1.20 Display impulse response of H(z) and the impulse response of its continuous time approximation H(s)

Plot the impulse response of \(H(z)=z/(z^2-1.4 z+0.5)\) and using sampling period of \(T=0.5\) find continuous time approximation using zero order hold and show the impulse response of the system and compare both responses.

Mathematica

maxSimulationTime = 10; 
samplePeriod      = 0.5; 
tf = TransferFunctionModel[z/(z^2-1.4 z+0.5), 
          z, 
          SamplingPeriod->samplePeriod];
 

Find its impulse response

discreteResponse=First@OutputResponse[tf,DiscreteDelta[k], 
                                           {k,0,maxSimulationTime}]
 

{0.,1.,1.4,1.46,1.344,1.1516,0.94024,0.740536, 
0.56663,0.423015,0.308905,0.22096,0.154891, 
0.106368,0.0714694,0.0468732,0.0298878, 
0.0184063,0.0108249,0.00595175,0.00291999}
 

approximate to continuous time, use ZeroOrderHold

ctf = ToContinuousTimeModel[tf, s, Method -> "ZeroOrderHold"]
 

Find the impulse response of the continuous time system

continouseTimeResponse=Chop@First@OutputResponse[ctf,DiracDelta[t],t]
 

\[ -1.25559 e^{-0.693147 t} (-13.3012 \theta (t) \sin (0.283794 t)-1. \theta (t) \cos (0.283794 t)) \]

continouseTimeResponse=Chop@First@OutputResponse[ctf,DiracDelta[t], 
                                         {t,0,maxSimulationTime}]
 

Plot the impulse response of the discrete system

ListPlot[ 
discreteResponse, 
DataRange->{0,maxSimulationTime}, 
Filling->Axis, 
FillingStyle->{Red, 
  PlotMarkers->Graphics[{PointSize[0.03], 
  Point[{0,0}]}]},PlotRange->All, 
  ImageSize->300,ImagePadding->{{45,10},{35,10}}, 
  AspectRatio->0.9,AxesOrigin->{0,0},Frame->True,Axes->None, 
  ImageSize->300, 
  Frame->True 
  ]
 

Plot the impulse response of the continuous system

Plot[samplePeriod *continouseTimeResponse,{t,0,maxSimulationTime}, 
     PlotRange->All,Frame->True,ImageSize->300]
 

Plot both responses on the same plot

p = Show[ 
  ListPlot[ 
   discreteResponse, 
   Filling -> Axis, 
   FillingStyle -> {Red, 
     PlotMarkers -> Graphics[{PointSize[0.03], Point[{0, 0}]}]}, 
   PlotRange -> All, 
   DataRange -> {0, maxSimulationTime}, 
   ImageSize -> 300, 
   ImagePadding -> {{45, 10}, {35, 50}}, 
   AspectRatio -> 0.9, 
   AxesOrigin -> {0, 0}, 
   Frame -> True, 
   Axes -> None], 
  Plot[samplePeriod *continouseTimeResponse, {t, 0, maxSimulationTime}, 
   PlotRange -> All], 
  PlotRange -> All, 
  FrameLabel -> {{Style["response", 10], None}, 
     {Style["time (sec)", 10], 
    "impulse response of discrete system \ 
     with its continouse time approximatio"}} 
  ]
 

Do the same plot above, using stair case approximation for the discrete plot

Show[ 
ListPlot[ 
 discreteResponse, 
 Filling->None,PlotRange->All,DataRange->{0,maxSimulationTime}, 
 ImageSize->300,ImagePadding->{{45,10},{35,50}},AspectRatio->0.9, 
 AxesOrigin->{0,0},Frame->True,Axes->None, 
 Joined->True,InterpolationOrder->0,PlotStyle->Red], 
Plot[samplePeriod *continouseTimeResponse,{t,0,maxSimulationTime}, 
 PlotRange->All], 
 PlotRange->All,FrameLabel->{{Style["response",10],None}, 
 {Style["time (sec)",10],"impulse response"}}]
 

Matlab

clear all; 
maxSimulationTime = 15; 
samplePeriod = 0.5; 
 
z=tf('z',samplePeriod); 
tf=z/(z^2-1.4*z+0.5) 
impulse(tf,0:samplePeriod:maxSimulationTime) 
 
ctf=d2c(tf,'zoh') 
 
hold on 
impulse(ctf,0:maxSimulationTime) 
title(['impulse response of discrete system with',... 
       ' its continuous time approximation']); 
xlabel('time (sec)'); 
ylabel('response');