diskretizacija po vremenu zoran m....

Auditorne ve�be iz Raqunarskog uprava�aDiskretizacija po vremenu

Zoran M. Buqevac

Maxinski fakultet u Bgd.

oktobar 2011.

Z. B. (Maxinski fakultet u Bgd.) Auditorne ve�be iz RU oktobar 2011. 1 / 59


1. Odrediti izlazni signal x∗ idealnog odabiraqa periodeodabira�a T ako je �egov ulazni signal x:

a) x (t) = h (t)b) x (t) = th (t)v) x (t) = e−αth (t) , α ∈ ]0,+∞[g) x (t) = te−αth (t) , α ∈ ]0,+∞[d) x (t) = e−αt (sinωt)h (t) , α ∈ ]0,+∞[

Nacrtati �ihove grafike, grafike d iskretizovanih signala za

sve ove sluqajeve?

Rexe�e:

a) x∗ (t) =∞∑k=0

x (kT ) δ (t− kT )

x (kT ) = h (kT ) = 1

x∗ (t) =∞∑k=0

δ (t− kT ) = δ∗ (t)



T 2T 3T 5T 6T

1

0

t

7T4T

d (t)**

T 2T 3T 5T 6T

1

0

t

7T4T

x (t) x (t)=

b) x (kT ) = kTh (kT ) = kT

x∗ (t) =∞∑k=0

(kT ) δ (t− kT )



T 2T 3T0

t

4T

*

T 2T 3T0

t

4T

x (t)

x (t)

v) x (kT ) = e−αkTh (kT ) = e−αkT

x∗ (t) =∞∑k=0

e−αkT δ (t− kT )



T 2T 3T 5T 6T

1

0

t

7T4T

*

T 2T 3T 5T 6T

1

0

t

7T4T

x (t) x (t)

g) x (kT ) = (kT ) e−αkTh (kT ) = (kT ) e−αkT

T 2T 3T=

=

5T 6T0

t

7T4T

*

T 2T 3T=

=

5T 6T0

t

7T4T

x (t)x (t)

1ae

1a

1a


Auditorne ve�be iz Raqunarskog uprava�aDiskretizacija po vremenu, Xenonova teorema

d) x∗ (t) =∞∑k=0

e−αkT sin (kωT ) δ (t− kT )

T 2T 3T

5T 6T

0

t

7T

4T

*

T 2T 3T

5T 6T

0

t

7T

4T

x (t)x (t)

ω0 > 2ω ispu�ena Xenonova teorema


Auditorne ve�be iz Raqunarskog uprava�aDiskretizacija po vremenu, Xenonova teorema

T

2T0

t

*

T

2T0

t

x (t)x (t)

ω0 < 2ω nije ispu�ena Xenonova teorema


Auditorne ve�be iz Raqunarskog uprava�aDiskretni kompleksni i frekventni lik

2. Odrediti diskretni kompleksni i frekventni lik signala

x (t) = te−αth (t)?Rexe�e:

1. oblik:

X∗ (s) =∞∑k=0

x (kT ) e−kTs =⇒ X∗ (s) =∞∑k=0

(kT ) e−αkT e−kTs =

∞∑k=0

(kT ) e−(s+α)kT = T∞∑k=0

ke−(s+α)kT

X∗ (jω) =∞∑k=0

(kT ) e−(jω+α)kT = T∞∑k=0

ke−(jω+α)kT

2. oblik

X∗ (s) = 1T

∞∑k=−∞

X (s+ jkω0) =⇒ X (s) = L{te−αth (t)

}= 1

(s+α)2



X∗ (s) = 1T

∞∑k=−∞

X (s+ jkω0) =1T

∞∑k=−∞

1(s+jkω0+α)

2

X∗ (jω) = 1T

∞∑k=−∞

X [j (ω + kω0)] =1T

∞∑k=−∞

1[j(ω+kω0)+α]

2

3. oblik

X∗ (s) =∑iRes 1

1−epT e−sTX (p)

∣∣∣∣p=p∗i

=⇒ X (p) = 1(p+α)2

=⇒ p∗1 =

−α; ν∗1 = 2

Res 11−epT e−sTX (p)

∣∣∣p=p∗i

= Ri1 =

1

(ν∗i −1)!dν∗i −1

dpν∗i−1

[(p− p∗i )

ν∗i 11−epT e−sTX (p)

]∣∣∣∣p=p∗i

u vixestrukom polu p∗i



R11 =1

(2−1)!ddp

[(p− p∗1)

2 11−epT e−sTX (p)

]∣∣∣p=p∗1=−α

=

ddp

[(p+ α)2 1

1−epT e−sT1

(p+α)2

]∣∣∣p=p∗1=−α

= ddp

11−epT e−sT

∣∣∣p=p∗1=−α

=

−T(−epT e−sT )[1−epT e−sT ]2

= Te−(s+α)T

[1−e−(s+α)T ]2 =⇒

X∗ (s) =1∑i=1


∣∣∣p=p∗1

= Te−(s+α)T

[1−e−(s+α)T ]2 =⇒

X∗ (jω) = Te−(jω+α)T

[1−e−(jω+α)T ]2

3. Za signal x (t) = e−αt (sinβt)h (t) odrediti diskretni

kompleksni i frekventni lik?

Rexe�e:

1. oblik

x (t) = e−αt(ejβt−e−jβt

2j

)h (t) = C

(ewt − ewt

)h (t) ;C = 1

2j , w =

− (α− jβ) , w = − (α+ jβ)



X∗ (s) =∞∑k=0

x (kT ) e−kTs =⇒ X∗ (s) =

C

( ∞∑k=0

ekTwe−kTs −∞∑k=0

ekTwe−kTs)

=

C

( ∞∑k=0

ekT (w−s) −∞∑k=0

ekT (w−s))

= C 11−e−T (s−w) − C 1

1−e−T (s−w) =

12j

(1

1−e−Tse−(α−jβ)T − 11−e−Tse−(α+jβ)T

)=

12j

([1−e−Tse−(α+jβ)T ]−[1−e−Tse−(α−jβ)T ][1−e−Tse−(α−jβ)T ][1−e−Tse−(α+jβ)T ]

)=⇒ X∗ (s) =

e−(s+α)T sinβT1−2e−(s+α)T cosβT+e−2(s+α)T

X∗ (jω) = e−(jω+α)T sinβT1−2e−(jω+α)T cosβT+e−2(jω+α)T

2. oblik

X∗ (s) = 1T

∞∑k=−∞

X (s+ jkω0) ;x (t) = e−αt (sinβt)h (t) =⇒ X (s) =



β

(s+α)2+β2=⇒ X∗ (s) = 1

T

∞∑k=−∞

β

(s+jkω0+α)2+β2

3. oblik

X∗ (s) =∑iRes 1

1−epT e−sTX (p)∣∣∣p=p∗i

=⇒ X (p) = β

(p+α)2+β2=⇒ p∗1 =

−α+ jβ; p∗2 = −α− jβ

X∗ (s) =2∑i=1


∣∣∣p=p∗i

=2∑i=1

11−epT e−sT Res X (p)|p=p∗i =

2∑i=1

11−epT e−sT

β

[(p+α)2+β2]′

∣∣∣∣p=p∗i

=2∑i=1

11−epT e−sT

β2(p+α)

∣∣∣p=p∗i

=

1

1−e−T(s−p∗1)

β

2(p∗1+α)+ 1

1−e−T(s−p∗2)

β

2(p∗2+α)=

11−e−T [s−(−α+jβ)]

β2(−α+jβ+α) +

11−e−T [s−(−α−jβ)]

β2(−α−jβ+α) =

12j

11−e−T [s−(−α+jβ)] − 1

2j1

1−e−T [s−(−α−jβ)] =

12j

1−e−T [s−(−α−jβ)]−1−e−T [s−(−α+jβ)]

[1−e−T [s−(−α+jβ)]][1−e−T [s−(−α−jβ)]]= e−T (s+α) sinβT

1−2e−T (s+α) cosβT+e−2T (s+α)



X∗ (s) = e−T (s+α) sinβT1−2e−T (s+α) cosβT+e−2T (s+α) =⇒

X∗ (jω) = e−T (jω+α) sinβT1−2e−T (jω+α) cosβT+e−2T (jω+α)

4. Odrediti sva tri oblika kompleksnog lika izlaznog signala

x∗ (t) idealnog odabiraqa ako je �egov ulazni signal

x (t) =(2e−t + e−2t

)h (t). Odrediti sve nule i polove tog

kompleksnog lika?

Rexe�e:

1. oblik kompleksnog lika

x (t) =(2e−t + e−2t

)h (t) =⇒ X (s) = 2

s+1 + 1s+2 = 2(s+2)+(s+1)

(s+1)(s+2) =3s+5

(s+1)(s+2) =⇒ s∗1 = −1; s∗2 = −2; s01 = −53 = −1, 6̇; x (kT ) =(

2e−kT + e−2kT)h (kT ) =

(2e−kT + e−2kT

)X∗ (s) =

∞∑k=0

x (kT ) e−kTs; =⇒ X∗ (s) =∞∑k=0

(2e−kT + e−2kT

)e−kTs



2. oblik

X∗ (s) = 1T

∞∑k=−∞

X (s+ jkω0) ;X (s) = 3s+5(s+1)(s+2) =⇒ X∗ (s) =

1T

∞∑k=−∞

3(s+jkω0)+5(s+jkω0+1)(s+jkω0+2)

3. oblik

X∗ (s) =∑iRes 1


=⇒ X (p) = 3p+5(p+1)(p+2) =⇒ p∗1 =

−1; p∗2 = −2

X∗ (s) =2∑i=1


∣∣∣p=p∗i

=2∑i=1


2∑i=1

11−epT e−sT

3p+5[(p+1)(p+2)]′

∣∣∣p=p∗i

=2∑i=1

11−epT e−sT

3p+52p+3

∣∣∣p=p∗i

=

1

1−ep∗1T e−sT

3p∗1+52p∗1+3 + 1

1−ep∗2T e−sT

3p∗2+52p∗2+3 = 2

1−e−T e−sT + 11−e−2T e−sT

=2

1−e−T (s+1) +1

1−e−T (s+2)



s

jw

jw

2

0

j w0

jw

2

0

j w0

j2

w03

j2

w03

0s1s2 s1**


Auditorne ve�be iz Raqunarskih sistema Diskretnikompleksni i frekventni lik

5. Nacrtati frekventni spektar izlaznog signala x∗ (t) idealnogodabiraqa ako je �egov ulazni signal x (t) = 2e−th (t) za dvevrednosti uqestanosti odabira�a: ω0 = 1rad/s i ω0 = 5rad/s.Prodiskutovati dobijeno rexe�e?

Rexe�e:

x (t) = 2e−th (t) =⇒ X (s) = 2s+1 =⇒ X (jω) = 2

1+jω1−jω1−jω =

21+ω2 − j 2ω

1+ω2 =⇒ |X (jω)| =√(

21+ω2

)2+(− 2ω

1+ω2

)2= 2√

1+ω2



-5 -4 -3 -2 -1 0 1 2 3 4 5

0.5

1.0

1.5

2.0



ww

2

0 1

2=

| ( )|X jw*

-5 -4 -3 -2 -1 0 1 2 3 4 5

1

2

3

4

5



p| X (jω)|



-5 -4 -3 -2 -1 0 1 2 3 4 5

1

2

3



p| X (jω)|



6. Posmatra se sistem prikazan na slici. Odrediti tre�i oblik

kompleksnog lika diskretnog izlaznog signala sistema S za

sluqaj da je X (s) = 1s a prenosna funkcija sistema S i perioda

odabira�a, W (s) = 1s+1 ;T = 0, 1 sec?

T x (t)*x(t) x (t)i

S

Rexe�e:

T

T

x (t)

x (t)

*

*

x(t) x (t)i

S

i

•

X z( ) X z( )

W ( )zi



3. oblik W ∗ (s)

W ∗ (s) =∑iRes 1

1−epT e−sTW (p)∣∣∣p=p∗i

=⇒ W (p) = 11+p =⇒ p∗1 = −1

W ∗ (s) =1∑i=1

Res 11−epT e−sTW (p)

∣∣∣p=p∗i

=1∑i=1

11−epT e−sT ResW (p)|p=p∗i =

1∑i=1

11−epT e−sT

1(p+1)′

∣∣∣p=p∗i

=1∑i=1

11−epT e−sT

∣∣∣p=p∗i

=

1

1−ep∗1T e−sT

= 11−e−T e−sT = 1

1−e−T (s+1)

X∗ (s)

X∗ (s) =∞∑i=0

x (kT ) e−kTs =∞∑i=0

h (kT ) e−kTs =∞∑i=0

e−kTs =

1 + e−Ts + e−2Ts + · · ·+ e−kTs + · · ·+ =⇒X∗ (s) = 1

1−q = 11−e−sT

3. oblik X∗i (s)X∗i (s) =W ∗ (s)X∗ (s) = 1

1−e−T (s+1)1

1−e−sTZ. B. (Maxinski fakultet u Bgd.) Auditorne ve�be iz RU oktobar 2011. 23 / 59

Prvi kolokvijum iz Raqunarskog uprava�a

1. Grafiqki odrediti izlazni signal predajnika koji sadr�i

nelinearnost prikazanu na slici, a za ulazni signal

x (t) = e−th (t):

0,05

0,1

Rexe�e:



0 1 2 3 4 5 6 7 8

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

t

x




x∗ (t) idealnog odabiraqa ako je �egov ulazni signal x:

x (t) = (5 sin 2t)h (t) .

Tako�e, odrediti sve nule i polove tog kompleksnog lika?

Rexe�e:

1. oblik:

x (t) = (5 sin 2t)h (t) = 5(ej2t−e−j2t

2j

)h (t) = C

(ej2t − e−j2t

)h (t) ;

C = 52j

X∗ (s) =∞∑k=0

x (kT ) e−kTs =⇒ X∗ (s) =

∞∑k=0

C(ej2kT − e−j2kT

)e−kTs =

C

( ∞∑k=0

ej2kT e−kTs − e−j2kT e−kTs)

=



C

( ∞∑k=0

ej2kT e−kTs −∞∑k=0

e−j2kT e−kTs)

=

C

( ∞∑k=0

ekT (j2−s) −∞∑k=0

ekT (−j2−s))

= C 11−e−T (s−2j) − C 1

1−e−T (s+2j) =

52j

(1

1−e−Tsej2T −1

1−e−Tse−j2T

)= 5

2j

([1−e−Tse−j2T ]−[1−e−Tsej2T ][1−e−Tsej2T ][1−e−Tse−j2T ]

)=

5e−Ts(ej2T−e−j2T

2j

)1−2e−Ts (e

j2T+e−j2T )2

+e−2Ts= 5e−Ts sin 2T

1−2e−Ts cos 2T+e−2Ts

=⇒ X∗ (s) = 5e−Ts sin 2T1−2e−Ts cos 2T+e−2Ts

2. oblik

X∗ (s) = 1T

∞∑k=−∞

X (s+ jkω0) =⇒ X (s) = L{(5 sin 2t)h (t)} = 10s2+4

X∗ (s) = 1T

∞∑k=−∞

X (s+ jkω0) =1T

∞∑k=−∞

10(s+jkω0)

2+4



3. oblik

X∗ (s) =∑iRes 1


=⇒ X (p) = 10p2+4

=⇒ p∗1 =

+2j, p∗2 = −2j

X∗ (s) =2∑i=1


∣∣∣p=p∗i

=2∑i=1


2∑i=1

11−epT e−sT

10[p2+4]′

∣∣∣p=p∗i

=2∑i=1

11−epT e−sT

5p

∣∣∣p=p∗i

=

1

1−ep∗1T e−sT

5p∗1

+ 1

1−ep∗2T e−sT

5p∗2

= 51−ej2T e−sT

12j +

51−e−j2T e−sT

1−2j =

−j2,51−ej2T e−sT + j2,5

1−e−j2T e−sT =−j2,5(1−e−j2T e−sT )+j2,5(1−ej2T e−sT )

(1−ej2T e−sT )(1−e−j2T e−sT ) =j2,5(1−ej2T e−sT−1+e−j2T e−sT )

(1−ej2T e−sT )(1−e−j2T e−sT ) =

−j2,5e−sT 2j (ej2T−e−j2T )

2j

1−ej2T e−sT−e−j2T e−sT+ej2T e−sT e−j2T e−sT =



5e−sT sin 2T

1−2e−sT (ej2T+e−j2T )2

+e−2sT= 5e−sT sin 2T

1−2e−sT cos 2T+e−2sT

s

jw

jw

2

0

j w0

jw

2

0

j w0

j2

w03

j2

w03

0

s = 2j2*

s =- 2j1*



3. Nacrtati frekventni spektar signala x iz predhodnog zadatka.

Na osnovu toga izvrxiti izbor periode odabira�a i nacrtati

frekventni spektar signala x (t) na ulazu u idealniniskopropusni priguxivaq i frekventni spektar �egovog

izlaznog signala?

Rexe�e:

x (t) = (5 sin 2t)h (t) =⇒ X (s) = 10s2+4

=⇒ X (jω) = 104−ω2 =⇒

|X (jω)| =√(

104−ω2

)2+ 02 = 10√

(4−ω2)2=∣∣∣ 104−ω2

∣∣∣ .



| X(j )|ω

ω

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

10

20

30

0ω 10=

2 2



ω

∗X (j )ω

-18 -16 -14 -12 -10 -8 -6 -4 -2 0 2 4 6 8 10 12 14 16 18

1

2

3

4

5

6

7

8

0ω 10=

2 2



pX (j )ω

1 2 3 4 5 6 7 8

-5

-4

-3

-2

-1

0

1

2

3

4

5

tT 2T

ω

-4 -2 0 2 4

1

2

3

4

5

6

7

8

0ω 10=

2 2

( )x t

3T





sluqaj da je X (s) = 1s a prenosna funkcija sistema S i perioda

odabira�a W (s) = 1;T = 0, 1s.

T x (t)*x(t) x (t)i

S

Rexe�e:

T

T

x (t)

x (t)

*

*

x(t) x (t)i

S

i

•

X z( ) X z( )

W ( )zi



3. oblik W ∗ (s)

W ∗ (s) =∑iRes 1

1−epT e−sTW (p)∣∣∣p=p∗i

=⇒ W (p) = 1 =⇒ W ∗ (s) = 1

X∗ (s) =∞∑k=0

x (kT ) e−kTs =∞∑k=0

h (kT ) e−kTs =∞∑k=0

e−kTs =

1 + e−Ts + e−2Ts + · · ·+ e−kTs + · · ·+ =⇒X∗ (s) = 1

1−q = 11−e−sT

3. oblik X∗i (s)X∗i (s) =W ∗ (s)X∗ (s) = 1

1−e−sT



1. Grafiqki odrediti izlazni signal predajnika koji sadr�i

nelinearnost prikazanu na slici, a za ulazni signal

x (t) = (10− t)h (t):

0,5

1

Rexe�e:



1 2 3 4 5 6 7 8 9 10 11 12 13 14

-4

-3

-2

-1

0

1

2

3

4

5

6

7

8

9

10

t

( ) ( )ˆx t , x t
