Hi,
For even order of active filter, 2nd order filters are cascaded. How about the odd order? For example, 5th order Butterworth LPF.
1st order-->2nd order-->2nd order

My question is about the 1st order. I know that it's simply RC, but should it be buffered with an op amp? Or I can use only RC before the first 2nd order?

Thanks

__________________
Superman returns..

 

THere are many ways to implement a filter. One RC is a first order passive filter, two cascaded is a second order passive filter. An op-amp wired to be a filter in the simplest way is a first-order active filter (and there are even multiple ways to implement the same filter on an op-amp, especially for more complicated filters like elliptical, butterworth, bessel, etc). Other implementations include switched capacitor and digital. All can be single order, or multi-order.

Passive filters (like just a single RC) are small and cheap but have impedance problems in that you have to know the impedance of the source and load and take them into account for it to function properly. This also includes cascading. With active filters you don't have to worry so much about the load (or filter) that comes before and after the current filter stage while in a passive filter you must take everything that is along the input path and output path.

Sure you can use an op-amp buffer with an passive filter to do this...but why not just wire an op-amp as an active filter?

If I cascade 7 first-order filters (properly) I have a 7th order filter. It is sometimes possible to wire up a single op-amp 2nd (or higher) order filter, but it's still the same thing if I cascade 3 first order filters + 2 second order filters. I still get a 7th order filter. THere's no difference between even or odd.

1st-1st-1st = 3rd order
2nd-2nd-2nd = 6th order
1st-8rd-2nd = 11th order

DOesn't matter.

 

Yeah, I'm using op amp for the second order for second and third stage. For the first stage, I'm wondering a passive RC without op amp can work the same or not. It is because I'm using dual op amp to form 5th order LPF.

__________________
Superman returns..

 

Does the output of the RC filter go straight into the "infinite" impedance input of the op-amp without branching off anywhere else (like into the first filter's feedback loop)? If it does, then the whole it would probably work assuming the source can drive the RC filter properly.

If it doesn't then you have to take a look at things more carefully.

 

I mean.. something like this:

Attached Images

LPF.GIF (10.9 KB, 19 views)

__________________
Superman returns..

 

You can make an N-th order filter with cascaded 1st order filters, but it is not possible to make any sort of canonical filter such as Butterworth, Bessel, etc. This is because the poles of single-pole RC networks always lie on the real axis, while in a Butterworth, etc., most of the poles are complex. Feedback is required to move the poles off the real axis.
See this page for a 5-pole Butterworth example.

__________________
Ron

 

Hi Banana,
Notice that the second-order filter needs to be fed from the very low output impedance of an opamp that buffers the single RC first stage.

__________________
Uncle $crooge

 

Alright, I got what you mean. So I need 3 op amps. But is there any reason behind? Why must the second order filter be fed from low output impedace source?

Thanks

__________________
Superman returns..

 

The synthesis for each section assumes a zero impedance source. The single pole analysis assumes an infinite impedance load. If either of these conditions is not met, the filter will not work as designed, because the transfer function will be modified.

__________________
Ron

 

Without buffered, the RC at the first stage is included in the transfer function of the 2nd order in the second stage am I right?

__________________
Superman returns..

 

I would say that without the buffer it removes the transfer function of the first stage from the equation and instead uses those components to modify the transfer function of the second state. I wouldn't say it is included because then there would be no difference in behaviour.