How to calculate gain of two cascaded stages low pass filter (passive)?

I built the following circuit and I would like to calculate the gain. I would like to predict the output voltage by knowing the input voltage. Also, I would like to predict the phase shift.

I got the transfer function from this question:Deriving 2nd order passive low pass filter cutoff frequency

I use this website to do math quickly instead of using my own calculator. The website provide the same transfer function of the previous question: http://sim.okawa-denshi.jp/en/CRCRtool.php

Here is how I use the website: I set the range of frequency from 999 to 1001 to get the gain and phase shift precisely at the frequency of 1000 Hz.

My practical measurement of gain does not match the transfer function:

The transfer function says: The gain at 10 Hz would be -0.00125 dB that means the ratio between the output voltage and input voltage is 0.999 but mine is 0.5959 !!

Vout and Vin are in volts.

Gain = Vout / Vin

dT is the phase shift in seconds.

Phase shift is in degrees.

• Why does Vin vary with frequency (in particular for the 10Hz case, other cases have reasonable variations)? Have you measured the actual resistance and capacitance of the components? This is something which I have done myself in the past and it matches the theoretical results reasonable well as far as gain is concerned. – helloworld922 Aug 2 '16 at 2:49
• @helloworld922 I use a mobile phone application as a function generator. The output is taken from the socket of hands-free (speaker). I don't have a real function generator. At 100 Hz, the wave form was not sine wave it was similar to square wave or a clipped sine wave. So, I worked with lower volume (lower voltage) and so that the wave form become a good sine wave again. – Michael George Aug 2 '16 at 12:58

The transfer function is: $$\small G(s)=\frac{1}{(RC)^2s^2+3RCs+1}$$

Hence $\omega_n=\frac{1}{RC}=\small10^4\:rad/s\:(=1592\:Hz)$, and $\small\zeta=1.5$, and it can be seen that the DC gain ($\small s=0$) is unity.

Converting this to the frequency domain, using $s\rightarrow j\omega$:

$$\small G(j\omega)=\frac{1}{1-(\omega RC)^2+j3\omega RC}$$

At $\small 10\:\small Hz$, $\small \omega RC=0.00628$, hence the gain is almost unity and the phase angle is almost zero. At $\small 1\:\small kHz$, $\small \omega RC=0.628$, giving a gain of $\small 0.505$, and phase angle of $\small \phi=-72^o$.

So it seems that there's a problem with your experimental set-up. What's the input impedance of the instrument measuring Vout?

Let's do some detective work:

If the input impedance of the instrument were $\small 3 \: k\Omega$ resistive, then (i) the gain and phase at DC would be $\small 0.6$ and zero, respectively (i.e. same as your results); and (ii) the gain and phase at $\small 1590 \:\small Hz$ would be $\small 0.29$ and $\small -79^o$, which compares with your measurement of $\small 0.31$ and $\small -73^o$.

• I use a mobile phone application as a function generator. The output is taken from the socket of hands-free (speaker). Also, My oscilloscope is not real. It is a sound card based oscilloscope. It the wave form is displayed on the monitor of my PC. Could this be the reason? Measurements are not accurate? – Michael George Aug 2 '16 at 12:54
• If the output and input impedances affect the transfer function.. Can I use an op amp as a buffer? – Michael George Aug 2 '16 at 12:55
• Try a simple test - connect two 1k resistors in a voltage divider and measure the gain at one frequency. If it comes out at around 0.43 then the measuring device has 3k input resistance. – Chu Aug 2 '16 at 13:02
• If that turns out to be the problem then you could use an op-amp buffer. – Chu Aug 2 '16 at 13:04
• I measured the input resistance using a multi-meter. It was about 3k as you said. – Michael George Aug 2 '16 at 14:27

Signals and Systems is the topic you're looking for... it's a big field of study. If I recall correctly, a lot depends on being able to prove a system is "linear" and "time-invariant", but any system with an input and an output can be fully characterized by a "transfer function" - which is a function that varies with frequency of the input signal, and is essentially V_out(frequency) / V_in(frequency) and it's usually called 'H'.

If you have to systems and you chain them together, and system 1 has transfer function H1, and system 2 has transfer function H2, then the composed system has transfer function H1 * H2. In the time domain, systems are characterized by an analogous function called the "impulse response", usually called 'h(t)' and the composition operation is a nasty mathematical construct called "convolution."

This stuff is all wrapped up in Fourier and Laplace transforms, and I'm about out of steam remembering the nuts and bolts of it all. But that's the gist.

• The product H1*H2 gives the correct overall transfer function only if both parts are decoupled to each other. If this is not the case, the calculation is a bit more involved. – LvW Aug 2 '16 at 7:08
• @LvW as I said, I'm a bit rusty - please can you provide a reference that describes this limitation, as well as what you mean by "decoupled to each other" – vicatcu Aug 5 '16 at 23:56
• It is rather logical: Calculating the function H1=(1/sC1)/[(1/sC1)+R1] assumes that there is no other part in parallel to R1. However, this is not true for the shown circuit with 4 elements - unless R2 and C2 are decoupled with a suitable buffer amplifier. – LvW Aug 6 '16 at 7:39