Z-transform for a signal with an unknown sampling rate

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SUMMARY

The discussion centers on deriving the z-transform for a delayed ramp function x(t) = (t-2)heaviside(t-2) sampled at an unknown rate T. The original z-transform is given as X(z) = z / (z-1)^2. The user proposes a modified z-transform equation X(z) = T*(z / (z-1)^2)*(z^-ceil(2/T)), which incorporates the delay and sampling rate. The community seeks to confirm the mathematical correctness of this formulation.

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mtarek16
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Homework Statement


Given a ramp function x(t) = t*heaviside(t) with the known z-transform ( z / (z-1)^2 ).
If the time-domain signal has been delayed by 2 seconds, then sampled with an unknown sampling rate (T). How could I get the z-transform (one-sided) for the discrete signal, for any values of T ?
I guess my question is really how to express that delay in the Z-domain. I have searched a lot and couldn't find any good resources.

Homework Equations


original signal : x(t) = t*heaviside(t) - X(z) = z / (z-1)^2
delayed signal: x(t) = (t-2)*heaviside(t-2) - X(z) = ?

The Attempt at a Solution


I have been trying a lot with this problem. The last (and seemingly to me) the closest I've got is :
X(z) = T*( z / (z-1)^2 )*(z^-s)
where :
s = 2/T if T < 1
s = 2*T if 1<= T < 2
s = T if T >= 2
Of course, if s turns out to be a non-integer value, the solution would be wrong.

Any help on that is greatly appreciated.

Regards,
MT
 
Last edited:
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Another take :
X(z) = T*( z / (z-1)^2 )*(z^-ceil(s))
where :
s = 2/T if T <= 1
s = T if T > 1

Seems right to me, but is the equation is in an acceptable/correct mathematical form ??
 
Uh ! I'm sorry, the conditions I wrote are again wrong, here's my final approach :
X(z) = T*( z / (z-1)^2 )*(z^-ceil(2/T))

Again, it seems to me that it's functionally correct, although I'm not sure if it's syntactically correct too.
 

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