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If a system has the following step response:

$$\begin{align} y(t) &= 1 - \Big(A e^{−t/T_1} + (1-A) e^{−t/T_2}\Big) \\ &= 1 - A e^{−t/T_1} - (1-A) e^{−t/T_2} \\ \end{align}$$

and I know the values of $A$, $T_1$ and $T_2$, can I transform this to a discrete-time IIR filter analytically? If there is no analytical solution, is it possible to say how many poles/zeros the IIR filter should have?

I've done it numerically by using the Prony's method on the impulse response, but I'm hoping there is a more direct way.

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You can use the step-invariant transformation, which makes sure that the step response of the discrete-time system matches the step response of the continuous-time system at the sampling instants. I suppose this is what you mean by "exact solution".

Sampling the given continuous-time step response at $t=nT$ gives the desired step response of the discrete-time system:

$$g[n]=y(nT)=\left[1-Ae^{-nT/T_1}-(1-A)e^{-nT/T_2}\right]u[n]\tag{1}$$

where $u[n]$ is the unit step. The $\mathcal{Z}$-transform of $(1)$ is

$$G(z)=\frac{1}{1-z^{-1}}-\frac{A}{1-e^{-T/T_1}z^{-1}}-\frac{1-A}{1-e^{-T/T_2}z^{-1}}\tag{2}$$

The frequency response of the desired discrete-time system is then given by

$$H(z)=(1-z^{-1})G(z)\tag{3}$$

After performing some basic algebra and using $a=e^{-T/T_1}$ and $b=e^{-T/T_2}$ we obtain

$$H(z)=\frac{[1-b+A(b-a)]z^{-1}+[a(b-1)+A(a-b)]z^{-2}}{1-(a+b)z^{-1}+abz^{-2}}\tag{4}$$

As a result, the IIR filter with transfer function $(4)$ has a step response that equals the given continuous-time step response $y(t)$ at the sample instants $t=nT$:

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  • $\begingroup$ That's great! The denominator factors easily too. $\endgroup$ Commented Sep 4, 2018 at 7:51
  • $\begingroup$ I rearranged the numerator a bit for myself: $b_1 = 1 - A \cdot a - (1 - A)b$ and $b_2 = a \cdot b - A \cdot b - (1 - A)a$. Many thanks! $\endgroup$ Commented Sep 4, 2018 at 11:11
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Here is a messy way it can be done. I started it and then it got too involved for me to finish, but I can see it is solvable.

First I simplified the equation to make the algebra less messy:

$$ y(t) = C + A e^{\alpha t} + B e^{\beta t} $$

You should be able to see the equivalence. Then I repeated it for two different domain values:

$$ y(t-1) = C + A e^{\alpha t} e^{-\alpha} + B e^{\beta t}e^{-\beta} $$

$$ y(t-2) = C + A e^{\alpha t} e^{-2\alpha} + B e^{\beta t}e^{-2\beta} $$

Next, I used the first equation to get:

$$ A e^{\alpha t} = y(t) - C - B e^{\beta t} $$

You can then substitute that into the second two equations. The two resulting equations can then be combined to get rid of the $ B e^{\beta t} $ term. In the mess you have left, you should be able to solve for $y(t)$ as a linear combination of $y(t-1)$ and $y(t-2)$ and a constant term, all with known values.

$$ y(t) = X + Y y(t-1) + Z y(t-2) $$

If that doesn't work out for you, I'll take another stab at it. There is probably a slicker way.

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