Soluzione d’esame – Fondamenti di segnali e trasmissione (08/06/2023)

PARTE PRATICA

Esercizio 1

h(t) (Bt)

Bsime (f(B)

H(f) rect

=

= =

h(t) H(f) reut (f(30)

(30t) 2

60 >

= =

simc =

Hd(f) j2nf

=

or(f)

H H(f) Ho(f) jaf rect(f(30)

= =

+ · = -

4cs(2nf) cos(2π ) 2cs(2π 20t)

+

x(t) 10 +

+

= -

-

(H(1))cos(2nt (2)

(0)

((i))

y(t) (H(20))cos(40πt

(H(20))cos(2 +(2) cos(20

+ π(z)

t

+ 1 cos(2πt +

4 + + 4 4π 4π 10

+

= +

+ 10 + +

+ =

=

. -

.

cos(2πt (s(20πt

π(2) /2)

16T 40 +

+

= + +

(164) (1600

Py (4π)) 256)

+ 92842

= 9159

+ =

= =

+ =

(

+* IHtoT(fll No :

Pn 16 f

Hot(f)l"

Sun(f)of No

(

Sww(f)

Sun(fl (fls

I

rect

con = =

= = .

· .

L

- ! 32 No (d 25=

2/ 16 .

32 N

af

No

Pm 10.103

33536 635

=

=

= . ,

. .

922577

Py 2411

=> =

=

Esercizio 2

M (x (t)) max(sz(t))

max(s

xz(t))

(t)

a) 10

12

= 40 6+

max 6

= =

+

+ =

+ =

,

,

D 2]

E A

2 = =

2

-

= =

,

b) A2

SNRin Psz)

(Psi Psi Psz

+

= +

=

A2 (Pr Pri

(

Prz Prz

+

+

.

Psi sinusoidale

(segnale

=6 = 10

Psz a a

=

= Psz

Psi 27

= + =

Pri Psi 18 4

18 :

10

= =

= .

104

SNR ,

Prz 10

Psz 10-4

a 90

9

= = = .

.

= 3

10

SNR2 10-2

10-4

Pri Prz (10 90) 108

+ 1 08

=

15

= +

+ = = .

.

. , dB

SNRimaB

102

SNRim 98

27 33

2500

25

=> =

= = =

.

= ,

10

1 08 .

,

SNRadB SNRinpB-zolB

c) dB SNRu

98

32 = 1986

=

= =

,

A2(Psi

SNRm Psz) Psi PS2

+

+

= = Pri

(Ps1 Pe/A

2 Pr2

A Psz) Pe + +

+

+

Pe 2-2

1 I

16

= = = .

23

12 .

SNRn 27 1986

= = =

-2 23

4

10

08

2 81 10 -

+ .

. .

, 2- =27 2 (1 1

102 00)

108 100

00 36-1

1 + =

.

. .

. ,

, .

22B 10-4 d)

log(5910 it

bit-GofcB

10 235Bbit

28

0 2B

= -15 230400

= = =

=

.

= - =

=

. ,

108