z transform properties

8/2/2019 Z Transform Properties

1/7

Problem 5.16 Repeat problem 5.9 for the signal.

1 n. 1tx[n] =("3) sm (-:4n) u[n]

ROC: Izi >1/3.

The ROC and pole-zero locations are plotted in fig P 5.16.

The Z-transform has several properties that can be used in the study of discrete-time

signals and systems. They are,

v1: Linearity.

~. Time shifting.

~. Scaling in the Z-domain.

Jk/ Time reversal.

5. Time Expansion.

6.....,Conjugation.

v 1 ':" Convolution.

: . ir : " " " - Differentiation in the Z-domain.

9. Initial value theorem.

10. Final value theorem.


2/7

and xin) ~ Xz(z) with ROC = Rz

zthen a1x1(n) + bxin) ~ aX1(z) + bXz(z) with ROC atleast R1nRz

Proof: We know that,

Z[x(n)) = X(z) = Lx[n). z-n

Zx(n) ~X(z) with ROC=R

zx(n-no) ~ z-I'oX(z) with ROC = R except for the possible

addition or deletion of the origin or inftnity.

Proof: Z[x(n)] = X(z) = LX(ll) z-n

=Lx(I). z-(l+no)1 = - 0 0


3/7

=Lx(l). Z-l. z-n o1 = - 0 0

= z -n oL

x(l) Z-l

1 = - 0 0

then

where

zx(n) ~ X(z) with ROC = R

anx(ll) AX(~) with ROC = laiR

a is a complex number.

Proof: Z[X(ll)] = X(z) = LX(ll) z-n

Z[ anx(n)] = Lan X(ll) z-n

= X(a-1 z)z

Z[anx(ll)] =X(a)

ejOonx(n) ~ X(e -jOoz)

zx(n) ~X(z)

z 1x(-n) ~X(z)

with ROC =R

with ROC = ~ .

Proof:' Z[X(ll)] = X(z) = Lx(n) z-n

Z[X(-ll)] =Lx(-n)z-n


4/7

Z(x(-n) = Lx(l). Zl

1 = - 0 0

= Lx(/). (Z - l )-11 = - 0 0

=X(Z-l)

1Z(x(-n)] = X(-)

z

zx(n) ~X(z)

x(k)(n) ~ X(Zk)

nX(k/n) = x( k)

= 0

X(z) =L

x(n) z-n

Similarly, X(Zk) =Lx(n) Z-kll.n=-oo

o I o f " . I I f 'k' d I (tll) - f ' , .t.e. z-m equa s zero 1 m IS not a mu tIp e 0 an equa to xk1 HI IS amultiple ofk. Thus the inverse transform is xk(n).

zx(n) ~X(z)

zx*(n) ~ X*(z*)

Thus ifX(z) has a pole (or zero) at z=zo, it must have a pole (or zero) at the

complex conjugate point z= Zo*.


5/7

zxt(n) ~ Xt(z)

zX2(n)~~(Z)

ZXt(n) * xin) ~ Xt(z) ~(z)

with ROC = R1

with ROC = R2

with ROC atleast Rll~.

xt(n) * x2(n) =~xt(k) xin-k).k=_oo

=i: [i:Xt (k) xin-k) 1 z -nn =-o o k =_ oo

00 [00 1

=~Xt(k) ~x2(n-k) z-nj

k= _oo n= -oo

=f..X\(k) ~i:X2(l)Z-I. Z - k \k= _oo 1= _00

=~ Xl (k) Z-k ~ x2( 1 ) z - 1

k= _oo 1= _00

zx(n)~X(z)

,then nx(n) A-z dX(z)dz

Proof: Z[x(n)] =X(z) =L,x(n) z-n


6/7

dX(z) = Lx(n) (-n) Z-n-ldz

= -Z-IL[n x(n)]z-n

= -Z-I. Z[n x(n)]

dX(z):. z[n x(n)] = -z dz

5.4.9 .nitia. Va'ue Theorem

If x(n) = 0 for n < 0 (i.e. x(n) is causal)

Lt limthen, n-70 x(n) = x [O ) = Z-7OO X(z)

Proof: Z[x(n)) = X(z) = Lx(n) z-n

n=O

= 2,x(n) z -nn=O

=x(O) + x(1 )Z-I + x(2) Z-2+ .

Make limit Z-7OO on both the sides,

Lt X(z) =x(O) + 0 + 0 + .Z-700

:. Lt x(n) = x(O) = Lt X(z)Il-70 Z-7OO

5.4.10 Fina. Va'ue Theorem

Ifx(n)AX(z) and ifX(z) exists and no poles outside the unit circle and it has nodouble or higher order poles on the unit circle centered at the origin of the

Z-plane, then

Lt x(n) = x(oo) = Lt (z-1) X(z)n-7OO Z-71

Proof: Z[x(n)) =X(z)

Z[x(n+1)) = zX(z) - zx(O)

Substractingeqn. (5.11) from eqn. (5.12),

(5.11)

(5.12)


7/7

.L,x(n+l) Z-ll- .L,x(n) Z-ll= (z-l) X(z) - zx(O)

n=O n=O

Taking limit z~ 1on both the sides, we get

Lt Lt ~ }z~ 1 [(z-l) X(z) - z x(O)] = z~ 1 """'-' {x(n +1) - x(n) Z-ll

n= O

Lt Lt

n~oo x(n) = x(oo) = 1 (z-l) X(z)z~

Table 5.1 Som e com m on Z t ra ns fo rm pa i r s

Signal Traniform

1. 8(n) 1

2. u(n) 1l-z-1

3. -u(-n-l) 1l_z.1

4. 8(n-k) Z-k

ROC

All z

I z l >1

5. ex ,"u(n) 1l-e x ,z-1

6 . _ex,"u( -n-l) 1l-ex,z-I

7. nex,llu(n) ex,Z-1

(l-ex,zlf

8. -nex,"u( -n-l) ex,Z-1

(l-ex,z-lf

9. cos non. u(n) l-(cosno)zol1-(2 cosno) Z - I + Z-2

10. sin non. u(n) (sin no)z-1

1-(2 cos no) Z-I + Z-2

11. ex,"cos no n. u(n) 1- (ex,cos no)z-1

1- (2 e x , cos no) Z-1+ e x ,2 Z-2

12. < X " sin non. u(n) ( < x sin no)z.11-(2 ex,cos no) Z-1+ < X 2z-i

All z except

o (ifk>O) or00 (ifk I e x ,I

z transform properties

Documents