Frequency Response Magnitude and Gain in Decibels

Hello. In the previous lesson, you moved from a transfer function $H(s)$ to the complex frequency response $H(j\omega )$ by substituting $s=j\omega$. For example, the RC low-pass response at one frequency was found to be

$$H(j5000)=0.8-j0.4$$

That complex number contains two distinct pieces of information: how much the signal amplitude changes, and how much its phase changes. This lesson extracts the first piece: magnitude, then expresses it on the logarithmic decibel scale used in filter plots.

By the end, you should be able to calculate $|H(j\omega )|$, convert it to gain in dB, and interpret what that value says about the output sinusoid’s amplitude.

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Magnitude: the amplitude part of a complex response

For a sinusoidal input at angular frequency $\omega$,

$${\tilde{V}}_o=H(j\omega ){\tilde{V}}_i$$

Taking magnitudes on both sides gives

$$|{\tilde{V}}_o|=|H(j\omega )||{\tilde{V}}_i|$$

Therefore,

$$\boxed{|H(j\omega )|=\frac{|{\tilde{V}}_o|}{|{\tilde{V}}_i|}}$$

The magnitude of the frequency response is a nonnegative, dimensionless voltage-amplitude ratio.

• $|H|=1$: output and input have equal amplitudes.
• $|H|<1$: the output is attenuated.
• $|H|>1$: output amplitude exceeds input amplitude.
• $|H|=0$: no output remains at that frequency.

Use the same amplitude convention for input and output: both peak values, both RMS values, or both peak-to-peak values. Since the ratio is taken, the convention cancels out.

If a frequency response has rectangular form

$$H(j\omega )=a+jb$$

then its magnitude is found using the complex-number modulus:

$$\boxed{|H(j\omega )|=\sqrt{a^2+b^2}}$$

Here $a$ and $b$ are real numbers. The sign of $b$ does not affect the magnitude, because it is squared.

For a quotient of complex expressions,

$$H(j\omega )=\frac{N(j\omega )}{D(j\omega )}$$

use the useful property

$$\boxed{\left|\frac{N}{D}\right|=\frac{|N|}{|D|}}$$

This often avoids unnecessary complex-conjugate algebra.

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Worked example: magnitude of the response from the previous lesson

Continue with the RC low-pass result already obtained:

$$H(j5000)=0.8-j0.4$$

Apply the modulus formula:

$$|H(j5000)|=\sqrt{0.8^2+(-0.4{)}^2}$$ $$|H(j5000)|=\sqrt{0.64+0.16}$$ $$\boxed{|H(j5000)|\approx 0.894}$$

The interpretation is direct: at $5000\text{rad/s}$, the output amplitude is $0.894$ times the input amplitude. For a $2\text{V}$ RMS input, for example, the output magnitude would be

$$|V_o|=(0.894)(2\text{V})$$ $$|V_o|\approx 1.79\text{V RMS}$$

Notice that we have deliberately not calculated the phase angle. The magnitude tells us the vertical coordinate on a magnitude plot; phase will be handled in the next lesson.

Direct magnitude calculation for an RC low-pass expression

For the standard RC low-pass response, with output taken across the capacitor,

$$H(j\omega )=\frac{1}{1+j\omega RC}$$

the numerator has magnitude $1$. The denominator has real part $1$ and imaginary part $\omega RC$, so its magnitude is

$$|1+j\omega RC|=\sqrt{1+(\omega RC{)}^2}$$

Thus,

$$\boxed{|H(j\omega )|=\frac{1}{\sqrt{1+(\omega RC{)}^2}}}$$

This is an important pattern: calculate the magnitude of the numerator and denominator separately, then divide.

At the frequency used in the previous lesson,

$$RC=100\mu \text{s}$$

and

$$\omega =5000\text{rad/s}$$

so

$$\omega RC=(5000)(100\times 10^{-6})=0.5$$

Hence,

$$|H(j5000)|=\frac{1}{\sqrt{1+0.5^2}}$$ $$|H(j5000)|=\frac{1}{\sqrt{1.25}}$$ $$|H(j5000)|\approx 0.894$$

This agrees with the magnitude found from $0.8-j0.4$. Both routes are valid:

• convert the completed response to rectangular form, then take its modulus;
• take numerator and denominator magnitudes directly.

The second method is usually quicker when the response is already in factored or quotient form.

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From an amplitude ratio to decibels

A raw magnitude such as $0.894$ is physically meaningful, but it becomes inconvenient when a filter must be described over frequencies ranging from a few hertz to megahertz. Decibels compress these ratios into a practical scale.

For a voltage or current amplitude ratio, define gain in decibels as

$$\boxed{G_{\mathrm{dB}}(\omega )=20{\log }_{10}\left(|H(j\omega )|\right)}$$

The factor $20$, rather than $10$, is used because power is proportional to the square of voltage when the relevant impedances are the same. A power ratio would instead use

$$10{\log }_{10}\left(\frac{P_o}{P_i}\right)$$

For filter voltage responses, use the first formula: $20{\log }_{10}|H|$.

Returning to the example,

$$|H(j5000)|=0.894$$

Therefore,

$$G_{\mathrm{dB}}=20{\log }_{10}(0.894)$$ $$\boxed{G_{\mathrm{dB}}\approx -0.97\text{dB}}$$

A gain of $-0.97\text{dB}$ means a modest reduction in output amplitude. It does not mean that the output voltage is negative, nor that the output is zero.

These values are worth recognizing immediately:

Magnitude ratio $|H|$	Gain in dB	Meaning
$2$	$+6.02\text{dB}$	Amplitude doubled
$1$	$0\text{dB}$	No amplitude change
$1/\sqrt{2}\approx 0.707$	$-3.01\text{dB}$	Half-power point
$0.5$	$-6.02\text{dB}$	Amplitude halved
$0.1$	$-20\text{dB}$	Amplitude reduced to one tenth

The $-3\text{dB}$ row will become especially important in the next lessons, because it defines the conventional cutoff point of many passive filters.

To reverse a dB calculation and recover the ordinary magnitude ratio, use

$$\boxed{|H(j\omega )|=10^{G_{\mathrm{dB}}/20}}$$

For instance, a response value of $-20\text{dB}$ corresponds to

$$|H|=10^{-20/20}=10^{-1}=0.1$$

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Reading Bode magnitude plots

A Bode magnitude plot displays frequency on the horizontal axis, usually logarithmically, and $G_{\mathrm{dB}}$ on the vertical axis. Each point gives the output-to-input amplitude ratio at that frequency, expressed in dB.

On the plotted low-pass response:

• Near $0\text{dB}$, $|H|$ is close to $1$, so low-frequency signals retain nearly their full amplitude.
• At $-3\text{dB}$, $|H|\approx 0.707$.
• At $-20\text{dB}$, $|H|=0.1$.
• More negative dB values mean stronger attenuation.

The logarithmic frequency axis helps display many decades of frequency, while the dB axis makes multiplicative changes in amplitude easy to compare. The detailed asymptotic construction of the $-20\text{dB}$ per decade line comes later; for now, focus on treating any point on the curve as a magnitude ratio written in dB.

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A dependable calculation routine

When asked for the magnitude gain and dB gain at a specified frequency, use this sequence:

1. Convert the stated frequency to angular frequency if necessary:

   $$\omega =2\pi f$$

2. Evaluate the response at that frequency:

   $$H(j\omega )$$

3. Calculate the ordinary magnitude. For rectangular form, use

   $$|a+jb|=\sqrt{a^2+b^2}$$

   For a quotient, take numerator and denominator magnitudes separately.

4. Convert the magnitude to decibels:

   $$G_{\mathrm{dB}}=20{\log }_{10}\left(|H(j\omega )|\right)$$

5. State both forms in words. For example: “The gain is $0.316$, or $-10\text{dB}$; the output amplitude is about $31.6\%$ of the input amplitude.”

Two errors account for many incorrect results:

• Applying $20{\log }_{10}$ to a complex number rather than to its magnitude.
• Treating a dB value as a voltage. A value such as $-6\text{dB}$ is a logarithmic ratio, not $-6\text{V}$.

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Key takeaways

The magnitude of a frequency response is the amplitude ratio

$$|H(j\omega )|=\frac{|V_o|}{|V_i|}$$

For a complex number $a+jb$,

$$|a+jb|=\sqrt{a^2+b^2}$$

and for a quotient, calculate the numerator and denominator magnitudes separately.

Express the result in decibels with

$$G_{\mathrm{dB}}=20{\log }_{10}\left(|H(j\omega )|\right)$$

A magnitude of $1$ corresponds to $0\text{dB}$; magnitudes below $1$ produce negative dB values and indicate attenuation. In particular, $1/\sqrt{2}$ corresponds to approximately $-3\text{dB}$.

Next, you will calculate the phase angle $\angle H(j\omega )$, completing the two-part interpretation of a complex frequency response.