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 [346], [365, 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.^9.3

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

$$\Gamma_a(s) \isdefs \frac{B_a(s)}{A_a(s)} \isdefs \frac{b_a(0) s^{N-1} + b_a(1) s^{N-2} + \cdots + b_a(N-2)s + b_a(N-1)}{s^N + a_a(1) s^{N-1} + \cdots + a_a(N-1)s + a_a(N)} \protect$$ (9.1)

and perform a partial fraction expansion (PFE) down to first-order terms [452]:^9.4

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

where $s_i$ is the $i$ th pole of the analog system, and $K_i$ is its residue [452]. Assume that the system is at least marginally stable [452] so that there are no poles in the right-half plane ( $\mbox{re\ensuremath{\left\{s_i\right\}}}\le 0$ ). Such a PFE is always possible when $\Gamma (s)$ is a strictly proper transfer function (more poles than zeros [452]).^9.5 Performing the inverse Laplace transform on the partial fraction expansion we obtain the impulse response in terms of the system poles and residues:

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

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

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

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

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

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

$$\zbox {z_i \isdefs e^{s_iT}} \protect$$ (9.2)

and the residues have remained unchanged. Clearly we must have $-\pi < \mbox{im\ensuremath{\left\{s_i\right\}}} T < \pi$ , i.e., the analog poles must lie within the bandwidth spanned by the digital sampling rate $f_s=1/T$ . Otherwise, the pole angle $\mbox{im\ensuremath{\left\{s_i\right\}}} T$ will be aliased into the interval $[-\pi,\pi)$ . Stability is preserved since $\mbox{re\ensuremath{\left\{s_i\right\}}} \le 0 \;\Leftrightarrow\; \vert z_i\vert \le 1$ .

Note that the series combination of two digital filters designed by the impulse-invariant method is not impulse invariant. In other terms, the convolution of two sampled signals is not the same as the sampled convolution of those two (continuous-time) signals. This is easy to see when aliasing is considered. For example, let the signal $h(t)=2f_c$sinc$(2f_ct)=2f_c\sin(\pi 2f_c t)/(\pi 2f_c t)$ be the impulse-response of an ideal lowpass-filter having cut-off frequency $f_c$ tuned below half the sampling rate $f_s=1/T$ , i.e., $f_c < f_s/2$ . Then the sampled version of this signal $h(nT)$ contains no aliasing. Let a second signal be defined as $x(t) = \cos(2\pi f_s t)$ . Then $x(nT)=1$ for all $n$ , and we obtain the discrete-time convolution $(x\ast h)(n) = \sum_n h(nT)$ . On the other hand, the continuous-time convolution $(h\ast x)(t) = 0$ for all $t$ , and the sampling of that yields only zeros.