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I have this circuit here which is a low pass chebyshev filter, I have been asked to find it's cutoff frequency. I have used matlab to plot an output response against angular frequency graph here. But I'm unsure of how to find the cut off frequency from this, is there an equation to do this? enter image description here

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  • \$\begingroup\$ Do you know the exact value of the passband ripple? (This will help in finding the cut-off frequency). \$\endgroup\$ – Matt L. Mar 9 '16 at 18:35
  • \$\begingroup\$ How are you sure that's a Chebyshev?? \$\endgroup\$ – Scott Seidman Mar 9 '16 at 21:44
  • \$\begingroup\$ I believe the curve that has been plotted matches the shape of a Chebyshev. Also I don't know the passband ripple, is there a way of calculating this? \$\endgroup\$ – ohkneel Mar 9 '16 at 21:47
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The usual definition of the cut-off frequency of a (type I) Chebyshev filter is shown in the figure below:

enter image description here

The common practice of defining the cutoff frequency at −3 dB is usually not applied to Chebyshev filters; instead the cutoff is taken as the point at which the gain falls to the value of the ripple for the final time.

Knowing the characteristics of a Chebyshev filter helps in computing the cut-off frequency (as defined above) without explicitly solving the equation \$|H(j\omega_0)|=c\$, where the constant \$c\$ is chosen according to the definition of the cut-off frequency.

The squared magnitude of the frequency response of an \$n^{th}\$ order type I Chebyshev filter is given by

$$|H(j\omega)|^2=\frac{1}{1+\epsilon^2T^2_n(\frac{\omega}{\omega_0})}\tag{1}$$

where \$T_n(\omega)\$ is the \$n^{th}\$-order Chebyshev polynomial of the first kind, \$\omega_0\$ is the cut-off frequency as defined above, and the constant \$\epsilon\$ specifies the pass band ripple, as shown in above figure. You should know the exact value of \$\epsilon\$ from your design specifications, but I can estimate it from your figure: \$\epsilon\approx 0.23403\$ (note that you need to take into account that the maximum of your transfer function is \$\frac12\$ instead of \$1\$, so the smallest (linear) pass band value is given by \$0.5/\sqrt{1+\epsilon^2}\$).

In order to find \$\omega_0\$ we need to compare the expression for the actual transfer function to the one given by (1). It's a basic exercise to show that the transfer function of your filter is

$$H(s)=\frac{\frac12}{\frac{L^2C}{2R}s^3+LCs^2+(\frac{L}{R}+\frac{RC}{2})s+1}\tag{2}$$

Knowing that \$T_3(x)\$ is given by

$$T_3(x)=4x^3-3x\tag{3}$$

we can compare the factors of the highest power of \$\omega\$ (which is \$\omega^6\$) of the denominators of (1) and of the squared magnitude of (2) for \$s=j\omega\$:

$$\frac{16\epsilon^2}{\omega_0^6}=\left(\frac{L^2C}{2R}\right)^2\tag{4}$$

From (4) \$\omega_0\$ can be expressed as

$$\omega_0=\sqrt[3]{\frac{8R\epsilon}{L^2C}}\approx 3829.7\text{ rad/s}\tag{5}$$

where I've used the approximate value of \$\epsilon\$ given above.

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It doesn't matter how the filter was designed. Once you have a specific design, you write down the equation for its frequency response, and solve for the frequency at which that response drops to half power (-3 dB), which is the definition of cutoff frequency.

Matlab is very good at finding the roots of polynomial equations.

If you don't need a lot of precision, just read the value off the graph. Since the peak voltage response is 0.5, look for the point at which the curve crosses the 0.35 level (0.5/√2), which is just shy of 5000 rad/s.

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  • \$\begingroup\$ Thank you very much! unfortunately I do need high accuracy. So I will be needing to calculate the frequency response. The input is just sin(wt) how would I form an equation for it's frequency response? \$\endgroup\$ – ohkneel Mar 9 '16 at 17:33
  • \$\begingroup\$ Is this a homework question? If so, you need to show us what you do understand about the circuit and how to analyze it, and what specific point you're stuck on. The Q and A format used here is not suitable for providing you a complete education in circuit analysis. It should have been covered before you were given this problem. \$\endgroup\$ – Dave Tweed Mar 9 '16 at 17:38
  • \$\begingroup\$ yes homework question, and everything I have been given is in this post. Sorry I can't be of anymore help on this. Thanks for the advice \$\endgroup\$ – ohkneel Mar 9 '16 at 17:56
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    \$\begingroup\$ This is not how the cut-off frequency is generally defined for Chebyshev filters: "the cutoff is taken as the point at which the gain falls to the value of the ripple for the final time" (from here). \$\endgroup\$ – Matt L. Mar 9 '16 at 18:32
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    \$\begingroup\$ @MattL. that isn't 100% true - if the passband ripple is less than 3 dB then the proper 3 dB point is chosen. If pass band ripple is greater than 3 dB then you choose the final excusion point as the pass band definition. Most designs I've ever done focus on low ripple hence a real 3 dB point definition. \$\endgroup\$ – Andy aka Mar 9 '16 at 19:04
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The cut-off frequency is slightly below 5 kHz: -

enter image description here

That's the point where the output signal falls to the half power point i.e. 3 dB below the input signal. If you need a more accurate answer then do the math on the filter and find the half power point that way.

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  • \$\begingroup\$ For Chebyshev filters (type I) the cut-off frequency is commonly defined in a different way. See my comment under Dave Tweed's answer. \$\endgroup\$ – Matt L. Mar 9 '16 at 18:34
  • \$\begingroup\$ See this for an explanation that both methods are valid (depending on the amount of pass band ripple: matheonics.com/Tutorials/Chebyshev.html#Paragraph_4.2 - go to the section entitled Natural cutoff frequency determination \$\endgroup\$ – Andy aka Mar 9 '16 at 19:02

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