On Z-transform and Its Applications
In the fifth chapter applications of Z-transform in digital signal processing such as analysis of linear shift-invariant systems
The z-Transform and Its Application
Manolakis Digital Signal Processing: Principles
z-Transforms: basics difference equations
https://nucinkis-lab.cc.ic.ac.uk/HELM/HELM_Workbooks_21-25/WB21-all.pdf
Untitled
03-Jan-2012 The Z-Transform and Its. Application to the. Analysis of LTI Systems. CHAPTER. 3. Transform techniques are an important tool in the analysis ...
Introduction to the z-transform
Therefore for the examples and applications considered in this book we can restrict the z-transforms to be rational functions. This restriction is emphasized
Chapter 6 - The Z-Transform
06-Apr-2011 Characteristics • Z-Transform and Discrete Fourier Transform ... Although applications of Z-transforms are relatively new the essential ...
Chapter 3 - The z-transform and Analysis of LTI Systems
u[n] . Applying previous example with “p = e± ?0 ” and linearity: X(z) = 1.
Discrete - Time Signals and Systems Z-Transform-FIR filters
n output y n by applying inverse Z transform to Y yn. Y z z. -. ?--?. - ! -. Calculating the output of a FIR filter using Z transforms.
Chapter 5 - z-transform
z-transform sum. For our example we have. ?. ? n=0. (2 z. )n.
X(s)estds;
X(s) =1
1 x(t)estdt:X(s) =1
0 x(t)estdt:X(z) =1X
n=0x[n]zn?????X(z) =1X
n=02 nzn;X(z) =1X
n=0 2z n S=1X n=0a n R;?? ????? ??? ??? ??? ?? ??? ?????? ??z???? ????jzj> R:S= limN!1SN= limN!1N
X n=0a n: SN= (1 +a+a2+:::+aN):
aSN= (a+a2+:::+aN+aN+1):
SNaSN= (1aN+1)
SN(1a) = (1aN1):
???? ??a= 1;?? ???? ????SN=N+ 1:????a6= 1;??? ?????? ???? ????? ??(1a)?? ?????? SN=1aN+11a
S= limN!1SN= limN!11aN+11a;
????? ???? ???? ?? ????? ????jaj<1;??? ????? ?? ????S=11a:
S=N 2X n=N1a n= (aN1+aN1+1+:::+aN2) aS= (aN1+1+aN1+2+:::+aN2+aN2+1);S(1a) = (aN1aN2+1)
S=aN1aN2+11a
N lim N2!1S= limN
2!1aN1aN2+11a=aN11a;
2X n=N1a n=aN1aN2+11a;???a6= 1????? ???1 X n=N1a n=aN11a;???jaj<1:?????X(1) =1X
n=02 n1n 1X n=02 n;Z(3) =1X
n=02 n3n 1X n=0 23n 1123
= 3: n=0 2z n
X(z) =1X
n=0 2z n 112z;2z <1 zz2;jzj>2;
X(z) =1X
n=0a nzn 1X n=0 az n;X(z) =zza:
az <1,jajjzj<1, jajX(z) =zza;jzj>jaj:
X aX X3(z) =1X
n=0(ax1[n] +bx2[n])zn 1X n=0ax1[n]zn+1X
n=0bx2[n]zn
=a1X n=0x1[n]zn+b1X
n=0x2[n]zn
=aX1(z) +bX2(z):X(z) =1X
n=0a nzn=zza;jzj>jaj ddzX(z) =ddz
1X n=0a nzn! ddz zza ;jzj>jaj =1X n=0na nzn1=a(za)2;jzj>jaj: zddzX(z) =1X
n=0na nzn=az(za)2;jzj>jaj: 1X n=0na nzn=az(za)2;jzj>jaj: nx[n], zddz X(z) 12 n(n1)x[n],a2z(za)3;jzj>jaj;1m!n(n1)(nm+ 1)an,amz(za)m+1;jzj>jaj:
X(z) =1X
n=0cos(!n)zn 1X n=012 (ej!n+ej!n)zn 12 1 X n=0(ej!z1)n+12 1 X n=0(ej!z1)n 1211ej!z1+12
11ej!z1;jzj>jej!j= 1
12 zzej!+12 zzej!;jzj>1 12 z(zej!)z2z(ej!+ej!) + 1+z(zej!)z
2z(ej!+ej!) + 1
;jzj>1 z2zcos(!)z22zcos(!) + 1;jzj>1:
x[n] =12jej!nej!n12jej!n12jej!n
X(z) =12jzzej!12jzzej!;jzj>1
12jz(zej!)z(zej!)z
22zcos(!) + 1;jzj>1
zsin(!)z22zcos(!) + 1;jzj>1:
[n],1: [nk],zk; u[n],zz1;jzj>1:Z(ay1[n] +by2[n]) =aY1(z) +bY2(z):
1 X n=0(ay1[n] +by2[n])zn=a1X n=0y1[n] +b1X
n=0y 2[n] =aY1(z) +bY2(z):Z(y[nk]u[nk]) =zkY(z):
Z(y[nk]u[nk]) =1X
n=0y[nk]u[nk]zn 1X n=ky[nk]zn 1X m=0y[m]z(m+k) =zkY(z); ????? ??? ?????? ???? ??????? ????u[nk]????? ???? ???n < k??? ??? ????? ???? ??????? ???? ?????? ???Z(y[nk]) =zk"
Y(z) +kX
m=1y[m]zm# y[nk] =y[nk]u[nk] +kX m=1y[m][nk+m]Z(y[nk]) =zkY(z) +kX
m=1y[m]z(km) =zk"Y(z) +kX
m=1y[m]zm#Z(y[n+k]u[n]) =zk"
Y(z)k1X
m=0y[m]zm#Z(y[n+k]u[n]) =1X
n=0y[n+k]zn =zk1X n=0y[n+k]z(n+k) =zk1X m=ky[m]zm =zk 1X m=0y[m]zmk1X m=0y[m]zm! =zkY(z)k1X
m=0y[m]zm! y[n] =1X m=1h[m]x[nm]:Y(z) =1X
n=0y[n]zn 1X n=01 X m=1h[m]x[nm]zn 1X m=1h[m]1X n=0x[nm]zn 1X m=1h[m]X(z)zm =X(z)1X m=1h[m]zm =X(z)H(z); 1 X k=1x[nk]y[nk]h[k]u[k] =nX k=0x[nk]h[k] !X(z)H(z); y[n] =12jY(z)zn1dz
Y(z) =b0+b1z+:::+bMzMa
0+a1z+:::+aNzN:
Y(z) =zza
zza=1 + az +a2z 2 za)z za0 +a+ 00 +aa2z
0 + 0 +
a2z0 + 0 +a2z
a3z 2Y(z) = 1 +az1+a2z2+a3z3+
=y[0] +y[1]z1+y[2]z2+y[3]z3+; ???? ????? ?? ??? ????? ????y[n] =an;n0: ???????x[n+ 1]u[n],z(X(z)x[0]) nX k=0x[k]y[nk],X(z)Y(z) k=1x[k]y[nk],X(z)Y(z) ??????? ??k y[n+k]u[n],zk"Y(z)k1X
m=0y[m]zm#Y(z) +kX
x[n],X(z) =1X n=0x[n]znROCX [n] =(1; n= 0
0; n6= 0,1???z
[nk],( zk; k00; k <0z6= 1
a n,zzajzj>jaj na n,az(za)2jzj>jaj a nsin(!n),azsin(!)z22azcos(!) +a2jzj>jaj
a ncos(!n),1azcos(!)z22azcos(!) +a2jzj>jaj
u[n] =(1; n= 0
0; n6= 0,zz1jzj>1
1,zz1jzj>1
X(z) =b0+b1z+bMzMa
0+a1z++aNzN=b0+b1z+bMzM(zr1)(zr2)(zrN);
X(z) =NX
k=1A kzzk; A k= (zrk)X(z)jz=rk;X(z) =NX
k=1A kzzk=NX k=1z1Akzzrk,x[n] =NX
k=1A krn1 ku[n1];Y(z) =z1(z2)(z3):
Y(z) =z1(z2)(z3)=A1z2+A2z3;
quotesdbs_dbs21.pdfusesText_27[PDF] application of z transform with justification
[PDF] application pour apprendre l'anglais gratuit sur pc
[PDF] application security risk assessment checklist
[PDF] applications of composite materials
[PDF] applications of dft
[PDF] applications of exponential and logarithmic functions pdf
[PDF] applications of therapeutic drug monitoring
[PDF] applied environmental microbiology nptel
[PDF] applied environmental microbiology nptel pdf
[PDF] applied information and communication technology a level
[PDF] applied information and communication technology a level notes
[PDF] applied microbiology in the field of environment
[PDF] applied robotics with the sumobot
[PDF] apply for scholarship canada