# Impulse invariance

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. $\displaystyle{ H_c(j\Omega) \lt \delta }$ when $\displaystyle{ |\Omega| \ge \pi/T }$), 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 $\displaystyle{ |\omega| \le \pi\, }$

### 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

$\displaystyle{ h_c(t) = \begin{cases} \sum_{k=1}^N{A_ke^{s_kt}}, & t \ge 0 \\ 0, & \mbox{otherwise} \end{cases} }$

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

$\displaystyle{ h[n] = Th_c(nT)\, }$
$\displaystyle{ h[n] = T \sum_{k=1}^N{A_ke^{s_knT}u[n]}\, }$

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

$\displaystyle{ H(z) = T \sum_{k=1}^N{\frac{A_k}{1-e^{s_kT}z^{-1}}}\, }$

Thus the poles from the continuous-time system function are translated to poles at z = eskT. 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 = sk transform to poles in the discrete-time system at z = exp(skT), 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. 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 $\displaystyle{ h }$.

Making this correction gives

$\displaystyle{ h[n] = T \left( h_c(nT) - \frac{1}{2} h_c(0_+)\delta [n] \right) \, }$
$\displaystyle{ h[n] = T \sum_{k=1}^N{A_ke^{s_knT}} \left( u[n] - \frac{1}{2} \delta[n] \right) \, }$

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

$\displaystyle{ H(z) = T \sum_{k=1}^N{\frac{A_k}{1-e^{s_kT}z^{-1}} - \frac{T}{2} \sum_{k=1}^N A_k}. }$

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