IIR Filter Design: Impulse Invariance & Aliasing

Hello! Welcome back to our module on Digital Filter Design.

In our previous lesson, we established the historical context for IIR and FIR filter design. We saw that early IIR filter design was rooted in the mature field of analog electronics, relying on methods to transform a well-understood analog filter, $H_{a} (s)$ , into a digital equivalent, $H (z)$ .

Today, we will dive deep into the first of these major transformation techniques. Our goal is to formally design IIR digital filters using the impulse invariance method and analyze its potential for aliasing. We will derive the method from first principles, understand its effect on the filter's poles and stability, and critically examine its primary limitation, which will set the stage for our next topic.

1. The Principle of Impulse Invariance

The name "impulse invariance" perfectly describes the method's goal: to create a digital filter whose impulse response, $h [n]$ , is simply a sampled version of the analog prototype's impulse response, $h_{a} (t)$ . Mathematically, the core principle is:

h [n] = h_{a} (n T_{s})

where $T_{s}$ is the sampling period. By preserving the shape of the impulse response, the hope is to preserve the filter's characteristics.

The derivation follows a logical path that leverages concepts you are already familiar with from your studies of Laplace and Z-transforms:

Start with the analog transfer function $H_{a} (s)$ .
Find its impulse response $h_{a} (t)$ using the inverse Laplace transform.
Sample $h_{a} (t)$ to get the discrete-time impulse response $h [n]$ .
Find the Z-transform of $h [n]$ to arrive at the digital transfer function $H (z)$ .

2. Deriving the s-to-z Transformation

To make the derivation manageable, we assume the analog transfer function $H_{a} (s)$ has distinct poles and can be expressed using partial fraction expansion:

H_{a} (s) = k = 1 \sum N \frac{A _{k}}{s - p _{k}}

Here, $p_{k}$ represents the $k$ -th pole of the analog filter and $A_{k}$ is the corresponding coefficient from the expansion.

The following video provides a clear, step-by-step mathematical derivation that takes us from this starting point to the final digital transfer function.

IIR FILTERS ( DERIVATIONS)

This video from Prof. Barapate's Tutorials provides a concise mathematical derivation of the impulse invariance method. It follows the exact four-step process outlined above.

Watch the segment from 00:45 to 09:20. Follow the derivation closely, paying attention to how the standard formulas for the inverse Laplace transform and the Z-transform of an exponential sequence are used to arrive at the final result.

As the video demonstrates, by applying the inverse Laplace transform, sampling, and then the Z-transform, we find a direct mapping. A single pole term in the s-domain corresponds to a single pole term in the z-domain:

\frac{A _{k}}{s - p _{k}} Impulse Invariance \frac{A _{k}}{1 - e ^{p_{k} T_{s}} z ^{- 1}}

This gives us the rule for designing the filter:

Decompose the analog $H_{a} (s)$ into its partial fractions.
Replace each term using the mapping rule above.
Sum the resulting terms to get $H (z)$ .

Notice that the coefficients $A_{k}$ remain unchanged. The transformation only affects the poles.

3. Mapping Poles from the s-plane to the z-plane

The relationship between the analog poles $p_{k}$ and the resulting digital poles is profound. The video you just watched derived that a pole at $s = p_{k}$ in the s-plane is mapped to a pole at $z_{k} = e^{p_{k} T_{s}}$ in the z-plane. This is a specific instance of the general mapping $z = e^{s T_{s}}$ . Let's analyze what this means for stability and frequency response.

The next segment of the same video explains this mapping visually.

IIR FILTERS ( DERIVATIONS)

Now, let's analyze the properties of this s-to-z mapping. This part of the video explains how different regions of the s-plane are transformed into the z-plane.

Watch from 09:20 to 17:36. Focus on these key points: The derivation of the relationships for magnitude and phase: r = e^{\sigma T_s} and \omega = \Omega T_s. How the left-half of the s-plane (\sigma < 0) maps to the interior of the unit circle in the z-plane (r < 1). How the imaginary axis of the s-plane (\sigma = 0) maps to the unit circle itself (r = 1).

Let's summarize the critical consequences of the mapping $z = e^{s T_{s}}$ , where $s = σ + j Ω$ and $z = r e^{j ω}$ :

Magnitude Mapping: $r = e^{σ T_{s}}$ .
Frequency Mapping: $ω = Ω T_{s}$ .

From these two simple equations, we can draw two important conclusions:

Stability is Preserved: A stable analog filter has all its poles in the left-half of the s-plane, meaning $σ < 0$ for all poles. According to the magnitude mapping, this means $r = e^{σ T_{s}} < 1$ . All poles of the resulting digital filter will be inside the unit circle, guaranteeing that the digital filter is also stable. This is an excellent property.
Frequency Axis Mapping is Linear: The digital frequency $ω$ is directly proportional to the analog frequency $Ω$ . This means the frequency response shape should, in theory, be preserved—a low-pass analog filter should become a low-pass digital filter. However, this is where the main problem lies.

Test your understanding!

An unstable analog filter has a pole in the right-half plane, at $s = 0.5 + j 2$ . If you use the impulse invariance method with a sampling period $T_{s} = 0.1$ s, where will the corresponding pole of the digital filter be located in the z-plane? Will the digital filter be stable?

Show answer

The analog pole is $p_{k} = 0.5 + j 2$ . The digital pole will be at $z_{k} = e^{p_{k} T_{s}} = e^{(0.5 + j 2) (0.1)} = e^{0.05} e^{j 0.2}$ .

The magnitude of this pole is $r = e^{0.05} \approx 1.051$ . Since the magnitude is greater than 1, the pole lies outside the unit circle. Therefore, the resulting digital filter will be unstable, just as the original analog filter was. The impulse invariance method preserves both stability and instability.

4. The Inherent Flaw: Aliasing

The core operation of impulse invariance is sampling $h_{a} (t)$ . As we learned in Module 4, sampling in the time domain corresponds to a periodic repetition in the frequency domain. The spectrum of the sampled signal, $H (e^{j ω})$ , is related to the analog spectrum, $H_{a} (j Ω)$ , by the formula:

H (e^{j ω}) = \frac{1}{T _{s}} k = - \infty \sum \infty H_{a} (j \frac{ω + 2 π k}{T _{s}})

where $ω = Ω T_{s}$ . This equation shows that the digital filter's frequency response is a sum of infinitely many shifted and scaled copies of the original analog filter's frequency response.

Aliasing occurs when these shifted copies overlap.

resource_id

Frequency response of an analog filter (left) and the corresponding digital filter designed via impulse invariance (right). The dashed lines show the shifted replicas of the analog spectrum that sum together, causing aliasing in the digital spectrum. Source: M. Soleymani, ELEC442 Lecture 7, Concordia University.

The ideal scenario is that $H_{a} (j Ω)$ is perfectly band-limited to the Nyquist interval, i.e., $H_{a} (j Ω) = 0$ for $∣Ω∣ \geq π / T_{s}$ . If this were true, the replicas wouldn't overlap, and the digital frequency response would be a perfect copy of the analog one.

However, standard analog prototype filters (like Butterworth, Chebyshev, etc.) are not band-limited. Their frequency responses decrease but never become exactly zero. This means that with the impulse invariance method, there will always be some amount of aliasing.

The following reading explains this problem and shows a practical example.

Lecture 7

These lecture notes from Concordia University clearly illustrate the problem of aliasing in the impulse invariance method.

Please read the section titled 'Impulse Invariance Techniques: Aliasing'. Focus on the formula relating the digital and analog spectra and the example plot that compares the filter response for T=0.1 and T=0.5. Note how the aliasing is worse for the larger sampling period (smaller sampling rate).

This unavoidable aliasing has a major practical consequence:

The impulse invariance method is generally unsuitable for designing high-pass or band-stop filters. The high-frequency response of these filters would alias down and distort the low-frequency portion of the spectrum.
The method is best suited for designing low-pass or band-pass filters whose frequency response is already small at high frequencies, thus minimizing the effect of aliasing. You can reduce aliasing by using a very high sampling rate (small $T_{s}$ ), but you cannot eliminate it entirely.

Conclusion

We have now formally derived the impulse invariance method and analyzed its properties. You can see how it builds directly on fundamental signal processing principles and why, as a historical first step, it was a logical approach.

Key Takeaways:

Principle: The digital filter's impulse response $h [n]$ is a sampled version of the analog prototype's impulse response $h_{a} (n T_{s})$ .
Design Process: Involves partial fraction expansion of $H_{a} (s)$ and applying the pole mapping rule: $\frac{A _{k}}{s - p _{k}} \to \frac{A _{k}}{1 - e ^{p_{k} T_{s}} z ^{- 1}}$ .
Stability: The method is stability-preserving. A stable analog filter will always yield a stable digital filter.
Primary Limitation: Aliasing. Because analog filter prototypes are not band-limited, sampling their impulse response inevitably causes aliasing in the frequency domain. This makes the method unsuitable for high-pass or band-stop filter designs.

Preview of the Next Lesson:

The aliasing problem of the impulse invariance method was a significant motivation for finding a better transformation technique. In our next lesson, we will derive the bilinear transformation. We will see how it cleverly avoids aliasing entirely but, in doing so, introduces its own unique characteristic: a non-linear relationship called "frequency warping," which we must learn to manage.

Recommended resources

1. The Principle of Impulse Invariance

2. Deriving the s-to-z Transformation

3. Mapping Poles from the s-plane to the z-plane

4. The Inherent Flaw: Aliasing

Conclusion