How to derive the Z-transform of a low-pass filter(LPF)

Here we describe the steps on how to derive the Z-transform of a low-pass filter based on its continuous-time transfer function or impulse response.  

Here's a breakdown of how to approach low pass filter from Z-Transform:

1. Start with a desired frequency response The usual method to design an IIR filter is to start with a frequency response you want. A low-pass filter passes signals with frequencies lower than a cutoff frequency and attenuates higher frequencies.
2. Continuous-Time Transfer Function: You often begin with the transfer function of a continuous-time low-pass filter. A common example is the first-order RC low-pass filter, which has the transfer function $H(s)=\frac{a}{s+a}$, where $a=\frac{1}{RC}$.
3. Discretization: You need to convert the continuous-time transfer function $H(s)$ into a discrete-time transfer function $H(z)$ using a discretization method. One common method is to use the Z-transform with a zero-order hold: $H(z)=\frac{z-1}{z}Z\left(L^{-1}\right(\frac{H(s)}{s}\left)\right)$. You can use the online Z-Transform calculator for this step.                                                                                                      
4. Example Discretization: The search results show the discretization of $H(s)=\frac{{\omega }_c}{s+{\omega }_c}$ which leads to $H(z)=\frac{1-e^{-{\omega }_{c}T}}{z-e^{-{\omega }_{c}T}}$.
5. Inverse Z-Transform: After obtaining $H(z)$, you can express it as a difference equation by taking the inverse Z-transform1. This difference equation describes how to compute the output $y[n]$ based on past outputs $y[n-1]$ and current/past inputs $x[n]$.
6. Z-Transform of the Input Signal: The Z-transform of the input signal $x[n]$ is represented as $X(z)$. For example, if $x(n)=120,132,144,155,163,\mathrm{.}\mathrm{.}\mathrm{.}$, then $X(z)=120+132z^{-1}+144z^{-2}+155z^{-3}+\mathrm{.}\mathrm{.}\mathrm{.}$.
7. Output Response: The output response in the Z-domain, $Y(z)$, is the product of the input $X(z)$ and the filter's transfer function $H(z)$: $Y(z)=X(z)H(z)$.

In summary, there isn't a single Z-transform sequence input for a low-pass filter. The filter itself has a Z-transform (its transfer function, $H(z)$, and the input signal has a Z-transform ($X(z)$. The output $Y(z)$ is the product of these. The specific form of $H(z)$ depends on the design choices for the low-pass filter (cutoff frequency, order, etc.).