trivial digital filter

[Hugo Coolens]

I'm trying to simulate the following digital filter in octave:
sys=tf([1 1],[1],1e-3)
However this gives an error message:
# of poles (0) < # of zeros (1)
Which is of course clear enough but I wonder whether there is a way to use
a "similar method" as the one above in octave for this kind of filter?

regards,
Hugo


To demonstrate the filter I already simulated its behaviour in the time
domain as follows:
a=[1];
b=[1 1];
fs=1000;
time=0:1/fs:40e-3;
x=sin(2*pi*200*time);
y=filter(b,a,x);
stem(time,y)



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