Solving an Example of a Low Pass Filter: RC Circuit Exam Walkthrough

When you are staring down an AC circuits exam or trying to prototype an audio crossover on the bench, the passive RC low pass filter is the foundational building block you must master. It is the most common example of a low pass filter encountered in both university coursework and practical RF/audio design. Rather than just memorizing the cutoff formula, you need to understand how to navigate component tolerances, phasor domain math, and real-world verification.

This walkthrough dissects a classic exam problem, showing every algebraic step, the standard component table you actually need at the bench, and the specific traps that cost students points.

The Core Problem: Designing a 1 kHz RC Filter

Before we touch the algebra, we need to acknowledge a reality of the physical workbench: exact theoretical resistor values do not exist in your component bins. You must map your theoretical target to standard E12 or E24 series values. Below is a data-dense reference table for hitting a ~1 kHz target using common capacitor values and their corresponding nearest E24 resistors.

Step-by-Step Algebraic Solution & Method Justification

Which method applies and why? We use the Complex Voltage Divider Rule in the phasor (frequency) domain. Because the circuit contains a reactive component (the capacitor), its impedance changes with frequency. We cannot use simple DC Ohm's law; we must use complex impedance ($Z$) to account for both magnitude and phase shift. As detailed in Electronics Tutorials' passive filter guide, the transfer function $H(j\omega)$ is derived directly from this divider.

Part A: Finding the Resistance

1. State the cutoff frequency formula: The -3dB cutoff frequency for a first-order RC filter is defined as:
   f_c = 1 / (2 * π * R * C)
2. Isolate R algebraically: Multiply both sides by R, then divide by f_c:
   R = 1 / (2 * π * f_c * C)
3. Substitute the given values: $f_c = 1000$ Hz, $C = 10 \text{ nF} = 10 \times 10^{-9}$ F.
   R = 1 / (2 * π * 1000 * 10e-9)
4. Calculate the exact theoretical resistance:
   R = 1 / (0.0000628318) = 15,915.5 Ω
5. Select the standard E24 value: Looking at our table above, the nearest E24 value is 16 kΩ (16,000 Ω).
6. Recalculate the actual $f_c$ with the chosen resistor:
   f_c_actual = 1 / (2 * π * 16000 * 10e-9) = 994.7 Hz

Part B: Calculating Output Voltage at 10 kHz

1. Find the capacitive reactance ($X_C$) at 10 kHz:
   X_C = 1 / (2 * π * f * C)
X_C = 1 / (2 * π * 10000 * 10e-9) = 1,591.5 Ω
2. Calculate the total circuit impedance magnitude ($|Z_{total}|$): Since R and C are in series, their impedances add vectorially.
   |Z_{total}| = √(R² + X_C²)
|Z_{total}| = √(16000² + 1591.5²) = √(256,000,000 + 2,532,872) = 16,079.1 Ω
3. Apply the magnitude voltage divider rule:
   |V_{out}| = |V_{in}| * (X_C / |Z_{total}|)
|V_{out}| = 2.0 Vpp * (1591.5 / 16079.1)
4. Final calculation:
   |V_{out}| = 2.0 * 0.09898 = 0.198 Vpp

The Exam Trap & Independent Verification

Answer Sanity Check

How do we know 0.198 Vpp is correct without re-running the calculator? We use Bode plot asymptotes. The input frequency (10 kHz) is exactly one decade above our cutoff frequency (~1 kHz). A first-order low pass filter attenuates high frequencies at a rate of -20 dB per decade. A -20 dB attenuation corresponds exactly to a voltage ratio of 0.1 (or 1/10th). Therefore, a 2.0 Vpp input should yield roughly 0.2 Vpp at the output. Our calculated 0.198 Vpp aligns perfectly with this order-of-magnitude check.

How to Verify Independently

If you are building this on a bench, do not trust the math blindly. Ceramic capacitors (especially X7R and Y5V dielectrics) exhibit severe capacitance droop under DC bias and have wide manufacturing tolerances (often ±10% to -20%).

• Simulation: Run an AC analysis in LTspice. Plot V(out)/V(in) in decibels and verify the trace crosses the -3.01 dB line at exactly 994.7 Hz.
• Physical Bench Test: Connect a function generator (e.g., Siglent SDG1032X) to the filter input. Set it to a 1 kHz sine wave at 2.0 Vpp. Probe the output with an oscilloscope. Adjust the generator frequency upward until the measured Vpp drops to 1.414 V (which is $2.0 \times 0.707$). The frequency displayed on the generator is your true physical $f_c$. For deeper design theory, refer to Analog Devices' application notes on first-order filters.

Frequently Asked Filter Exam Questions

Why is the output voltage not exactly 0V at high frequencies?

An ideal first-order filter approaches 0V asymptotically but never truly reaches it. In the real world, however, you will hit a noise floor. Furthermore, parasitic capacitance across the resistor and the Equivalent Series Inductance (ESL) of the capacitor will eventually cause the filter to behave like a band-pass or even resonate at VHF/UHF frequencies, ruining the high-frequency attenuation.

Does the order of the resistor and capacitor matter?

For a standard ungrounded signal path, no. The voltage divider math remains identical whether the signal hits the resistor first (series R, shunt C to ground) or the capacitor first (series C, shunt R to ground). However, in practical PCB layout, placing the resistor first (closest to the source) is preferred. This isolates the driving op-amp or microcontroller GPIO from the purely capacitive load of the filter, preventing phase-margin degradation and high-frequency oscillation in the driving stage.

What happens to the phase shift at the cutoff frequency?

At exactly $f_c$, the resistive and reactive voltage drops are equal in magnitude. This results in a phase shift of exactly -45° (the output lags the input by 45 degrees). At frequencies well below $f_c$, the phase shift approaches 0°, and well above $f_c$, it approaches -90°.