Electronic – Is a low pass RC circuit the same thing as a first order Butterworth filter

The key point is that it is 0 in steady state. That is, as \$t\rightarrow\infty\$, \$V_{out}\rightarrow0\$. When you take the differential equation, you're looking at output voltage with respect to time.

Let's try taking a look at an example of this system. I'm going to simplify \$K=1\$ and \$RC=1\$. Then, we are left with: $$ \frac{V_{out}}{V_{in}}\left(j\omega\right)=T\left(j\omega\right)=\frac{1}{1-j/\omega} $$ To simplify our calculations, let's change the form of it and replace \$j\omega\$ with \$s\$. Then, we get: $$ T\left(s\right)=\frac{s}{s+1} $$ Taking the step response of this (that is, \$V_{out}\$ when the input \$V_{in}=u\left(t\right)\$ where \$u\left(t\right)\$ is the Heaviside step function) we get this:

Note that this is in time domain, and in steady state, our output voltage is zero. The bode plot comes out as you'd expect (first order high-pass filter):

For years I tutored the very first electrical engineering course at my uni, so this comes up a lot - don't worry :)

The best way to understand this is to think about all 4 cases:

• An AC (or, you could think about it as a "high frequency") signal through the low-pass filter
• An AC (or, you could think about it as a "high frequency") signal through the high-pass filter
• An DC (or, you could think about it as a "low frequency") signal through the low-pass filter
• An DC (or, you could think about it as a "low frequency") signal through the low-high filter

Key thing is to substitute in what the impedance of the capacitor would be (the resistor has the same impedance as it does resistance, so that doesn't matter). For a capacitor:

• High-frequency would mean the capacitor has a very low impedance (i.e. high-frequency can pass through it)
• Low-frequency would mean the capacitor has a very high impedance (i.e. just like the symbol for it - two open plates - the low-f or DC can't pass through easily)

With these things in mind, let's think about the path the current takes in all 4 cases:

Note: (and for those who are going to call me out) the "output" will actually be some resistor it is going towards, not an open-circut like we always draw when designing filters.

You can see here for the 4 cases:

• For AC, the low-pass filter blocks it from reaching the output as it goes to ground first, because the capacitor offers a low-impedance path to ground (remember: all signals, DC and AC, want to go to ground. That's their end goal)
• For AC in the high-pass, it passes through as the capacitor is an easy path to get through, and the resistor is just there doing it's thing - not offering any easy path to ground
• For DC, the low-pass is just like having a resistor in the way, as it can't pass through the capacitor easily and so the path to ground is not an easy option
• For DC, the high-pass blocks it almost before it even starts flowing, because the capacitor is present and as I mentioned earlier: for DC/low-frequency signals, the capacitor is just like it's symbol - it blocks the signal, and presents a high-impedance path.

I hope this helps. Happy to clarify if my explanation isn't clear enough :)