Question

For a RC Low Pass filter, what is the approximate amplitude (as a decimal proportion of input amplitude) at 1/4 the cutoff frequency?

Answer 1

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:

$V_{in$ = the input signal  

$V_o$ = the output signal

Since; $V_o$  is used across the capacitor C,

By using the potential divider equation we have:

$V_o =V_{in} \times \dfrac{X_C}{\sqrt{ R^2+X_C^2} }$

From above; $X_C$ = capacitive reactance ;

and The total impedance Z is illustrated as $Z = \sqrt{R^2+X_C^2}$

Thus;

$V_o =V_{in} \times \dfrac{X_C}{Z}$

Recall that;

$X_C = \dfrac{1}{2 \pi fC}$

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:

$F_C = \dfrac{1}{2 \pi RC}$

Also;

the frequency of input signal $f = \dfrac{F_c}{4}$

$f = \dfrac{1}{8 \pi RC}$

Hence;

$X_C = \dfrac{1}{2 \pi fC}$

$X_C = \dfrac{1}{2 \pi \times \dfrac{1}{8 \pi RC} \times C}$

$X_C = 4R$

Finally;

From $Z = \sqrt{R^2+X_C^2}$

$Z = \sqrt{R^2+(4R)^2}$

$Z = \sqrt{17R^2}$

Z $\simeq$ 4.12R

As such, the output will be:

$V_o =V_{in} \times \dfrac{X_C}{\sqrt{ R^2+X_C^2} }$

$V_o =V_{in} \times \dfrac{4R}{4.12R^2} }$

$V_o =0.97V_{in}$

So, if we regard $A_{in}$ to be the input amplitude, then $A_{out}$ i.e the output amplitude will also be $A_{out}$ = $0.97 A_{in}$