Please help me about active filter!! thanks

[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 out there, 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

 

[Active8]

On 4 May 2004 03:12:44 -0700, Liu Ju wrote:

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.

YOu mean the resistor is in series and the cap is a shunt to ground.
You can omit the op amp in this case to do the transfer function as
it appears to be used as a unity gain follower.

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

Thats the complex frequency. **Lets call it s for ease of
calculation**

(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!

If you mean w (omega break - WTF?), then

w = 2*pi*f where f is the cutoff freq. You'd probably do better by
reading Don Lancasters Active filter cookbook. It explains all that.
Starting with a normalized 1 rad/s filter, IIRC. Figure that RC is a
time constant. 1/RC would be frequency in rad/s and multiplying by
2*pi gives you hertz.

Now consider the frequency selective parts


Ei ___ Eo
------|___|-----+------
|
|
|
---
---
|
|
===
GND

Look at the circuit as a frequency dependant voltage divider.

Eo/Ei = Xc/(Xc + R)

Xc = 1/(2*pi*f*C) but as I prefer (see **...** above)

Xc = 1/Cs so

Cs/(R + Cs) = 1/(1 + RCs)

There's yer xfer function.

To go High pass, swap the R and C leaving all else the same.

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.

Get you r money back. We can't order somethings anymore. THey were
replaced long ago by nothings.

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?

Eh, I think you just transform the components. In this case, it's
easier to just swap them than calculating new and most likely
non-standard values.

By the way, would anyone out there, 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.

Get a real ISP that carries alt.binaries.schematics.electronic and
post there.
--
Best Regards,
Mike

 

[Active8]

On Tue, 4 May 2004 11:25:24 -0400, Active8 wrote:

<snip>

Now consider the frequency selective parts

Ei ___ Eo
------|___|-----+------
|
|
|
---
---
|
|
===
GND

Look at the circuit as a frequency dependant voltage divider.

Eo/Ei = Xc/(Xc + R)

Xc = 1/(2*pi*f*C) but as I prefer (see **...** above)

Xc = 1/Cs so

Cs/(R + Cs) = 1/(1 + RCs)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Sorry, it's (1/Cs)/(R + 1/Cs) = 1/(1 + Cs)

snip

--
Best Regards,
Mike

 

[Genome]

"Liu Ju" <chinabooter@hotmail.com> wrote in message
news:8cda960d.0405040212.466d4cb0@posting.google.com...
| 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 out there, 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

You are right about the transfer function being K=1/(1+jwRC) which, if
you substitute P for jw ends up as being K=1/(1+PRC) which is very nice
for the PRC. In the west we use s instead of P and end up with
K=1/(1+sRC).

I think you have misunderstood the book when you say it gives the
transfer function as K=1/(1+P(omega break)RC) where K=1/SQRT(2).

At this point it is stating the particular frequency, omega break, when
the transfer function is 3dB down from unity gain. 20log(1/SQRT(2))
= -3dB.

DNA

 

[Tom Bruhns]

Hello LiuJu,

Yes, it sounds like the book either has a mistake or is just confusing
in the way they have written the transfer function. For the simple
circuit you describe, the transfer function should just be the same as
the RC filter. The op amp is simply buffering the output, so that a
following circuit does not load the filter and thus change its
response. I would usually write the transfer function as H(s) =
1/(1+s*R*C) -- really the same as you suggest. So "Omega break" is
just R*C, but you should not use both at the same time.

The classic way to change a low-pass transfer function to a high-pass
is to replace s by 1/s, or to keep the cutoff frequency the same, I
suppose replace s by 1/(s*(omega_cutoff)^2). That transforms the the
lowpass equation above to H(s)=s*R*C/(1+s*R*C). In the circuit you
can do that by replacing the C with an inductor: the capacitor was a
reactance of 1/(s*C) and gets replaced by an inductor whose reactance
is s*L. The value of L must be such that L=1/((omega_cutoff)^2*C).
The resistor does not change, because its impedance does not depend on
s. But of course inductors are not as nice to work with at low
frequencies as are capacitors. So you can then divide all the parts
by s, and see that the resistor must change to a capacitor (R --> R/s)
and the inductor to a resistor (s*L --> s*L/s = L), and you may scale
the impedances to something reasonable to implement. That may be a
little more detailed way to do it than you want, but if you realize
that's what's underneath the part swapping, you can always figure out
what to do, just remembering the one simple transformation.

(You can even transform one step further, and make the resistor a
capacitor, and the capacitor a 'super-capacitor' with impedance
1/(k*s^2) -- it's possible to synthesize the 'super-capacitor' with
op amp circuits.)

There are similar transformations from lowpass to band-pass and to
band-stop.

Welcome to contact me at the email which should show in the header.
Be sure to put a good subject line on any message so I don't ignore
it.

Cheers,
Tom


chinabooter@hotmail.com (Liu Ju) wrote in message news:<8cda960d.0405040212.466d4cb0@posting.google.com>...

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 out there, 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

 

[Tom Bruhns]

Active8 <reply2group@ndbbm.net> wrote in message news:<16wq7e2hr2fw3$.dlg@news.individual.net>...

On Tue, 4 May 2004 11:25:24 -0400, Active8 wrote:
....

Cs/(R + Cs) = 1/(1 + RCs)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Sorry, it's (1/Cs)/(R + 1/Cs) = 1/(1 + Cs)

Third time's a charm?

= 1/(1+RCs)



Some days are like that, I know.

Cheers,
Tom

 

[Active8]

On 4 May 2004 14:16:28 -0700, Tom Bruhns wrote:

Active8 <reply2group@ndbbm.net> wrote in message news:<16wq7e2hr2fw3$.dlg@news.individual.net>...
On Tue, 4 May 2004 11:25:24 -0400, Active8 wrote:
...

Cs/(R + Cs) = 1/(1 + RCs)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Sorry, it's (1/Cs)/(R + 1/Cs) = 1/(1 + Cs)

Third time's a charm?

= 1/(1+RCs)

Some days are like that, I know.

Cheers,
Tom

Yeah, and that must have been one of 'em.
--
Best Regards,
Mike