Bode Plot — How to Draw

To draw a Bode plot, first factor the transfer function, then add the effect of each gain, pole, and zero on a log-frequency axis. You usually sketch two graphs: magnitude in dB and phase in degrees for $G(j\omega)$.

For a transfer function $G(s)$, the standard quantities are

$$\text{magnitude in dB} = 20 \log_{10} |G(j\omega)|$$

and

$$\text{phase} = \angle G(j\omega).$$

The key simplification is that multiplication in the transfer function becomes addition on the Bode plot. That is why a complicated expression can still be sketched by hand.

How To Draw A Bode Plot Fast

Use this order:

1. Rewrite the transfer function as simple factors.
2. Mark each break frequency on a logarithmic frequency axis.
3. Add the magnitude contribution of each factor in dB.
4. Add the phase contribution of each factor.

A common factored form is

$$G(s) = K \frac{\prod (1 + s / \omega_z)}{s^m \prod (1 + s / \omega_p)}.$$

Here, $K$ is a constant gain, each $\omega_z$ is a zero frequency, and each $\omega_p$ is a pole frequency.

What Each Factor Does

Constant Gain $K$

• Magnitude: add $20 \log_{10}|K|$ dB everywhere.
• Phase: add $0^\circ$ if $K > 0$. If $K < 0$, the phase differs by $180^\circ$ up to your angle convention.

Zero At $\omega_z$

For a factor $(1 + s / \omega_z)$:

• Magnitude: about $0$ dB before $\omega_z$, then slope $+20$ dB/decade after $\omega_z$.
• Phase: rises from about $0^\circ$ to about $+90^\circ$ across the transition around $\omega_z$.

Pole At $\omega_p$

For a factor $(1 + s / \omega_p)$ in the denominator:

• Magnitude: about $0$ dB before $\omega_p$, then slope $-20$ dB/decade after $\omega_p$.
• Phase: falls from about $0^\circ$ to about $-90^\circ$ across the transition around $\omega_p$.

Pole Or Zero At The Origin

For a factor $s$ in the denominator:

• Magnitude: slope $-20$ dB/decade at all frequencies.
• Phase: constant $-90^\circ$.

For a factor $s$ in the numerator:

• Magnitude: slope $+20$ dB/decade at all frequencies.
• Phase: constant $+90^\circ$.

These straight lines are asymptotes, not exact curves. Near a break frequency, the real graph bends smoothly.

Worked Example: Draw $G(s) = \frac{10}{s(1 + s / 10)}$

This example has one constant gain, one pole at the origin, and one first-order pole at $\omega = 10$. That is enough to show the full sketching process without extra algebra.

Step 1: Substitute $s = j\omega$

$$G(j\omega) = \frac{10}{j\omega(1 + j\omega / 10)}.$$

Step 2: Sketch The Magnitude Plot

The exact magnitude is

|G(j\omega)| = \frac\{10\}\{\omega \sqrt\{1 + (\omega / 10)^2\}}.

So the exact magnitude in decibels is

$$20 \log_{10}|G(j\omega)| = 20 - 20 \log_{10}\omega - 10 \log_{10}\left(1 + (\omega / 10)^2\right).$$

For a hand sketch, it is faster to add factor-by-factor contributions:

• Gain $10$: $+20$ dB everywhere.
• Pole at the origin: slope $-20$ dB/decade everywhere.
• Pole at $10$: no extra slope before $\omega = 10$, then another $-20$ dB/decade after it.

So the total slope is:

• $-20$ dB/decade for $\omega < 10$
• $-40$ dB/decade for $\omega > 10$

Use one anchor point to place the line. At $\omega = 1$,

|G(j1)| \approx \frac\{10\}\{1 \cdot \sqrt\{1 + 0.1^2\}} \approx 9.95,

so

$$20 \log_{10}(9.95) \approx 20 \text{ dB}.$$

That places the straight-line sketch near $20$ dB at $\omega = 1$. It reaches about $0$ dB at $\omega = 10$, then falls with slope $-40$ dB/decade after the break.

At the corner frequency, the exact curve is lower than the asymptote. For a first-order pole, the difference is about $3$ dB, so here

|G(j10)| = \frac\{10\}\{10\sqrt\{2\}} = \frac\{1\}\{\sqrt\{2\}},

which is about $-3$ dB.

Step 3: Sketch The Phase Plot

The phase is the sum of the phase contributions:

• pole at the origin: $-90^\circ$
• pole at $10$: $-\tan^{-1}(\omega / 10)$

So the exact phase is

$$\angle G(j\omega) = -90^\circ - \tan^{-1}(\omega / 10).$$

That gives three clean checkpoints:

• At very low frequency, the phase is close to $-90^\circ$.
• At $\omega = 10$, the phase is $-135^\circ$.
• At very high frequency, the phase approaches $-180^\circ$.

For a quick sketch, use the usual first-order approximation: start the phase change around $\omega_p / 10$, pass through $-45^\circ$ at $\omega_p$, and finish near $10\omega_p$. Here the extra phase drop happens roughly from $\omega = 1$ to $\omega = 100$.

What The Finished Bode Plot Tells You

Once the sketch is done, you can read the behavior quickly.

• High frequencies are attenuated more strongly than low frequencies in this example.
• The break at $\omega = 10$ is where the roll-off becomes steeper.
• The phase lag grows as frequency increases.

That combination is typical of a low-pass response with an integrator.

Common Bode Plot Mistakes

• Using a linear frequency axis instead of a logarithmic one.
• Multiplying magnitudes on the graph instead of adding them in dB.
• Using $10 \log_{10}$ for amplitude ratios. For transfer-function magnitude, use $20 \log_{10}|G(j\omega)|$.
• Forgetting a pole or zero at the origin, which changes the slope everywhere.
• Treating the straight-line sketch as exact near a corner frequency.

When Bode Plots Are Used

Bode plots are useful when you care about how a system responds to different frequencies.

• In electronics, they describe filters and amplifiers.
• In control, they help estimate bandwidth, crossover behavior, and phase lag.
• In signal processing, they show which frequencies are passed or suppressed.

They are especially helpful when the system is linear and time-invariant and the transfer function can be written as poles and zeros.

Try A Similar Sketch

Try your own version with

$$G(s) = \frac{5(1 + s / 2)}{s(1 + s / 20)}.$$

Mark the break frequencies first, then add the slope and phase changes one factor at a time. If you want to go one step further, compare your sketch with a graphing tool and check where the straight-line approximation differs most.