# Understanding the gain in an RC circuit's transfer function

I am learning about 1st order transfer functions, and I am trying to understand it with practical examples of systems, because it still looks weird to me how a same function can describe various physical systems. I am currently looking at the transfer function of a simple RC circuit.

It is denoted as

lookig at the standard function, we can tell we have a gain of 1 in this function, as explained in this post

I am ok with the fact that the time constant tau is equal to RC in this particular real world example, as a resistance multiplied by a capacitance can be simplified to time units, a little bit like a speed multiplied by a distance.

however, I am not sure how to look at this and, set apart all the math, intuitively understand why the gain should be 1, rather than anything else. I need help understanding what is the concept behind the gain here, because I am completely lost.

Thank you so much!

• The gain is 1 at dc. The gain is approximately 1 as frequency rises but, at the cut off frequency, the gain is approx 0.71 and falls as frequency increases more. Feb 10 '21 at 19:42
• For gain you need an active component, or amplifier. Without that the best you can get is 1. Feb 10 '21 at 19:42
• @Andyaka I am sorry, I do not understand. Where does the 0.71 comes from? and what would be the frequency and the cutoff frequency in the case of the RC circuit, since it is DC?
– JCSB
Feb 10 '21 at 19:47
• Look up what s is and then look up what the magnitude of a complex number is. If you know that then it should become very clear. Feb 10 '21 at 20:08
• @JCSB Do you follow why the frequency domain description uses 's' and how it relates to time domain, Laplace as a handy way of turning solutions involving complex differential equations into a simpler algebra process, how Euler's operates here, and only need to know how to compute a magnitude of a complex number? Or is there a perch above the fray you are still searching for, too?
– jonk
Feb 10 '21 at 20:17

The gain in the transfer function shows what gain you'll get in infinite time after input stabilizes. And this is indeed 1, because after enough time Vout always equals Vin.

To understand what happens with real (changing) input, you shall examine frequency response of the system, that is use Fourier transform instead of Laplace transform. For your transfer function you may just replace $$\s\$$ with $$\j\omega \$$.

In this case frequency response is

$$H(j\omega)=\frac{1}{1+j\omega RC}$$

And if you are interested in amplitude gain, you shall consider amplitude frequency response

$$|H(j\omega)|=\frac{1}{1+(\omega RC)^2}$$

As you can see, gain is 1 when frequency is zero and less than one otherwise.

The shown common simple RC lowpass filter has the mentioned transfer function Vout/Vin = K/(1+sT) . Factor K happens to be 1 because at 0Hz (i.e. at s=0) the transfer function must get value =1. Factor T happens to be =RC. All this is calculated easily with elementary s-domain circuit analysis.

By inserting an amplifier in front of the filter one can get a version which has K>1.

Or inserting passive attenuating resistor one can get a version which has K<1. The next version has transfer function Vout/Vin = 0.5/(1 + sRC) The resistances are chosen to keep the multiplier of s equal with RC:

Other circuits with no capacitors can well have transfer function of the same general 1st order form Vout/Vin = K/(1+sT)

K is still =1, but T = L/R. If you set R = 1 Ohm and L = 1 Henry you get T = 1 second. The filtering effect is the same as with the simple RC filter which has R = 1 Ohm and C = 1 Farad.

The transfer function doesn't define the structure of the circuit, the same transfer function is possible with infinite many different circuits. But if you see that a LRC circuit has N inductors and M capacitors the order of the transfer function must be M+N or less. It can be less because some of the capacitors or inductors can be redundant, for. ex 2 capacitors in series affect only as one.

If it happens that you want to know how the transfer functions are derived from the circuits I suggest learning circuit analysis. To describe the calculations shortly I tell that all shown circuits are voltage dividers which obey the usual voltage divider rule. The upper and lower halves of the voltage dividers are s-domain impedances.