1st/2nd order Bode plots

In this section, we will derive the Bode plots for first and second-order systems. Just like understanding the time-domain response of first and second order systems allows you to predict the behavior of more complex systems (via partial fraction decomposition), understanding the frequency response of first and second order systems allows you to predict the frequency response of more complex systems. However, it’s even easier with frequency response because no partial fraction decomposition is needed!

We will start with the simplest systems (constant gains and integrators) and then move on to first and second-order systems.

Constant gains¶

Let’s start with the simplest of transfer functions, a constant gain $G(s) = K$. The frequency response is given by $G(j\omega) = K$, which is a complex number with magnitude $M(\omega) = |K|$ and phase $\phi(\omega) = \angle K$. The frequency response is constant for all frequencies, which means that the Bode plot will be a horizontal line in both the magnitude and phase plots. However, the value of the magnitude and phase will depend on whether $K$ is positive or negative. Here are the plots:

Shifting property¶

Multiplying a system by a constant gain simply adds a constant offset to its Bode magnitude plot. This is because:

So we can just shift the Bode magnitude plot of $G$ up or down by $20\log_{10} |K|$ dB, depending on whether $K$ is greater than or less than 1. For the phase plot, we have:

So if $K$ is positive, the phase plot is unchanged, while if $K$ is negative, the phase plot is shifted by 180°.

Integrator¶

An integrator has the transfer function $G(s) = \frac{1}{s}$, which means that its frequency response is given by $G(j\omega) = \frac{1}{j\omega}$. The magnitude of the frequency response is $M(\omega) = \frac{1}{\omega}$, which means that:

So on a log scale, the magnitude plot will be a straight line with a slope of -20 dB/decade. The phase of the frequency response is

As with the constant gain, the phase is constant at all frequencies. This time it’s a 90° phase lag. Here is the corresponding Bode plot:

Notice that the DC gain of the integrator is infinite. This is because $G(s)=\frac{1}{s}$ so $G(0) \to \infty$. We can see why this is the case by looking at the time-domain response. Integration of a sinusoid produces:

The output amplitude is $\frac{1}{\omega}$, which gets larger the smaller we make $\omega$. In the limit $\omega\to 0$, the input becomes $u(t)=1$ and the output becomes $y(t)=t$, which grows without bound as $t\to\infty$. This is why the DC gain is infinite. That being said, integrators still possess high-frequency roll-off, so high-frequency inputs are attenuated/smoothed.

Differentiator¶

A differentiator has the transfer function $G(s) = s$, which means that its frequency response is given by $G(j\omega) = j\omega$. This is the inverse of what we had for the integrator. This time, $M_\text{dB}(\omega) = 20\log_{10}(\omega)$ and $\phi(\omega) = 90\degree$. So our magnitude plot is a constant slope of +20 dB/decade and the phase plot is a constant 90° of phase lead. Here is the Bode plot:

This time, the DC gain is zero (the derivative of a constant is zero), while the high-frequency gain is infinite. We can see this again with the time-domain response:

The output amplitude is $\omega$, which gets larger the larger we make $\omega$. In the limit $\omega\to \infty$, the output has infinite amplitude.

First-order systems¶

Let’s consider a first-order system with transfer function $G(s) = \frac{1}{\tau s + 1}$. The frequency response is given by $G(j\omega) = \frac{1}{j\omega\tau + 1}$. Let’s treat the magnitude and phase separately.

Magnitude (first-order)¶

The magnitude of the frequency response is

Let’s look at what happens at low, medium, and high frequencies:

• At low frequencies ($\omega\tau \ll 1$), we can neglect $\omega\tau$ and the magnitude is approximately 1 (0 dB).

• At high frequencies ($\omega\tau \gg 1$), we can neglect the 1 in the denominator and the magnitude is approximately $\frac{1}{\omega\tau}$ Converting to dB, we have:

This is a -20 dB/decade slope, since every time we multiply $\omega$ by 10, $M$ decreases by 20 dB. This is the same as what happened with the integrator.

• When both denominator terms are equal, ($\omega\tau = 1$), we have $M(\omega) = \frac{1}{\sqrt{2}} \approx -3$ dB. This is a corner frequency, and it marks the transition between the low-frequency and high-frequency behavior.

Phase (first-order)¶

The phase of the frequency response is

Again, we can look at low, medium, and high frequencies:

• At low frequencies ($\omega\tau \ll 1$), we can let $\omega\tau \approx 0$ and the phase is approximately 0°.

• At high frequencies ($\omega\tau \gg 1$), we can let $\omega\tau \to \infty$ and the phase becomes approximately -90°.

• In the middle, at the corner frequency ($\omega\tau = 1$), the phase is exactly -45°.

The phase plot also has “point symmetry” about the corner frequency. This means that the phase plot makes an S-shaped curve that looks the same if you rotate it by 180° around the corner frequency. To see why, expand the explanation below.

Why point symmetry?

To see why, imagine two frequencies $\omega_1$ and $\omega_2$ equidistant from the corner frequency on a log scale, so $\omega_1 \tau = \alpha$ and $\omega_2\tau =\frac{1}{\alpha}$ for some $\alpha > 0$. Then,

$$\underbrace{\atan(\alpha)}_{\phi_1} + \underbrace{\atan\bigl(\tfrac{1}{\alpha}\bigr)}_{\phi_2} = 90\degree.$$

(10)

This follows from the right-angle triangle below.

It follows that $\phi\bigl(\frac{\alpha}{\tau}\bigr) + \phi\bigl(\frac{1}{\alpha\tau}\bigr) = -90\degree$, which means that the phase plot has point symmetry about the corner frequency $\frac{1}{\tau}$.

Bode sketch (first-order)¶

We can sketch the Bode magnitude by plotting the two asymptotes (low frequency constant and high-frequency slope) and then connecting them smoothly around the corner frequency.

For the Bode phase plot, the S-curve starts and ends roughly one decade before and one decade after the corner frequency. We can approximate the shape by connecting these points with straight lines and drawing a smooth curve on top.

Here is the bode plot for our first-order system, showing the approximation and the true plot overlaid:

First-order demo¶

The interactive demonstration below shows the frequency response of a first-order system: a water heating device where the input is the flame intensity and the output is the water temperature. You can adjust the frequency of oscillation $\omega = \frac{1}{\tau}$ and observe how the response changes. You can also click the boxes to animate the plot and reveal the Bode plot. Notice that as you increase the frequency, the magnitude decays (high-frequency roll-off) and the phase lag increases up to 90°.

Summary¶

Second-order systems¶

The derivation of the Bode plot for second-order systems is similar to the first-order case, but more involved. Let’s start with a standard second-order system with DC gain 1:

Again, we shall treat the magnitude and phase separately.

Magnitude (second-order)¶

The magnitude of the frequency response is

• At low frequencies ($\omega \ll \omega_n$), we can neglect the $\omega$ terms in the denominator and the magnitude is approximately 1 (0 dB).

• At high frequencies ($\omega \gg \omega_n$), and the $\bigl(\tfrac{\omega}{\omega_n}\bigr)^2$ term dominates, so $M(\omega) \approx \bigl(\tfrac{\omega}{\omega_n}\bigr)^{-2}$. Every time $\omega$ increases by a factor of 10, $M(\omega)$ decreases by a factor of 100, which corresponds to a slope of -40 dB/decade.

At the corner frequency, we don’t have a constant drop of -3 dB like in the first-order case. Instead, the magnitude at $\omega = \omega_n$ depends on the damping ratio $\zeta$. This leads to the possibility of resonance.

Resonance phenomenon¶

To see whether there is a resonance peak, we can see if $M(\omega)$ reaches a local maximum for some $\omega > 0$. This happens when the denominator is minimized, which occurs when

Using the chain rule and simplifying, we obtain:

The solution $\omega = 0$ corresponds to the low-frequency extremum, which is not what we are interested in. The other solution satisfies

This solution is only valid if $1-2\zeta^2 > 0$ because the left-hand side must be positive, which means that $\zeta < \frac{1}{\sqrt{2}} \approx 0.707$. In this case, the maximizing frequency (called the resonance frequency) is the solution of Eq. (15), which is $\omega_r = \omega_n \sqrt{1 - 2\zeta^2}$.

We can plug this back into the expression for $M(\omega)$ in Eq. (12) to find the peak height:

For very small $\zeta$, the peak gain becomes very large, which is why we say that the system is “resonant”. In such cases, a useful approximation is to set $\zeta^2 \approx 0$ in the denominator, which gives $M(\omega_r) \approx \frac{1}{2\zeta}$. Remember that to read this on a Bode plot, you need to convert this value of $M(\omega_r)$ to decibels! Here is a plot showing the peak size and its approximation as a function of $\zeta$:

Phase (second-order)¶

The phase of the frequency response is

Again, we can look at low, medium, and high frequencies:

• At low frequencies ($\omega \ll \omega_n$), we can neglect the $\bigl(\frac{\omega}{\omega_n}\bigr)$ terms and we obtain

  $$\phi(0) = -\angle(1+0j) = 0\degree$$

  (22)

• At high frequencies ($\omega \gg \omega_n$), the $\bigl(\frac{\omega}{\omega_n}\bigr)^2$ term dominates, so we obtain

  $$\phi(\infty) = -\angle(-\infty+0j) = -180\degree$$

  (23)

• At the corner frequency ($\omega = \omega_n$), we obtain

  $$\phi(\omega_n) = -\angle(0 + 2\zeta j) = -90\degree$$

  (24)

Just like the first-order case, the second-order case also exhibits point symmetry about the corner frequency. The main difference is that the transition between the low and high-frequency behavior gets steeper as $\zeta$ gets smaller. We wil see this in the Bode sketch below.

Bode sketch (second-order)¶

We can sketch the Bode magnitude by plotting the two asymptotes (low frequency constant and high-frequency slope) similarly to how we did for the first-order system, remembering that the roll-off is now -40 dB/decade instead of -20 dB/decade. For the resonance peak, as covered above:

• If $\zeta < \frac{1}{\sqrt{2}} \approx 0.707$, there will be a peak. As $\zeta \to 0$, the peak becomes pronounced, growing roughly like $\frac{1}{2\zeta}$, and occurring at a frequency slightly before the corner frequency $\omega_n$.

• If $\zeta \geq \frac{1}{\sqrt{2}}$, there is no peak and the magnitude plot looks more like that of a first-order system. The larger $\zeta$ is, the more flat the plot.

For the phase plot, the shape is a similar point-symmetric S-curve to the first-order case, but the curve goes from 0° to -180° instead of -90°. The transition region gets steeper as $\zeta$ gets smaller. You can approximate it as three straight lines like in the first-order case, but this time the middle segment should range $\pm \zeta$ decades around the corner frequency $\omega_n$ instead of $\pm 1$ decade. Here is an example for $\zeta=0.4$:

The dashed guidelines break at $\pm \zeta$ decades around the corner frequency $\omega_n$, which means that the bend occurs $\pm 0.4$ fraction of the distance to the next decade. For the magnitude plot, the two asymptotes always meet at $\omega_n$ regardless of $\zeta$, but the resonance peak (if it exists) occurs at $\omega_r$, which is slightly before $\omega_n$.

Here is what happens for different values of $\zeta$ (dashed lines removed for clarity).

Second-order demo¶

The interactive demonstration below shows the frequency response of a second-order mass-spring-damper system. You can adjust $m$, $k$, and $b$ and the derived quantities $\omega_n$ and $\zeta$ are displayed. As with the first-order demo, you can change the input frequency and observe the output. Notice that as you increase the input frequency, the magnitude first increases as we approach resonance, and then decreases as we move past it. Phase lag also increases with frequency, up to 180°. Try decreasing the damping $b$ to see how the resonance peak becomes more pronounced and the phase transition gets steeper.

Summary¶

Test your knowledge¶