# LTI system - RC low-pass filter

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Linear time-invariant (LTI) systems can be represented by the transfer function $H(f) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamisaiaacI cacaWGMbGaaiykaaaa@3908@$ . It determines the output signal of an LTI system for a given input signal in the frequency domain:

$Y(f)=H(f)·X(f)$

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

$H(f)= 1 j2πC R+ 1 j2πC = 1 1+j2πRC MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamisaiaacI cacaWGMbGaaiykaiabg2da9maalaaabaWaaSGaaeaacaaIXaaabaGa amOAaiaaikdacqaHapaCcaWGdbaaaaqaaiaadkfacqGHRaWkdaWcca qaaiaaigdaaeaacaWGQbGaaGOmaiabec8aWjaadoeaaaaaaiabg2da 9maalaaabaGaaGymaaqaaiaaigdacqGHRaWkcaWGQbGaaGOmaiabec 8aWjaadkfacaWGdbaaaaaa@4E46@$
or
$H(f)= 1 1+j(f/ f C ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamisaiaacI cacaWGMbGaaiykaiabg2da9maalaaabaGaaGymaaqaaiaaigdacqGH RaWkcaWGQbGaaiikamaalyaabaGaamOzaaqaaiaadAgadaWgaaWcba Gaam4qaaqabaGccaGGPaaaaaaaaaa@41A8@$ where $f c MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaaBa aaleaacaWGdbaabeaaaaa@37D5@$ is the cutoff frequency $f c = 1 2πRC MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaaBa aaleaacaWGdbaabeaakiabg2da9maalaaabaGaaGymaaqaaiaaikda cqaHapaCcaWGsbGaam4qaaaaaaa@3DC8@$.

The rearrangement in polar coordinates leads to:

$H(f)= 1 1+(f/ f C ) 2 ⋅ e −j⋅arctan(f/ f C ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamisaiaacI cacaWGMbGaaiykaiabg2da9maalaaabaGaaGymaaqaamaakaaabaGa aGymaiabgUcaRiaacIcadaWcgaqaaiaadAgaaeaacaWGMbWaaSbaaS qaaiaadoeaaeqaaOGaaiykamaaCaaaleqabaGaaGOmaaaaaaaabeaa aaGccqGHflY1caWGLbWaaWbaaSqabeaacqGHsislcaWGQbGaeyyXIC TaciyyaiaackhacaGGJbGaaiiDaiaacggacaGGUbGaaiikamaalyaa baGaamOzaaqaaiaadAgadaWgaaadbaGaam4qaaqabaWccaGGPaaaaa aaaaa@531B@$

Thus the absolute value $| H(f) | MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca WGibGaaiikaiaadAgacaGGPaaacaGLhWUaayjcSdaaaa@3C29@$ yields the amplitude response, and the argument $arg(H(f)) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciyyaiaack hacaGGNbGaaiikaiaadIeacaGGOaGaamOzaiaacMcacaGGPaaaaa@3D28@$ yields the phase response.

$| H(f) |= 1 1+(f/ f C ) 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca WGibGaaiikaiaadAgacaGGPaaacaGLhWUaayjcSdGaeyypa0ZaaSaa aeaacaaIXaaabaWaaOaaaeaacaaIXaGaey4kaSIaaiikamaalyaaba GaamOzaaqaaiaadAgadaWgaaWcbaGaam4qaaqabaGccaGGPaWaaWba aSqabeaacaaIYaaaaaaaaeqaaaaaaaa@44D4@$

$φ(f)=−arctan(f/ f C ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqOXdOMaai ikaiaadAgacaGGPaGaeyypa0JaeyOeI0IaciyyaiaackhacaGGJbGa aiiDaiaacggacaGGUbGaaiikamaalyaabaGaamOzaaqaaiaadAgada WgaaWcbaGaam4qaaqabaGccaGGPaaaaaaa@45C1@$

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
$y ^ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaaja aaaa@3705@$ (output amplitude) [V]             0.707             0.447
$φ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqOXdOgaaa@37B4@$ (phase shift) [ $° MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiSaalaaa@37E3@$ ] 45.00 63.44
Example output amplitude and phase shift for $f c MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaaBa aaleaacaWGdbaabeaaaaa@37D5@$ = 10 MHz and $x ^ =1V MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmiEayaaja Gaeyypa0JaaGymaiaadAfaaaa@39A0@$

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
$y ^ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaaja aaaa@3705@$ (output amplitude) [V]             0.707
t (phase delay time) [ns] 12.5
$φ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqOXdOgaaa@37B4@$ (phase shift) [ $° MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiSaalaaa@37E3@$ ] 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.

$φ= t⋅360° T MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqOXdOMaey ypa0ZaaSaaaeaacaWG0bGaeyyXICTaaG4maiaaiAdacaaIWaGaeyiS aalabaGaamivaaaaaaa@4109@$

## 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.