# Representing a continuous LTI system as a discrete one

I am aware that there are different ways to represent a continuous time system in discrete domain (e.g. bilinear transform, impulse invariance transform).

But my problem is as follows: Given an arbitrary LTI system $$h_c(t)$$, its output is given as $$y(t) = (x * h_c)(t)$$.

Now suppose $$x(t)$$ is given as:

$$x(t) = \sum_{n=-\infty}^{\infty} x[n] \operatorname{sinc}\left(\frac{t-n T}{T}\right)$$

i.e., Whittaker interpolation formula. I can then write a discrete-time system:

$$y[n] = \sum_{q=-\infty}^{\infty} h[q] x[n-q] ,$$

such that $$y[n]$$ is identical to $$y(n T)$$.

My question: What is the relation between $$h[n]$$ and $$h_c(t)$$?

There must be an exact one and it is neither given by impulse invariance nor bilinear transform. I am aware that when $$H_c(f)$$ is bandlimited to $$1/(2B)$$, the answer is just impulse invariance transform, i.e., $$h[n] = h_c(nT)$$. If this is not satisfied, it will result in aliasing but this should not prevent such a relationship from existing.

The sequence $$h[n]$$ consists of samples of a low pass filtered version of $$h_c(t)$$. Define that low pass filtered impulse response as

$$\tilde{h}_c(t)=\int_{-\infty}^{\infty}h_c(\tau)\frac{\sin(\pi(t-\tau)/T)}{\pi(t-\tau)/T}d\tau\tag{1}$$

and

$$h[n]=\tilde{h}_c(nT)\tag{2}$$

The output signal is

\begin{align}y(t)&=(x\star h_c)(t)\\&=\int_{-\infty}^{\infty}x(t-\tau)h_c(\tau)d\tau\\&=\int_{-\infty}^{\infty}\sum_{k=-\infty}^{\infty}x[k]\frac{\sin(\pi(t-kT-\tau)/T)}{\pi(t-kT-\tau)/T}h_c(\tau)d\tau\\&=\sum_{k=-\infty}^{\infty}x[k]\int_{-\infty}^{\infty}h_c(\tau)\frac{\sin(\pi(t-kT-\tau)/T)}{\pi(t-kT-\tau)/T}d\tau\\&=\sum_{k=-\infty}^{\infty}x[k]\tilde{h}_c(t-kT)\tag{3}\end{align}

Consequently, we have

\begin{align}y(nT)&=\sum_{k=-\infty}^{\infty}x[k]\tilde{h}_c((n-k)T)\\&=\sum_{k=-\infty}^{\infty}x[k]h[n-k]\end{align}\tag{4}

This results is also obvious from the fact that $$(x\star h_c)(t)=(x\star\tilde{h}_c)(t)$$ must hold, because $$x(t)$$ is band-limited. And since $$\tilde{h}_c(t)$$ is also band-limited, we can simply take samples from $$x(t)$$ and $$\tilde{h}_c(t)$$ to obtain the result.