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I'm reading through an electronics book to teach myself, and I'm in the section about filters. I've been following along so far, but now I'm confused about how the author came to a conclusion in his math. Here is an excerpt from the book (Practical Electronics for Inventors, 4th edition, page 213): a low-pass filter with a resistive load

He is computing the transfer function for the circuit in the image, and he says that: $$ H=\frac{V_{out}}{V_{in}}=\frac{1/(j\omega C)||R_L}{R+[1/(j\omega C)||R_L]} $$ This makes sense to me, since this is just computing \$V_{out}\$ using the voltage divider equation. The next step is what confuses me, where he says that the equation above is equivalent to: $$ = \frac{R'}{1+j(\omega R'C)}V_{in} \text{ where } R'=R||R_L $$

How do I come to this conclusion? I've tried simplifying the circuit using Thevenin's theorem by combining \$R\$ and \$C\$, which gives me: $$ R_{THEV} = R||\frac{1}{j\omega C} = \frac{R/(j\omega C)}{R+1/(j\omega C)} = \frac{R}{1+j\omega RC} $$

That's pretty close to the author's answer, except I have \$R\$ instead of \$R'\$. I've searched around and haven't found other resources making this leap. I'm a little lost! The book continues making use of this throughout the filters section, so I really want to understand it. Any help here would be greatly appreciated.

Edit: The answer by Paul below solved the problem. The numerator of the equation in red should be \$R'/R\$ and not \$R\$. I expanded both the green equation and the corrected red equation and got a matching result.

Expansion of green equation: $$ H=\frac{V_{out}}{V_{in}}=\frac{1/(j\omega C)||R_L}{R+[1/(j\omega C)||R_L]}V_{in} = \frac{(R_L/(j\omega C))/(R_L + 1/(j\omega C))}{R+(R_L/(j\omega C))/(R_L + 1/(j\omega C))}\\ = \frac{R_L/(R_L j\omega C+1)}{R+R_L/(R_L j\omega C+1)} = \frac{R_L}{R(R_L j\omega C+1)+R_L} = \frac{R_L}{R+R_L+j\omega CRR_L} $$

Expansion of corrected red equation (with \$R'\$ replaced with \$R||R_L\$): $$ \frac{(R||R_L)/R}{1+j\omega (R||R_L)C} = \frac{\frac{RR_L}{R+R_L}/R}{1+j\omega C(\frac{RR_L}{R+R_L})} = \frac{R_L}{(R+R_L)(1+j\omega C\frac{RR_L}{R+R_L})} = \frac{R_L}{R+R_L+j\omega CRR_L} $$

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There is a glitch in the book answer. The expression framed in red should have R'/R in the numerator not R'.

Remember also that Vout /Vin has no unit and the expression framed in red is in ohm.

This wont affect the book result for the cutoff frequency so it makes sense to mention this analogy.

And if you want to apply Thevenin do not forget to replace Vin by Vth also.

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  • \$\begingroup\$ Thank you! This gave me a matching result, and really helps clear things up. I'll edit my original post to include the math for others. \$\endgroup\$
    – Chris
    Jan 28 at 4:38

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