# Amplitude ratio is larger than 0.5 on a RC-filter if RC = 1.7684e-03

I have two signals

>> t = linspace(0, 10, 10000); % 0 to 10 seconds
>> u1 = 30*sin(2*pi*5*t); % Sine signal 1
>> u2 = 10*sin(2*pi*90*t); % Sine signal 2
>> u = u1 + u2; % Combine
>> spa(u, t); % Spectral analysis
>> grid on


And the spectral analysis shows these two signals u1 and u2

Now I want to filter the signal u by using a RC-filter

$$G(s) = \frac{1}{RCs + 1}$$

I say that at 90 Hz, I'm going to reduce the amplitude of the u2 sine signal with 50%, also half. 90 Hz is $$s = 2\pi90$$ radians.

So I put in the $$s$$ inside $$G(s)$$

$$G(2\pi90) = 0.5 = \frac{1}{RC2\pi90 + 1}$$

And now trying to solve $$RC$$

$$RC = \frac{1-0.5}{0.5*2\pi90} = 1.7684*10^{-3}$$

Assume that we are using $$R = 1 kOhm$$ and $$C = 1.7684 uF$$.

So I created my transfer function i MALTAB / Octave and plotted the bode diagram

>> R = 1000;
>> C = 1.7684e-06;
>> G = tf(1, [R*C 1]) % Transfer function
G =

scalar structure containing the fields:

num = 1
den =

1.7684e-03   1.0000e+00

delay = 0
tfnum =  1
tfdash = ---------------
tfden =  0.0017684s + 1
type = TF
sampleTime = 0

>> bode(G, 0, 600); % From 0 rad/s to a little bit over 2*pi*90 rad/s. As you can see, at  2*pi*90 = 565.49 rad/s, I have -3 dB, that's 0.7079 in magnitude.


I do linear simulation of my transfer function and the spectral analysis. As you can see, why has not the amplitude of the peak of 90 Hz become half?

>> y = lsim(G, u, t);
>> spa(y, t); % Spectral analysis


• For starters: $G(s)$ is a complex function that you need to evaluate on the imaginary axis, i.e. $s = j\omega$, $G(s) = \frac{1}{jRC\omega + 1)}$. If you want to cut the amplitude in half, you need solve for $|G(s)| = .5$, not $G(s) = 0.5$ . The more standard definition is the solve for $\omega = 1/RC$ which reduces the amplitude by $\sqrt{2}$ and the power by a factor of $2$ That's the -3dB point Sep 7 at 20:43
• @Hilmar I understand! I solved it now. Thank you so much. Sep 7 at 21:20