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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.

 $Y\left(f\right)=H\left(f\right)\cdot X\left(f\right)$
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 $\stackrel{^}{x}=1V$