20
\$\begingroup\$

I'm reading Practical Electronics for Inventors, 4th edition, to freshen up my long forgotten knowledge of electronics. I love the book so far, but I just came across an equation there which seems incorrect. I don't want to immediately assume that the error is there, so I just want to make sure I'm not missing something.

It's about the transfer function of a loaded RC filter (page 213 if you have the book). This is the circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

The transfer function, according the book is this:

Transfer function according to book

R and RL in parallel

This jump from the formula with the parallel notation to the one with R' didn't make much sense to me, and it's not explained how we get there. After converting the equation, I think there's a 1/R missing.

Here are the steps:

  1. First I expanded and simplified the parallel CR part.

    Enter image description here

  2. Then I plugged that back in the original equation.

    Enter image description here

    Enter image description here

So the same thing, except with an extra 1/R.

I did an other round using the Thevenin equivalent circuit and got the same result.

schematic

simulate this circuit

So if I switch RL and C there and cut the circuit before C as seen above, then the equivalent resistance should be equal to R' and the equivalent voltage should be the voltage across RL.

Enter image description here

schematic

simulate this circuit

So with that:

Enter image description here

And the transfer function is:

Enter image description here

Do you all see any errors in this process? It makes sense to me, but I'm a bit rusty regarding all this, so I'm doubting myself a bit.

\$\endgroup\$
0

3 Answers 3

17
\$\begingroup\$

Standard Form

Over some time, standard forms have been developed for these equations. There's a reason why. But I'll get to that in a moment.

The usual 1st order low-pass standard forms for your case is:

$$G_s=K\frac{\omega_{_0}}{s+\omega_{_0}}=K\frac{1}{1+\frac{s}{\omega_{_0}}}$$

Here, \$K\$ is the gain and \$\omega_{_0}\$ is the cutoff frequency. Either form is often used, though at this point the right form has \$K\$ in exactly one place and \$\omega_{_0}\$ in exactly one place, so that may be nicer. However, when you get to a 2nd order standard form then there becomes a reason to hold more towards the left form. But that's for another time.

Whichever you use, if you set \$\omega_{_0}=1\$ and \$K=1\$ then you get:

$$G_s=\frac1{1+s}$$

If you study this one for all values of \$s=j\,\omega\$ (setting \$\sigma=0\$ to ignore the exponential part and to instead focus only on the frequency behavior), then you know the behavior of all such 1st order low-pass filters. All you have to do is scale the plot for \$\omega_{_0}\$, when you decide to care about it. But you do not have to study it for all \$\omega_{_0}\$ because there's no real difference except a shift. The curves all look identical. (Similarly for \$K\$, which is just a gain factor -- the curves are essentially the same when you study the \$K=1\$ case.)

So your goal is to put your transfer function into standard form.

Your Case

I like the fact that you attempted this several different ways. That's an important practice. Keep that up. I liked your approach using the two resistors as a divider to help simplify the problem. But let me take a somewhat different approach just to add to what you already did.

I prefer to not use \$j\,\omega\$ but to always use \$s\$. It's just less writing. But it is also more general. I can always decide that \$\sigma=0\$ and then it reduces to what you did. But why go to all the trouble of writing out two symbols when one is good enough and is less work?

Let's just treat \$R_\text{L}\$ in parallel with \$C\$ to create \$Z_\text{L}\$. Then we can still use the divider approach. We have \$Z_\text{L}=R_\text{L}\mid\mid C=\frac{R_\text{L}\,\frac1{s\,C}}{R_\text{L}+\frac1{s\,C}}\$. So:

$$\begin{align*} G_s&=\frac{Z_\text{L}}{Z_\text{L}+R}=\frac{\frac{R_\text{L}\,\frac1{s\,C}}{R_\text{L}+\frac1{s\,C}}}{\frac{R_\text{L}\,\frac1{s\,C}}{R_\text{L}+\frac1{s\,C}}+R}\cdot\frac{R_\text{L}+\frac1{s\,C}}{R_\text{L}+\frac1{s\,C}}=\frac{R_\text{L}\,\frac1{s\,C}}{R_\text{L}\,\frac1{s\,C}+R\left(R_\text{L}+\frac1{s\,C}\right)}\\\\ &=\frac{R_\text{L}}{R_\text{L}+R\left(R_\text{L}\,C\,s+1\right)}\cdot\frac{\frac1{R_\text{L}\,R\,C}}{\frac1{R_\text{L}\,R\,C}}=\frac{\frac1{R\,C}}{s+\frac1{C}\left(\frac1{R_\text{L}}+\frac1{R}\right)}\\\\ &=\frac{\frac1{R\,C}}{s+\frac1{\left(R_\text{L}\mid\mid R\right)\,C}} \end{align*}$$

You can set \$\alpha_{_0}=\frac1{R\,C}\$ and \$\omega_{_0}=\frac1{\left(R_\text{L}\mid\mid R\right)\,C}\$ to get \$G_s=\frac{\alpha_{_0}}{s+\omega_{_0}}\$. But \$K\$ is missing there and we'd like to extract that because it is really nice to know what it is. Since we know that \$\alpha_{_0}=K\,\omega_{_0}\$, it follows that \$K=\frac{\alpha_{_0}}{\omega_{_0}}=\frac{R_\text{L}}{R_\text{L}+R}\$, which is just that resistor divider network!

So,

$$G_s=K\frac{\omega_{_0}}{s+\omega_{_0}}=\left[\frac{R_\text{L}}{R_\text{L}+R}\right]\frac{\omega_{_0}}{s+\omega_{_0}}=\left[\frac{R_\text{L}}{R_\text{L}+R}\right]\frac{1}{1+\frac{s}{\omega_{_0}}}$$

You can recover your form (which is definitely not standard) by replacing \$\omega_{_0}\$ and multiplying by \$\frac{R}{R}\$:

$$\begin{align*} G_s&=\left[\frac{R}{R}\right]\left[\frac{R_\text{L}}{R_\text{L}+R}\right]\frac{1}{1+\frac{R_\text{L}\,R}{R_\text{L}+R}\,C\,s}\\\\ &=\left[\frac{1}{R}\right]\left[\frac{R_\text{L}\,R}{R_\text{L}+R}\right]\frac{1}{1+\frac{R_\text{L}\,R}{R_\text{L}+R}\,C\,s}\\\\ &=\left[\frac{1}{R}\right]\frac{R^{'}}{1+R^{'}\,C\,s} \end{align*}$$

So you are right.

But you should get into the practice of putting things into standard form. It's more quickly readable for meaning.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Thank you for this approach! I definitely have to get comfortable with standard forms again, a PID controller design is on the roadmap for later. \$\endgroup\$
    – bnorb
    Commented Feb 12, 2021 at 19:07
8
\$\begingroup\$

The book is wrong. You are right. You can verify by punching in random sets of sets of values into the left-hand side and right-hand side. They are off by a factor of 1/R.

It doesn't even need to be complex values in this case. Just replace jwC with some real integer (i.e., treat C as just a resistor if you have trouble calculating the complex result).

\$\endgroup\$
2
  • 9
    \$\begingroup\$ Even more simple is to look at units. H has to be in per units, but appears to be in Ohms in the book. \$\endgroup\$
    – AlexVB
    Commented Feb 12, 2021 at 15:52
  • \$\begingroup\$ Checking the units is a great idea to quickly verify it! Didn't occur to me now, but I'll definitely utilize that moving forward. \$\endgroup\$
    – bnorb
    Commented Feb 12, 2021 at 18:58
2
\$\begingroup\$

I simplified the transfer function (1) differently and also defined \$ R^\prime \$ (2) differently before substituting. $$ H=\frac{V_{out}}{V_{in}} =\frac{R^\prime}{R+R^\prime} =\frac{1}{1+\frac{R}{R^\prime}} \tag{1} $$

Where: $$ R^\prime=\frac{1}{j\omega C}||R_L =\frac{R_L}{1+j\omega C R_L} =\frac{1}{\frac{1}{R_L}+j\omega C} \tag{2} $$

Therefore: $$ \frac{R}{R^\prime}=\frac{R}{\frac{1}{\frac{1}{R_L}+j\omega C}} =R\left(\frac{1}{R_L}+j\omega C\right) =\frac{R}{R_L}+j\omega CR \tag{3} $$

Substituting for \$ \frac{R}{R^\prime} \$ in (1) gives: $$ H=\frac{1}{1+R\left(\frac{1}{R_L}+j\omega C\right)} =\frac{1}{1+\frac{R}{R_L}+j\omega CR} \tag{4} $$

I hope you find this different approach helpful.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.