Revisiting the lowpass filter, in which you get the opposite frequency behavior

Question

Revisiting the lowpass filter, in which you get the opposite frequency behavior by interchanging R and C (Figure 1.90, repeated here as Figure 1.102), we find the accurate result V_out = 1/(1 + omega^2 R^2 C^2)^1/2 V_in as seen in Figure 1.103 (and earlier in Figure 1.91). The 3 dB point is again at a frequency^45 f = 1/2 pi RC. Lowpass filters are quite handy in real life. For instance, a lowpass filter can be used to eliminate interference from nearby radio and television stations (0.5-800 MHz), a problem that plagues audio amplifiers and other sensitive electronic equipment. Show that the preceding expression for the response of an RC lowpass filter is correct.^44 One often omits the minus sign when referring to the - 3dB point.^45 As mentioned in $1.7.1A, we often like to define the breakpoint frequency omega_0=1/RC, and work with frequency ratios omega/omega_0. Then a useful form for the denominator in eq'n 1.36 is squareroot 1 + (omega/omega_0)^2. The same applies to eq'n 1.35, where the numerator becomes omega/omega_0.

Explanation / Answer

according to the definition of low pass filter it will only pass the frequencies which are lower than the cut-off frequency. The ideal case determines passing all frequencies lower than cutoff. In this case for the RC response we will check both sinusoidal and step-voltage case.

lets consider an input sinusoidal voltage V1 is applied to the RC circuit to get output V2.

considering i as complex plane identity we can write for this RC circuit

V2=V1(-iXc/R-iXc)

Here R=resistance and Xc=capacitive reactance

lets consider -iXc<<R then the equation becomes V2=V1(1/R) or V2 nearly gives a value 0.

from this relation we can say that t low frequencies with condition -iXc>>R we get output voltage same as input.

2nd case:

now we will take step voltage as input. For step voltage we get output voltage as the function of exponential term of the time constant RC. Again, we consider output voltage V2 accross the capacitor C.

then V2=V1(1-e-1/RC)

for simplicity we consider that response comes from an fully discharged capacitor C. If there are some initial charge in C then we get the expression as V2=V1-[(V1-Vc)e^-1/RC]

so in both cases we get that the preceding expression is correct.

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