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Physical Audio Signal Processing
    Transfer Function Models
       Sampling the Impulse Response
          Impulse Invariant Method

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Impulse Invariant Method

The impulse-invariant method converts analog filter transfer functions to digital filter transfer functions in such a way that the impulse response is the same (invariant) at the sampling instants [353], [371, pp. 216-219]. Thus, if $ \gamma(t)$ denotes the impulse-response of an analog (continuous-time) filter, then the digital (discrete-time) filter given by the impulse-invariant method will have impulse response $ \gamma(nT)$, where $ T$ denotes the sampling interval in seconds. Moreover, the order of the filter is preserved, and IIR analog filters map to IIR digital filters. However, the digital filter's frequency response is an aliased version of the analog filter's frequency response.R.3

To derive the impulse-invariant method, we begin with the analog transfer function

$\displaystyle \Gamma_a(s) \isdefs \frac{B_a(s)}{A_a(s)} \isdefs \frac{b_a(0) ... ... b_a(N-2)s + b_a(N-1)}{% s^{N} + a_a(1) s^{N-1} + \cdots + a_a(N-1)s + a_a(N)} $

and perform a partial fraction expansion (PFE) down to first-order terms [460]:R.4

$\displaystyle \Gamma_a(s) \eqsp \sum_{i=1}^N \frac{K_i}{s-s_i}, $

where $ s_i$ is the $ i$th pole of the analog system, and $ K_i$ is its residue [460]. The PFE is always possible when $ \Gamma (s)$ is a strictly proper transfer function (more poles than zeros [460]).R.5 We now perform the inverse Laplace transform on the partial fraction expansion to obtain the impulse response in terms of the system poles and residues:

$\displaystyle \gamma_a(t) \eqsp \sum_{i=1}^N K_i e^{s_i t}, \quad t\ge 0. $

We now sample at intervals of $ T$ seconds to obtain the digital impulse response

$\displaystyle \gamma_d(n) \isdefs \gamma_a(nT) \eqsp \sum_{i=1}^N K_i e^{s_i nT}, \quad n= 0,1,2,\ldots\,. $

Taking the z transform gives the digital filter transfer function designed by the impulse-invariant method:

$\displaystyle \Gamma_d(z) \eqsp \sum_{i=1}^N \frac{K_i}{1 - e^{s_iT}z^{-1}} \isdefs \frac{B_d(z)}{A_d(z)}. $

We see that the $ s$-plane poles $ s_i$ have mapped to the $ z$-plane poles

$\displaystyle \zbox {z_i \isdefs e^{s_iT}} \protect$ (R.1)

and the residues have remained unchanged. Clearly we must have $ -\pi < \omega_a T < \pi$, i