Analog Signals and Systems - Digital Signal Processing: An Introduction with MATLAB and Applications - page 46

From ( 1.32 ) one obtains:

2

j s o ð t Þj 2 ¼ Z

1

e j2pft o df H ð f Þ

|{z}
g 1 ð f Þ

S ð f Þ
|{z}
g 2 ð f Þ

: ð 1 : 35 Þ

1

Now the Schwartz inequality (Tables) states that:

0
@

1
A

0
@

1
A ;

2 2 2

Z Z Z

1 1 1

g 1 ð f Þ g 2 ð f Þ df g 1 ð f Þ df g 2 ð f Þ df ð 1 : 36 Þ

1 1 1

where equality holds in ( 1.36 ) only if:

g 1 ð f Þ¼ kg 2 ð f Þ

ð 1 : 37 Þ

and where k is a constant. * means complex conjugation. Using ( 1.35 ) and ( 1.36 ), a
maximum value for |s o (t)| 2 is obtained as follows:

0 1 0 1

Z

1

Z

1

j s o ð t o Þj 2 max ¼

@

j H ð f Þj 2 df

A @

j S ð f Þj 2 df

A :

ð 1 : 38 Þ

1 1

Using ( 1.31 ), ( 1.34 ), and ( 1.38 ) it follows that:

R 1 1 j H ð f Þj 2 df R 1 1 j S ð f Þj 2 df

¼ E

SNR o ð t o Þ¼

2 R 1

g = 2 :

ð 1 : 39 Þ

1 j H ð f Þj 2 df

g

where E ¼ R 1

1 j s ð t Þj 2 dt ¼ R 1

1 j S ð f Þj 2 df is the symbol energy. Using ( 1.37 ), this
maximum is obtained only when one chooses H(f) to be:

2 3

4 5

¼ kS ð f Þ e j2pft o
| {z }
g 2 ð f Þ

¼ kS ð f Þ e j2pft o :

H ð f Þ
|{z}
g 1 ð f Þ

ð 1 : 40 Þ

Hence, the impulse response of this filter is given by:

¼ kp ð t t o Þ:

h ð t Þ¼F 1 H ð f f ¼F 1 kS ð f Þ e j2pft o

(using the time-shift property of the Fourier transform (see Tables). Note also that
p ð t Þ¼F 1 S ð f f g ). From the conjugation property of the Fourier transform:
p(t) = s*(-t) = s(-t) [since s(t) is real]. Hence, the impulse response of the
matched filter is:

h ð t Þ¼ ks ð t t o Þ: ð 1 : 41 Þ