RC Low Pass Filter with Sine Wave Input

Thread Starter

jaydnul

Joined Apr 2, 2015
155
Hey Everyone,

I am trying to gain a level of fundamental understanding of an RC circuit sine wave response through the mathematics and was wondering if someone could help me work it out.

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Fundamentally a sine wave is represented by the equation y=-ky'' . When a sine wave is used as the input of an RC low pass filter, you get a sine wave out (just phase and amplitude shifted). So I know that when the input waveform has the form Vin=-kVin'' then somehow the output also retains that relationship, Vout=-kVout''. I am having trouble proving this with the equation that describes an RC circuit response: Vout' = 1/RC*(Vin-Vout).

Any help would be great, thanks!
 

Papabravo

Joined Feb 24, 2006
16,160
I'd be surprised if you could prove it. The simple answer is that you cannot make a 1st order system behave like a 2nd order system. To to that you need to add an inductor to the RC circuit and then you will have a second order system
 

Thread Starter

jaydnul

Joined Apr 2, 2015
155
Just for an intuitive visualization I am more looking for an equation in the form of a differential equation (instead of the solutions to the differential equation), just to show that the relationship between acceleration and voltage remains the same without converting to sines or exp functions. I know that getting a sine wave at the output when there is sine wave at the input must mean the the output voltage retains this voltage/acceleration information.

I understand the phasor/exponential/sinusoidal explanations, but is there a way to prove this without exp or sine functions, just keeping it in the form of the differential equations?

My attempt here:

$$(1) Vin = -k\frac {d^2 Vin} {dt^2}$$
Plugged into the RC circuit equation
$$(2) \frac {d Vout} {dt} = \frac {Vin - Vout}{RC}$$
$$(3) \frac {d Vout} {dt} = \frac {-k\frac {d^2 Vin} {dt^2} - Vout}{RC}$$

Differentiating equation (2)
$$(4) \frac {d^2 Vout} {dt^2} = \frac {\frac {d Vin} {dt} - \frac {d Vout} {dt}} {RC}$$

Then plugging equation (3) into equation (4)

$$(5) \frac {d^2 Vout} {dt^2} = \frac {\frac {d Vin} {dt} - \frac {-k\frac {d^2 Vin} {dt^2} - Vout}{RC}} {RC}$$

Not sure where to go from there, or if im going in completely the wrong direction...
 

Papabravo

Joined Feb 24, 2006
16,160
That's all fine and dandy, but the RC circuit does not behave according to 2nd Order ODE (Ordinary Differential Equation).
It behaves according to a 1st Order ODE. so all your fine machinations mean nothing when it comes to understanding what is going on if you start with the wrong premise.

What is will do for a given frequency is provide attenuation in the magnitude and a shift in the phase.
If you calculate and plot the magnitude and phase you will see immediately the low pass characteristic in the magnitude, and the shift in phase at the corner frequency given by the values of R and C.

Here is a simulation of a typical RC circuit. The dotted blue lone is the phase of the output with respect to the input and it starts a 0° for low frequency AC inputs. As you approach the corner frequency the phase begins to lag and at high frequency it reach 90°. In order for a sign change to happen, as is the case in a 2nd Order system the phase needs to shift by 180°,

In case you are interested the corner frequency is:

\( f_c\;=\;\cfrac{1}{2\pi RC}\;\approx\;159.15\;\text {Hertz} \)

Edit: One more thing the homogeneous ODE you should be looking at is:

\( R\cfrac{di}{dt}\;+\;\cfrac{1}{C}i\;=\;0 \)

You can transform this to a differential equation with voltage as the dependent variable because the current is the same in every part of a series circuit. Can you do that?
 

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Papabravo

Joined Feb 24, 2006
16,160
@jaydnul
There are potentially solutions in the complex domain as well. The differential equation represents a much larger class of functions than you think. The solution to a differential equation consists of two parts:
  1. The first part is the solution to the homogeneous equation which you keep trying to misuse.
  2. The second part is the particular solution, given a forcing function which you have neglected.
You have also neglected to specify any initial conditions which are required for the general solution to any ODE. Your attempted definition of a sine function as the solution to a homogeneous ODE is doomed. It won't work because it is not unique. AFAIK, to be useful a definition must be unique.

So riddle me this:
The sine function is a odd function. That is \( \;\;sin(-x)\;=\;-sin(x) \)​
The cosine function is an even function. That is \( \;\;cos(-x)\;=\;cos(x) \)​
Your definition is not able to create a single function that is both odd and even.
 
Last edited:

bogosort

Joined Sep 24, 2011
674
There are potentially solutions in the complex domain as well. The differential equation represents a much larger class of functions than you think. The solution to a differential equation consists of two parts:
  1. The first part is the solution to the homogeneous equation which you keep trying to misuse.
  2. The second part is the particular solution, given a forcing function which you have neglected.
You have also neglected to specify any initial conditions which are required for the general solution to any ODE. Your attempted definition of a sine function as the solution to a homogeneous ODE is doomed. It won't work because it is not unique.
Why are you telling me this, I've done no such things. :)
 

Tesla23

Joined May 10, 2009
445
Hey Everyone,

I am trying to gain a level of fundamental understanding of an RC circuit sine wave response through the mathematics and was wondering if someone could help me work it out.

Fundamentally a sine wave is represented by the equation y=-ky'' . When a sine wave is used as the input of an RC low pass filter, you get a sine wave out (just phase and amplitude shifted). So I know that when the input waveform has the form Vin=-kVin'' then somehow the output also retains that relationship, Vout=-kVout''. I am having trouble proving this with the equation that describes an RC circuit response: Vout' = 1/RC*(Vin-Vout).

Any help would be great, thanks!
I think you will have trouble as I don't think it is true. Whilst it is true that the response of the RC network to a sinusoid is a sinusoid (shifted in amplitude and phase), your assumption that "Fundamentally a sine wave is represented by the equation y=-ky" needs the added note that k > 0. Without this you admit all the exponentials with real exponents.

Restricting yourself to steady-state sinusoidal excitation, you can analyze the circuit in the Fourier transform domain - and here the relationship is clear - so it should be possible to prove it in the time domain from differential equations.

Without the requirement k>0, you can no longer rely on a fourier representation and you are trying to prove it in the Laplace domain, and at first glance it appears that the circuit introduces extra exponentials in response to non-steady state excitation,
 
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