Transfer function for a low pass filter

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SUMMARY

The discussion focuses on calculating the output voltage expression for a first-order low-pass filter with a transfer function defined as H(f) = 1/(1+j*(f/f_B)), where the half-power frequency (f_B) is 200 Hz. The input signal is v_in(t) = 3 + 2sin(800*PI*t + 30) + 5cos(20000*PI*t). The user successfully decomposes the input signal and applies the transfer function to the frequency components, resulting in H(400) = 1/sqrt(5) phase(-arctan(2)) and H(1000) = 1/sqrt(26) phase(-arctan(5). The discussion concludes with a clarification that the DC component (3) remains unchanged as it corresponds to zero frequency.

PREREQUISITES
  • Understanding of transfer functions in signal processing
  • Familiarity with first-order low-pass filter characteristics
  • Knowledge of frequency domain analysis and Fourier series
  • Proficiency in complex numbers and phasor representation
NEXT STEPS
  • Study the implications of DC components in transfer functions
  • Learn about the frequency response of first-order low-pass filters
  • Explore the concept of phase shift in signal processing
  • Investigate the application of Fourier transforms in analyzing signals
USEFUL FOR

Electrical engineers, signal processing students, and anyone involved in designing or analyzing low-pass filter circuits will benefit from this discussion.

cseanm
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Homework Statement



The input signal of a first-order lowpass filter with a transfer function given by eqn 1 below and a half power frequency of 200 Hz is eqn 2 below. Find an expression for V_out

Homework Equations



eqn1:
H(f) = 1/(1+j*(f/f_B) where f is the frequency and f_B is the half-power frequency
eqn2:
v_in (t) = 3 + 2sin(800*PI*t + 30) + 5cos(20000*PI*t)

v_out/v_in = H(f)

The Attempt at a Solution



I have already gotten that v_in = 3 + 2cos(800*PI*t - 60) + 5cos(20000*PI*t)
then I would multiply each component by H(f) which will give me:
H(400) = 1/sqrt(5) phase(-arctan(2))
H(1000) = 1/sqrt(26) phase(-arctan(5))

v_out = 3*H(?) + 2 phase(-60)*H(400) + 5 phase(0)*H(1000)

pretty much I just do not know what I do about the 3. Do I simply leave it alone?

Thanks!
 
Last edited:
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cseanm said:
pretty much I just do not know what I do about the 3. Do I simply leave it alone?

It's a DC component, frequency is zero. What does your transfer function do to a signal with zero frequency?
 

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