Half-Power Cutoff: Understanding the 1/√2 Magnitude Ratio and −3 dB Point

Hello. In the previous lesson, you calculated the phase of a frequency response and saw that a first-order RC low-pass filter has a phase of $-45^{\circ }$ at its cutoff frequency. We now connect the other familiar cutoff marker, $-3\text{dB}$, to a physical statement about power.

The key idea is simple but easy to misstate: half power does not mean half voltage. At a half-power cutoff, the voltage magnitude has fallen only to $1/\sqrt{2}$, about $0.707$, of its passband value. This lesson derives that relationship, expresses it in decibels, and connects it to the cutoff point of an RC filter.

---

Voltage magnitude and power are related by a square

Your transfer function magnitude,

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

is a voltage-amplitude ratio. It tells you what fraction of the input voltage amplitude appears at the output at a particular frequency.

For a sinusoidal voltage across a resistor $R$, the average power is

$$P=\frac{V_{\text{rms}}^2}{R}$$

Thus, provided we compare voltages across the same resistance (or the same reference impedance), power is proportional to voltage squared:

$$P\propto V^2$$

Suppose that the passband voltage is $V_{\text{PB}}$, and that at some frequency the output voltage magnitude is a fraction $A$ of the passband value:

$$A=\frac{|V_o|}{|V_{\text{PB}}|}$$

The corresponding power ratio is therefore

$$\frac{P_o}{P_{\text{PB}}}=\frac{|V_o{|}^2/R}{|V_{\text{PB}}{|}^2/R}$$ $$\boxed{\frac{P_o}{P_{\text{PB}}}={\left(\frac{|V_o|}{|V_{\text{PB}}|}\right)}^2=A^2}$$

At the half-power point,

$$\frac{P_o}{P_{\text{PB}}}=\frac{1}{2}$$

so the voltage magnitude ratio must satisfy

$$A^2=\frac{1}{2}$$

Taking the positive square root, since magnitude cannot be negative,

$$\boxed{A=\frac{1}{\sqrt{2}}\approx 0.707}$$

Therefore, the half-power cutoff condition is

$$\boxed{\frac{|V_o|}{|V_{\text{PB}}|}=\frac{1}{\sqrt{2}}}$$

For a unity-gain passive filter whose passband magnitude is approximately one, this is often written more simply as

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

A numerical picture

Imagine that a passband output is $10{\text{V}}_{\text{rms}}$ across a $100\Omega$ resistor. Its power is

$$P_{\text{PB}}=\frac{10^2}{100}=1\text{W}$$

At the half-power frequency, the voltage is not $5{\text{V}}_{\text{rms}}$. It is

$$V_o=\frac{10}{\sqrt{2}}=7.07{\text{V}}_{\text{rms}}$$

The resulting power is

$$P_o=\frac{(7.07{)}^2}{100}\approx 0.5\text{W}$$

Condition	Voltage ratio	Power ratio
Passband reference	$1$	$1$
Half-power cutoff	$1/\sqrt{2}\approx 0.707$	$1/2$
Half voltage	$1/2$	$1/4$

So a half-voltage point would be a quarter-power point, not a half-power point.

In introductory filter analysis, “power” at cutoff is often shorthand for this squared-magnitude comparison. For a voltage measured across a reactive component, such as the capacitor in an unloaded RC low-pass filter, the capacitor itself does not dissipate average real power. The half-power terminology remains useful because $|H{|}^2$ is the normalized power-transfer quantity when the voltage is referenced to an equal resistive impedance or a specified load.

---

Why half power is called $-3\text{dB}$

There are two equivalent decibel formulas:

$$G_P(\text{dB})=10{\log }_{10}\left(\frac{P_o}{P_{\text{ref}}}\right)$$

and, when the compared voltages have equal reference resistance,

$$G_V(\text{dB})=20{\log }_{10}\left(\frac{|V_o|}{|V_{\text{ref}}|}\right)$$

The factor of $20$, rather than $10$, in the voltage form comes directly from the square relation between power and voltage.

At half power,

$$\frac{P_o}{P_{\text{PB}}}=\frac{1}{2}$$

Therefore,

$$G_P=10{\log }_{10}\left(\frac{1}{2}\right)$$ $$\boxed{G_P=-3.0103\text{dB}}$$

Using the associated voltage ratio gives exactly the same result:

$$G_V=20{\log }_{10}\left(\frac{1}{\sqrt{2}}\right)$$ $$\boxed{G_V=-3.0103\text{dB}}$$

In normal filter work, engineers round this value:

$$\boxed{\text{half power}⟺\frac{|V_o|}{|V_{\text{PB}}|}=\frac{1}{\sqrt{2}}⟺-3\text{dB}}$$

Be precise about the wording:

• A drop of $-3\text{dB}$ means half the power.
• A drop of $-3\text{dB}$ means the voltage magnitude is $70.7\%$ of its reference value.
• A voltage magnitude of $50\%$ corresponds to

$$20{\log }_{10}\left(\frac{1}{2}\right)=-6.02\text{dB}$$

and hence one-quarter power.

---

The cutoff point on a real filter response

An ideal low-pass filter would change instantly from full transmission to zero transmission. Physical passive filters do not: their magnitude response changes gradually. The cutoff frequency gives a standardized, reproducible location within that transition.

For a normalized low-pass response, the passband level is

$$|H|=1$$

which is

$$20{\log }_{10}(1)=0\text{dB}$$

The cutoff level is $3.0103\text{dB}$ below that passband level:

$$0\text{dB}-3.0103\text{dB}=-3.0103\text{dB}$$

More generally, if a filter has passband gain $G_{\text{PB}}$ in decibels rather than $0\text{dB}$, the half-power cutoff level is

$$\boxed{G_{\text{cutoff}}=G_{\text{PB}}-3.0103\text{dB}}$$

This “relative to the passband” wording matters. A passive filter often has a unity passband gain in simplified analysis, but component loading can make the actual passband gain lower than one. The cutoff is still $3\text{dB}$ below the relevant maximum or nominal passband level, not necessarily at an absolute level of $-3\text{dB}$.

---

Connection to the first-order RC low-pass filter

From the previous lesson, the standard RC low-pass transfer function is

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

Its magnitude is

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

The cutoff angular frequency is

$${\omega }_c=\frac{1}{RC}$$

Substitute $\omega ={\omega }_c$:

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

Because

$${\omega }_{c}RC=1$$

we obtain

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

That directly proves that the RC cutoff frequency is the half-power frequency. Expressed in decibels,

$$20{\log }_{10}|H(j{\omega }_c)|=20{\log }_{10}\left(\frac{1}{\sqrt{2}}\right)$$ $$\boxed{20{\log }_{10}|H(j{\omega }_c)|=-3.0103\text{dB}}$$

For this particular first-order RC low-pass circuit, cutoff has two useful simultaneous signatures:

$$\boxed{|H(j{\omega }_c)|=\frac{1}{\sqrt{2}}}$$ $$\boxed{20{\log }_{10}|H(j{\omega }_c)|\approx -3\text{dB}}$$ $$\boxed{\angle H(j{\omega }_c)=-45^{\circ }}$$

The first two define the half-power magnitude condition; the $-45^{\circ }$ phase value is a characteristic of this specific first-order low-pass topology.

---

Key takeaways

For equal reference resistances,

$$P\propto V^2$$

Therefore, at half power,

$$\frac{P_o}{P_{\text{PB}}}=\frac{1}{2}$$

the corresponding magnitude ratio is

$$\boxed{\frac{|V_o|}{|V_{\text{PB}}|}=\frac{1}{\sqrt{2}}\approx 0.707}$$

In decibels, the same condition is

$$\boxed{10{\log }_{10}\left(\frac{1}{2}\right)=20{\log }_{10}\left(\frac{1}{\sqrt{2}}\right)=-3.0103\text{dB}}$$

Thus, “cutoff,” “half-power point,” and “$-3\text{dB}$ point” refer to the same standard magnitude condition. In the next lesson, you will use magnitude-response plots to identify passbands, stopbands, and cutoff frequencies directly from the graph.