LTI system example: RC low-pass filter

Linear time-invariant (LTI) systems can be represented by the transfer function $H (f)$ . It determines the output signal of an LTI system for a given input signal in the frequency domain:

Y (f) = H (f) \cdot X (f)

This diagram illustrates how the output for a sine wave input signal is determined:

Transform the input signal into the frequency domain.
Weight the spectral components with the respective gain and phase of the transfer function.
Transform the output signal's spectrum into the time domain.

LTI system's transfer function determines the output for a given input signal. The response of a LTI system to a sine wave is a sine wave at the same frequency. Amplitude and phase are determined by the transfer function's gain and phase at this frequency.

An RC low-pass filter is a LTI system. It shall be used to examine amplitude and phase of a complex valued frequency response. Let us start deriving the RC low-pass filter's transfer function.

An RC low-pass filter is a potential divider circuit containing a resistor and a capacitor. It implements a first order low-pass.

Recalling the capacitor impedance the transfer function results to

$H (f) = \frac{\frac{1}{j 2 π C}}{R + \frac{1}{j 2 π C}} = \frac{1}{1 + j 2 π R C}$
or
$H (f) = \frac{1}{1 + j (f / f_{C})}$ where $f_{c}$ is the cutoff frequency $f_{c} = \frac{1}{2 π R C}$ .

The rearrangement in polar coordinates leads to:

$H (f) = \frac{1}{\sqrt{1 + (f / f_{C})^{2}}} \cdot e^{- j \cdot \arctan (f / f_{C})}$

Thus the absolute value $| H (f) |$ yields the amplitude response, and the argument $\arg (H (f))$ yields the phase response.

$| H (f) | = \frac{1}{\sqrt{1 + (f / f_{C})^{2}}}$

$φ (f) = - \arctan (f / f_{C})$

Amplitude response |H(f)| of RC low-pass with fc = 10MHz

Calculate the missing values in the table!

f (frequency) [MHz]	5	10	15	20	40
$\hat{y}$ (output amplitude) [V]		0.707		0.447
$φ$ (phase shift) [ $°$ ]		45.00		63.44

Example output amplitude and phase shift for

f_{c}

= 10 MHz and

\hat{x} = 1 V

In this laboratory experiment you will analyze the amplitude and phase response of a RC low-pass filter. In a sine-wave analysis the amplitude and phase response is measured.

LTI system example: RC low-pass filter

Start

The simulation starts with a sine input signal at amplitude 1V and frequency of 10MHz. The oscilloscope shows both the input and output signal.

LTI system sine input and output phase shift oscilloscope

Sine input signal (blue) at 1V amplitude and output signal (black). The amplitude, delay and thus the phase shift of the output signal can be measured. The input frequency of 10MHz causes an output amplitude of 0.707 V, hence marking the cutoff frequency.

Experiment

Adjust the input frequency and measure the respective amplitudes and phase delay times of the output signal. Compare the results with the values calculated in the tutorial.

f (frequency) [MHz]	5	10	15	20	40
$\hat{y}$ (output amplitude) [V]		0.707
t (phase delay time) [ns]		12.5
$φ$ (phase shift) [ $°$ ]		45

Note that, with the input signal starting at zero, the phase delay time can be metered at the zero crossing of the output signal (black). The phase shift is the phase delay time divided by the cycle duration.

φ = \frac{t \cdot 360 °}{T}

Note

Stop the simulation after a complete oscilloscope sweep by pressing pause on your keypad and click on the point to measure. Ctrl + click yields the local maximum or zero crossing of the measured curve.

Next steps

To change the filter itself, click on the block diagram to open the properties.
The cutoff-frequency can be changed and the filter is re-initialized.
A right-click on the filter offers to open the transfer function or the impulse response according to the cutoff frequency setting.
Beyond the input properties frequency and amplitude, the waveform can be changed as well.

Square input signal (blue) and output signal (black) at cutoff frequency.

LTI system - RC low-pass filter

Start

Experiment

Note

Next steps