LTI system - RC low-pass filter

Peak-to-average power ratio (PAPR) of OFDM systems

LTI system example: RC low-pass filter

Linear time-invariant (LTI) systems can be represented by the transfer function. It determines the output signal of an LTI system for a given input signal in the frequency domain. An RC low-pass filter serves as example to examine amplitude and phase of this complex valued frequency response. $Y\left(f\right)=H\left(f\right)\cdot X\left(f\right)$
LTI system's output is related to the input by the transfer function $H\left(f\right)$ .

In this briefing and the subsequent experiment an RC low-pass filter serves as example for an LTI system. The transfer function, amplitude response and phase response are derived. 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

$H\left(f\right)=\frac{1}{j2\pi C}}{R+1}{j2\pi C}}=\frac{1}{1+j2\pi RC}$
or
$H\left(f\right)=\frac{1}{1+j\left(f/{f}_{C}\right)}$ where ${f}_{C}$ is the cutoff frequency ${f}_{C}=\frac{1}{2\pi RC}$.

The rearrangement in polar coordinates leads to:

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

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

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

$\phi \left(f\right)=-\mathrm{arctan}\left(f/{f}_{C}\right)$ 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
$\stackrel{^}{y}$ (output amplitude) [V]             0.707             0.447
$\phi$ (phase shift) [ $°$ ] 45.00 63.44
Example output amplitude and phase shift for fc =10MHz and $\stackrel{^}{x}=1V$
LTI system example: low-pass filter - labAlive experiment

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. 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
$\stackrel{^}{y}$ (output amplitude) [V]             0.707
t (phase delay time) [ns] 12.5
$\phi$ (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.

$\phi =\frac{t\cdot 360°}{T}$

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.