step() giving a smooth curve for a discrete function

I am trying to work out the error between the a function and a discrete ZOH version of itself as shown in the image. When I use (R=step(Z,0:0.1:4); plot(x,R,'*') at the end it brings up a non ZOH held version of the TF. For some reason the output data (G(HS)) remains the same as H(s) yet the graph is ZOH.

%Method one

x=0:0.1:4;

hold on

H=tf(5,[1 5],'InputDelay',1); %http://au.mathworks.com/help/control/examples/specifying-time-delays.html

figure(1)

% step(H,0:0.1:4,'*');

% y=5*heaviside(x-1).*(1-exp(-5.*(x-1)));

%Method two

syms s c t z n

F=5*exp(-1*s)/(s+5)*(1/s); %with step responce

pretty(F)

assume(t > 0) %assume t>0 for the conversion

f=ilaplace(F,s,t); %inverse laplace

simplify(f)

figure(1)

o=ezplot(f,[0,4]); %plot for time domain from 0s to 4s

set(o,'Color',[1,0,0])

set(o,'LineStyle', '- -','LineWidth',2)

%%

% hold off

% figure(2)

% bode(H)

%%

Z=c2d(H,0.1,'zoh');

hold on

figure(1)

R=step(Z,0:0.1:4);

plot(x,R,'*')

Answer 1

[Arkadiy Turevskiy]

It seems you are getting tripped up by the fact that the code below results in two identical lines

y1=step(H,x);
y2=step(c2d(H,0.1),x);
plot(x,y1); 
hold;
plot(x,y2,'r')

but this code does not

close;
step(H,x);
hold;
step(c2d(H,0.1),x,'r');

There is no error here. As you noticed y1 and y2 are identical, as they should be, because your discretization step of 0.1 is the same step you use for defining x.

The reason why the first batch of code does not produce the same graph as second batch of code is that function plot always interpolates between data points. It is not the right function to use for plotting the step response of the discrete-time transfer function. For that, use the function stairs:

close;
plot(x,y1); 
hold;
stairs(x,y2,'r')

Hope this helps.