Show the use of the inverse Z transform

Clear["Global`*"]; 
 
x[n_]:= InverseZTransform[z/(z-1),z,n] 
 
ListPlot[Table[{n,x[n]},{n,0,10}], 
  Frame->True, 
  FrameLabel->{{"x[n]",None}, 
      {"n","Inverse Ztransform of z/(z-1)"}}, 
  FormatType->StandardForm, 
  RotateLabel->False, 
  Filling->Axis, 
  FillingStyle->Red, 
  PlotRange->{Automatic,{0,1.3}}, 
  PlotMarkers->{Automatic,12}]

pict

function e19() 
 
nSamples = 10; 
z = tf('z'); 
h = z/(z-1); 
[num,den] = tfdata(h,'v'); 
[delta,~] = impseq(0,0,nSamples); 
xn = filter(num,den,delta); 
 
stem(xn); 
title('Inverse Z transform of z/(z-1)'); 
xlabel('n'); ylabel('x[n]'); 
ylim([0 1.3]); 
set(gcf,'Position',[10,10,400,400]); 
 
end 
 
%function from Signal Processing by Proakis 
%corrected a little by me for newer matlab version 
function [x,n] = impseq(n0,n1,n2) 
% Generates x(n) = delta(n-n0); n1 <= n,n0 <= n2 
% ---------------------------------------------- 
% [x,n] = impseq(n0,n1,n2) 
% 
if ((n0 < n1) || (n0 > n2) || (n1 > n2)) 
    error('arguments must satisfy n1 <= n0 <= n2') 
end 
n = n1:n2; 
x = (n-n0) == 0; 
end

pict

1.13.2 example 2

In Mathematica analytical expression of the inverse Z transform can be generated as well as shown below

Clear["Global`*"]; 
x[n_]:= InverseZTransform[(5 z)/(z-1)^2,z,n]; 
 
ListPlot[Table[{n,x[n]},{n,0,10}], 
   Frame->True, 
   FrameLabel->{{"x[n]",None}, 
  {"n","Inverse Ztransform of (5 z)/(z-1)^2"}}, 
   FormatType->StandardForm, 
   RotateLabel->False, 
   Filling->Axis, 
   FillingStyle->Red, 
   PlotRange->{Automatic,{0,60}}, 
   PlotMarkers->{Automatic,12}, 
   ImageSize->350]

pict

function e19_2() 
 
nSamples = 10; 
z = tf('z'); 
h = (5*z)/(z-1)^2; 
[num,den] = tfdata(h,'v'); 
[delta,~] = impseq(0,0,nSamples); 
xn = filter(num,den,delta); 
 
stem(xn); 
title('Inverse Z transform of 5z/(z-1)^2'); 
xlabel('n'); ylabel('x[n]'); 
ylim([0 60]); 
set(gcf,'Position',[10,10,400,400]); 
 
end

pict

1.13 Show the use of the inverse Z transform

1.13.1 example 1

1.13.2 example 2