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Study of Frequency Response of Low-Pass and High-Pass Filters Using Op-Amp

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Aim

To study and analyze the frequency response of low-pass and high-pass filters using an operational amplifier (op-amp).

Apparatus & Software

ComponentQuantity
Function Generator1
Oscilloscope1
DC Supply (+15V, -15V)1
Bread Board1
5.6 kΩ Resistor2
2.8 nF Capacitor1
Connecting WiresAs required
Op-Amp 071

Theory

1. Low-Pass Filter: A low-pass filter is an electronic circuit that lets signals with a frequency lower than a certain cutoff frequency pass through, while reducing higher-frequency signals. It is often built using a resistor-capacitor (RC) network or an operational amplifier (op-amp) in active filter setups. The output voltage is taken across the capacitor, which charges and discharges in response to changes in input frequency.
When an op-amp is used, it increases the gain and prevents loading effects on the input signal. The cutoff frequency fc is determined by the values of the resistor and capacitor as:
fc=12πRCf_c = \frac{1}{2\pi RC}
2. High-Pass Filter: A high-pass filter lets signals with frequencies above a certain cutoff frequency pass through, while it reduces those below that level. The simplest type uses an RC network. In this setup, the capacitor is in series with the input signal, and the resistor connects to ground. In active filters that use op-amps, we can add gain while keeping the signal clear.
The cutoff frequency fc in a high-pass filter is also calculated using the same formula. These filters are commonly used to eliminate low-frequency noise or DC components from signals.
fc=12πRCf_c = \frac{1}{2\pi RC}

Pre-Lab / Circuit Diagram

Fig (a): Theoretical Diagram of Low-Pass Filter

Fig (a): Theoretical Diagram of Low-Pass Filter (R1 = Rf = 5.6 kΩ, C = 2.8 nF, fc = 10 kHz).

Fig (c): Theoretical Diagram of High-Pass Filter

Fig (c): Theoretical Diagram of High-Pass Filter (R1 = Rf = 5.6 kΩ, C = 2.8 nF, fc = 10 kHz).

Procedure

Part 1 — Low-Pass Filter:
  1. Design and realize a first-order active low-pass filter with R1 = Rf = 5.6 kΩ and C = 2.8 nF for a cutoff frequency of 10 kHz.
  2. Connect the circuit on a breadboard with Op-Amp 07 powered by ±15V DC supply.
  3. Apply a 1 Vpp AC sinusoidal input from the function generator.
  4. Vary the input frequency from 10 Hz to 500 kHz.
  5. Record the output voltage (Vout) at each frequency step using the oscilloscope.
  6. Calculate gain (Vout/Vin) at each frequency and plot gain vs log(frequency) to obtain the frequency response curve.
Part 2 — High-Pass Filter:
  1. Design and realize a first-order active high-pass filter with R1 = Rf = 5.6 kΩ and C = 2.8 nF for a cutoff frequency of 10 kHz.
  2. Apply a 1 Vpp AC sinusoidal input and vary frequency from 10 Hz to 1 MHz.
  3. Record Vout at each frequency step using the oscilloscope.
  4. Calculate gain (Vout/Vin) at each frequency and plot gain vs log(frequency) to obtain the frequency response curve.
  5. Verify the −3 dB point by checking the gain at the cutoff frequency (expected ≈ 0.707).

Simulation / Execution

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Observations

1. Low-Pass Filter (R1 = Rf = 5.6 kΩ, C = 2.8 nF, Vin = 1 Vpp)
The oscilloscope output at the cutoff frequency (10 kHz) showed an output voltage of approximately 696 mV (≈ 700 mV Vpp), while the input signal was 1 V Vpp. The measured gain at cutoff is 0.7, which closely matches the theoretical value of 0.707, confirming the −3 dB behavior of the low-pass filter.
Output waveform of Low-Pass Filter at Cutoff Frequency

Output waveform of Low-Pass Filter at Cutoff Frequency (10 kHz). Measured Vout ≈ 696 mV, Vin = 1 V.

Frequency (Hz)log(f)Vout (mV)
101.001.16
1002.001.14
5002.701.12
10003.001.08
25003.401.04
50003.701.00
75003.880.86
100004.000.70
125004.100.64
150004.180.52
200004.300.35
250004.400.16
500004.700.14
750004.880.12
1000005.000.08
5000005.700.08
Frequency Response of the Low-Pass Filter

Frequency Response of the Low-Pass Filter — Gain (Vout/Vin) vs Frequency (Hz) on log scale.

2. High-Pass Filter (R1 = Rf = 5.6 kΩ, C = 2.8 nF, Vin = 1 Vpp)
The oscilloscope output at the cutoff frequency (10 kHz) showed an output voltage of 752 mV Vpp, while the input signal was 1 V Vpp. The measured gain at cutoff is 0.752, which approximately matches the theoretical value of 0.707, confirming the −3 dB response of the high-pass filter.
Output waveform of High-Pass Filter at Cutoff Frequency

Output waveform of High-Pass Filter at Cutoff Frequency (10 kHz). Measured Vout = 752 mV, Vin = 1 V.

Frequency (Hz)log(f)Vout (mV)
101.0014
1002.0016
5002.7064
10003.00114
25003.40270
50003.70492
75003.88644
100004.00752
125004.10860
150004.18888
250004.40976
500004.701020
1000005.001060
2500005.401080
5000005.701100
10000006.001140
Frequency Response of the High-Pass Filter

Frequency Response of the High-Pass Filter — Gain (Vout/Vin) vs Frequency (Hz) on log scale.

Calculations

For both the low-pass and high-pass filters, the desired cutoff frequency (fc) is 10 kHz. The standard formula for cutoff frequency is:
fc=12πRCf_c = \frac{1}{2\pi RC}
Rearranging to solve for capacitance C:
C=12πRfcC = \frac{1}{2\pi R f_c}
Substituting R = 5.6 kΩ and fc = 10 kHz:
C=12π×5.6×103×10×103≈2.84 nFC = \frac{1}{2\pi \times 5.6 \times 10^3 \times 10 \times 10^3} \approx 2.84\,\text{nF}
Hence, the required capacitor value is approximately 2.8 nF. Since a single 2.8 nF capacitor was unavailable in the lab, two 5.6 nF capacitors were connected in series to achieve the equivalent value:
Ceq=C2=5.6 nF2=2.8 nFC_{eq} = \frac{C}{2} = \frac{5.6\,\text{nF}}{2} = 2.8\,\text{nF}
Since R1 = Rf = 5.6 kΩ and the circuit is an inverting amplifier, the voltage gain is:
Av=−RfR1=−5.6 kΩ5.6 kΩ=−1A_v = -\frac{R_f}{R_1} = -\frac{5.6\,\text{k}\Omega}{5.6\,\text{k}\Omega} = -1
At the cutoff frequency, the gain drops by a factor of 1/√2, so the magnitude of gain at cutoff is:
∣Avcutoff∣=12≈0.707|A_{v_{\text{cutoff}}}| = \frac{1}{\sqrt{2}} \approx 0.707
The negative sign indicates a 180° phase shift, which is expected in inverting amplifier configurations.
Measured gain verification at cutoff (LPF):
∣Avmeasured∣=VoutVin=700 mV1 V=0.7|A_{v_{\text{measured}}}| = \frac{V_{out}}{V_{in}} = \frac{700\,\text{mV}}{1\,\text{V}} = 0.7
Measured gain verification at cutoff (HPF):
∣Avmeasured∣=VoutVin=752 mV1 V=0.752|A_{v_{\text{measured}}}| = \frac{V_{out}}{V_{in}} = \frac{752\,\text{mV}}{1\,\text{V}} = 0.752

Results & Analysis

Filter TypeDesign fcMeasured Gain at fcTheoretical Gain at fc
Low-Pass Filter10 kHz0.7000.707
High-Pass Filter10 kHz0.7520.707
  • The low-pass filter passed frequencies below 10 kHz and attenuated higher frequencies, as expected.
  • The high-pass filter passed frequencies above 10 kHz and suppressed lower frequencies, as expected.
  • The measured gain at the cutoff frequency for the LPF (0.700) and HPF (0.752) closely match the theoretical −3 dB value of 0.707.
  • The plotted frequency response curves clearly illustrate the intended filter behavior for both circuits.

Conclusion

The frequency response characteristics of low-pass and high-pass filters were studied and confirmed using op-amp-based setups. The cutoff frequency for both filters was set to 10 kHz. The gain measured at this frequency closely matched the theoretical value of 1/√2 ≈ 0.707, confirming the expected −3 dB point. The plotted frequency responses clearly illustrated the attenuation of higher frequencies in the low-pass filter and suppression of lower frequencies in the high-pass filter, validating their intended functionality.

Post-Lab / Viva Voce

  1. Q: Why is the gain of the filter at the cutoff frequency equal to 1/√2 (≈ 0.707) and not 0.5?

    A: At the cutoff frequency, the impedance of the capacitor equals the resistance (Xc = R), so the reactive and resistive components are equal. The output is taken across one of two equal-magnitude impedances, but since they are 90° out of phase, the total impedance is √2 times each individual impedance. This results in a voltage division by √2, giving a gain of 1/√2. The power at this point is half the passband power (−3 dB), which is why it is called the half-power point, not the half-voltage point.
  2. Q: In an inverting op-amp filter, the DC gain is set by −Rf/R1. If you wanted to make a unity-gain low-pass filter, what constraint would that impose on the components, and what trade-off would you face?

    A: For unity gain, Rf = R1 is required. However, the cutoff frequency of the LPF is fc = 1/(2πRfC), so fixing Rf = R1 ties the gain and cutoff frequency together through the same resistor. You cannot independently adjust gain and cutoff frequency — changing R1 or Rf to tune gain will simultaneously shift fc unless the capacitor value is re-calculated. This lack of independent control is a limitation of the simple first-order inverting topology.
  3. Q: What happens to the phase relationship between input and output as frequency increases in a first-order low-pass filter?

    A: At very low frequencies, the capacitor impedance is very high and contributes little, so the phase shift is approximately 180° (due to the inverting configuration). As frequency approaches the cutoff frequency fc, the phase contribution of the RC network adds an additional −45°, giving a total of −225° (or equivalently −45° relative to the ideal inverting output). At very high frequencies, the additional phase lag approaches −90°, giving a total of −270°. So the RC network adds a phase lag that varies from 0° to −90° as frequency increases.
  4. Q: If the observed cutoff frequency is lower than the designed value, what are the most likely causes and how would you correct each?

    A: The most likely causes are: (1) The actual capacitance is higher than the nominal value due to component tolerance — verify with an LCR meter and replace if needed. (2) Breadboard parasitic capacitance adds to the circuit capacitance — use a PCB or minimize wire lengths. (3) The actual resistance is higher than nominal — measure with a multimeter. The cutoff frequency can be restored by reducing either R or C proportionally, since fc = 1/(2πRC).
  5. Q: Why does the frequency response of the practical filter deviate from the ideal 'brick wall' response, and why can a linear circuit never achieve a true brick wall?

    A: A brick wall filter would require an instantaneous, discontinuous transition between passband and stopband, which mathematically corresponds to an infinite-order transfer function with infinitely many poles. A physical linear circuit can only implement a finite number of poles (each real passive or active component contributes at most one pole), so the roll-off is always gradual. Furthermore, the Paley–Wiener theorem from signal processing proves that any causal, physically realizable filter cannot have a perfectly sharp frequency cutoff.
  6. Q: How would you modify the first-order low-pass filter circuit to double its cutoff frequency without changing the capacitor value?

    A: Since fc = 1/(2πRfC), doubling fc requires halving Rf. However, halving Rf would also halve the passband gain (Av = −Rf/R1). To maintain the same gain while doubling fc, both Rf and R1 must be halved simultaneously — this keeps the ratio Rf/R1 unchanged (preserving gain) while halving Rf to double the cutoff frequency.

References & Resources (Not Applicable)

This section is not required for this experiment.