# Finding the bandwidth of second order lowpass filter

I have a lowpass filter with the transfer function $$\H(s) = -\frac{3}{3s^2+s+3} \$$. Converting it to standard form yields $$H(s) = -\frac{1}{s^2+\frac{1}{3}s+1}$$ Comparing this to the standard equation $$\\frac{\omega_n^2}{s^2+2\zeta\omega_n+\omega_n^2} \$$ we see that $$\\omega_n = 1 \text{rad/s} \$$ and $$\\zeta=\frac{1}{6} \$$. We can also see that the cut-off frequency is $$\\omega_{cutoff} = 1\text{rad/s} \$$.

I want to find the quality factor and bandwidth, and for that, I have found these two formulas given to us by our instructor (where d is $$\\zeta \$$):-

$$Q=\frac{1}{2\zeta} = \frac{1}{2/6} = 3 = 9.54\text{dB}$$ $$B=\frac{\omega_{cutoff}}{Q} = \frac{1 \text{rad/s}}{3} = \frac{1}{3}\text{rad/s}$$

However, when using MATLAB it tells me that the bandwidth is 1.52

>> H = -tf([0,1],[1,1/3,1])

H =

-1
------------------
s^2 + 0.3333 s + 1

Continuous-time transfer function.

>> bandwidth(H)

ans =

1.5226


Can anyone spot a mistake I have made, and is the formula given by my instructor even valid?

One of the reasons I want to know the cutoff frequency is to draw a straight line approximation of the bode amplitude plot. I believe it should look like this, without the peak because it's an approximation:

According to their online help, bandwidth() returns:

the first frequency where the gain drops below 70.79% (-3 dB) of its DC value

This makes the function put in the same bowl all the 2nd order transfer functions, by treating them, all, for their -3 dB point. I suppose it maks sense in a generic way, however, you should know that a Chebyshev filter, for example, has its passband defined until the end of the ripples, and a Bessel until its natural frequency (e.g. $$\\sqrt{3}\$$ for a normalized transfer function). That's not to say that one cannot define its own bandwidth. After all, a filter is made with requirements in mind, and if those requirements call for a different definition, there's nothing stopping you from considering it so.

However it is, the formula that you have there makes more sense for bandpass/bandstop filters. If you think about it, that formula considers a center frequency and the bandwidth that falls on both sides. Plotting the magnitude of your filter will show a large peak, but the filter is still a lowpass. Which means it makes more sense to consider the -3 dB point than the bandwidth that the peak that it makes.

The phase being half its final asymptotic point is not a definition, but a useful property, for example in this case, where the passband is hardly flat. For lowpass/highpass, the bandwidth should correspond to the corner frequency, but this, as LvW also shows, varies depending on the Q, which makes the -3 dB point vary. To make things more confusing, it's not mandatory to consider the -3 dB point as the limit for the bandwidth. But, since that function needs to have a reference specified, it chose the -3 dB point, so that's what it reports. Otherwise, $$\\omega_c\$$ is what bandwidth() reports, while the $$\\omega_n\$$ is 1.

The conclusion remains: the bandwidth formula is not fit for a lowpass/highpass, but it is for a bandpass/bandstop. The proof is easy, here shown for a bandpass:

$$H(s)=\dfrac{s}{s^2+\dfrac13s+1}\tag{1}$$

A simulator will show these:

m: MAX(mag(v(x)))=(3,0) FROM 0.1 TO 10
f1: mag(v(x))=m/2**.5 AT 0.847039
f2: mag(v(x))=m/2**.5 AT 1.18071
b: f2-f1=(0.333668,0)


where 0.333668 is 1/3 in disguse (roundings, only 101 points/dec, etc), while the mathematical proof is simple enough, too:

\begin{align} |H(j\omega)|^2&=\dfrac{\omega^2}{\Bigl(1-\omega^2\Bigr)^2+\dfrac{\omega^2}{9}}=\dfrac{|H(j\omega_n)|^2}{2} \tag{1} \\ |H(j\omega_n)|^2&=\dfrac{1}{(1-1^2)^2+\dfrac{1}{9}}=9 \tag{2} \\ \dfrac{\omega^2}{\Bigl(1-\omega^2\Bigr)^2+\dfrac{\omega^2}{9}}=\dfrac92\quad &\Rightarrow\quad 9\omega^4-19\omega^2+9=0 \tag{3} \end{align}

This gives you 4 solutions, out of which only two are real positive:

\begin{align}\left\{ \begin{array}{x} \omega_{1,2}=\pm 1.1805 \\ \omega_{3,4}=\pm 0.84713 \end{array}\right .\tag{4} \end{align}

and these correspond to the readings you see above, and also with the formula you give, but this became true only when the transfer function became a bandpass. It can be easily shown that it's the same for bandstop. Also, the way you want to represent the magnitude with a piecewise function is not applicable given the variation of $$\\omega_c\$$ with Q.

• I know the -3dB point definition for the cutoff frequency as well. But isn't there also a definition that for a second order filter, when the phase is shifted 90 degrees that is the cut-off frequency?
– Carl
Commented Apr 24, 2021 at 16:01
• I've updated my answer, see if it helps. Commented Apr 24, 2021 at 18:25
• Thank you for your great elaboration. It makes me wonder why there isn't an absolute definition for cut-off frequency. Didn't people primarily use the information about cut-off frequency to draw straight line approximations of bode plots, back in the day before software could do it for us?
– Carl
Commented Apr 24, 2021 at 19:03
• @Carl No, they actually computed those. Don't forget that handbooks and tables for functions existed for more than the age of computers. How do you think they drew those graphs? Try plotting your piecewise plot next to the real graph. Here is one version, see how large the errors are. This is not the kind of response that can be approximated with lines. Sure, you could, but the result will obscure the true response so much, that it would not be a valid choice. Commented Apr 24, 2021 at 19:49
• Carl, some general remarks about definitions for lowpass cut-off: (1) For 1st-order filters and for all filters with a BUTTERWORTH response there is a commonly agreed defintion: 3dB below the value at DC (for BUTTWORTH of 2nd order only, this is identical with -90 deg phase shift). (2) For other responses (Chebyshev, Cauer,...) there is another agreement: End of passband is defined by the allowed ripple within the passband. That means: At a frequency where the magnitude assumes (to the last time) the value for DC.
– LvW
Commented Apr 25, 2021 at 8:27

From the given transfer function we can derive (as you did):

• pole frequency wp=1 rad/s and pole quality factor Qp=1/2d=3 (d=1/6).

• The transfer function belongs to a second-order Chebyshev lowpass having a large peak in the passband (Amax=Qp/SQRT(1-1/4Qp²)=3.04 ...9.66 dB). Note that for such lowpass functions the pole frequency is not identical to the cut-off frequency. This applies to first order or second order Butterworth functions only.

• For Chebyshev functions the cut-off is defined (in most cases) at a frequency where the magnitude assumes again the value as is available for w=0.

• The last equation (marked in yellow) is applicable for bandpass responses only (wc is the mid-frequency).

• I am not assuming that $\omega_c = \omega_n$ but the transfer function with s = jw becomes $H(jw) = -\frac{1}{\Big(\frac{jw}{1}\Big)^2+ \frac{1}{6} \frac{jw}{1}+1}$ and I can read the cutoff frequency as the denominator of the jw-fractions (so 1rad/s). One of the reasons I want the cutoff frequency is to draw a straight line approximation of the amplitude bode plot.
– Carl
Commented Apr 24, 2021 at 16:11
• Carl, it is not correct that one can "read the cut-off frequency" directly from the denominator. As I have mentioned, this is possible for 1st-order or 2nd-order Butterworth responses only!
– LvW
Commented Apr 24, 2021 at 17:28

The Q for a LPF only extends the BW on the high side, whereas that formula is correct for a BPF which extends. On both sides equally for the full -3dB half power BW.

It is not exactly 1/2 of the BW but close for a LPF.

Well, the thing you need to solve is:

$$\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\left|\underline{\mathscr{H}}\left(\text{j}\omega_0\right)\right|\cdot\frac{1}{\sqrt{2}}\space\Longleftrightarrow\space\omega=\dots\tag1$$

Using that for your circuit, we first find:

$$\frac{\partial\left|\underline{\mathscr{H}}\left(\text{j}\omega_0\right)\right|}{\partial\omega_0}=0\space\Longrightarrow\space\omega_0=\frac{1}{3}\sqrt{\frac{17}{2}}\tag2$$

So, we also know:

$$\left|\underline{\mathscr{H}}\left(\text{j}\omega_0\right)\right|=\frac{18}{\sqrt{35}}\tag3$$

So:

$$\frac{3}{\sqrt{\omega^2\left(9\omega-17\right)+9}}=\frac{18}{\sqrt{35}}\cdot\frac{1}{\sqrt{2}}\space\Longrightarrow\space\omega=\frac{1}{3}\sqrt{\frac{17\pm\sqrt{35}}{2}}\tag4$$

And the bandwidth is given by:

$$\mathscr{B}=\frac{1}{3}\sqrt{\frac{17+\sqrt{35}}{2}}-\frac{1}{3}\sqrt{\frac{17-\sqrt{35}}{2}}=\frac{\sqrt{17-\sqrt{254}}}{3}\approx0.343612\space\text{rad/sec}\tag5$$