If you are analyzing a passive RC network with a 10 kΩ series resistor, a 10 nF capacitor to ground, and a 10 kΩ load resistor in parallel with the capacitor, the true -3dB cutoff frequency is 3183.1 Hz and the DC gain is 0.5 (-6.02 dB). Many students and junior engineers miss this because they calculate the filter in isolation and forget the downstream load. Below is the complete exam-style walkthrough showing every algebraic step, the underlying theorem, and how to verify your results on the bench.
The Practice Problem: Loaded RC Low Pass Filter Example
Problem Statement:
A passive low pass filter is constructed using a series resistor $R_1 = 10\text{ k}\Omega$ and a shunt capacitor $C_1 = 10\text{ nF}$. The output of this filter is connected to a measurement device that presents a purely resistive load of $R_L = 10\text{ k}\Omega$ in parallel with $C_1$.
Find:
- The DC voltage gain ($A_{DC}$) from input to output.
- The actual -3dB cutoff frequency ($f_c$) of the loaded circuit.
The Trap in This Problem
The most common mistake in this low pass filter example is ignoring $R_L$ and blindly applying the standard formula $f_c = \frac{1}{2\pi R_1 C_1}$. That formula only applies when the filter is unloaded (i.e., $R_L = \infty$). In real-world applications—like driving a 10 kΩ ADC input impedance or a microcontroller GPIO—the load resistor forms a voltage divider with $R_1$, altering both the passband gain and the RC time constant.
Which Theorem Applies and Why
We use Thevenin's Theorem. By treating the input voltage source ($V_{in}$), the series resistor ($R_1$), and the load resistor ($R_L$) as a linear network, we can replace them with a single Thevenin equivalent voltage source ($V_{th}$) and a single Thevenin series resistance ($R_{th}$) driving the capacitor. This reduces the complex loaded network back to a simple, standard RC low pass topology, making the transfer function trivial to solve.
Step-by-Step Solution: Applying Thevenin's Theorem
Step 1: Calculate the DC Gain ($A_{DC}$)
At DC (0 Hz), the capacitor $C_1$ acts as an open circuit. The circuit simplifies to a purely resistive voltage divider formed by $R_1$ and $R_L$.
- $V_{out(DC)} = V_{in} \times \frac{R_L}{R_1 + R_L}$
- $A_{DC} = \frac{10\text{ k}\Omega}{10\text{ k}\Omega + 10\text{ k}\Omega} = \frac{10}{20} = 0.5$
- In decibels: $20 \times \log_{10}(0.5) = -6.02\text{ dB}$
The Thevenin equivalent voltage is therefore $V_{th} = 0.5 \times V_{in}$.
Step 2: Calculate the Thevenin Equivalent Resistance ($R_{th}$)
To find the resistance 'seen' by the capacitor, we zero the independent voltage source (short $V_{in}$ to ground) and look into the network from the capacitor's terminals. $R_1$ and $R_L$ are now in parallel.
- $R_{th} = R_1 \parallel R_L = \frac{R_1 \times R_L}{R_1 + R_L}$
- $R_{th} = \frac{10\text{ k}\Omega \times 10\text{ k}\Omega}{10\text{ k}\Omega + 10\text{ k}\Omega} = \frac{100}{20}\text{ k}\Omega = 5\text{ k}\Omega$
Step 3: Calculate the New Time Constant ($\tau$)
The capacitor now charges and discharges through the Thevenin resistance, not just $R_1$.
- $\tau = R_{th} \times C_1$
- $\tau = 5,000\text{ }\Omega \times 10 \times 10^{-9}\text{ F} = 50 \times 10^{-6}\text{ seconds (50 }\mu\text{s)}$
Step 4: Calculate the Loaded Cutoff Frequency ($f_c$)
The -3dB cutoff frequency is defined by the new time constant.
- $f_c = \frac{1}{2\pi \tau}$
- $f_c = \frac{1}{2 \times \pi \times 50 \times 10^{-6}}$
- $f_c = \frac{1}{0.000314159} \approx 3183.1\text{ Hz}$
Comparison Table: Unloaded vs. Loaded Performance
| Parameter | Unloaded ($R_L = \infty$) | Loaded ($R_L = 10\text{ k}\Omega$) |
|---|---|---|
| DC Gain | 1.0 (0 dB) | 0.5 (-6.02 dB) |
| Effective Resistance | 10 kΩ | 5 kΩ |
| Time Constant ($\tau$) | 100 μs | 50 μs |
| -3dB Cutoff ($f_c$) | 1591.5 Hz | 3183.1 Hz |
Sanity Check and How to Verify Independently
Sanity Check: Order of Magnitude and Units
Before accepting the math, verify the physical reality. The load resistor is equal to the series resistor, meaning it pulls the Thevenin resistance down by exactly half (from 10 kΩ to 5 kΩ). Since resistance and cutoff frequency are inversely proportional in an RC filter network, halving the resistance must exactly double the cutoff frequency. Our unloaded baseline was ~1591 Hz; doubling it yields ~3183 Hz. The math holds. Furthermore, checking units: $\Omega \times \text{F} = \text{Seconds}$, and $\frac{1}{\text{Seconds}} = \text{Hertz}$. The dimensional analysis is correct.
Independent Verification Method
To verify this independently without relying on Thevenin algebra, use an AC sweep in a SPICE simulator like LTSpice or Micro-Cap. 1. Place a 1V AC source, 10k series resistor, 10nF capacitor to ground, and 10k load resistor in parallel with the capacitor. 2. Run an AC analysis from 10 Hz to 100 kHz. 3. Probe the node between the resistors. You will observe the low-frequency asymptote settling at -6.02 dB (0.5V). The -3dB point relative to that passband (which is -9.02 dB absolute on the plot) will intersect exactly at 3183.1 Hz. For a deeper look into how loading affects higher-order filter stages, refer to Analog Devices' MT-222 Tutorial on active filter design, which explains why buffer amplifiers are used to prevent this exact loading effect.
Frequently Asked Questions About Low Pass Filter Examples
How does a load resistor change a low pass filter example cutoff?
A load resistor placed in parallel with the shunt capacitor lowers the equivalent Thevenin resistance seen by the capacitor. Because the cutoff frequency formula is $f_c = \frac{1}{2\pi R_{eq} C}$, a lower equivalent resistance results in a higher cutoff frequency. The filter passes more high-frequency noise than originally designed, and the overall signal amplitude is attenuated due to the DC voltage divider effect.
When should I use an active op-amp low pass filter instead?
You should switch to an active topology (like a Sallen-Key or Multiple Feedback filter) when your downstream load has a low or unpredictable impedance, or when you need passband gain greater than 1. By placing an op-amp buffer between the RC network and the load, the op-amp's high input impedance prevents the load from altering your carefully calculated cutoff frequency, while its low output impedance can drive heavy loads like 50 Ω coaxial cables or low-impedance ADC multiplexers.
How to calculate low pass filter phase shift at the cutoff frequency?
In any standard first-order low pass filter example (loaded or unloaded, provided you use the correct Thevenin resistance), the phase shift at the exact -3dB cutoff frequency is always -45 degrees. The general phase formula is $\phi = -\arctan(\frac{f}{f_c})$. When the input frequency $f$ equals the cutoff frequency $f_c$, the ratio becomes 1, and $\arctan(1) = 45^\circ$. The negative sign indicates that the output voltage lags behind the input voltage.
