Impulse invariance is a technique for designing discrete-time infinite-impulse-response (IIR) filters from continuous-time filters in which the impulse response of the continuous-time system is sampled to produce the impulse response of the discrete-time system. The frequency response of the discrete-time system will be a sum of shifted copies of the frequency response of the continuous-time system; if the continuous-time system is approximately band-limited to a frequency less than the Nyquist frequency of the sampling, then the frequency response of the discrete-time system will be approximately equal to it for frequencies below the Nyquist frequency.

Discussion

The continuous-time system's impulse response, ${\displaystyle h_{c}(t)}$, is sampled with sampling period ${\displaystyle T}$ to produce the discrete-time system's impulse response, ${\displaystyle h[n]}$.

${\displaystyle h[n]=Th_{c}(nT)\,}$

Thus, the frequency responses of the two systems are related by

${\displaystyle H(e^{j\omega })={\frac {1}{T))\sum _{k=-\infty }^{\infty }{H_{c}\left(j{\frac {\omega }{T))+j{\frac {2{\pi )){T))k\right)}\,}$

If the continuous time filter is approximately band-limited (i.e. ${\displaystyle H_{c}(j\Omega )<\delta }$ when ${\displaystyle |\Omega |\geq \pi /T}$), then the frequency response of the discrete-time system will be approximately the continuous-time system's frequency response for frequencies below π radians per sample (below the Nyquist frequency 1/(2T) Hz):

${\displaystyle H(e^{j\omega })=H_{c}(j\omega /T)\,}$ for ${\displaystyle |\omega |\leq \pi \,}$

Comparison to the bilinear transform

Note that aliasing will occur, including aliasing below the Nyquist frequency to the extent that the continuous-time filter's response is nonzero above that frequency. The bilinear transform is an alternative to impulse invariance that uses a different mapping that maps the continuous-time system's frequency response, out to infinite frequency, into the range of frequencies up to the Nyquist frequency in the discrete-time case, as opposed to mapping frequencies linearly with circular overlap as impulse invariance does.

Effect on poles in system function

If the continuous poles at ${\displaystyle s=s_{k))$, the system function can be written in partial fraction expansion as

${\displaystyle H_{c}(s)=\sum _{k=1}^{N}{\frac {A_{k)){s-s_{k))}\,}$

Thus, using the inverse Laplace transform, the impulse response is

${\displaystyle h_{c}(t)={\begin{cases}\sum _{k=1}^{N}{A_{k}e^{s_{k}t)),&t\geq 0\\0,&{\mbox{otherwise))\end{cases))}$

The corresponding discrete-time system's impulse response is then defined as the following

${\displaystyle h[n]=Th_{c}(nT)\,}$

${\displaystyle h[n]=T\sum _{k=1}^{N}{A_{k}e^{s_{k}nT}u[n]}\,}$

Performing a z-transform on the discrete-time impulse response produces the following discrete-time system function

${\displaystyle H(z)=T\sum _{k=1}^{N}{\frac {A_{k)){1-e^{s_{k}T}z^{-1))}\,}$

Thus the poles from the continuous-time system function are translated to poles at z = e^s_kT. The zeros, if any, are not so simply mapped.^{[clarification needed]}

Poles and zeros

If the system function has zeros as well as poles, they can be mapped the same way, but the result is no longer an impulse invariance result: the discrete-time impulse response is not equal simply to samples of the continuous-time impulse response. This method is known as the matched Z-transform method, or pole–zero mapping.

Stability and causality

Since poles in the continuous-time system at s = s_k transform to poles in the discrete-time system at z = exp(s_kT), poles in the left half of the s-plane map to inside the unit circle in the z-plane; so if the continuous-time filter is causal and stable, then the discrete-time filter will be causal and stable as well.

Corrected formula

When a causal continuous-time impulse response has a discontinuity at ${\displaystyle t=0}$, the expressions above are not consistent.^[1] This is because ${\displaystyle h_{c}(0)}$ has different right and left limits, and should really only contribute their average, half its right value ${\displaystyle h_{c}(0_{+})}$, to ${\displaystyle h[0]}$.

Making this correction gives

${\displaystyle h[n]=T\left(h_{c}(nT)-{\frac {1}{2))h_{c}(0_{+})\delta [n]\right)\,}$

${\displaystyle h[n]=T\sum _{k=1}^{N}{A_{k}e^{s_{k}nT))\left(u[n]-{\frac {1}{2))\delta [n]\right)\,}$

Performing a z-transform on the discrete-time impulse response produces the following discrete-time system function

${\displaystyle H(z)=T\sum _{k=1}^{N}((\frac {A_{k)){1-e^{s_{k}T}z^{-1))}-{\frac {T}{2))\sum _{k=1}^{N}A_{k)).}$

The second sum is zero for filters without a discontinuity, which is why ignoring it is often safe.