questions about Basics of active filters (using op amp), please help/

[Liu Ju]

Dear members:

I am studying about applications of op amps in designing the active
filter in a book.
I have some confusion and would like to ask for explaination.

1. In the book they analyze an active low pass filter consisting of "
1 R, 1 C, and 1 op amp"
The R and C create a passive low pass RC filter and are connected to
the positive pin of the op amp in series.

The book provides the transfer function of the low pass filter as
K=1/(1+P*(omega break)*R*C)
Where (omega break) is the corner frequency (at which the
K=1/SQRT(2)).

I don't understand this function. In my understanding P=j*omega
(Omega: frequency of the input) and thus the transfer function should
be: K=1/(1+P*R*C).
Why is there the (omega break) in the function? Please explain it to
me!

2. In the book they said we can get the transfer function of a high
pass filter by changing somethings from the transfer function of a low
pass filter.

I guess that in the transfer function of a low pass filter, we replace
R by 1/C, C by 1/R, (omega break) by 1/(omega break) and P by 1/P, and
thereby we have the transfer function of the corresponding high pass
filter.


Is it correct?

By the way, would anyone, who know the basic theory of the active
filters and op amp, give me the contact email address. I have some
complex confusion and would like to ask but there are some figures
that I cannot present in the form of text in GOOGLE group. Thank you
very much.


Sincerely
LiuJu

 

[John Popelish]

Liu Ju wrote:
(snip)

1. In the book they analyze an active low pass filter consisting of "
1 R, 1 C, and 1 op amp"
The R and C create a passive low pass RC filter and are connected to
the positive pin of the op amp in series.

This is a basic 1 pole RC filter followed by a buffer amplifier that
keeps operation of the RC filter independent of any downstream load.

The book provides the transfer function of the low pass filter as
K=1/(1+P*(omega break)*R*C)
Where (omega break) is the corner frequency (at which the
K=1/SQRT(2)).

I don't understand this function. In my understanding P=j*omega
(Omega: frequency of the input) and thus the transfer function should
be: K=1/(1+P*R*C).
Why is there the (omega break) in the function? Please explain it to
me!

The break or corner frequency is the frequency at which the response
is switching from little attenuation versus frequency to an
attenuation proportional to frequency. If you draw a graph of the
amplitude response versus frequency on log log paper, the response is
a hyperbola that approaches constant (flat, horizontal) amplitude in
the low frequency side, and constant down slope to the high frequency
side with the intersection of those two asymptotes at the corner or
break frequency.

This frequency is most naturally described in radians per second,
instead of cycles per second. R*C describes that radian frequency.
Since there are 2*pi radians in a complete cycle, you have to multiply
by those factors to change the units of frequency from radians per
second to cycles per second.

2. In the book they said we can get the transfer function of a high
pass filter by changing somethings from the transfer function of a low
pass filter.

If you interchange the capacitor and resistor, the right (higher
frequency side) asymptote of the response graph becomes horizontal
(flat response versus frequency) and the left (lower frequency side)
slopes down proportional to frequency. The corner or break frequency
is still R*C radians per second.

Both the high and low pass filters are just voltage dividers made up
of one fixed impedance (the resisters impedance is not a function of
frequency) and one variable impedance (the impedance of a capacitor is
1/(2*pi*F*C), if F is is hertz or cycles per second, or 1/(omega*C) if
the frequency is in units of radians per second) with a 90 degree
phase shift. The attenuation versus frequency response is just the
result of dividing the voltage between these two impedances.

I guess that in the transfer function of a low pass filter, we replace
R by 1/C, C by 1/R, (omega break) by 1/(omega break) and P by 1/P, and
thereby we have the transfer function of the corresponding high pass
filter.

Is it correct?

Not quite. The break frequency still remains at R*C because that is
the frequency at which the magnitudes of the resistive and capacitive
impedances are equal.

By the way, would anyone, who know the basic theory of the active
filters and op amp, give me the contact email address. I have some
complex confusion and would like to ask but there are some figures
that I cannot present in the form of text in GOOGLE group. Thank you
very much.

You are welcome to email me if this post has been at all helpful to
you.

 

[CFoley1064]

Subject: questions about Basics of active filters (using op amp), please

help/
From: [email protected] (Liu Ju)
Date: 5/4/2004 5:12 AM Central Standard Time
Message-id: <[email protected]>

Dear members:

I am studying about applications of op amps in designing the active
filter in a book.
I have some confusion and would like to ask for explaination.

1. In the book they analyze an active low pass filter consisting of "
1 R, 1 C, and 1 op amp"
The R and C create a passive low pass RC filter and are connected to
the positive pin of the op amp in series.

The book provides the transfer function of the low pass filter as
K=1/(1+P*(omega break)*R*C)
Where (omega break) is the corner frequency (at which the
K=1/SQRT(2)).

I don't understand this function. In my understanding P=j*omega
(Omega: frequency of the input) and thus the transfer function should
be: K=1/(1+P*R*C).
Why is there the (omega break) in the function? Please explain it to
me!

2. In the book they said we can get the transfer function of a high
pass filter by changing somethings from the transfer function of a low
pass filter.

I guess that in the transfer function of a low pass filter, we replace
R by 1/C, C by 1/R, (omega break) by 1/(omega break) and P by 1/P, and
thereby we have the transfer function of the corresponding high pass
filter.


Is it correct?

By the way, would anyone, who know the basic theory of the active
filters and op amp, give me the contact email address. I have some
complex confusion and would like to ask but there are some figures
that I cannot present in the form of text in GOOGLE group. Thank you
very much.


Sincerely
LiuJu

If you're going to be working with active filters, you should look at "Active
Filter Cookbook" by Don Lancaster. All the basics of a tough subject are
covered there, in a manner which is slightly unorthodox but accessible to just
about everybody. The best part is that it has "instant design" sections which
walk you through component specification quickly. It's really oriented toward
actually making and using active filters, as well as analysis. I found the
book to be helpful, and would highly recommend it to any newbie.

The book is still in print, and is available from the library, amazon, or Mr.
Lancaster's website.

http://www.tinaja.com/
http://www.tinaja.com/books/bkdons.asp

Good luck
Chris