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

Step-by-step 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) * (X_C)/(√( R^2+X_C^2) )`

From above;
`X_C` = capacitive reactance ;

and The total impedance Z is illustrated as
`Z = √(R^2+X_C^2)`

Thus;

`V_o =V_(in) * (X_C)/(Z)`

Recall that;

`X_C = (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 = (1)/(2 \pi RC)`

Also;

the frequency of input signal
`f = (F_c)/(4)`

`f = (1)/(8 \pi RC)`

Hence;

`X_C = (1)/(2 \pi fC)`

`X_C = (1)/(2 \pi * (1)/(8 \pi RC) * C)`

`X_C = 4R`

Finally;

From
`Z = √(R^2+X_C^2)`

`Z = √(R^2+(4R)^2)`

`Z = √(17R^2)`

Z
`\simeq` 4.12R

As such, the output will be:

`V_o =V_(in) * (X_C)/(√( R^2+X_C^2) )`

`V_o =V_(in) * (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)`