Deriving DC and HF Gain of Filter: Need Help with Calculations?

In summary: Thanks for helping me :) Do you think I'm making a mistake in there, or do you know of any other things we might be able to try?In summary, the book defines the gain as the G = 20 log | T(jw) | dB. However, for the HF gain, the book states the answer is \frac{C_1}{C_1 + C_2}. I can't seem to derive this.
  • #1
maxsthekat
55
0

Homework Statement


I need to find the DC and high frequency (HF) gain of this filter:
[PLAIN]http://www.flickr.com/photos/96575810@N00/4420162040/

(The url is [PLAIN]http://www.flickr.com/photos/96575810@N00/4420162040/ in case that doesn't work)

Homework Equations


The book defines the gain as the G = 20 log | T(jw) | dB
Where |T(jw)| = |Vo (jw)| / |Vi (jw) |

The Attempt at a Solution


For the DC gain, the book states that the gain is [tex]\frac{R_2}{R_1 + R_2}[/tex]. This seems straightforward if we apply traditional DC analysis, and assume the capacitors act as breaks in the circuit.

However, for the HF gain, the book states the answer is [tex]\frac{C_1}{C_1 + C_2}[/tex]. I can't seem to derive this. Here's what I've tried:

1) Began by combining R1 and C1, in the complex frequency domain (C = 1/jwC)
R1 || C1 = [tex]\frac{1}{\frac{1}{R_1}+\frac{1}{\frac{1}{jwC_1}}}[/tex]

= [tex]\frac{1}{\frac{1}{R_1}+ jwC_1}[/tex]
(multiply top and bottom by [tex]R_1[/tex]
= [tex]\frac{R_1}{1 + jwC_1R_1}[/tex] (call this expression A)

2) I can derive a similar expression for the combination of R2 and C2
= [tex]\frac{R_2}{1 + jwC_2R_2}[/tex] (call this expression B)

Now, to find Vo, I would just take Vi * [tex]\frac{B}{A + B}[/tex]. Since we're looking at the gain, the Vi term will drop out and I'll get:

[tex]\frac{V_o}{V_i} = \frac{B}{A + B}[/tex]

Finally, I need to find the magnitude of [tex]\frac{B}{A+B}[/tex], and then see what happens as w approaches infinity. This is where I'm getting stuck. How do I take the magnitude of this? B has a real term in the numerator, and a real and imaginary term in the denominator. I've tried just doing the complex division, but either I'm doing something wrong or it's a bad idea, since the expression doesn't seem to be correct.

Can anyone lend a hand?

Thanks!

-Max
 
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  • #2
You know, I tried to do some of the algebra too, and didn't find what I was looking for... I got close, but I still had a mix of R1 and R2 terms. So I decided to look at it another couple ways:
Usually, when we start talking about the HF extreme, we start looking at things in the extreme opposite from DC... so the Caps become shorts and the resistors become breaks in the circuit. At that extreme, if you ignore the Rs and just look at the Cs, then I'm sure you would agree with the book's answer.

But what if you don't feel comfortable with just ignoring the Rs?

Well, take a look at your two terms... A, and B. What happens if you take the limit on those? Don't you find the Rs just disappear?

I'll work a little more on the algebra and see if I can't find something a little more rigorous.
 
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  • #3
OK, got it. ... How about you try to do it over, but this time, before you create your expression "A", try NOT multiplying top and bottom by R1 (or R2 for "B")

see if that works for you.
 
  • #4
Well, I totally agree with you: intuitively, it seems that we should be able to ignore the resistors for w -> infinity.

What bugs me is the notion of "shorting" the capacitors is really just an application of looking at the capacitor's impedance as w-> infinity. In other words, it's a result of 1/(jwC). So, by doing this calculation in complex, I should be able to derive the same behaviors by looking at what happens to w as it approaches zero and infinity.

Looking at the limit of the functions: say I take the limit of B as w-> infinity. Then, the denominator becomes large, taking the whole expression to zero. If I apply this to A as well, and plug back into A / (A + B), then I have 0 / 0, which doesn't work.

Thanks for helping me :) Do you think I'm making a mistake in there, or do you know of any other things we might be able to try?

Edit: just saw your second post. I'll give that a try and let you know :)
 
  • #5
Or maybe I was just solving the problem that I ran into and didn't read closely about the problem you were having.

To answer your question, "How do I take the magnitude of this?", well, to find the magnitude of a complex term, you multiply the term by its complex conjugate and then take the square root.

(*the complex conjugate is where you replace every j with -j)
 
  • #6
maxsthekat said:
Looking at the limit of the functions: say I take the limit of B as w-> infinity. Then, the denominator becomes large, taking the whole expression to zero. If I apply this to A as well, and plug back into A / (A + B), then I have 0 / 0, which doesn't work.

And as long as I'm not being rigorous, I feel I can wink and say it's not 0/0 or even 0/(0+0),
but it's more like almostzeroC1/(almostzeroC1+almostzeroC2). Factor out the almostzero and there you go. ;)
 
  • #7
Can I see how you got this to work for the "new" A and B expressions?

Without multiplying by R, the expression is

[tex]\frac{1}{ \frac{1}{R} + \frac{1}{jwC} }[/tex]

Meaning the whole thing would look like:

[tex]
\frac{\frac{1}{ \frac{1}{R_2} + \frac{1}{jwC_2} }}{\frac{1}{ \frac{1}{R_1} + \frac{1}{jwC_1}} + \frac{1}{ \frac{1}{R_2} + \frac{1}{jwC_2}}}[/tex]

Which doesn't look at all like it's going to end up in the form of [tex]\frac{C_1}{C_1+C_2}[/tex]

Thanks again :)
 
  • #8
I think I steered you wrong on that... it only helped me because it meant that I did the algebra over and did not repeat an algebraic mistake the first time.

But make no mistake that I do end up with a proper expression whose limit is C1/(C1+C2).

Also, your terms look funny - you should have [tex]\frac{1}{\frac{1}{R} + j \omega C}[/tex] , not [tex]\frac{1}{\frac{1}{R} + \frac{1}{j \omega C}}[/tex]

Meaning the whole thing would look like:

[tex]
\frac{\frac{1}{ \frac{1}{R_2} + jwC_2 }}{\frac{1}{ \frac{1}{R_1} + jwC_1} + \frac{1}{ \frac{1}{R_2} + jwC_2}}[/tex]

Now you need to multiply top and bottom by both [tex](\frac{1}{R_1} + j \omega C_1)[/tex] and [tex](\frac{1}{R_2} + j \omega C_2)[/tex]

So how about you go through the algebra and gather your terms and show me what you have. (Don't take the magnitude - let's see it before that point)
 
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  • #9
Damn you, algebra! I shall defeat you one day ;)

Thanks! That trick worked. After I multiplied by those terms, I then multiplied top and bottom by 1/(jw). Taking the limit of this led to C1 / (C1 + C2).

Oh, and you were completely right-- that was a typo earlier-- should've been 1/(1/jwC) = jwC.

Thanks again for your help-- I owe you one :)
 

1. How do I calculate the DC gain of a filter?

The DC gain of a filter can be calculated by setting the frequency in the filter's transfer function to 0Hz. This will give you the ratio of the output DC voltage to the input DC voltage.

2. What is the HF gain of a filter?

The HF (high frequency) gain of a filter is the ratio of the output voltage to the input voltage at high frequencies. It is typically measured at the cutoff frequency of the filter.

3. What is the difference between DC gain and HF gain in a filter?

DC gain refers to the ratio of the output voltage to the input voltage at 0Hz, while HF gain refers to the same ratio at high frequencies. The DC gain is more relevant for low-frequency signals, while the HF gain is more important for high-frequency signals.

4. How does the type of filter affect the DC and HF gain?

The type of filter (e.g. low-pass, high-pass, band-pass) will affect the DC and HF gain values. For example, a low-pass filter will have a higher DC gain and a lower HF gain compared to a high-pass filter. This is because a low-pass filter allows low frequencies to pass through and attenuates high frequencies, while a high-pass filter does the opposite.

5. Can I adjust the DC and HF gain of a filter?

Yes, the DC and HF gain of a filter can be adjusted by changing the components (e.g. resistors, capacitors) in the filter circuit. This can be done by selecting different values for these components or by adding additional components to the circuit to modify the gain. However, it is important to note that changing the gain of a filter may also affect its other characteristics, such as its bandwidth and frequency response.

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