1. Jul 22, 2010

### hasan_researc

I do not understand section 8.6 of my lecture notes (see the attachment).

1. "If two low pass filters are cascaded the output voltage is not simply the product of
two transfer functions." : Why not? Why would we even think of multiplying two transfer functions? Which transfer functions are being referred to? Why?

2. "If you try to analyse such a circuit using phasors and an Argand diagram you will soon realise the shortcomings of that technique. This is because the second filter acts as a load for the first." : Could someone please explain how the second statement leads to the first statement?

3. "However, we shall later that a simple operational amplifier circuit called the unity gain buffer acts as an impedance matcher so that loading of the first filter by the second does not occur. " : What is an impedance matcher? Why does the loading not occur due to an impedance matcher?

Thanks in advance for any help.

2. Jul 22, 2010

### The Electrician

Where's the attachment?

3. Jul 22, 2010

### hasan_researc

Here it is.

#### Attached Files:

• ###### gtfhhhhhhhhhhhhhhh.pdf
File size:
130.2 KB
Views:
1,180
4. Jul 22, 2010

So a low pass works by being a two resistance voltage divider that is frequency dependent, and a low pass has a complex frequency dependent total resistance. If you put one in parallel to the capacitor of another low pass (which is what you do when chaining then) the voltage divider looks different, does it not? Because the second resistance is not formed by a pure capacitor but by a capacitor in parallel to a low pass.

The transfer function tells you how much of the input reaches the output. So it seems natural that you can multiply them, if the first low pass lets 50% through and the second is the same you expect 25% in the output, but this is normally not correct due to the above reasons.

You can imagine a unity gain buffer like a voltmeter that drives a power supply. If it reads 10V on the input, then it supplies 10V on the output. The nice thing is, that the input impedance is so high, that there is almost no current running through it, so the input circuit doesn't see it, and the output doesn't change the voltage when you draw current from it (which is what happens when you draw current from a low pass). In other words it has low output impedance.

Your questions are very elementary, maybe you should talk with other students about them.

5. Jul 22, 2010

### Antiphon

0xDEADBEEF, you are totally correct. However, it should be pointed out that it is quite possible to build higher order lowpass filters that doesn't use buffer amplifiers. The buffer is not the only way to do it.

6. Jul 22, 2010

### The Electrician

The cascaded filter shown in section 8.6 of the pdf file is already a higher order filter (higher order than a single R-C stage). And, if additional R-C stages are cascaded, the filter order will increase by one for each additional stage added, without buffers.

So, it's already apparent that higher order filters (R-C filters in this case) can be built without buffers; it didn't need to be pointed out.

You must have meant to say something more than that, didn't you? :-)

7. Jul 24, 2010

### Antiphon

Yes, I meant that the thread (not necessarily your post) seems to imply that cascading filters is somehow difficult. In fact it's one of the best established ways to make filters. Usually the stages are LC, not RC.

8. Oct 3, 2012