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John D Cook
RC values for a Pink Noise Filter

In our Creating Noise book we show how to use RC ladder networks to create pink noise. An \(n^{th}\) order network is shown below.

The R and C values are given by the following (equations 37 and 38 in the book).

\[C(n,i) = \frac{4i-1}{(2(n-i)+1)16^i}\frac{\binom{2(n+i)}{n+i}}{\binom{2(n-i)}{n-i}}\] \[R(n,i) = \frac{(4i-3)(2(n-i)+1)16^i}{4(n-i+1)(n+i)}\frac{\binom{2(n-i)}{n-i}}{\binom{2(n+i)}{n+i}}\]

Using these equations is a bit cumbersome so here are some asymptotic versions of the equations which will go into the next edition of the book.

\[C(n,i) = \frac{4i-1}{(2(n-i)+1)}\sqrt{\frac{n-i}{n+i}}\] \[R(n,i) = \frac{(4i-3)(2(n-i)+1)}{4(n-i+1)\sqrt{(n+i)(n-i)}}\] \[C(n,i)R(n,i) = \frac{(4i-1)(4i-3)}{4(n-i+1)(n+i)}\]

These equations obviously only work for \(i<n\). For \(i=n\) the asymptotic equations are

\[C(n,n) = \frac{4n-1}{\sqrt{2\pi n}}\] \[R(n,n) = \frac{(4n-3)\sqrt{2\pi n}}{8n}\]

The last resistor in the chain has the asymptotic formula \[R_L(n)=\sqrt{2\pi n}\]

In the limit \(n\rightarrow\infty\) with \(i/n\rightarrow x\) both the R and C equations have the following form

\[\frac{2x}{\sqrt{1-x^2}}\]

where \(x\) has the range \(0<x<1\). In this limit the ladder becomes a continuous transmission line and the total capacitance or resistance (measured from the beginning of the line) is found by integrating the above equation

\[C(x)=R(x)=2\left(1-\sqrt{1-x^2}\right)\]

This post as a pdf


A Bridge Circuit Puzzle

In the bridge circuit shown below, find the value of the resistance \(c\) such that the equivalent resistance connected across the voltage source is also equal to \(c\). Show that for this value of \(c\), the voltage across \(c\) is equal to \((\sqrt{b}-\sqrt{a})/(\sqrt{b}+\sqrt{a})\). Try to solve this problem yourself before looking at the solution below.

Start by finding the equivalent resistance. The easiest way to do this is to replace the voltage source with a current source \(I\), and then find the voltage \(V_1\) at node 1. The equivalent resistance is then \(V_1/I\). This involves solving 3 equations for nodes 1, 2 and 3. The equations are Kirchoff’s current law applied to the nodes. Leaving out the details, the equivalent resistance is

\[R = \frac{2ab+(a+b)c}{a+b+2c}\]

If you set \(R=c\) and solve for \(c\), you get \(c=\sqrt{ab}\). In other words, when \(c\) is equal to the geometric mean of \(a\) and \(b\), the equivalent resistance is equal to \(c\).

To solve the second part of the problem we need an expression for the voltage across \(c\). Going back to the voltage source and writing the node equations for nodes 2 and 3, we get

\[(1/a + 1/b + 1/c)V_2 - (1/c)V_3 = (1/a)V_1\]

\[-(1/c)V_2 + (1/a + 1/b + 1/c)V_3 = (1/b)V_1\]

Solving these 2 equations for \(V_2/V_1\) and \(V_3/V_1\) we get the following expression for the voltage across resistor \(c\).

\[\frac{V_2-V_3}{V_1} = \frac{(b-a)c}{(b+a)c+2ab}\]

If you substitute the value \(c=\sqrt{ab}\) into this equation, you find that

\[\frac{V_2-V_3}{V_1} = \frac{\sqrt{b}-\sqrt{a}}{\sqrt{b}+\sqrt{a}}\]

This simple circuit can calculate the ratio of the sum and difference of the square roots of 2 numbers.

This post as a pdf


Measuring Voltmeter Input Impedance

Sometimes you have to measure a voltage that has a high output impedance. To do this accurately you need to know the input impedance of your voltmeter. Ideally it should be much much larger than the impedance of what you’re trying to measure. The following figure illustrates the situation.

The voltage we’re trying to measure is \(V_x\) and it has an output resistance of \(R_x\). The meter has an input resistance of \(R_m\). The voltage measured by the meter will then be

\[V = \frac{V_xR_m}{R_x+R_m} = \frac{V_x}{1+R_x/R_m}\]

So you can see that \(V\approx V_x\) only when \(R_m>>R_x\). If you happen to know the values of \(R_x\) and \(R_m\) then you can determine what \(V_x\) is even when the two resistances are of the same magnitude. How do you measure the input resistance of your voltmeter? If you apply a known \(V_x\) using a known \(R_x\) then you can solve for \(R_m\) in the above equation.

\[R_m = \frac{R_x}{\frac{V_x}{V}-1}\]

We did this for the old Scope multimeter you see in the following picture using \(V_x=12v\), \(R_x=1M\Omega\).

The meter read \(10.98v\) which gave us \(R_m=10M\Omega\). This is pretty typical for most handheld multimeters.

Some really cheap multimeters such as the one shown below that we got at Harbor Freight Tools showed an input resistance of only \(1M\Omega\). You have to be careful about what you measure with a meter like that. In some cases it can give you very inaccurate readings.

This post as a pdf


Book Review: The Refrigerator and the Universe

The Refrigerator and the Universe: Understanding the Laws of Energy” by Martin Goldstein and Inge F. Goldstein is gold. The authors know the subject of thermodynamics very well and take us on an epic journey through the far reaches of the subject’s domain, as well as cover its history. While this book doesn’t contain problems, it is an inspiration for problems and further investigations. I couldn’t recommend it more highly. My only criticism is really just a personal preference that the book be at the level of a university physics textbook, with calculus, not just algebra. But as such, it appeals to a wider audience. The extensive annotated bibliography at the end serves as a roadmap for further reading.



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