Frequency response and Bode plots

Here is the gain of an RC circuit with $RC=1$ s, drawn against frequency. Watch the dots at 0.1, 1 and 10 rad/s as the ordinary axis turns into one where every tenfold step of frequency gets the same width.

A frequency axis that multiplies

Fourier series and LTI systems (7.4) found the RC circuit’s answer to a handful of harmonics. With $RC=1$ s the answer is $H(j\omega)=1/(1+j\omega)$, with gain $\lvert H\rvert=1/\sqrt{1+\omega^2}$ and lag $\arctan\omega$. This page asks for the answer at every frequency, and for a way to draw it that an engineer can read at a glance.

The trouble starts with the axis. The interesting part of this circuit’s response spans 0.1, 1, 10 and 100 rad/s. Draw the gain against $\omega$ on an ordinary axis from 0 to 100 rad/s and the first three of those frequencies fall in the first tenth of the axis. The rest of it is a flat line near zero.

The cure is to give each tenfold range of frequency the same width. A piano keyboard does this for pitch: every octave, a doubling of frequency, takes the same width. On a logarithmic axis the same holds for steps of ten, and one such step, say from 1 to 10, is a decade. A mark at frequency $\omega$ sits at a distance proportional to $\log_{10}\omega$ from the start, so multiplying $\omega$ by 10 always moves the mark by one decade.

That is what the picture at the top of the page does. On the ordinary axis, the dots at 0.1 and 1 rad/s sit almost on top of each other near zero. Stretch the low end and squeeze the high end, and the dots read $0.995$, $0.707$ and $0.0995$ on the gridlines at 0.1, 1 and 10 rad/s. The dot on the mark “2” is 2 rad/s, gain 0.447. The corner, 1 rad/s, sits in the middle.

Those four values are $1/\sqrt{1+\omega^2}$ at four frequencies: $1/\sqrt{1.01}=0.995$, $1/\sqrt2=0.707$, $1/\sqrt{101}=0.0995$ and $1/\sqrt5=0.447$. The corner at $\omega=1/RC=1$ rad/s is the one that 7.4 called $\omega_c$, and it sits in the middle because 1 is as many decades above 0.01 as it is below 100.

Between two gridlines, for example between 1 and 10, the small marks are 2, 3, 4, up to 9 times the first. They crowd toward the high-frequency end of the decade. The reason is that equal distances mean equal ratios: the step from 1 to 2 doubles the frequency, while the step from 9 to 10 multiplies it by only $10/9=1.11$.

Zero is never reached on this axis, because each step toward lower frequency divides by 10.

The sweep: points landing on the Bode plot

Gains are usually quoted in decibels, as in How big is a signal (1.3). The size of $H$ in decibels is $20\log_{10}\lvert H\rvert$, relative to the input. A gain of one tenth is $-20$ dB and a gain of one hundredth is $-40$ dB. A Bode plot draws two panels on one logarithmic frequency axis: the gain in dB, and the phase of $H$ in degrees.

I want you to measure the circuit the way a lab does. Feed in a sine, wait until the start-up wobble has died away, then read the size and the lag of the output. That is one point of the frequency response. Then move to the next frequency. An audio engineer sweeping a sine through an amplifier is doing this.

The next instrument does it for you over 14 s, from 0.01 to 100 rad/s. Before you watch, predict: as the input gets faster, does the output shrink, lag, or both? Watch the output sine against the input, and the dots it leaves on the Bode plot.

Very slow, at 0.01 rad/s, the output follows the input: $0.00$ dB and $-0.57°$. At the corner, 1 rad/s, it is 3 dB down and 45° behind: $-3.01$ dB and $-45.0°$, which is $-\pi/4$ rad. On the way up the stamps read $-0.04$ dB and $-5.7°$ at 0.1 rad/s, and $-20.04$ dB and $-84.3°$ at 10 rad/s. At 100 rad/s the output is $-40.00$ dB and $-89.4°$. Every tenfold step past the corner makes the output about 20 dB smaller, and the lag heads for 90°. When the clip stops, drag along the Bode plot to move the test frequency yourself.

So the answer to the prediction is both. The dots lie on one smooth curve, which is why a handful of measurements is enough to draw it.

What number is $H(j\omega)$, exactly? Send the arrow $x(t)=e^{j\omega t}$ into a system with impulse response $h$. By Continuous-time convolution (5.3) the output is

Here $\tau$ is only the dummy variable of the integral, as in 5.3. The arrow comes out unchanged except for the factor in front, so that factor is the frequency response:

This is the Fourier transform of $h$ from From series to transform (8.1). For the RC circuit, $h(t)=\tfrac1{RC}e^{-t/RC}u(t)$, which is $e^{-t}u(t)$ when $RC=1$ s. Its transform is $1/(1+j\omega)$, the pair of 8.1, and that is the number 7.4 found by trying $y=He^{j\omega t}$ in the equation. The sweep has been measuring the transform of $h$ all along.

Two straight lines

Look at the end of the sweep again. Far above the corner, $\lvert H\rvert\approx1/\omega$ when $RC=1$ s, because $1+\omega^2$ is nearly $\omega^2$. A tenfold step in $\omega$ then makes the gain ten times smaller, which is $-20$ dB for every decade.

That gives two rules for sketching the gain: flat at 0 dB up to the corner, then down 20 dB for every tenfold step. An engineer can draw them on the back of an envelope.

The true curve is never far from them. In decibels it is exactly $-10\log_{10}(1+\omega^2)$. Below the corner the lines say 0, so the gap is $10\log_{10}(1+\omega^2)$. Above it the lines say $-20\log_{10}\omega$, so the gap is $10\log_{10}(1+1/\omega^2)$. Both are largest at $\omega=1$, where each is $10\log_{10}2=3.01$ dB.

The next instrument draws the sketch first and then checks it. Watch the gap between the true curve and the lines.

The lines say $-20$ dB at 10 rad/s and $-40$ dB at 100 rad/s. At half the corner, at the corner and at twice the corner, the true curve is $0.97$ dB, $3.01$ dB and $0.97$ dB from them. It is never more than 3.01 dB from the lines, and that is at the corner.

At twice the corner, for example, the exact gain is $-6.99$ dB and the line says $-6.02$ dB, a gap of $0.97$ dB.

I call this one RC stage with corner $\omega_c=1/RC$: flat, then $-20$ dB per decade, with $-3.01$ dB and $-45°$ at the corner. It is the first-order system of First- and second-order systems (6.3).

Two stages in a row

Put a second RC stage after the first, with its corner at 10 rad/s, and keep a buffer between them so that the second does not load the first. Each stage multiplies the arrow by its own number, so the pair has $H=H_1H_2$ (convolution becomes multiplication, Properties of the Fourier transform (8.2)). Here

In decibels the product turns into a sum. The logarithm of a product is the sum of the logarithms, $\log_{10}(ab)=\log_{10}a+\log_{10}b$, so the two stages’ dB curves add. The slopes add as well. Above 10 rad/s both stages are falling, and the line falls 40 dB per decade.

The sketch is still good. At 1 rad/s the pair is $3.054$ dB below the lines: the 3.01 dB of the first stage, plus a small $0.043$ dB from the second. At 100 rad/s the exact gain is $-60.04$ dB against the line’s $-60$ dB. Worked example 3 gives the rest.

Second order: the gain at every rhythm

First- and second-order systems (6.3) pushed a mass on a spring with a sine at three rhythms and quoted the steady gains, $1.3216$, $5$ and $0.3304$, without deriving them. You now have what is needed. The equation of that page is

Feed in the arrow $x=e^{j\omega t}$ and try $y=He^{j\omega t}$. By the differentiation rule of 8.2, each derivative multiplies by $j\omega$, so $\dot y=j\omega\,He^{j\omega t}$ and $\ddot y=(j\omega)^2He^{j\omega t}=-\omega^2He^{j\omega t}$. Cancel the common factor $e^{j\omega t}$:

Divide top and bottom by $\omega_n^2$ and write $r=\omega/\omega_n$:

This is the steady gain of 6.3 for every rhythm. For $\zeta=0.1$, half the natural rate gives $1/\sqrt{0.5625+0.01}=1.3216$, which is $2.42$ dB. At $r=1$ the denominator of $H$ is $j2\zeta$, so $\lvert H\rvert=1/(2\zeta)=5$, the quality factor $Q$ of 6.3, and in decibels $20\log_{10}5=13.98$ dB. At twice the natural rate the gain is $1/\sqrt{9+0.16}=0.3304$, which is $-9.62$ dB.

On a Bode plot this system is flat at 0 dB for slow rhythms, rises to a peak of 13.98 dB at $\omega_n$, and then falls. Far above $\omega_n$ the gain is about $1/r^2$, which is $-40$ dB per decade, the slope of two stages.

Bandwidth

An amplifier’s datasheet does not quote a whole Bode plot. It quotes one frequency: where the gain has fallen to the maximum divided by $\sqrt2$, a drop of 3.01 dB. This is the 3 dB bandwidth $f_c$, and the circuit lets frequencies below it through almost unchanged.

For a first-order system the corner and the 3 dB point are the same. With the time constant $\tau=RC$, the 3 dB point is $\omega_c=1/\tau$ rad/s, or $f_c=1/(2\pi\tau)$ Hz. For $\tau=1$ s that is $0.15915$ Hz.

Bandwidth is tied to speed. 6.3 §1 gave the rise time, the time from 10 % to 90 % of a step, as $t_r=\tau\ln9=2.1972$ s for $\tau=1$ s. Multiply the two:

The $\tau$ cancels, so the product is the same for every first-order system. A circuit with twice the bandwidth rises in half the time.

Why a brick wall cannot be built

So far the filters have been gentle slopes. What would the best possible low-pass filter look like? An ideal filter keeps a band of frequencies exactly and removes everything else. There are four kinds: low-pass (keeps everything below a cutoff), high-pass (keeps everything above it), band-pass (keeps a band) and band-stop (removes a band).

The ideal low-pass is the one to know, because the others can be built from it. The high-pass is the low-pass subtracted from 1. A band-pass slides the low-pass bump up to $\pm\omega_0$ by multiplying $h$ by $2\cos\omega_0t$, the shift of 8.2. The band-stop is the band-pass subtracted from 1.

Transforms of common signals (8.3) gave the ideal low-pass pair. For cutoff $\omega_c=\pi$ rad/s it is $\lvert H\rvert=1$ for $\lvert\omega\rvert<\pi$ and 0 elsewhere, with

Look at $h$ for negative time. At $t=-0.5$ s it is $\sin(-\pi/2)/(-\pi/2)=0.637$. A system whose impulse response is nonzero before $t=0$ would respond before the input arrives, so it is not causal, as Properties of LTI systems (5.4) showed. A brick wall cannot be built.

Can a delay help? Delaying $h$ by $D$ seconds multiplies $H$ by $e^{-j\omega D}$, which only turns each frequency by a straight-line angle, as in 8.2. The size stays a brick wall. Delaying moves most of $h$ after 0, and what remains before 0 can be cut off.

The next instrument shows this. It measures the share of $h$‘s energy that comes before $t=0$. The energy is the area under $h^2$ (1.3), and for this $h$ the total is 1. Watch that share as the delay grows.

With no delay, half of this filter’s response comes before the input arrives. The share is a half because $h$ is symmetric about 0. Delayed by 3 s, most of it comes after 0, and only 1.68 % is left before. At 6 s the share is 0.84 %, and what is left before 0 is cut off. The cut part shrinks, but never to nothing.

The shares fall as $D$ grows: $0.0486$ at $D=1$ s, $0.0250$ at 2 s, $0.0168$ at 3 s and $0.0084$ at 6 s. Roughly, the share is $0.05/D$. It never reaches zero, because $h^2$ has a tail that never stops. And the cut changes the response a little. The result is causal, but its frequency response has ripples near the cutoff, of the kind Convergence and the Gibbs phenomenon (7.3) showed. More delay gives a better wall, and costs a longer wait.

Two delays

A delay can be measured in two ways. Take a burst: a tone under a smooth envelope that rises and falls. The tone’s crests and the envelope’s peak are two different things to follow through a circuit.

The phase delay is $\tau_p=-\angle H/\omega$. It is how late the crests inside arrive. The group delay is $\tau_g=-d\angle H/d\omega$, the negative slope of the phase against $\omega$. It is how late the envelope arrives. For a pure delay of $t_0$ the phase is the straight line $-\omega t_0$ (8.2), so both are $t_0$. The shape passes through unchanged, which is called distortionless.

The RC circuit’s phase is not a straight line. With $RC=1$ s, $\angle H=-\arctan\omega$, so $\tau_p=\arctan(\omega)/\omega$ and $\tau_g=1/(1+\omega^2)$. At $\omega=0.1$ rad/s they are $0.9967$ s and $0.9901$ s, nearly equal. At 1 rad/s they are $0.7854$ s and $0.5$ s. At 10 rad/s they are $0.1471$ s and $0.0099$ s.

The figure sends a burst with carrier 1 rad/s and envelope $e^{-t^2/200}$ through the circuit. That envelope has fallen to $0.61$ of its peak 10 s from the centre, so the burst is long compared with one crest, and only frequencies near 1 rad/s matter.

In the figure the dashed curves are the envelopes. A square marks an envelope peak and a dot marks a crest. Simulated, the envelope peaks are $0.505$ s apart and the chosen crests $0.782$ s apart. The theory at 1 rad/s says $0.5$ s and $0.785$ s, and the small differences come from the burst’s bandwidth.

Worked example

1. RC low-pass, $RC=1$ s. With $H=1/(1+j\omega)$:

   $\omega$ (rad/s)	$\lvert H\rvert$	dB	$\angle H$
   0.01	0.99995	$-0.0004$	$-0.573°$
   0.1	0.99504	$-0.0432$	$-5.711°$
   1	0.70711	$-3.0103$	$-45.000°$
   10	0.099504	$-20.0432$	$-84.289°$
   100	0.0099995	$-40.0004$	$-89.427°$

2. Line error of one stage at $0.5\omega_c$, $\omega_c$ and $2\omega_c$ is $0.969$, $3.010$ and $0.969$ dB. At $2\omega_c$ the exact gain is $-6.990$ dB and the line is $-6.021$ dB.

3. Two stages, corners 1 and 10 rad/s. The line errors at 0.5, 1 and 2 rad/s are $0.980$, $3.054$ and $1.139$ dB. At 10 rad/s it is $3.054$ dB. At 100 rad/s the exact gain is $-60.04$ dB against the line’s $-60$ dB.

4. Delays, $RC=1$ s. $\tau_p=\arctan(\omega)/\omega$ and $\tau_g=1/(1+\omega^2)$ give $0.99669$ s and $0.99010$ s at 0.1 rad/s, $0.78540$ s and $0.5$ s at 1 rad/s, and $0.14711$ s and $0.00990$ s at 10 rad/s. A simulated burst with carrier 1 rad/s and envelope $e^{-t^2/50}$ gives $0.519$ s and $0.770$ s. With $e^{-t^2/200}$ it gives $0.505$ s and $0.782$ s.

5. Bandwidth and rise time, $\tau=1$ s. $\omega_c=1$ rad/s, $f_c=0.15915$ Hz, $t_r=\ln9=2.1972$ s, and $t_rf_c=0.3497$.

6. 6.3’s rig on a Bode plot. For $\zeta=0.1$ the gain at $\omega_n$ is $Q=5$, which is $13.98$ dB. At $0.5\omega_n$ it is $1.3216$ ($2.42$ dB) and at $2\omega_n$ it is $0.3304$ ($-9.62$ dB).

7. Brick wall, $\omega_c=\pi$ rad/s. $h(t)=\sin(\pi t)/(\pi t)$, $h(0)=1$ and $h(-0.5)=0.63662$. It is zero at every nonzero integer, and its total energy is 1. The share of the energy before $t=0$ after a delay $D$ is $0.5$ at $D=0$, then $0.0486$, $0.0250$, $0.0168$ and $0.0084$ at $D=1$, 2, 3 and 6 s.

Where you’ll meet this

Every datasheet that quotes a bandwidth, a corner or a roll-off in dB per decade is quoting this page. An audio equaliser is a Bode plot you can move with sliders. A loudspeaker, a microphone and a tuned circuit have the resonant peak of the second-order system. The question of how much delay a filter adds, and whether it adds the same to every frequency, is the question of the last section.

Later pages read the same plot from other directions. Poles, zeros and the s-plane (9.3) shows how to read the bends of a Bode plot off a map of the system, and Frequency response of discrete-time systems (12.4) draws the same plot for sampled systems. Linear-phase systems (17.3) builds filters whose phase is a straight line, All-pass systems (17.1) and Minimum phase (17.2) separate size from phase, and Analog prototype filters (20.1) designs filters that approach the brick wall.

Reference card