RC Low-Pass Filter: Transfer Function and Frequency Response

What an RC Low-Pass Filter Does

The RC low-pass filter is the single most common building block in analog electronics. It is nothing more than one resistor followed by one capacitor, yet that humble pair quietly does the work of cleaning up power rails, smoothing sensor readings, taming the harshness of audio, and guarding the input of an analog-to-digital converter against aliasing. The filter earns the name "low-pass" because it lets slow, low-frequency signals through almost untouched while progressively swallowing the fast, high-frequency content. Once you understand how a single pole shapes a frequency response, every more elaborate filter you meet later will feel like a variation on this one idea.

The behavior is easiest to grasp through the capacitor's impedance, which depends on frequency. At zero frequency the capacitor is an open circuit: no current flows through the resistor, no voltage is dropped across it, and the output equals the input. As the frequency climbs, the capacitor's impedance shrinks, it begins to short the output node toward ground, and the output collapses. Somewhere between these two extremes the filter makes a graceful transition, and the frequency at which that transition is centered is set entirely by the product $RC$.

Deriving the Transfer Function

Put the resistor $R$ in series with the signal path and connect the capacitor $C$ from the output node to ground. The output is taken across the capacitor. In the $s$-domain the resistor has impedance $R$ and the capacitor has impedance $1/(sC)$, so the two elements form a frequency dependent voltage divider. Applying the divider rule:

Multiplying numerator and denominator of the intermediate step by $sC$ clears the fraction and yields the compact form on the right. This is a first-order transfer function with a single pole at $s = -1/(RC)$ and no finite zero. The time constant $\tau = RC$ ties together the transient speed and the frequency response: the same number that tells you how fast a step settles also tells you where the filter starts to roll off. If you want to see how this derivation generalizes, the transfer functions tutorial walks through poles and zeros in detail.

The Cutoff Frequency

The cutoff frequency is the frequency at which the output power has dropped to half the input power, which corresponds to the output voltage magnitude falling to $1/\sqrt{2} \approx 0.707$ of the input. That is why it is also called the −3 dB point or the half-power frequency. Evaluating the transfer function on the imaginary axis, $s = j\omega$, the magnitude is:

Setting this equal to $1/\sqrt{2}$ forces $\omega RC = 1$, so $\omega_c = 1/(RC)$. Converting from radians per second to hertz gives the formula every student memorizes:

Worked Example

Suppose you want a low-pass filter with a cutoff near 1 kHz, and you have a 1 kΩ resistor on hand. Rearranging $f_c = 1/(2\pi RC)$ to solve for the capacitance gives $C = 1/(2\pi R f_c)$. Plugging in numbers:

Now run the check in the forward direction with $R = 1\;\text{k}\Omega$ and $C = 159\;\text{nF}$:

So a 1 kΩ resistor with a 159 nF capacitor gives a clean 1 kHz cutoff. To feel the slope, evaluate the magnitude one decade above cutoff at 10 kHz. There $\omega RC = 10$, so $|H| = 1/\sqrt{1 + 100} \approx 0.0995$, which is about $20\log_{10}(0.0995) \approx -20\;\text{dB}$. That confirms the textbook −20 dB-per-decade roll-off. At the cutoff itself the magnitude is $1/\sqrt{2} = 0.707$, or −3 dB, and the phase has slipped to exactly −45°.

Reading the Bode Plot

The Bode magnitude plot has two straight-line asymptotes. Below cutoff it sits flat at 0 dB (unity gain); above cutoff it falls at −20 dB per decade, equivalently −6 dB per octave. The two asymptotes meet at $f_c$, where the true curve sags 3 dB below the corner. The phase response is just as orderly: it begins at 0° far below cutoff, passes through −45° exactly at $f_c$, and asymptotes to −90° far above. The negative phase tells you the output lags the input, and the lag grows with frequency. In the time domain this same single pole shows up as an exponential step response with time constant $\tau = RC$: the output reaches 63% of its final value after one time constant and settles within 1% after five.

It is worth dwelling on why the magnitude and phase asymptotes work so well as design tools. Engineers rarely need the exact value of the response at every frequency; they need to know roughly where a signal sits relative to the corner and how much attenuation to expect. Because the magnitude curve hugs its straight-line asymptotes everywhere except within about a decade of the corner, you can sketch the entire response with a pencil and a ruler in seconds. The asymptote crossing the corner at 0 dB, the −20 dB/decade slope beyond it, and the single 3 dB correction at the corner together capture everything a first-order low-pass does. This is the practical payoff of recognizing the single-pole structure: the math collapses into a picture you can reason about by inspection.

Choosing Component Values in Practice

When you design a real filter you usually start from the cutoff you want and the impedance level that suits the surrounding circuit. The resistor sets the impedance: too small and the filter loads down the previous stage and wastes current; too large and it forms a divider with the input bias currents and noise of the following stage. A resistor in the range of a few hundred ohms to a few hundred kilohms is usually sensible. With the resistor fixed, the capacitor follows directly from $C = 1/(2\pi R f_c)$. Standard capacitor values are coarsely spaced, so in practice you pick the nearest standard part and accept a cutoff that lands a few percent off target — which is almost always fine, since the response is gentle and the exact corner rarely matters to better than ten percent. If precision does matter, trim the cutoff with the resistor, which comes in much finer value steps than capacitors do.

Frequency Response Summary

Common Mistakes

• Forgetting the factor of $2\pi$. The pole lives at $\omega_c = 1/(RC)$ in radians per second, but the cutoff in hertz is $f_c = 1/(2\pi RC)$. Mixing the two off by a factor of about 6.28 is the single most frequent error.
• Confusing the −3 dB point with where the output dies. At cutoff the filter still passes 70.7% of the signal, not a tiny remnant. Real attenuation only sets in well above $f_c$.
• Ignoring loading by the next stage. The clean $1/(1 + sRC)$ result assumes the output drives an ideal high-impedance load. A low-impedance load in parallel with $C$ shifts the cutoff and lowers the gain.
• Swapping R and C and expecting low-pass. Taking the output across the resistor instead of the capacitor turns the very same parts into a high-pass filter. Topology, not component choice, decides the filter type.
• Treating the roll-off as a brick wall. A first-order filter rolls off gently at −20 dB/decade; it does not sharply block everything past $f_c$. Steeper skirts require higher-order designs.

Build One in CircuitMath

To analyze an RC low-pass filter, place a resistor and a capacitor in the voltage-divider configuration above, add an input voltage source and a ground reference, and run the analysis. CircuitMath returns the transfer function $H(s)$ in LaTeX, confirms the cutoff formula, and shows the loop equation. You can open the editor and sweep $R$ and $C$ to watch the cutoff slide, or browse the full tutorial library for related topics.

Related Tutorials

• RL High-Pass Filter — the complementary first-order filter built around an inductor.
• RLC Resonant Circuit — add an inductor for resonance and sharper selectivity.
• Transfer Functions in Circuit Analysis — poles, zeros, and how to read $H(s)$.