I'm solving the following problem:

Consider the design of a Chebyshev lowpass filter with 3 dB DC gain, a passband limited to 1 kHz with a maximum ripple of 3 dB, and a stopband characterized by AS = 60 dB and fS = 4 kHz. Show that the necessary filter order is n = 4.

Ok so we have $$f_p= 1 kHz $$ $$ A_p=3 dB $$

with this we can calculate the filter \$ \epsilon\$

$$ \epsilon = \sqrt{10^{0.1A_p}-1} = 0.998$$

which we will round up to one. So \$ \epsilon = 1\$.

Now my professor told me that now we have to consider two different situations for n even and n odd.

For n odd (ripple starts at the DC gain), we have to consider:

$$f_s= 4 kHz $$ $$ A_s=63 dB $$


We obtain


So, we would take n=5 (the next odd value).

For n even (ripple starts at the DC gain + ripple value):

$$f_s= 4 kHz $$ $$ A_s=66 dB $$

We obtain


Now here we have a problem: Should I round up this to n=4, or should I consider the next n value, n=6?

Because now my professor told us to take the least value of n (the minimum value of n that works): if we consider n=6 for the even case then the order of the filter should be n=5. Otherwise it should be n=4? Or am I rounding up the value wrong: I should round up the value of n odd to n=4 and the value of n even to n=5 and therefore considering n=4?

Can someone help me?

Thank you!


1 Answer 1


Your calculations are right, only your presumption is off: you are required to have a 3 dB gain, which means passive filters are out of the question, which also means you are no longer restricted to odd orders, because only passive filters suffer from attenuation at DC. If it's an active filter, the DC will have a 3 dB amplification, the ripple will contribute with an additional 3 dB, and the attenuation is considered (unless specified) at max peak minus ripple, thus +3 dB. The solution requires you to prove that n=4, so you got it right when you calculated that \$n\ge3.85\$. Congratulations!

A quick check proves it -- the amplification is 3 dB (cursor 1 is at DC), the same cursor shows that fc is at 1 kHz and has the same amplification, 3 dB, and the 2nd cursor shows -60 dB attenuation at 4 kHz which, added with the 3 dB in the passband, make the almost 63 dB difference that is shown:


There isn't much else that can be said, except to clear out the possible confusion regarding the difference between the gain of the passive and active filters.

For one, if it's passive, the DC response can never be 0 dB or higher (well, technically it can be, but that's more of an exception), and that makes sense since there are an input and an output impedance which, considering the filter lossless, would mean that the highest output voltage would be the one given by \$\frac{Z_{out}}{Z_{in}+Z_{out}}\$.

However, in the case of the filters with passband ripple, such as Chebyshev, there is a peculiarity at even orders, because the transfer function causes the alternances of the passband ripple to start at the DC value of \$1-\delta_p\$ (relative to the maximum in the passband). This also means that the output impedance must always be less than the input. There is a whole passive filter synthesis that I won't repeat here (@jonk also has a very nice document linked in the comments), but I'll say that the load needs to be the input divided by \$\small\tanh\left[\frac12\mathrm{asinh}\left(\frac{1}{\epsilon_p}\right)\right]^2\$, if the load is fed by a series inductor, or the reciprocal if the load is fed by a parallel capacitor (a sweep for orders from 2 to 7):


Nothing will blow up if you'll use a different value, but the response will not be Chebyshev, anymore. For example, calculating for, and using unity I/O impedances:


Active filters can have any DC level and, thus, they can compensate for the loss of the passive filters. The attenuation at the corner frequency now will shift to \$1+\delta_p\$ for even orders, meaning the ripples will be over unity, and that the problem of compensating for the ripples is now inverted:


This is the case for your problem, where you need an active filter in order to have an over-unity gain, and thus choose an even order. If your teacher told you that you should compare the results with an odd order, then you should account for the \$1+\delta_p\$.

I'll add a final note. Suppose you actually end up with \$n=4.01\$ (for your pronlem, as shown, it's not the case), but if you do, then that requires a choice:

  • if the design is strict, that is, the minimum attenuation must be preserved at all costs, then there must be a rounding up
  • if, however, the tolerances account for a slight amplification that can result in rounding down (there is, after all, only a tiny fraction in there that's extra), then the lesse order can be chosen.

Mathematically, and since this is an exercise, you would need to use the rounding up (unless your teacher says it can be otherwise). To what order? That, again, depends on the constraints. Here, it would have been 5, since there is no passive filter.

  • 2
    \$\begingroup\$ Simple and direct. +1. I always like to see more written, so that others reading the question and answer, but who otherwise are somewhat ignorant of Chebyshev/Tschebysheff (and its varieties with respect, for example, to a passband tolerance factor) might learn to have more of an interest and a better sense. Here, all they can tell is that there's an idea with a name to it and some math and that the OP got it right. But they could walk away otherwise none the wiser and with perhaps no further interest. \$\endgroup\$
    – jonk
    Commented Aug 30, 2020 at 21:40
  • 1
    \$\begingroup\$ @jonk Thank you. I would have added more but, honestly, OP has already put on the table what was necessary. Re-reading your comment, you're right, I'll add a few things, but tomorrow, now it's almost the end of the day, I'll let the neuron rest. \$\endgroup\$ Commented Aug 30, 2020 at 21:43
  • 1
    \$\begingroup\$ Agreed. So I'll just add this from Louis Weinberg. I was really looking for a ready copy of J G Linvill's earlier July 1950 publication, but couldn't find it on the web, sadly. If I find a copy of it, I may add it here. \$\endgroup\$
    – jonk
    Commented Aug 30, 2020 at 21:47
  • 1
    \$\begingroup\$ @LvW And that's what I would normally do: specify both Ap and As as a function of the filter's general response (thus independent of the gain), values which would come from the general requirements. But here I interpreted As to be the attenuation at 4 kHz (as in: the output must be so, not the filter must have so) because of the way the teacher's question is written, and because OP seems to account for the gain, too. I could be wrong, of course, feel free to blame my English, but whether it is so, or not, only OP can tell at this point. I'll gladly modify, accordingly. \$\endgroup\$ Commented Aug 31, 2020 at 11:33
  • 1
    \$\begingroup\$ @a concerned citizen..I do completely agree with you. Sorry, I have forgotten to make clear that my comment was primarily directed to the OP and NOT to you and your fine answer. \$\endgroup\$
    – LvW
    Commented Aug 31, 2020 at 12:08

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