Analog Filter Design: Type, Order, and Pole-Zero Placement

1. The Analog Filter Design Workflow

Designing an analog filter is not an ad-hoc process. It follows a structured procedure that translates a set of desired frequency-domain behaviors into a concrete transfer function. The general steps are:

1. Analyze Specifications and Choose Filter Type: Interpret the requirements—passband edge(s) ${\omega }_p$, stopband edge(s) ${\omega }_s$, maximum passband attenuation ${\alpha }_p$ (or ripple $r$), and minimum stopband attenuation ${\alpha }_s$. Based on these, select an appropriate filter approximation (e.g., Butterworth, Chebyshev, Elliptic).
2. Determine Lowpass Prototype Specifications: If the target is not a lowpass filter, use the frequency transformation rules to map the target specifications onto an equivalent set of specifications for a normalized lowpass prototype.
3. Calculate the Required Filter Order ($n$): Use the prototype specifications and the formulas specific to your chosen filter type to calculate the minimum integer order $n$ that will meet the requirements.
4. Determine the Prototype Transfer Function $H_{LP}(S)$: Find the poles (and zeros, if applicable) for the normalized lowpass prototype of order $n$. This gives you $H_{LP}(S)$. For Butterworth and Chebyshev filters, this is an all-pole filter.
5. Apply Frequency Transformation and De-normalization: Substitute the appropriate transformation $S=g(s)$ into $H_{LP}(S)$ to derive the final, un-normalized transfer function $H(s)$ that meets the original specifications.

Let's begin by refreshing our memory on the key differences between the filter types, which is the first crucial decision in our workflow.

As the video highlights, the Chebyshev filter offers a "sharper" transition band for the same order compared to a Butterworth filter, at the cost of ripples in the passband. An Elliptic filter (which has ripples in both the passband and stopband) is even more efficient in this regard. Your choice depends on the application's tolerance for passband distortion versus the need for a narrow transition band.

2. A Comprehensive Design Example: Bandpass Filter

Let's put the full workflow into practice. We will design a bandpass filter to meet the following specifications:

• Passband: 1000 Hz to 2000 Hz.
• Passband Ripple ($r$): No more than 1 dB. (We'll use this for the Chebyshev design).
• Passband Attenuation (${\alpha }_p$): No more than 1 dB at the passband edges. (We'll use this for the Butterworth design).
• Stopband: Attenuation of at least 20 dB (${\alpha }_s=20$ dB) at 450 Hz and 4000 Hz.

First, let's convert our specifications into the standard units (radians/second):

• ${\omega }_{p1}=2\pi (1000)=2000\pi$ rad/s
• ${\omega }_{p2}=2\pi (2000)=4000\pi$ rad/s
• ${\omega }_{s1}=2\pi (450)=900\pi$ rad/s
• ${\omega }_{s2}=2\pi (4000)=8000\pi$ rad/s
• Passband ripple $r=1$ dB
• Stopband attenuation ${\alpha }_s=20$ dB

We will now design this filter using two different approximations—Chebyshev and Butterworth—to compare the results.

2.1 Chebyshev Bandpass Filter Design

Step 1: Determine Lowpass Prototype Specifications

For a bandpass filter, the stopband specifications ${\omega }_{s1}$ and ${\omega }_{s2}$ map to a single stopband edge ${\omega }_s$ for the normalized lowpass prototype. As shown in Section 7.7-2 of the Lathi textbook (and your previous lesson), we take the smaller of the two possible transformed values. The prototype passband edge is always ${\omega }_p=1$.

The stopband edge for the prototype, ${\omega }_s$, is given by Eq. (7.56):
${\omega }_s=\min \left(\frac{{\omega }_{p1}{\omega }_{p2}-{\omega }_{s1}^2}{{\omega }_{s1}({\omega }_{p2}-{\omega }_{p1})},\frac{{\omega }_{s2}^2-{\omega }_{p1}{\omega }_{p2}}{{\omega }_{s2}({\omega }_{p2}-{\omega }_{p1})}\right)$

Plugging in our values:
$\frac{(2000\pi )(4000\pi )-(900\pi {)}^2}{(900\pi )(4000\pi -2000\pi )}=\frac{8\times 10^6-0.81\times 10^6}{900\times 2000}\approx 3.99$
$\frac{(8000\pi {)}^2-(2000\pi )(4000\pi )}{(8000\pi )(4000\pi -2000\pi )}=\frac{64\times 10^6-8\times 10^6}{8000\times 2000}=3.5$
So, our prototype lowpass filter must have ${\omega }_p=1$, ${\omega }_s=3.5$, $r=1$ dB, and ${\alpha }_s=20$ dB.

Step 2: Calculate Filter Order ($n$)

Using the Chebyshev order formula from Eq. (7.49b) in the Lathi textbook:
$n\ge \frac{{\cosh }^{-1}\left(\sqrt{\frac{10^{{\alpha }_s/10}-1}{10^{r/10}-1}}\right)}{{\cosh }^{-1}({\omega }_s)}$
$n\ge \frac{{\cosh }^{-1}\left(\sqrt{\frac{10^{20/10}-1}{10^{1/10}-1}}\right)}{{\cosh }^{-1}(3.5)}=\frac{{\cosh }^{-1}\left(\sqrt{\frac{99}{0.2589}}\right)}{{\cosh }^{-1}(3.5)}=\frac{{\cosh }^{-1}(19.58)}{1.925}=\frac{3.667}{1.925}\approx 1.905$
We must choose the next highest integer, so $n=2$.

Step 3: Determine the Prototype Transfer Function $H_{LP}(S)$

We need the transfer function for a 2nd-order, normalized Chebyshev lowpass filter with 1 dB ripple. From Table 7.5 in the Lathi textbook, the poles are at $-0.5489\pm j0.8951$. The denominator is:
$D(S)=(S-(-0.5489+j0.8951))(S-(-0.5489-j0.8951))=S^2+1.0978S+1.1025$
For a Chebyshev filter of even order, the DC gain is $1/\sqrt{1+{ϵ}^2}$, and we need to scale the numerator $K_n$ to achieve this. From Eq. (7.53), for even $n$, $K_n=a_0/\sqrt{1+{ϵ}^2}$. A simpler way is to ensure the magnitude at $S=0$ is correct. Here, $a_0=1.1025$.
$H_{LP}(S)=\frac{K_n}{S^2+1.0978S+1.1025}$
For an even-order Chebyshev filter, the numerator constant $K_n$ is chosen so that the gain at $S=0$ is $1/\sqrt{1+{ϵ}^2}$. For $r=1$ dB, ${ϵ}^2=0.2589$. The gain should be $1/\sqrt{1.2589}\approx 0.891$.
So, $K_n/1.1025=0.891⟹K_n\approx 0.9826$.
$H_{LP}(S)=\frac{0.9826}{S^2+1.0978S+1.1025}$

Step 4: Apply Frequency Transformation

We use the lowpass-to-bandpass transformation from Eq. (7.57):
$S=\frac{s^2+{\omega }_0^2}{Bs}$
where ${\omega }_0=\sqrt{{\omega }_{p1}{\omega }_{p2}}=\sqrt{(2000\pi )(4000\pi )}=2000\pi \sqrt{2}$ is the center frequency, and $B={\omega }_{p2}-{\omega }_{p1}=2000\pi$ is the bandwidth.
$S=\frac{s^2+(2000\pi \sqrt{2}{)}^2}{(2000\pi )s}=\frac{s^2+8\times 10^6{\pi }^2}{2000\pi s}$
Substituting this into $H_{LP}(S)$ yields the final transfer function $H(s)$. The algebra is tedious, but it results in a 4th-order bandpass filter. Each pole of the prototype maps to two poles in the final filter.

2.2 Butterworth Bandpass Filter Design

Now let's try to meet the same specifications with a Butterworth filter. The passband spec is ${\alpha }_p=1$ dB at the band edges.

Step 1: Determine Lowpass Prototype Specifications

This is identical to the Chebyshev case. The prototype must have ${\omega }_p=1$, ${\omega }_s=3.5$, ${\alpha }_p=1$ dB, and ${\alpha }_s=20$ dB.

Step 2: Calculate Filter Order ($n$)

Using the Butterworth order formula from Eq. (7.39):
$n\ge \frac{{\log }_{10}\left(\frac{10^{{\alpha }_s/10}-1}{10^{{\alpha }_p/10}-1}\right)}{2{\log }_{10}({\omega }_s)}$
$n\ge \frac{{\log }_{10}\left(\frac{10^{20/10}-1}{10^{1/10}-1}\right)}{2{\log }_{10}(3.5)}=\frac{{\log }_{10}(382.3)}{2\times 0.544}=\frac{2.582}{1.088}\approx 2.37$
We must choose the next highest integer, so $n=3$. Notice that the Butterworth filter requires a higher order (n=3) than the Chebyshev (n=2) to meet the same specs.

Step 3: Determine the Prototype Transfer Function $H_{LP}(S)$

This is a two-part step for Butterworth filters.
First, we calculate the prototype's 3-dB cutoff frequency, ${\omega }_c$, using Eq. (7.40):
${\omega }_c=\frac{{\omega }_p}{(10^{{\alpha }_p/10}-1{)}^{1/(2n)}}=\frac{1}{(10^{1/10}-1{)}^{1/6}}=\frac{1}{(0.2589{)}^{1/6}}=1.258\text{ rad/s}$
Next, we find the normalized (${\omega }_c=1$) 3rd-order Butterworth transfer function from Table 7.2:
$H_{norm}(S^{\prime })=\frac{1}{(S^{\prime }+1)(S^{\prime 2}+S^{\prime }+1)}$
To get our prototype $H_{LP}(S)$ with ${\omega }_c=1.258$, we substitute $S^{\prime }=S/{\omega }_c=S/1.258$:
$H_{LP}(S)=\frac{1}{(\frac{S}{1.258}+1)((\frac{S}{1.258}{)}^2+\frac{S}{1.258}+1)}=\frac{1.258^3}{(S+1.258)(S^2+1.258S+1.258^2)}$
$H_{LP}(S)=\frac{1.992}{(S+1.258)(S^2+1.258S+1.583)}$

Step 4: Apply Frequency Transformation

The transformation is the same as before. Substituting $S=\frac{s^2+8\times 10^6{\pi }^2}{2000\pi s}$ into $H_{LP}(S)$ gives the final transfer function $H(s)$. Since the prototype order was $n=3$, the final bandpass filter will be of 6th order.

This comparative example clearly illustrates the fundamental trade-off: the Chebyshev filter is more efficient (lower order), but the Butterworth filter provides a maximally flat passband, which may be preferable in many applications.

3. An Intuitive Look at Pole-Zero Placement

We've now successfully determined the filter type, order, and pole locations needed to construct the transfer function. But what is the intuitive link between these pole locations in the $s$-plane and the final frequency response?

The magnitude of the frequency response, $|H(j\omega )|$, at any frequency $\omega$, can be thought of as a geometric relationship:
$|H(j\omega )|=K\times \frac{\text{product of distances from all zeros to the point }j\omega }{\text{product of distances from all poles to the point }j\omega }$

This means:

• Poles near the $j\omega$ axis create peaks (resonance) because the distance in the denominator becomes very small as $j\omega$ passes by the pole.
• Zeros on or near the $j\omega$ axis create nulls (attenuation) because the distance in the numerator becomes very small.

This perspective helps explain the behavior of our filter types:

• Butterworth: The poles are arranged on a circle, far from the $j\omega$ axis in the passband region, resulting in a smooth, flat response.
• Chebyshev: The poles are on an ellipse, pushed closer to the $j\omega$ axis in the passband, causing the characteristic ripples.
• Elliptic: These filters have zeros in the stopband (on the $j\omega$ axis) to create deep nulls and achieve a very sharp transition, in addition to poles that create passband ripples.

This geometric view is a powerful tool for qualitatively understanding and even sketching the frequency response of a system, given its pole-zero plot—a skill that is invaluable in control systems and signal processing.

Conclusion

In this lesson, we have integrated all the components of analog filter design into a single, cohesive methodology. You are now equipped to tackle a filter design problem from start to finish.

Key Takeaways:

• Analog filter design is a systematic process of translating specifications into a transfer function.
• The choice of filter type (Butterworth, Chebyshev, etc.) is a critical engineering trade-off between passband flatness, transition sharpness, and filter complexity (order).
• The design process typically involves designing a normalized lowpass prototype and then using frequency transformations to achieve the final filter type (highpass, bandpass, bandstop).
• The final pole and zero locations in the $s$-plane geometrically determine the filter's frequency response. Poles near the $j\omega$ axis boost gain, while zeros suppress it.

Preview of the Next Lesson:
With our study of analog filters complete, we now turn our attention to the bridge between the analog and digital worlds. In the first lesson of the next module, we will explore one of the most important ideas in all of signal processing: the Sampling Theorem. We will trace its historical development from Nyquist to Shannon and derive the fundamental conditions required to convert a continuous-time signal into a discrete sequence of numbers without losing information.