# Applying a time delay with Z-Transform

I am trying to take a signal $F(t)$ that has been sampled at some time DelT, I then wish to pass this signal through a channel $H(s)$. To do this I am sampling my signal $H(s)$ at DelT time intervals then finding the Z-transform $H(z)$. How would I go about applying this in matlab?

Currently I am using c2d to convert s-domain to z-transform for my channel but am unsure how I can go about the convolution with my sampled signal which is just an array of numbers. I have tried to take the fft of sampled $F(t)$ then use freqz of $H(z)$. To begin with I am just trying to apply a simple time delay $e^{-st}$ to the signal.

• Do you understand the physical meaning of $z$ (or $z^{-1}$) in a transfer function in the Z-domain? – fibonatic May 14 '16 at 10:32

There's a lot of confusion in your question. You refer to "a channel $H(s)$", then to "my signal $H(s)$", so for the sake of clarity, let's properly define:

1. Input signal - continuous time $f(t)$, impulse sampled $f[n]=f(n\Delta T)$, Fourier transform of continuous time signal $F(s)$, Z-transform of impulse sampled signal $F(z)=F(e^{s\Delta T})$.
2. Continuous time channel - impulse response $h(t)$, transfer function (Laplace transform of impulse response) $H(s)$.
3. Discrete time channel - discrete impulse response $h[n]$, transfer function (Z-transform of discrete impulse response) $F(z)$.

Note the definitions above follow a convention:

1. time-domain signals/functions are lower case ($f, h$); frequency-domain signals/functions are upper case ($F, H$).
2. arguments of continuous signals/functions are enclosed in parentheses $(t)$; arguments of discrete signals/functions are enclosed in brackets $[n]$.

Then, you write:

(...) z-transform for my channel but am unsure how I can go about the convolution with my sampled signal which is just an array of numbers

It seems you do not understand the basics of LTI systems. Note that $H(s)$ and $H(z)$ are frequency-domain and $f(t)$ and $f[n]$ are time-domain.

Convolution makes sense when the two convolved functions are in the same exact domain (continuous/discrete time/frequency), although for the usual purposes the two convolved functions should both be in the continuous/discrete time domain.

If you're dealing with a discrete-time signal ("signal which is just an array of numbers"), then you might want to convolve $f[n]$ with $h[n]$. If you already have $H(z)$, then an inverse Z-transform on $H(z)$ would produce $h[n]$.