Soluzione d’esame – Fondamenti di segnali e trasmissione (21/12/2018)

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Esercizio 1

As2

Psr 3 4 5

= =

= .

2

Az 42

PSz 8

= =

=

2 PS1/SNR1

Pos

Psz/Pr1 10

SNRc 15 4

= 5

= = = = , .

PS2/PRz Psz/SNRz

PRz

SNR2 10

= 8 8

= =

= = .

Ps1

SNRi Psz 8

4 5 +

a) 2350 49

+

= ,

= ,

=

10-4

5.10-3

PR24

Pre 8

+ .

+ ,

logso

SNRig SNRi dB

73

10 33

=

= ,

b) A = 1

I

= =

Max(se(t)) max(sc(t))

+

SNRm SNRi-10B dB

e) 32 73 2874

=

=

= , +Be

Pe

=1 1273

=

Ap

28 5

- 0 -

2 394 10

= .

.

10g10 10 - 5

2B B bit

53

934

= 0 -13 7

=

- =

=

. .

.

logro 2 fc bit/s

fe bit B

400Hz

9) al secondo

= 20000

=

= : .

Esercizio 2

H(z) 3}

1 10 /z1

z2

z ROC

1 27

+ + <

2 <

= = = ,

z2

27-2

3 1

1 27 6

3 0

+

0 -

67

+

- -

, .

,

.

H(z) Bor

=

= +

s

ta

H(z)

A -3)

(z 785

2

z+ 1

= = = ,

. z-0 2

3

z ,

=

B H(z) (7-0 2)

= 785

2 E 22 -0

= =

· ,

-

, z 0

= .

H(z) 705 E 785

. z

1 0

= .

, - , 0 2

z - ,

hImT 24

34uI-m-1]

a) uIm]

-25 11 0

= - .

. .

14

14

(z)

H Hac(z) Hc(z)

+

=

b) Hae(z) Yac(z)

. XL

25 2

= E.

= - =

3

147 - Yc(z)

He(z) X(z)

11

-2 1

=

= >

- .

14 27-1

0

1

2

0 - ,

1

Yac(z) (z 3) X(z)

25 z

=

· - .

14

Yae(z) zYazz) 2X(

3 z

= - -

↓ 1] +

yacInT In 15 XIM

Eyec +

= - .

Ye(z) (1 27-2) 11X(z)

0 =

- , 14

Y

c(z) Yc(z) 11X(z)

1

2 -

0 z

= -

-

. 14

↓ XIn]

YeInT 15

In

2

0 yc +

= .

YInT EyInt1]-XIMt]

yacInT ycInT 2yIM-1] I

t

+

= = ,

H(z)

9) = 2

0

,

Y(z) ROCy Roch

H(z) Ulz) ROCu

zCE 1

=

= z

=

· 2) .

0

, 20 33

/(2) 1z1

Scompongo <

<

: 41zk

Y(z) 13

A 23

+0

= + < 33

(

ROCy /21

=

z(z

A 2) 240 15

875

+ 1

= = = -

-

= , J

8

2) z 1

=

(2-3)(z 0 -

- ,

,

B z(z 2)

= 196 =

+ 0

= .

(z 3)(z 1) z 2

0

=

-

- .

C zz 26

= =

Y(z)

27uInT-34uI-m

XIn] (0 1

In]

15 +

= - -

,