Filters and Bode plots
Introduced| English | Meaning |
|---|---|
| filter/ˈfɪltə/ | A circuit that passes some frequencies and reduces others. |
| transfer function | The ratio of the Laplace transform of a system output to that of its input, with zero initial conditions. |
| decibels/ˈdesɪblz/ | decibels |
| Bode plot | A pair of graphs of a system's gain in decibels and its phase against log frequency. |
| corner frequency | A pole or zero break frequency where an asymptotic Bode slope changes; an isolated first-order low-pass pole is 3 dB down there. |
The bass knob
- Turn up the bass on a speaker and the low notes get louder while the voices stay the same. That knob controls a filter.
- Filters are everywhere: in power supplies, radios and sensors. This lesson shows how to draw a filter's response with a few straight lines.
What a filter does
- A filter 滤波器 passes one band of frequencies and weakens the rest. Its transfer function 传递函数 $H(j\omega) = \dfrac{\mathbf{V}_{out}}{\mathbf{V}_{in}}$ has a size and a phase at each frequency.
- The RC low-pass, output across C: $H = \dfrac{1}{1 + j\omega RC}$. Low frequencies pass ($H \approx 1$); high ones are shorted out by the capacitor.
An RC low-pass with its output across C passes DC unchanged.
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At $\omega = 0$ the capacitor is an open circuit, so no current flows and $H = 1$: the output equals the input.
Decibels and the Bode plot
- Gain in decibels 分贝 is $20\log_{10}|H|$: ×10 is +20 dB, ×0.1 is −20 dB, and $\dfrac{1}{\sqrt{2}}$ is −3 dB. Multiplied gains become added decibels.
- A Bode plot 波特图 draws gain in dB and phase against frequency on a log axis. On these axes, the response is nearly straight lines.
An amplifier has a voltage gain of 100. What is that in dB?
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$20\log_{10}100 = 20 \times 2 = 40$ dB. Two stages of 40 dB each give 80 dB, a gain of 10 000.
The corner frequency
- The corner frequency 转折频率 is $\omega_{c} = \dfrac{1}{RC}$. Below it the low-pass gain is flat at 0 dB; above it, it falls 20 dB per decade.
- At the corner itself the gain is −3 dB (half power) and the phase −45°. The phase runs from 0° at low frequency to −90° at high.
One pole above its corner costs how much gain per decade?
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$20\log_{10}10 = 20$ dB per decade per pole. At $10\omega_{c}$ the RC low-pass reads −20.04 dB exactly, against the asymptote's −20.
Worked example: a 1 kHz low-pass
- $R = 1.6\,\text{k}\Omega$, $C = 0.1\,\mu\text{F}$: $\omega_{c} = \dfrac{1}{1600 \times 10^{-7}} = 6250\,\dfrac{\text{rad}}{\text{s}}$, which is 995 Hz.
- At $10\omega_{c}$ the straight line says −20 dB; the exact value is $20\log_{10}\dfrac{1}{\sqrt{101}} = -20.04$ dB. At $100\omega_{c}$: about −40 dB.
The exact curve rounds the Bode corner
$H_{total}=H_{stage}^{N}$
The 1.6 kΩ and 0.1 µF section has a 6250 rad/s corner. Compare exact gain and phase with straight-line approximations, swap the output to make a high-pass, and see how two buffered stages multiply gain and shift the overall half-power corner.
R = 10 kΩ and C = 10 nF in an RC low-pass. What is the corner frequency, in Hz?
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$\omega_{c} = \dfrac{1}{RC} = \dfrac{1}{10^{4} \times 10^{-8}} = 10\,000$ rad/s, and $f_{c} = \dfrac{10\,000}{2\pi} = 1592$ Hz.
A first-order RC low-pass at 10 times its corner frequency reads —
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One pole rolls off 20 dB per decade past the corner. −3 dB is the corner itself; −40 dB is two decades past it, or two poles.
The high-pass mirror
- Take the output across R instead: $H = \dfrac{j\omega/\omega_{c}}{1 + j\omega/\omega_{c}}$, a high-pass with the same corner.
- It is the mirror image: −20.04 dB at $0.1\omega_{c}$, −3 dB at the corner, almost 0 dB at $10\omega_{c}$. Its phase leads, from +90° down to 0°, passing +45° at the corner.
A high-pass RC at its corner frequency has gain and phase of
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The high-pass mirrors the low-pass: −3.01 dB at the corner, but the phase leads by +45° instead of lagging.
Steeper filters
- Each pole adds −20 dB per decade above its corner; each zero adds +20 dB per decade. Two poles roll off at 40 dB per decade.
- Cascading two sections multiplies their responses, so their decibels add. A second RC loads the first, so a buffer between them keeps the sum exact.
Match each filter to its slope well above the corner.
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Each pole contributes −20 dB per decade and each zero +20 dB per decade, added together above their corners.
Three common slips
- Hertz and rad/s on one axis. $\omega_{c} = \dfrac{1}{RC}$ is in rad/s; $f_{c} = \dfrac{\omega_{c}}{2\pi}$.
- Misreading −3 dB. It means half the mid-band power reaches the load, not that the filter burns half.
- Ignoring phase. A flat magnitude can still distort a pulse if the phase shifts unevenly.
Recap
- $20\log_{10}|H|$ on a log frequency axis turns the response into straight lines.
- Each pole: −20 dB per decade above its corner $\omega_{c} = \dfrac{1}{RC}$; −3 dB and ±45° at the corner.
- The high-pass mirrors the low-pass; cascaded sections add in dB.