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Is there a difference between a band-pass filter (BPF) and tying a low-pass filter's (LPF) output to the input of a high-pass filter (HPF)?

It seems to me that they should do the same thing---or perhaps they are algebraically equiveilent (are they?). If not, what is the benefit of using a BPF instead of a serial LPF+HPF?

For example:

Serial LPF+HPF:

Serial High-pass+Low-pass filter

Is that effectively the same as a BPF?

Band-pass filter

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    \$\begingroup\$ Yes. There is a time where you'd chose one approach and a time where you'd chose another. I have another post elsewhere here on the topic. That may be enough for you. But if not, just say so and I'll write something up for you. \$\endgroup\$
    – jonk
    Jul 9, 2020 at 4:14
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    \$\begingroup\$ No. For radios you would not consider these equivalents due to the difference in input and output impedance. \$\endgroup\$ Jul 9, 2020 at 4:30

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Your intuition is correct, the 2nd is derived from the first through frequency transformations applied to the transfer function. Usually, the first step is to calculate the lowpass prototype, which has unity, non-scaled frequency, and which has the form:

$$H(s)=\frac{\omega^2}{s^2+\frac{\omega}{Q}s+\omega^2}\tag{1}$$

This form is reached after calculating the roots of the filter. Transforming this into a bandpass is done by subsituting \$s\$ into the real-valued transfer function. For simplicity, let's consider all the terms \$1\$ (\$BW=\omega_2-\omega_1\$ is the bandwidth, \$\omega_0=\sqrt{\omega_1\omega_2}\$ is the center frequency):

$$\begin{align} H_{LP}(s)&=\frac{1}{s^2+s+1}\tag{2} \\ H_{BP}(s)&=H_{LP}(s)\biggr|_{s\to\frac{s^2+\omega_0^2}{BW\,s}} \\ &=\frac{1}{\left(\frac{s^2+\omega_0^2}{BW\,s}\right)^2+\left(\frac{s^2+\omega_0^2}{BW\,s}\right)+1} \\ &=\frac{BW^2s^2}{s^4+BW\,s^3+(BW^2+2\omega_0^2)s^2+BW\omega_0^2\,s+\omega_0^4}\tag{3} \end{align}$$

Equation \$(3)\$ gives a new set of roots which can be used to group \$(3)\$ into two transfer functions:

$$\begin{align} H_{BP}(s)&=H_{LP}(s)\cdot H_{HP}(s) \\ &=\frac{a_0}{s^2+b_1s+b_0}\cdot\frac{s^2}{s^2+c_1s+c_0} \end{align}$$

which are nothing but a lowpass and a highpass. The connection to the topology you are showing is that \$s\$ represents one reactive element (depends whether it's series or parallel, normally \$s=L\$ and \$\frac1s=C\$), while \$\frac{s^2+\omega_0^2}{BW\,s}\$ represents a resonant bandpass \$LC\$: if it's in series with the load then it's a series \$LC\$ (G1 and G3 in your 2nd picture), if it's shunt then it's a parallel \$LC\$ (G2).

For notch lowpass/highpass filters, the resultants will be the same combination of two lowpass/highpass filters, but each of them notch.

As @Tony Stewart shows, you cannot calculate the two resultant lowpass/highpass filters and build them as separate filters, as in your 1st picture, since 1) as @jonk links in the comments, there is a loading effect, and 2) these two resulant filters are meant to be used in their composed form (your 2nd picture).

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  • \$\begingroup\$ They are not algebraically equivalent as shown. With no load the top reduces to a 5th order and the bottom to a 4th order polynomial, yet with load may be similar for s21with a 6th order polynomial. \$\endgroup\$ Jul 10, 2020 at 16:20
  • \$\begingroup\$ Great answer, I'm still wrapping my head around the math. Thanks! \$\endgroup\$
    – KJ7LNW
    Jul 10, 2020 at 18:40
  • \$\begingroup\$ @TonyStewartSunnyskyguyEE75 That's part of the conclusion in the last paragraph. \$\endgroup\$ Jul 10, 2020 at 18:51
  • \$\begingroup\$ It wasn’t clear from the math that they are not equivalent. But they can be loaded and chosen to get similar but not identical results. \$\endgroup\$ Jul 10, 2020 at 18:59
  • \$\begingroup\$ @TonyStewartSunnyskyguyEE75 Sure, provided the loading effect is minimized (which loses power), but why would anyone complicate like that when the tried-and-tested method is OP's 2nd picture? :-) \$\endgroup\$ Jul 10, 2020 at 19:01
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rough Simulation You need a load R to obtain an impedance ratio on the output and so looking at the s11 s22 impedance of the in,out , they are different, but similar but not the same s21 transfer function.

This configuration is sub-optimal for a BPF. Here a double tuned 3~8.5 MHz BPF.

The left and right images just switch the output response. enter image description here

Below is a narrow BPF 8th order Cheb. Filter Chosen here for 50 Ohm input and load impedance using the Large passive Filter option.

enter image description here

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  • \$\begingroup\$ Thanks for the graphs, it is neat to see the response of the two filters in an example. \$\endgroup\$
    – KJ7LNW
    Jul 10, 2020 at 18:40

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