Introduction

Importance of normalization and denormalization

Normalization means designing the filter with convenient reference values, usually a cut-off (or half-power) frequency of $\omega_o = 1$ rad/s and a termination of $R = 1\ \Omega$. Denormalization (scaling) converts the normalized element values to the actual frequency and impedance level required.

Importance:

• Filter tables (Butterworth, Chebyshev, Bessel poles and ladder element values) are published only in normalized form; one table serves every frequency and impedance level.
• Calculations use simple numbers like 1, 1.414, 2 instead of values like $10^{-9}$ and $10^{4}$, so errors are fewer.
• Designs can be compared and reused: a 1 rad/s prototype becomes a 1 kHz or 1 MHz filter by simple scaling.
• Scaling lets the designer choose practical element values (e.g. capacitors in nF–μF, resistors in kΩ) without changing the shape of the response.

Magnitude (impedance) scaling

Every impedance in the network is multiplied by $k_m$, at the same frequency. A voltage transfer function is a ratio of impedances, so it does not change.

$$\begin{aligned} R' &= k_m R \\ sL' = k_m (sL) &\Rightarrow L' = k_m L \\ \frac{1}{sC'} = k_m \frac{1}{sC} &\Rightarrow C' = \frac{C}{k_m} \end{aligned}$$

Frequency scaling

The response at the old frequency $\omega$ must appear at the new frequency $k_f\omega$, where $k_f = \omega_{new}/\omega_{old}$. So each element must have at $k_f\omega$ the same impedance that the old element had at $\omega$:

$$\begin{aligned} j(k_f\omega)L' = j\omega L &\Rightarrow L' = \frac{L}{k_f} \\ \frac{1}{j(k_f\omega)C'} = \frac{1}{j\omega C} &\Rightarrow C' = \frac{C}{k_f} \\ R' &= R \quad (\text{frequency independent}) \end{aligned}$$

The new transfer function is $T'(s) = T(s/k_f)$: the response keeps its shape but is shifted along the frequency axis.

Combined scaling

Applying both:

$$R' = k_m R, \qquad L' = \frac{k_m}{k_f}L, \qquad C' = \frac{C}{k_m k_f}$$

Example: the normalized third-order Butterworth ladder (1 Ω, 1 H, 2 F, 1 H, 1 Ω, $\omega_o$ = 1 rad/s) scaled to $\omega_o = 10^4$ rad/s ($k_f = 10^4$) and 1 kΩ terminations ($k_m = 1000$) gives $R$ = 1 kΩ, $L = 1\times1000/10^4 = 0.1$ H and $C = 2/(1000\times10^4) = 0.2\ \mu$F.

Significance of normalization and denormalization

Normalization means designing the filter with convenient reference values, usually a cut-off (or half-power) frequency of $\omega_o = 1$ rad/s and a termination of $R = 1\ \Omega$. Denormalization (scaling) converts the normalized element values to the actual frequency and impedance level required.

Importance:

• Filter tables (Butterworth, Chebyshev, Bessel poles and ladder element values) are published only in normalized form; one table serves every frequency and impedance level.
• Calculations use simple numbers like 1, 1.414, 2 instead of values like $10^{-9}$ and $10^{4}$, so errors are fewer.
• Designs can be compared and reused: a 1 rad/s prototype becomes a 1 kHz or 1 MHz filter by simple scaling.
• Scaling lets the designer choose practical element values (e.g. capacitors in nF–μF, resistors in kΩ) without changing the shape of the response.

Changing the half-power frequency from $\omega_0$ to $\omega_n$

Let the frequency scaling factor be

$$k_f = \frac{\omega_n}{\omega_0}$$

The new filter must show at $\omega_n = k_f\omega_0$ the same response the old one had at $\omega_0$. So each element's impedance at the new frequency must equal the old element's impedance at the old frequency.

Inductor:

$$j\omega_n L_n = j\omega_0 L_0 \;\Rightarrow\; L_n = \frac{\omega_0}{\omega_n}L_0 = \frac{L_0}{k_f}$$

Capacitor:

$$\frac{1}{j\omega_n C_n} = \frac{1}{j\omega_0 C_0} \;\Rightarrow\; C_n = \frac{\omega_0}{\omega_n}C_0 = \frac{C_0}{k_f}$$

Resistor: the impedance of a resistor does not depend on frequency, so $R_n = R_0$.

Thus $T_n(s) = T_0(s/k_f)$ and the half-power point moves from $\omega_0$ to $\omega_n$ with the same shape of response.

If the impedance level is also changed by $k_m$ (to get practical values):

$$R_n = k_m R_0, \qquad L_n = \frac{k_m L_0}{k_f}, \qquad C_n = \frac{C_0}{k_m k_f}$$

Example: the normalized third-order Butterworth ladder (1 Ω, 1 H, 2 F, 1 H, 1 Ω, $\omega_o$ = 1 rad/s) scaled to $\omega_o = 10^4$ rad/s ($k_f = 10^4$) and 1 kΩ terminations ($k_m = 1000$) gives $R$ = 1 kΩ, $L = 1\times1000/10^4 = 0.1$ H and $C = 2/(1000\times10^4) = 0.2\ \mu$F.

Ideal vs practical filters

An ideal filter passes all frequencies in the passband with constant gain and zero phase distortion and completely blocks all frequencies in the stopband, with a vertical edge between them. It cannot be built, because its impulse response is non-causal and infinitely long. A practical filter only approximates this.

 |T|   ideal              |T|   practical
 1 +------+               1 +~~~~~.  <- ripple (amax)
   |      |                 |      \
   |      |                 |       \  transition
   |      |                 |        `--..__ (amin)
 0 +------+------> w      0 +------+---+------> w
          wc                       wp  ws

Magnitude (impedance) scaling

Every impedance in the network is multiplied by $k_m$, at the same frequency. A voltage transfer function is a ratio of impedances, so it does not change.

$$\begin{aligned} R' &= k_m R \\ sL' = k_m (sL) &\Rightarrow L' = k_m L \\ \frac{1}{sC'} = k_m \frac{1}{sC} &\Rightarrow C' = \frac{C}{k_m} \end{aligned}$$

Frequency scaling

The response at the old frequency $\omega$ must appear at the new frequency $k_f\omega$, where $k_f = \omega_{new}/\omega_{old}$. So each element must have at $k_f\omega$ the same impedance that the old element had at $\omega$:

$$\begin{aligned} j(k_f\omega)L' = j\omega L &\Rightarrow L' = \frac{L}{k_f} \\ \frac{1}{j(k_f\omega)C'} = \frac{1}{j\omega C} &\Rightarrow C' = \frac{C}{k_f} \\ R' &= R \quad (\text{frequency independent}) \end{aligned}$$

The new transfer function is $T'(s) = T(s/k_f)$: the response keeps its shape but is shifted along the frequency axis.

Combined scaling

Applying both:

$$R' = k_m R, \qquad L' = \frac{k_m}{k_f}L, \qquad C' = \frac{C}{k_m k_f}$$

Example: the normalized third-order Butterworth ladder (1 Ω, 1 H, 2 F, 1 H, 1 Ω, $\omega_o$ = 1 rad/s) scaled to $\omega_o = 10^4$ rad/s ($k_f = 10^4$) and 1 kΩ terminations ($k_m = 1000$) gives $R$ = 1 kΩ, $L = 1\times1000/10^4 = 0.1$ H and $C = 2/(1000\times10^4) = 0.2\ \mu$F.

Attenuation (loss) is $\alpha(\omega) = -20\log_{10}|T(j\omega)|$ dB, so a large $\alpha$ means a small output.

 alpha(dB)
     |                     ______________
amin |- - - - - - - - - - |
     |                   /
     |                  /  transition
amax |~~~~~~~~~~~~~~~~ /
   0 +---------------+---+---------------> w
       passband      wp  ws   stopband

• $\alpha_{max}$ ($\alpha_p$): the maximum attenuation allowed anywhere in the passband (e.g. 0.5 dB). It sets the allowed ripple or droop.
• $\alpha_{min}$ ($\alpha_s$): the minimum attenuation required everywhere in the stopband (e.g. 40 dB).
• $\omega_p$: passband edge frequency, the last frequency at which $\alpha \le \alpha_{max}$.
• $\omega_s$: stopband edge frequency, from which $\alpha \ge \alpha_{min}$.
• Half-power frequency $\omega_{3dB}$: the frequency at which output power falls to half of its maximum, i.e. $|T(j\omega)| = |T|_{max}/\sqrt{2}$, an attenuation of 3.01 dB.
• Bandwidth: the width of the passband. For a low-pass filter it is $0$ to $\omega_p$ (or to $\omega_{3dB}$ for the half-power bandwidth); for a band-pass filter $BW = \omega_2 - \omega_1$, where $\omega_1, \omega_2$ are the lower and upper half-power frequencies.

Insertion gain and insertion loss compare the load voltage with and without the filter between a source $V_1$ (resistance $R_s$) and load $R_L$.

 Without filter:  Rs                 With filter:  Rs   +--------+
  V1 o--/\/\--+-- RL  (V20)          V1 o--/\/\--| filter |-- RL (V2)
                                                 +--------+

$$V_{20} = \frac{R_L}{R_s + R_L}V_1, \qquad \text{Insertion gain} = 20\log_{10}\left|\frac{V_2}{V_{20}}\right|\ \text{dB}$$ $$\text{Insertion loss} = 20\log_{10}\left|\frac{V_{20}}{V_2}\right| = -(\text{insertion gain})\ \text{dB}$$

A passive filter normally has a positive insertion loss; an active filter can have insertion gain.

Filter

A filter is a frequency-selective two-port network that passes signals in a chosen band of frequencies (passband) with little attenuation and attenuates signals at other frequencies (stopband). Filters may be passive (R, L, C), active (R, C, op-amp), switched-capacitor or digital.

Applications in communication

• Channel selection: IF and RF band-pass filters in radio, TV and mobile receivers select one channel and reject adjacent ones.
• Frequency division multiplexing: band-pass filters separate channels in FDM telephony and cable systems.
• Anti-aliasing and reconstruction: low-pass filters before an ADC and after a DAC.
• Noise and interference removal: removing hum (50 Hz notch), out-of-band noise and harmonics from transmitters.
• Modulation/demodulation: removing the carrier after detection, SSB generation (sideband filters), pulse shaping (raised-cosine) in digital links.
• Duplexers and diplexers: separating transmit and receive bands sharing one antenna.

Ideal vs practical response

An ideal filter passes all frequencies in the passband with constant gain and zero phase distortion and completely blocks all frequencies in the stopband, with a vertical edge between them. It cannot be built, because its impulse response is non-causal and infinitely long. A practical filter only approximates this.

 |T|   ideal              |T|   practical
 1 +------+               1 +~~~~~.  <- ripple (amax)
   |      |                 |      \
   |      |                 |       \  transition
   |      |                 |        `--..__ (amin)
 0 +------+------> w      0 +------+---+------> w
          wc                       wp  ws

The same idea applies to HP, BP and BS filters: the ideal response has rectangular edges, while the practical one has finite slopes, passband ripple up to $\alpha_{max}$ and stopband attenuation of at least $\alpha_{min}$.

Filter

A filter is a frequency-selective two-port network that passes signals in a chosen band of frequencies (passband) with little attenuation and attenuates signals at other frequencies (stopband).

Significance of normalization and denormalization

Normalization means designing the filter with convenient reference values, usually a cut-off (or half-power) frequency of $\omega_o = 1$ rad/s and a termination of $R = 1\ \Omega$. Denormalization (scaling) converts the normalized element values to the actual frequency and impedance level required.

Importance:

• Filter tables (Butterworth, Chebyshev, Bessel poles and ladder element values) are published only in normalized form; one table serves every frequency and impedance level.
• Calculations use simple numbers like 1, 1.414, 2 instead of values like $10^{-9}$ and $10^{4}$, so errors are fewer.
• Designs can be compared and reused: a 1 rad/s prototype becomes a 1 kHz or 1 MHz filter by simple scaling.
• Scaling lets the designer choose practical element values (e.g. capacitors in nF–μF, resistors in kΩ) without changing the shape of the response.

New element values for changing $\omega_{old}$ to $\omega_{new}$

Let $k_f = \omega_{new}/\omega_{old}$. For the same response at the new frequency, each element's reactance at $\omega_{new}$ must equal its old reactance at $\omega_{old}$:

$$\begin{aligned} \omega_{new} L_{new} = \omega_{old} L_{old} &\Rightarrow L_{new} = \frac{L_{old}}{k_f} \\ \frac{1}{\omega_{new} C_{new}} = \frac{1}{\omega_{old} C_{old}} &\Rightarrow C_{new} = \frac{C_{old}}{k_f} \\ R_{new} &= R_{old} \end{aligned}$$

The transfer function becomes $T_{new}(s) = T_{old}(s/k_f)$. Combined with impedance scaling by $k_m$: $R' = k_m R$, $L' = k_m L/k_f$, $C' = C/(k_m k_f)$.

Example: the normalized third-order Butterworth ladder (1 Ω, 1 H, 2 F, 1 H, 1 Ω, $\omega_o$ = 1 rad/s) scaled to $\omega_o = 10^4$ rad/s ($k_f = 10^4$) and 1 kΩ terminations ($k_m = 1000$) gives $R$ = 1 kΩ, $L = 1\times1000/10^4 = 0.1$ H and $C = 2/(1000\times10^4) = 0.2\ \mu$F.

Characteristics of an ideal filter

• Constant (flat) gain over the whole passband, with no ripple.
• Zero output (infinite attenuation) over the whole stopband.
• Zero-width transition band: a vertical, brick-wall edge at the cut-off frequency.
• Linear phase in the passband, so constant group delay and no phase distortion.
• It is non-causal and needs infinite order, so it can only be approximated.

Importance of scaling

Filter tables and prototype designs are given for 1 rad/s and 1 Ω. Scaling turns these into the actual frequency and impedance needed, and lets the designer pick practical element values (nF–μF capacitors, kΩ resistors) without changing the shape of the response.

Magnitude (impedance) scaling

Every impedance in the network is multiplied by $k_m$, at the same frequency. A voltage transfer function is a ratio of impedances, so it does not change.

$$\begin{aligned} R' &= k_m R \\ sL' = k_m (sL) &\Rightarrow L' = k_m L \\ \frac{1}{sC'} = k_m \frac{1}{sC} &\Rightarrow C' = \frac{C}{k_m} \end{aligned}$$

Frequency scaling

The response at the old frequency $\omega$ must appear at the new frequency $k_f\omega$, where $k_f = \omega_{new}/\omega_{old}$. So each element must have at $k_f\omega$ the same impedance that the old element had at $\omega$:

$$\begin{aligned} j(k_f\omega)L' = j\omega L &\Rightarrow L' = \frac{L}{k_f} \\ \frac{1}{j(k_f\omega)C'} = \frac{1}{j\omega C} &\Rightarrow C' = \frac{C}{k_f} \\ R' &= R \quad (\text{frequency independent}) \end{aligned}$$

The new transfer function is $T'(s) = T(s/k_f)$: the response keeps its shape but is shifted along the frequency axis.

Combined scaling

Applying both:

$$R' = k_m R, \qquad L' = \frac{k_m}{k_f}L, \qquad C' = \frac{C}{k_m k_f}$$

Example: the normalized third-order Butterworth ladder (1 Ω, 1 H, 2 F, 1 H, 1 Ω, $\omega_o$ = 1 rad/s) scaled to $\omega_o = 10^4$ rad/s ($k_f = 10^4$) and 1 kΩ terminations ($k_m = 1000$) gives $R$ = 1 kΩ, $L = 1\times1000/10^4 = 0.1$ H and $C = 2/(1000\times10^4) = 0.2\ \mu$F.

Normalization means designing the filter with convenient reference values, usually a cut-off (or half-power) frequency of $\omega_o = 1$ rad/s and a termination of $R = 1\ \Omega$. Denormalization (scaling) converts the normalized element values to the actual frequency and impedance level required.

Importance:

• Filter tables (Butterworth, Chebyshev, Bessel poles and ladder element values) are published only in normalized form; one table serves every frequency and impedance level.
• Calculations use simple numbers like 1, 1.414, 2 instead of values like $10^{-9}$ and $10^{4}$, so errors are fewer.
• Designs can be compared and reused: a 1 rad/s prototype becomes a 1 kHz or 1 MHz filter by simple scaling.
• Scaling lets the designer choose practical element values (e.g. capacitors in nF–μF, resistors in kΩ) without changing the shape of the response.

Scaling relations used for denormalization

With frequency scaling factor $k_f = \omega_{actual}/\omega_{normalized}$ and magnitude scaling factor $k_m = R_{actual}/R_{normalized}$:

$$R' = k_m R, \qquad L' = \frac{k_m}{k_f}L, \qquad C' = \frac{C}{k_m k_f}$$

Example

A normalized second-order Butterworth low-pass filter ($\omega_o$ = 1 rad/s, 1 Ω terminations) has $L = 1.414$ H and $C = 1.414$ F (doubly terminated ladder). Required: $f_o$ = 1 kHz and 600 Ω terminations.

$$\begin{aligned} k_f &= 2\pi\times1000 = 6283.2, \quad k_m = 600 \\ L' &= \frac{600 \times 1.414}{6283.2} = 0.135\ \text{H} \\ C' &= \frac{1.414}{600 \times 6283.2} = 0.375\ \mu\text{F} \end{aligned}$$

 600R    0.135 H
 o-/\/\/--UUUU--+------+
 Vs             |      |
             0.375uF  600R  Vo
                |      |
 o--------------+------+

The design was done once with simple numbers (1.414) and then adapted to the real requirement only by scaling.

Scaling

Scaling is the change of element values of a filter so that its response moves to a new frequency (frequency scaling) or a new impedance level (magnitude scaling) while the shape of the response stays the same. It is used to convert normalized designs (1 rad/s, 1 Ω) into practical ones.

Frequency scaling

The response at the old frequency $\omega$ must appear at the new frequency $k_f\omega$, where $k_f = \omega_{new}/\omega_{old}$. So each element must have at $k_f\omega$ the same impedance that the old element had at $\omega$:

$$\begin{aligned} j(k_f\omega)L' = j\omega L &\Rightarrow L' = \frac{L}{k_f} \\ \frac{1}{j(k_f\omega)C'} = \frac{1}{j\omega C} &\Rightarrow C' = \frac{C}{k_f} \\ R' &= R \quad (\text{frequency independent}) \end{aligned}$$

The new transfer function is $T'(s) = T(s/k_f)$: the response keeps its shape but is shifted along the frequency axis.

Example: a 1 F capacitor in a 1 rad/s prototype becomes $1/10^4 = 100\ \mu$F when $k_f = 10^4$; combined with $k_m = 1000$ it becomes 0.1 μF.

Basic steps of filter design

1. Specifications: state the filter type (LP, HP, BP, BS) and $\alpha_{max}, \alpha_{min}, \omega_p, \omega_s$, gain, impedance levels.
2. Normalization: convert to a normalized low-pass prototype ($\omega_p$ = 1 rad/s, 1 Ω), using frequency transformation for HP/BP/BS.
3. Approximation: choose a response (Butterworth, Chebyshev, inverse Chebyshev, elliptic, Bessel), find the order $n$ and the transfer function $T(s)$ that meets the specifications.
4. Realization (synthesis): build a circuit for $T(s)$: a passive LC ladder, or cascaded active biquads (Sallen–Key, MFB, Tow–Thomas), or switched-capacitor sections.
5. Denormalization (scaling): apply frequency and magnitude scaling (and the inverse frequency transformation) to get practical element values.
6. Study of non-idealities: check sensitivity to element tolerances, op-amp limits and temperature; simulate.
7. Construction and testing: build, measure and tune the filter.

Analog filter

An analog filter is a frequency-selective circuit that works on continuous-time signals, built from resistors, inductors, capacitors and active devices such as op-amps. It passes the wanted band of frequencies and attenuates the others.

Applications of filter networks

• Communication receivers and transmitters: channel selection, IF filtering, harmonic suppression.
• Anti-aliasing filters before ADCs and smoothing (reconstruction) filters after DACs.
• Audio: equalizers, tone controls, loudspeaker crossover networks.
• Power systems: harmonic filters, EMI/RFI line filters, DC power supply ripple filters.
• Biomedical instruments: ECG/EEG filters, 50 Hz notch filters.
• Control and instrumentation: removing sensor noise.

Importance of element scaling equations

Filter tables give element values for $\omega_o$ = 1 rad/s and 1 Ω terminations, such as 1 H or 2 F, which are not practical. The scaling equations

$$R' = k_m R, \qquad L' = \frac{k_m}{k_f}L, \qquad C' = \frac{C}{k_m k_f}$$

($k_f = \omega_{new}/\omega_{old}$, $k_m = R_{new}/R_{old}$) convert them to real values without redesigning. They follow from keeping each impedance $k_m$ times larger at a frequency $k_f$ times higher.

Example 1 (frequency and magnitude scaling): normalized third-order Butterworth LPF: 1 Ω, $L_1$ = 1 H, $C_2$ = 2 F, $L_3$ = 1 H, 1 Ω. Required $\omega_o = 10^4$ rad/s and 1 kΩ terminations ($k_f = 10^4$, $k_m = 10^3$):

$$L_1' = L_3' = \frac{10^3 \times 1}{10^4} = 0.1\ \text{H}, \qquad C_2' = \frac{2}{10^3 \times 10^4} = 0.2\ \mu\text{F}$$

Example 2 (choosing $k_m$ for a practical capacitor): for an active RC section with $R$ = 1 Ω and $C$ = 1 F at 1 rad/s, moving to $\omega_o$ = 10⁴ rad/s with $C$ = 10 nF needs $k_m = 1/(10^4 \times 10^{-8}) = 10^4$, so $R$ = 10 kΩ.

Thus scaling makes the normalized design reusable and lets the designer choose standard, available component values.

Filter

A filter is a frequency-selective two-port network that passes signals in a chosen band of frequencies (passband) with little attenuation and attenuates signals at other frequencies (stopband).

Insertion gain and insertion loss

Insertion gain and insertion loss compare the load voltage with and without the filter between a source $V_1$ (resistance $R_s$) and load $R_L$.

 Without filter:  Rs                 With filter:  Rs   +--------+
  V1 o--/\/\--+-- RL  (V20)          V1 o--/\/\--| filter |-- RL (V2)
                                                 +--------+

$$V_{20} = \frac{R_L}{R_s + R_L}V_1, \qquad \text{Insertion gain} = 20\log_{10}\left|\frac{V_2}{V_{20}}\right|\ \text{dB}$$ $$\text{Insertion loss} = 20\log_{10}\left|\frac{V_{20}}{V_2}\right| = -(\text{insertion gain})\ \text{dB}$$

A passive filter normally has a positive insertion loss; an active filter can have insertion gain.

Magnitude (impedance) scaling

Every impedance in the network is multiplied by $k_m$, at the same frequency. A voltage transfer function is a ratio of impedances, so it does not change.

$$\begin{aligned} R' &= k_m R \\ sL' = k_m (sL) &\Rightarrow L' = k_m L \\ \frac{1}{sC'} = k_m \frac{1}{sC} &\Rightarrow C' = \frac{C}{k_m} \end{aligned}$$

Frequency scaling

The response at the old frequency $\omega$ must appear at the new frequency $k_f\omega$, where $k_f = \omega_{new}/\omega_{old}$. So each element must have at $k_f\omega$ the same impedance that the old element had at $\omega$:

$$\begin{aligned} j(k_f\omega)L' = j\omega L &\Rightarrow L' = \frac{L}{k_f} \\ \frac{1}{j(k_f\omega)C'} = \frac{1}{j\omega C} &\Rightarrow C' = \frac{C}{k_f} \\ R' &= R \quad (\text{frequency independent}) \end{aligned}$$

The new transfer function is $T'(s) = T(s/k_f)$: the response keeps its shape but is shifted along the frequency axis.

Combined scaling

Applying both:

$$R' = k_m R, \qquad L' = \frac{k_m}{k_f}L, \qquad C' = \frac{C}{k_m k_f}$$

$\alpha_p$, $\alpha_s$, $\omega_p$, $\omega_s$

Attenuation (loss) is $\alpha(\omega) = -20\log_{10}|T(j\omega)|$ dB, so a large $\alpha$ means a small output.

 alpha(dB)
     |                     ______________
amin |- - - - - - - - - - |
     |                   /
     |                  /  transition
amax |~~~~~~~~~~~~~~~~ /
   0 +---------------+---+---------------> w
       passband      wp  ws   stopband

• $\alpha_p$ ($\alpha_{max}$): the maximum attenuation allowed anywhere in the passband (e.g. 0.5 dB). It sets the allowed ripple or droop.
• $\alpha_s$ ($\alpha_{min}$): the minimum attenuation required everywhere in the stopband (e.g. 40 dB).
• $\omega_p$: passband edge frequency, the last frequency at which $\alpha \le \alpha_{max}$.
• $\omega_s$: stopband edge frequency, from which $\alpha \ge \alpha_{min}$.

The smaller $\alpha_p$, the larger $\alpha_s$ and the narrower the transition band $\omega_s/\omega_p$, the higher the filter order required.

Scaling

Scaling changes element values so that the response moves to another frequency or impedance level without changing its shape. It converts normalized designs (1 rad/s, 1 Ω) into practical filters.

Magnitude (impedance) scaling

Every impedance in the network is multiplied by $k_m$, at the same frequency. A voltage transfer function is a ratio of impedances, so it does not change.

$$\begin{aligned} R' &= k_m R \\ sL' = k_m (sL) &\Rightarrow L' = k_m L \\ \frac{1}{sC'} = k_m \frac{1}{sC} &\Rightarrow C' = \frac{C}{k_m} \end{aligned}$$

Frequency scaling

The response at the old frequency $\omega$ must appear at the new frequency $k_f\omega$, where $k_f = \omega_{new}/\omega_{old}$. So each element must have at $k_f\omega$ the same impedance that the old element had at $\omega$:

$$\begin{aligned} j(k_f\omega)L' = j\omega L &\Rightarrow L' = \frac{L}{k_f} \\ \frac{1}{j(k_f\omega)C'} = \frac{1}{j\omega C} &\Rightarrow C' = \frac{C}{k_f} \\ R' &= R \quad (\text{frequency independent}) \end{aligned}$$

The new transfer function is $T'(s) = T(s/k_f)$: the response keeps its shape but is shifted along the frequency axis.

Combined scaling

Applying both:

$$R' = k_m R, \qquad L' = \frac{k_m}{k_f}L, \qquad C' = \frac{C}{k_m k_f}$$

Attenuation is $\alpha(\omega) = -20\log_{10}|T(j\omega)|$ dB. The figure shows the specification of a low-pass filter:

 alpha(dB)
     |                 :  ////////////////
amin |- - - - - - - - -:-+    stopband
     |    passband     : |
     |                 :/  <- transition
amax |=================+
   0 +-----------------+--+---------------> w
                       wp ws

• Passband: the band of frequencies the filter passes with little attenuation, $\alpha \le \alpha_{max}$. For a LPF it is $0 \le \omega \le \omega_p$.
• Stopband: the band the filter rejects, with $\alpha \ge \alpha_{min}$. For a LPF it is $\omega \ge \omega_s$.
• $\alpha_{max}$: the maximum attenuation allowed in the passband (passband ripple), e.g. 0.5 dB or 1 dB.
• $\alpha_{min}$: the minimum attenuation that must be achieved in the stopband, e.g. 40 dB.
• $\omega_p$: passband edge frequency, the highest frequency (for LPF) where attenuation is still within $\alpha_{max}$.
• $\omega_s$: stopband edge frequency, from where attenuation is at least $\alpha_{min}$.
• Bandwidth: the width of the passband. For a LPF it is $\omega_p$ (or $\omega_{3dB}$ for half-power bandwidth). For a BPF it is $BW = \omega_2 - \omega_1$, with centre frequency $\omega_o = \sqrt{\omega_1\omega_2}$.

 |T|   band-pass
  1 +      .-----.
    |     /       \
    |    /         \
    +---+---+---+---+---> w
       w1   wo     w2
        <--- BW --->

The region between $\omega_p$ and $\omega_s$ is the transition band; the ratio $\omega_s/\omega_p$ (selectivity) together with $\alpha_{max}$ and $\alpha_{min}$ fixes the filter order.

Terms

 |T|(dB)
   0 |~~~~~~~~~~~~~~.
     |  passband     \   transition band
     |                \  (roll-off slope)
 -As |- - - - - - - - -\_______________
     |                  :   stopband
     +-------------+----+--------------> w
                   wp   ws

• Passband: the range of frequencies passed with attenuation not more than $\alpha_{max}$ (LPF: $0$ to $\omega_p$).
• Stopband: the range attenuated by at least $\alpha_{min}$ (LPF: $\omega \ge \omega_s$).
• Transition band: the region between $\omega_p$ and $\omega_s$, where attenuation rises from $\alpha_{max}$ to $\alpha_{min}$. An ideal filter has zero transition width.
• Roll-off: the rate at which the gain falls beyond the cut-off, in dB/decade or dB/octave. An $n$th-order all-pole LPF rolls off at $20n$ dB/decade ($6n$ dB/octave); a steeper roll-off needs a higher order.
• Bandwidth: the width of the passband: $\omega_p$ (or $\omega_{3dB}$) for LPF, $\omega_2 - \omega_1$ between half-power points for BPF.

Basic steps in filter design

1. Specifications: state the filter type (LP, HP, BP, BS) and $\alpha_{max}, \alpha_{min}, \omega_p, \omega_s$, gain, impedance levels.
2. Normalization: convert to a normalized low-pass prototype ($\omega_p$ = 1 rad/s, 1 Ω), using frequency transformation for HP/BP/BS.
3. Approximation: choose a response (Butterworth, Chebyshev, inverse Chebyshev, elliptic, Bessel), find the order $n$ and the transfer function $T(s)$ that meets the specifications.
4. Realization (synthesis): build a circuit for $T(s)$: a passive LC ladder, or cascaded active biquads (Sallen–Key, MFB, Tow–Thomas), or switched-capacitor sections.
5. Denormalization (scaling): apply frequency and magnitude scaling (and the inverse frequency transformation) to get practical element values.
6. Study of non-idealities: check sensitivity to element tolerances, op-amp limits and temperature; simulate.
7. Construction and testing: build, measure and tune the filter.

Insertion gain and insertion loss

Insertion gain and insertion loss compare the load voltage with and without the filter between a source $V_1$ (resistance $R_s$) and load $R_L$.

 Without filter:  Rs                 With filter:  Rs   +--------+
  V1 o--/\/\--+-- RL  (V20)          V1 o--/\/\--| filter |-- RL (V2)
                                                 +--------+

$$V_{20} = \frac{R_L}{R_s + R_L}V_1, \qquad \text{Insertion gain} = 20\log_{10}\left|\frac{V_2}{V_{20}}\right|\ \text{dB}$$ $$\text{Insertion loss} = 20\log_{10}\left|\frac{V_{20}}{V_2}\right| = -(\text{insertion gain})\ \text{dB}$$

A passive filter normally has a positive insertion loss; an active filter can have insertion gain.

Steps in filter design

1. Specifications: state the filter type (LP, HP, BP, BS) and $\alpha_{max}, \alpha_{min}, \omega_p, \omega_s$, gain, impedance levels.
2. Normalization: convert to a normalized low-pass prototype ($\omega_p$ = 1 rad/s, 1 Ω), using frequency transformation for HP/BP/BS.
3. Approximation: choose a response (Butterworth, Chebyshev, inverse Chebyshev, elliptic, Bessel), find the order $n$ and the transfer function $T(s)$ that meets the specifications.
4. Realization (synthesis): build a circuit for $T(s)$: a passive LC ladder, or cascaded active biquads (Sallen–Key, MFB, Tow–Thomas), or switched-capacitor sections.
5. Denormalization (scaling): apply frequency and magnitude scaling (and the inverse frequency transformation) to get practical element values.
6. Study of non-idealities: check sensitivity to element tolerances, op-amp limits and temperature; simulate.
7. Construction and testing: build, measure and tune the filter.

Active vs passive filters

Filter

A filter is a frequency-selective two-port network that passes signals in a chosen band of frequencies (passband) with little attenuation and attenuates signals at other frequencies (stopband).

Need in communication systems

A channel and a receiver contain many signals and noise at different frequencies. Filters select the wanted channel, reject adjacent channels and noise, limit bandwidth before sampling (anti-aliasing), remove the carrier after demodulation and suppress harmonics in transmitters. Without filters, FDM and radio would be impossible.

Types of filters

• Low-pass (LPF): passes $0 \le \omega \le \omega_c$ and attenuates higher frequencies. Use: anti-aliasing, audio woofer feed.
• High-pass (HPF): passes $\omega \ge \omega_c$. Use: removing DC/low-frequency drift, tweeter feed.
• Band-pass (BPF): passes $\omega_1 \le \omega \le \omega_2$, centre $\omega_o = \sqrt{\omega_1\omega_2}$. Use: tuning a radio channel.
• Band-stop / notch (BSF): rejects $\omega_1 \le \omega \le \omega_2$. Use: removing 50 Hz mains hum.
• All-pass (APF): $|T(j\omega)| = 1$ for all $\omega$; changes only phase. Use: delay (phase) equalization.

 LPF   |-----\        HPF        /-----
       |      \___          ___/
       +-----------> w   +-----------> w
 BPF      /---\         BSF ----\   /----
      ___/     \___           \_/
       +-----------> w   +-----------> w
 APF   |--------------- |T| = 1
       +-----------> w

Filters can also be classified by technology (passive LC, active RC, switched-capacitor, digital, crystal/SAW) and by approximation (Butterworth, Chebyshev, elliptic, Bessel).

Why analog filters are still relevant

Digital filters need an ADC and DAC, and the analog world is always at the input and output, so analog filters remain essential:

• Anti-aliasing and reconstruction: every ADC needs an analog low-pass filter before it (to remove components above $f_s/2$), and every DAC needs a smoothing filter after it. Example: audio codecs, data acquisition cards.
• High frequencies: RF and microwave signals (GHz) in mobile phones, Wi-Fi and radar are filtered with LC, SAW/BAW or cavity filters, because ADCs at such rates are costly or not available.
• Low power and real time: analog filters have no sampling delay and use very little power, useful in hearing aids, wearable sensors and implanted medical devices.
• Power and EMI: mains EMI filters, harmonic filters in power systems and ripple filters in power supplies handle large currents that DSP cannot.
• Dynamic range: a front-end filter removes strong out-of-band interferers before they saturate the ADC (e.g. receiver preselectors, ECG front ends).

Active vs passive filters

Frequency scaling

Frequency scaling moves the frequency response of a filter to a new frequency without changing its shape or impedance level. It is used to convert a normalized (1 rad/s) design to the required cut-off frequency.

Let $k_f = \omega_{new}/\omega_{old}$. The new network must have at $k_f\omega$ the same impedance as the old one at $\omega$:

$$\begin{aligned} j(k_f\omega)L' = j\omega L &\Rightarrow L' = \frac{L}{k_f} \\ \frac{1}{j(k_f\omega)C'} = \frac{1}{j\omega C} &\Rightarrow C' = \frac{C}{k_f} \\ R' &= R \end{aligned}$$

and $T'(s) = T(s/k_f)$. Example: a 1 rad/s prototype with $L$ = 1 H and $C$ = 2 F, scaled to $10^3$ rad/s, becomes $L$ = 1 mH and $C$ = 2 mF.

Normalization and de-normalization

Normalization is designing a filter for a reference frequency of 1 rad/s and a reference impedance of 1 Ω, so standard tables and simple numbers can be used. De-normalization is converting the normalized element values to the actual cut-off frequency and impedance level by frequency scaling ($k_f$) and magnitude scaling ($k_m$):

$$R' = k_m R, \qquad L' = \frac{k_m L}{k_f}, \qquad C' = \frac{C}{k_m k_f}$$

Numerical

Given: $R_s = R_L = 1\ \Omega$, $L_1 = L_3 = 1$ H, $C_2 = 2$ F, $\omega_o$ = 1 rad/s. Required: $f_o$ = 100 Hz, capacitor = 0.01 μF.

Frequency scaling factor:

$$k_f = \frac{\omega_{new}}{\omega_{old}} = \frac{2\pi\times100}{1} = 628.32$$

Magnitude scaling factor (chosen so that the 2 F capacitor becomes 0.01 μF):

$$\begin{aligned} C' = \frac{C}{k_m k_f} &\Rightarrow k_m = \frac{C}{k_f C'} = \frac{2}{628.32 \times 0.01\times10^{-6}} \\ &= 3.183\times10^{5} \end{aligned}$$

New element values:

$$\begin{aligned} R_s' = R_L' &= k_m \times 1 = 318.3\ \text{k}\Omega \\ L_1' = L_3' &= \frac{k_m L}{k_f} = \frac{3.183\times10^5 \times 1}{628.32} = 506.6\ \text{H} \\ C_2' &= 0.01\ \mu\text{F} \end{aligned}$$

 318.3k   506.6 H      506.6 H
 o-/\/\/--UUUU----+----UUUU----+
 Vs               |            |
               0.01uF       318.3k  Vo
                  |            |
 o----------------+------------+

Answer: $R_s = R_L$ = 318.3 kΩ, $L_1 = L_3$ = 506.6 H, $C_2$ = 0.01 μF; half-power frequency 100 Hz (628.3 rad/s).

Note: the inductors are very large because the capacitor is small and the frequency is low. In practice such a filter would be built as an active RC filter or with inductors simulated by a GIC.

Normalization and denormalization

Normalization is designing the filter for $\omega_o$ = 1 rad/s and 1 Ω terminations, so that standard tables and simple numbers can be used. Denormalization is converting the normalized values to the actual frequency and impedance level using frequency scaling ($k_f$) and magnitude scaling ($k_m$):

$$R' = k_m R, \qquad L' = \frac{k_m L}{k_f}, \qquad C' = \frac{C}{k_m k_f}$$

Numerical

Given: $R_1 = R_2 = 1\ \Omega$, $L_1 = L_2 = 2.024$ H, $C_1 = 0.994$ F at $\omega_o$ = 1 rad/s. Required $\omega_o = 10^5$ rad/s.

Frequency scaling factor: $k_f = 10^5/1 = 10^5$.

Magnitude scaling factor: with $k_m = 1$ the capacitor would be $0.994/10^5 \approx 10\ \mu$F and the resistors 1 Ω, which are not practical. Choose $k_m = 1000$ (1 kΩ terminations).

$$\begin{aligned} R_1' = R_2' &= k_m R = 1000 \times 1 = 1\ \text{k}\Omega \\ L_1' = L_2' &= \frac{k_m L}{k_f} = \frac{1000 \times 2.024}{10^5} = 20.24\ \text{mH} \\ C_1' &= \frac{C}{k_m k_f} = \frac{0.994}{1000 \times 10^5} = 9.94\ \text{nF} \end{aligned}$$

 1k     20.24 mH      20.24 mH
 o-/\/\--UUUU----+----UUUU----+
 V1              |            |
              9.94 nF        1k   V2
                 |            |
 o---------------+------------+

Answer: $R_1 = R_2$ = 1 kΩ, $L_1 = L_2$ = 20.24 mH, $C_1$ = 9.94 nF (≈ 10 nF standard), with $\omega_o = 10^5$ rad/s (≈ 15.9 kHz).

Check: $L'C'$ product scales by $1/k_f^2$: $2.024\times0.994\times10^{-10} = 20.24\times10^{-3}\times9.94\times10^{-9}$, so the response shape is unchanged and only moved to $10^5$ rad/s. Any other $k_m$ (e.g. 10 kΩ) is also correct if it gives practical values.

Significance of normalization and de-normalization

Filter tables and prototypes are given for $\omega_p$ = 1 rad/s and 1 Ω. Normalization lets one table serve every filter and keeps calculations simple; de-normalization (scaling) converts the prototype to the actual frequency and impedance level with practical element values, without changing the response shape:

$$R' = k_m R, \qquad L' = \frac{k_m L}{k_f}, \qquad C' = \frac{C}{k_m k_f}$$

Numerical

Given prototype ($\omega_p$ = 1 rad/s): shunt $C_1$ = 1.3 F, series $L_2$ = 1.5 H, shunt $C_3$ = 1.2 F, series $L_4$ = 0.5 H, load $R_L$ = 1 Ω.

Required: $\omega_p$ = 1000 rad/s and $R_L$ = 75 Ω.

$$k_f = \frac{1000}{1} = 1000, \qquad k_m = \frac{75}{1} = 75$$ $$\begin{aligned} C_1' &= \frac{1.3}{75 \times 1000} = 17.33\ \mu\text{F} \\ L_2' &= \frac{75 \times 1.5}{1000} = 0.1125\ \text{H} = 112.5\ \text{mH} \\ C_3' &= \frac{1.2}{75 \times 1000} = 16\ \mu\text{F} \\ L_4' &= \frac{75 \times 0.5}{1000} = 0.0375\ \text{H} = 37.5\ \text{mH} \\ R_L' &= 75 \times 1 = 75\ \Omega \end{aligned}$$

        112.5 mH      37.5 mH
 o---+---UUUU----+----UUUU----+
 V1  |           |            |
  17.33uF      16 uF        75R   V2
     |           |            |
 o---+-----------+------------+

Answer: $C_1$ = 17.33 μF, $L_2$ = 112.5 mH, $C_3$ = 16 μF, $L_4$ = 37.5 mH, $R_L$ = 75 Ω; passband edge 1000 rad/s. (If the source has a 1 Ω resistance it also becomes 75 Ω.)

Significance of normalization and denormalization

Normalization means designing the filter with convenient reference values, usually a cut-off (or half-power) frequency of $\omega_o = 1$ rad/s and a termination of $R = 1\ \Omega$. Denormalization (scaling) converts the normalized element values to the actual frequency and impedance level required.

Importance:

• Filter tables (Butterworth, Chebyshev, Bessel poles and ladder element values) are published only in normalized form; one table serves every frequency and impedance level.
• Calculations use simple numbers like 1, 1.414, 2 instead of values like $10^{-9}$ and $10^{4}$, so errors are fewer.
• Designs can be compared and reused: a 1 rad/s prototype becomes a 1 kHz or 1 MHz filter by simple scaling.
• Scaling lets the designer choose practical element values (e.g. capacitors in nF–μF, resistors in kΩ) without changing the shape of the response.

The scaling relations are $R' = k_m R$, $L' = k_m L/k_f$, $C' = C/(k_m k_f)$.

Numerical

Attenuation in dB is defined as

$$\alpha = 20\log_{10}\frac{V_{in}}{V_{out}}\ \text{dB} \;\Rightarrow\; V_{out} = V_{in}\times10^{-\alpha/20}$$

At $f$ = 30 kHz, $\alpha$ = 90 dB and $V_{in}$ = 1 V:

$$\begin{aligned} V_{out} &= 1 \times 10^{-90/20} = 10^{-4.5} \\ &= 3.162\times10^{-5}\ \text{V} \end{aligned}$$

Answer: output amplitude at 30 kHz ≈ 31.6 μV.

(The 78 dB at 20 kHz applies only to a 20 kHz input; a 1 V signal at 20 kHz would come out as $10^{-78/20} = 126\ \mu$V.)

Definitions

Attenuation (loss) is $\alpha(\omega) = -20\log_{10}|T(j\omega)|$ dB, so a large $\alpha$ means a small output.

 alpha(dB)
     |                     ______________
amin |- - - - - - - - - - |
     |                   /
     |                  /  transition
amax |~~~~~~~~~~~~~~~~ /
   0 +---------------+---+---------------> w
       passband      wp  ws   stopband

• $\alpha_{max}$ ($\alpha_p$): the maximum attenuation allowed anywhere in the passband (e.g. 0.5 dB). It sets the allowed ripple or droop.
• $\alpha_{min}$ ($\alpha_s$): the minimum attenuation required everywhere in the stopband (e.g. 40 dB).
• Half-power frequency $\omega_{3dB}$: the frequency at which output power falls to half of its maximum, i.e. $|T(j\omega)| = |T|_{max}/\sqrt{2}$, an attenuation of 3.01 dB.
• Bandwidth: the width of the passband. For a low-pass filter it is $0$ to $\omega_p$ (or to $\omega_{3dB}$ for the half-power bandwidth); for a band-pass filter $BW = \omega_2 - \omega_1$, where $\omega_1, \omega_2$ are the lower and upper half-power frequencies.

Numerical

Attenuation in dB is defined as

$$\alpha = 20\log_{10}\frac{V_{in}}{V_{out}}\ \text{dB} \;\Rightarrow\; V_{out} = V_{in}\times10^{-\alpha/20}$$

At $f$ = 30 kHz, $\alpha$ = 90 dB and $V_{in}$ = 1 V:

$$\begin{aligned} V_{out} &= 1 \times 10^{-90/20} = 10^{-4.5} \\ &= 3.162\times10^{-5}\ \text{V} \end{aligned}$$

Answer: output amplitude at 30 kHz ≈ 31.6 μV.

(The 78 dB at 20 kHz applies only to a 20 kHz input; a 1 V signal at 20 kHz would come out as $10^{-78/20} = 126\ \mu$V.)