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#### Liu Ju

- Jan 1, 1970

- 0

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