The formula was generated with the LATEX feature provided in the post editor here in the forum. When you use the advance post editor you will see the Ʃ in the tool bar above the text box.

There is a helpful thread here in the forum that discusses Latex.

hgmjr

 

All right, here's a stab at what my 89 gave me, I hope it comes out right. Also I just wrote +- instead of repeating it all. Its a repeated root.
\(\frac{4x^2}{(2x-41569.2)(-+.57735j)}\)

And for my Quadratic Equation i found 12000 +-23987.5j

One of these gives me my a and b values correct? Which would be the values of my capacitors? Then I could divide by 2pi to get my frequency?

 

Ok, had a bit of a breakthrough. I realized my 5.76x10^8 was my w^2 term. So I just took the square root, and now I have my w term. But now I don't really know what to do with it, since every equation I'm looking at, needs my C or R values.

 

Look up the general transfer function for a unity-gain, high-pass sallen-key active filter. Then you can obtain the coefficients for each of the terms expressed in Rs and Cs.

hgmjr

 

Take a look at this webpage on high-pass unity-gain sallen-key.

hgmjr

 

Ok, that all makes sense, but how am I to implement these? I don't have enough values. I'm sorry I'm just not getting it. I know the A and B terms from the quadratic. I know my ω^2 term. but I still don't get how to calculate the C R or Q values since they depend on each other.

 

Since Rs and Cs in filters can be scaled, I think you will find that there is no unique set of resistor and capacitor values that will satisfy the actual values in your transfer function. For example there are any number of combinations of R1, R2, C1 and C2 that can end up yielding the coefficients in the given expression. I suggest you assume C1 and C2 are equal and substitute C into the formula. You will then end up being able to calculate R2 equal K*R1. You will then have a ratio of R1 to R2 that you can use to get you to the answer.

What you have is two non-linear equations that are produced when you assign the expressions for the coefficient in Rs and Cs to the actual value from your given transfer function.

for example:

\(\frac{1}{R_1R_2C_1C_2}\ =\ 5.67\times 10^8\)

I'm sure this is a bit confusing to one who is new to the process.

hgmjr

 

Ok, so setting my Capacitors equal to each other then equal to 1. I get the following: This website helped http://www.maxim-ic.com/appnotes.cfm/an_pk/1795

I get:

\(\frac{S^2R1R2}{S^2R1R2+S(2R1)+1}\)

So I figured since I have the coefficient of the S(2R1) term (2.4*10^4) I could set 2R1 equal to that and then solve for R1 getting 12000. But then that leaves me with \(\frac{1}{12000}\) for my R2, which seems strangely low. But, i guess it would make those terms have a coefficient of one..... But It just doesn't seem right. It can't be that easy.

 

Ok, so setting my Capacitors equal to each other then equal to 1. I get the following: This website helped http://www.maxim-ic.com/appnotes.cfm/an_pk/1795

I get:

\(\frac{S^2R1R2}{S^2R1R2+S(2R1)+1}\)

So I figured since I have the coefficient of the S(2R1) term (2.4*10^4) I could set 2R1 equal to that and then solve for R1 getting 12000. But then that leaves me with \(\frac{1}{12000}\) for my R2, which seems strangely low. But, i guess it would make those terms have a coefficient of one..... But It just doesn't seem right. It can't be that easy.

You're on the right track here. Since it looks like your using the Sallen-Key topology, using 50k resistors whenever you can is a good idea. This should give you reasonable values for your capacitors also.

 

Ok, so setting my Capacitors equal to each other then equal to 1...

You should realize that setting a capacitor value to 1, means setting it to 1 Farad, an extremely large value. That's why your resistor ends up being so small. Try a capacitor value like 1E-6, 1 microfarad instead. Or, maybe even smaller.

 

What is needed is a schematic to move this discussion along. I believe we are all in agreement that the filter topology should be a Sallen-Key unity-gain second-order highpass filter.

That being the case I offer the following schematic.

The transfer function given is:

\(\frac{s^2}{s^2+2.4\times10^4s+5.76\times10^8}\)

Based on the general transfer function expression for the assumed filter topology, the 2 non-linear equations are:

\(\frac{1}{R_1R_2C_1C_2}\ =\ 5.76\times 10^8\)

\(\frac{1}{R_1C_1}\ +\ \frac{1}{R_1C_2}\ =\ 2.4\times 10^4\)

This is a system of two non-linear equation with four unknowns. Since these two equations can only be solved if there are only two unknowns, some assumption must be introduced if the system of equations is to be made solvable.

Out of the universe of assumptions, I have chosen the following two:

\(C_1\ =\ C_2\ =\ C\)

AND

\(R_1\ =\ k\times R_2\)

Step one is to find the value of "k" (the ratio of R1 to R2). I will leave that exercise for you to ponder and hopefully to solve.

hgmjr

 

I'm trying to guess some good values for R1 and C1 that won't give me some really wierd numbers. My professor said it would be easier to pick them that way, instead of having C1=C2, but It's kinda hard.

 

I'm trying to guess some good values for R1 and C1 that won't give me some really wierd numbers. My professor said it would be easier to pick them that way, instead of having C1=C2, but It's kinda hard.

You can try the guessing game but I think you will be guessing long after the assignment is due. That is unless you have inordinate luck on your side.

A systematic mathematical attack of the problem will be sure to yield a prompt answer to this an all subsequent similar circuit problems.

hgmjr

 

Ok, so picking C1= 1uf and R1=10kΩ using the above equations,I get C2=4.18uf and R2=4.1Ω Does this sound reasonable?

 

Ok, I will take you a bit closer to the curtain behind which the answer lies.

First, for convenience I will restate the equations along with my assumption that I posted earlier.

This is a system of two non-linear equation with four unknowns. Since these two equations can only be solved if there are only two unknowns, some assumption must be introduced if the system of equations is to be made solvable.

Out of the universe of assumptions, I have chosen the following two:

AND

Familiarity with this problem and three decades of experience in circuit analysis inform me that the best way to approach the solution to the value of "k" given the assumptions I have chosen is to decide which of the two resistors I will use in the analysis. I can choose either one since once I know the value of "k", I will be able to ascertain the other resistor. I choose R1.

Step 1:

Plugging in C for C1 and C2 in the above equation and then solving for the product of R1 and C yields:

\(\normalsize{R_1C}\ =\ \Large{\frac{2}{2.4\times 10^4}}\)

Step 2:

From the equation relating R1 to R2, I solve for R2 in terms of R1:

\(\Large{\frac{1}{\frac{{R_1}^2C^2}{k}}}\ =\ \normalsize{5.76\times 10^8}\)

Step 2a:

\(\Large{\frac{k}{{R_1}^2C^2}}\ =\ \normalsize{5.76\times 10^8}\)


Step 2b:

\(\normalsize{{R_1}^2C^2}\ =\ \frac{\Large{k}}{\normalsize{5.76\times 10^8}}\)


Step 2c:

\(\normalsize{{R_1C}\ =\ \LARGE{(}{\frac{\Large{k}}{\normalsize{5.76\times 10^8}}}\LARGE{)}\ \Large{^{\frac{1}{2}}}\)

Step 3:

\(\Large{\frac{2}{2.4\times 10^4}}\ =\ \LARGE{(}{\frac{\Large{k}}{\normalsize{5.76\times 10^8}}}\LARGE{)}\ \Large{^{\frac{1}{2}}}\)

Step 3a:

\(\LARGE{(}\Large{\frac{2}{2.4\times 10^4}}\LARGE{)}\Large{^{^2}}\ =\ \LARGE{(}{\frac{\Large{k}}{\normalsize{5.76\times 10^8}}}\LARGE{)}\)

Step 3b:

\(\LARGE{(}\Large{\frac{4}{5.76\times 10^8}}\LARGE{)}\ =\ \LARGE{(}{\frac{\Large{k}}{\normalsize{5.76\times 10^8}}}\LARGE{)}\)

From Step 3b above, it should be obvious to even the most casual observer that the value for "k" is 4.

Do you think you can carry on from here? I am exhausted from the climb. LaTex is a powerful and indispensible tool for forming equations, however, user-friendly it is not.

hgmjr

 

Ok, so picking C1= 1uf and R1=10kΩ using the above equations,I get C2=4.18uf and R2=4.1Ω Does this sound reasonable?

It should be easy enough for you to plug these values back into the general transfer function equation and verify for yourself that the values you have chosen are correct.

Did you do that for these values?

hgmjr

 

No, I did not plug them back in. I was guessing some more. But I wouldn't think that they would since they aren't ratios of each other.

Using your method above I chose R1= 1k so R2 = 4k, which yields the C = .83uF

but when calculating them I get 24096.4 = 24000, which is obviously wrong. But not by much. perhaps due to the .03 repeating.

 

No, I did not plug them back in. I was guessing some more. But I wouldn't think that they would since they aren't ratios of each other.

Using your method above I chose R1= 1k so R2 = 4k, which yields the C = .83uF

but when calculating them I get 24096.4 = 24000, which is obviously wrong. But not by much. perhaps due to the .03 repeating.

Those values come close enough to satisfy one of the equations but not the other other. The challenge is to get the choice of R1, R2, C1, and C2 to satisfy both equations.

hgmjr

 

that's supposed to be .83 nF not uf.

but it's not clear to me what you mean by satisfies one equation but not the other. The equations you listed in step 1 and 2? But we set C1=C2 and such so you can't guess a value for C1 AND C2 since they are the same?

I am making this so much harder than I need to. I just can't get that AHA moment. Thanks for your patience.

 

\(\frac{1}{R_1R_2C_1C_2}\ =\ 5.76\times 10^8\)

\(\frac{1}{R_1C_1}\ +\ \frac{1}{R_1C_2}\ =\ 2.4\times 10^4\)

It is these two equations in the insert above that I was referring to when I spoke of the two equations.

It matters not whether you choose to make C1 = C2 = C as I did. You should be able to plug whatever values for C1, C2, R1, and R2 into the two equation and have both sides of the equation be equal. Until this can be done, the problem is not solved.

At this point you can take the second of the two equations above and choose a value for R1. You can then solve for C (remember I chose to set C1 equal to C2).

Next you can check the correctness of your values by plugging in the value for C (just calculated) and R1 (guessed) that you have at hand into the first equation. Remember that based on my calculation of "k", if you have a value for R1 you can calculate R2 by dividing R1 by "k" . If both equation are satisfied by your values then you have been gloriously successful.

hgmjr

P.S. As I look at this solution, I can now see that the calculation of "k" while interesting is needless. In fact the second equation above can be used to guess and calculate R1 and C respectively then it is possible to use the first equation to calculate R2. That means that the solution can be carried out in fewer intermediate calculations.