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I have an analog, continuous impulse response $$h_a(t)=u(t)\cdot\sum_{n=1}^4A_ne^{s_nt}$$, and by sampling it with the usual method I get $$h[n]=\Delta th_a(n\Delta t).$$ Now, that $\Delta t$ is a bit problematic for me, because upon performing the z-transform of $h[n]$ I get $$H(z)=\sum_{n=0}^\infty h[n]z^{-n}\\=\sum_{n=0}^\infty\sum_{k=1}^4\Delta tA_k\Big(\frac{e^{s_k\Delta t}}{z}\Big)^{n}\\=\Delta t\sum_{k=1}^4A_k\sum_{n=0}^\infty\Big(\frac{e^{s_k\Delta t}}{z}\Big)^{n}\\=\Delta t\sum_{k=1}^4A_k\sum_{n=0}^\infty\beta^n \\=\Delta t\sum_{k=1}^4A_k\frac{1}{1-\beta}$$ $$=\sum_{k=1}^4\frac{\mathbf{\Delta t}A_k}{1-e^{s_k\Delta t}z^{-1}}$$

with $\beta\triangleq\frac{e^{s_k\Delta t}}{z},|\beta|<1$. If I implement such transfer function in an IIR discrete-time filter, each coefficient in the difference equation contains $\Delta t$, except for the coefficients in the feedback part. This results in an incorrect amplitude, when compared to the plot of $h_a(t)$. Therefore, I wonder, since not including $\Delta t$ makes the discretized $h[n]$ practically equal to $h_a(t)$ (in the sampling points), why would I use that coefficient at all? Or better, am I supposed to use it and since I get different results am I doing something wrong? I remember it was necessary for some definition (DFT or autofunctions), but I don't see any practical use of it.

Sorry if it's a stupid question, I hope I haven't missed anything.

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