1

Eb

T T





2

=Eb T

whereas the noise power

P n = E


T

0

T

0

n(τ )n(v)dτ dv



=

T

0

T

0

E[n(τ )n(v)]dτ dv

= N0 2

T

0

T

0

δ(τ − v)dτdv = N0

2 T Hence, the output SNR is

SNR = Es

Pn =

2Eb

N0

2) The transfer function of the RC filter is

H(f ) = 1

1 + j2πRCf

Thus, the impulse response of the filter is

h(t) = 1

RC e

− t

RC u −1 (t)

and the output signal is given by

y(t) = 1

RC

t

−∞ r(τ )e

− t−τ

RC dτ

RC

t

−∞ (si(τ ) + n(τ ))e

− t−τ

RC dτ

RC e

− t RC

t

0

s i (τ )e RC τ dτ + 1

RC e

− t RC

t

−∞ n(τ )e

τ

RC dτ

At time t = T we obtain

y(T ) = 1

RC e

− T RC

T

0

s i (τ )e RC τ dτ + 1

RC e

− T RC

T

−∞ n(τ )e

τ

RC dτ

The signal energy at the output of the filter is

(RC)2e − 2T

RC

T

0

T

0

si(τ )si (v)e RC τ e RC v dτ dv

(RC)2e − 2T

RC Eb T


T

0

e RC τ dτ

2

= e − 2T

RC Eb T

e RC T − 12

= Eb T

1− e − T RC

 2

The noise power at the output of the filter is

(RC)2e − 2T

RC

T

−∞

T

−∞ E[n(τ )n(v)]dτ dv

(RC)2e − 2T

RC

T

−∞

T

−∞

N0

2 δ(τ − v)e τ +v RC dτ dv

(RC)2e − 2T

RC

T

−∞

N0

2 e

2τ

RC dτ

2RC e

− 2T

RC N0

2 e

2T

2RC

N0

2 Hence,

SNR = Es

Pn =

4EbRC

T N0

1− e − T RC

2

3) The value of RC that maximizes SNR, can be found by setting the partial derivative of SNR

with respect to RC equal to zero Thus, if a = RC, then

ϑSNR

ϑa = 0 = (1− e − T a)− T

a e

− T

a =−e − T a 1 +T

a

+ 1

Solving this transcendental equation numerically for a, we obtain

T

a = 1.26 = ⇒ RC = a = T

1.26

Problem 7.18

1) The matched filter is

h1(t) = s1(T − t) =

−1

T t + 1, 0≤ t < T

0 otherwise