Answer:
Explanation:
RC low Pass Filter is an electronic circuit that comprises of a resistor and capacitor and it functions to permit low-frequency signals depending on the design and reject the high-frequency signals above a given frequency known as the cutoff frequency.
From the diagram attached below:
[tex]V_{in[/tex] = the input signal
[tex]V_o[/tex] = the output signal
Since; [tex]V_o[/tex] is used across the capacitor C,
By using the potential divider equation we have:
[tex]V_o =V_{in} \times \dfrac{X_C}{\sqrt{ R^2+X_C^2} }[/tex]
From above; [tex]X_C[/tex] = capacitive reactance ;
and The total impedance Z is illustrated as [tex]Z = \sqrt{R^2+X_C^2}[/tex]
Thus;
[tex]V_o =V_{in} \times \dfrac{X_C}{Z}[/tex]
Recall that;
[tex]X_C = \dfrac{1}{2 \pi fC}[/tex]
Here; f denotes the frequency of the input signal
Since the cutoff frequency is related to the frequency at which the capacitive reactance and resistance are said to be the same, then:
The Cutoff frequency can be expressed as:
[tex]F_C = \dfrac{1}{2 \pi RC}[/tex]
Also;
the frequency of input signal [tex]f = \dfrac{F_c}{4}[/tex]
[tex]f = \dfrac{1}{8 \pi RC}[/tex]
Hence;
[tex]X_C = \dfrac{1}{2 \pi fC}[/tex]
[tex]X_C = \dfrac{1}{2 \pi \times \dfrac{1}{8 \pi RC} \times C}[/tex]
[tex]X_C = 4R[/tex]
Finally;
From [tex]Z = \sqrt{R^2+X_C^2}[/tex]
[tex]Z = \sqrt{R^2+(4R)^2}[/tex]
[tex]Z = \sqrt{17R^2}[/tex]
Z [tex]\simeq[/tex] 4.12R
As such, the output will be:
[tex]V_o =V_{in} \times \dfrac{X_C}{\sqrt{ R^2+X_C^2} }[/tex]
[tex]V_o =V_{in} \times \dfrac{4R}{4.12R^2} }[/tex]
[tex]V_o =0.97V_{in}[/tex]
So, if we regard [tex]A_{in}[/tex] to be the input amplitude, then [tex]A_{out}[/tex] i.e the output amplitude will also be [tex]A_{out}[/tex] = [tex]0.97 A_{in}[/tex]
