# Quick bandwidth question

Discussion in 'Homework Help' started by blah2222, Jun 7, 2011.

1. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
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Hello, I just have a quick question about finding the bandwidth of a circuit if its frequency response is known.

$H(s) = \frac{s}{s^2 + 3s + 1}$

From this I know that the center frequency is going to be 1 rad/s due to the quadratic pole, but I have no idea how to get the bandwidth of this response.

Thanks,
J

2. ### steveb Senior Member

Jul 3, 2008
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For reasonable high-Q filters (meaning bandwidth is small compared to center frequency), the bandwidth in rad/s is given by the coefficient of "s" in the denominator. This attachment may help. Note that I"m pretty sure that this is an approximation, but I'd have to work out the math to be 100% sure. But either way, this is a useful rule of thumb, and it does seem to agree with your answer.

• ###### Bandpass.pdf
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Last edited: Jun 9, 2011
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3. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I did a Bode plot and got this result which may be of interest ...

• ###### Bode_Plot.jpg
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4. ### steveb Senior Member

Jul 3, 2008
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So, I think this reveals that the above rule is, in fact, an approximation. In this case bandwidth is not small compared to center frequency, so we can see some error in the precise rule. It still seems to work as a rule of thumb, but it not precisely true. I do think it is exact in the limit as Q goes to infinity, but again I'd have to double check that to be sure.

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5. ### hgmjr Moderator

Jan 28, 2005
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It is my understanding that you can normalize your expression and get the gain of the circuit described by your expression.

$H(s) = \frac{s}{s^2 + 3s + 1}$

Multply the numerator by the unity factor, $\ \Large \frac{3}{3}$

$H(s) = \frac{\large \frac{3}{3}s}{s^2 + 3s + 1}$

Then rearrange the expression as follows.

$H(s) = \large \frac{1}{3}\ * \frac{3s}{s^2 + 3s + 1}$

Based on this calculation, the circuit gain at the center frequency of this bandpass filter is 1/3.

hgmjr

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6. ### steveb Senior Member

Jul 3, 2008
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Yep, I agree with that also. It looks correct on paper and seems to agree with t_n_k's plot too.

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7. ### hgmjr Moderator

Jan 28, 2005
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The technique I have used here has been helpful to me over my career. It is straightforward for a bandpass filter that is second order. This technique is not as easy to apply to bandpass filters of order greater than two.

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8. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
565
33
Great, thanks a lot for the replies guys, I really appreciate it!

One thing that still eludes me from this equation:

$H(s) = \frac{Bs}{s^2 + Bs + w_{0}^2}$

Say I am using a series RLC circuit with the output taken across R, this forms a simple passive bandpass filter. So solving for the response we get:

$H(w) = \frac{R}{R + jwL + \frac{1}{jwC}}$

$= \frac{jwRC}{jwRC + LC(jw)^2 + 1}$

$s = jw$

$= \frac{sRC}{s^2LC + sRC + 1}$

$= \frac{sRC}{s^2 + \frac{R}{L}s + \frac{1}{LC}}$

In this case, B = RC or R/L, and it seems to violate that rule of thumb equation stated above.

EDIT*** Oh, do you have to normalize it and then you get the gain factor?

$H(s) = LC \frac{\frac{R}{L}s}{s^2 + \frac{R}{L}s + \frac{1}{LC}}$

I am using values given in my textbook for a series RLC bandpass:

R = 2k, L = 40mH, and C = 1uF

Therefore B = R/L = 2k/40m = 50 krad/s AND w0 = 1/sqrt(LC) = 1/sqrt(1u*40m) = 5 krad/s

Or even the response in the very first post I made. B = 3 rad/s and w0 = 1 rad/s.

My question here is, how can you have a bandwidth greater than the center frequency, ie: a quality factor less than 1? Does this just mean that all frequencies lower than the center frequency will be passed and all signals up to w0 + B/2 will be passed, and the rest are filtered out?

Last edited: Jun 8, 2011
9. ### hgmjr Moderator

Jan 28, 2005
9,030
214
You have an error in going from the first equation above to the second equation above. You divided the terms in the denominator by LC but you neglected to divide the lone term in the numerator by the same factor. The second expression above should be

$= \frac{\frac{R}{L}s}{s^2 + \frac{R}{L}s + \frac{1}{LC}}$

You will notice that the expression is already normalized since the coefficient of the middle s-term in the numerator is already the same as the coefficient of the middle s-term in the denominator. For this particular transfer function, the gain of the circuit at the center frequency is 1.

hgmjr

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10. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
565
33
You're right, how silly of me haha, completely forgot to touch the numerator. Thanks for catching that!

Also, just going back to my second question: How can there be a bandwidth that is larger than the center frequency?

11. ### steveb Senior Member

Jul 3, 2008
2,433
469
There is no problem with that. The bandwidth just defines the range where the voltage response is 0.707 times the peak value. If the center is at 1, the low limit might be at 0.5 and the high limit might be at 5, hypothetically. Here you have a bandwidth of 4.5, but the center is at 1.

In other words the shape is not symmetrical on a linear scale. It looks symmetrical on a logarithmic (Bode) type of plot, but such plots don't show zero frequency, so there is no danger of hitting zero, no matter how wide the bandwidth. There is basically infinte space on both sides of the center frequency.

12. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
565
33
Ah, that make sense, I forgot I was dealing with a logarithmic plot haha thanks!

So, basically as Q decreases, the more asymmetric the passband will become and be less selective, and vice-versa for increasing Q.

13. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Steveb,

In your diagram in post #2 of this thread. I believe that the bandwidth intersection is at the magnitude Ao/√2 rather than Ao*√2. Isn't that correct?

hgmjr

14. ### steveb Senior Member

Jul 3, 2008
2,433
469
You are correct, sir!
Thank you for the correction.

I'll update the file a little later.

15. ### steveb Senior Member

Jul 3, 2008
2,433
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Note that I updated the above mentioned file to fix the sqrt(2) factor. I also bit the bullet, and finally derived out the correct bandwidth formulas. I've been meaning to do this out of my own curiosity, and this seems the appropriate time since it answers a question posed here.

Here I am attaching a derivation and formulas to calculate the bandwidth at the precise 3 dB down points even when the Q is low. The formula gives cutoff frequencies of 0.303 rad/s and 3.303 rad/s (or 0.0482 Hz and 0.526 Hz), for the example in this thread, which agrees with the plot that t_n_k posted previously.

Also, in this document is a derivation of the approximate formula for the case of high Q, which means that bandwidth is much less than center frequency. Here the following is confirmed:

1. That in the limit of high Q, the bandwidth (in rad/s) equals the coefficient of s in the denominator, if the standard form is used. Or, bandwidth is Wb if H(s)=Ao*Wb*s/(s^2+Wb*s+Wo^2).

2. That in the limit of high Q, the upper and lower cutoff frequencies are an equal distance from the center frequency. Or, cutoffs are Wo-Wb/2 and Wo+Wb/2 for the lower and upper limits of the band (at the 3 dB down points).

• ###### Bandpass2.pdf
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16. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
565
33
Wow, thanks for posting that derivation! I've gained a much better understanding from everyone's posts.

J

17. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Like Steveb, I set myself the challenge to derive the answer to blah2222's question on how to calculate the bandwidth of his second-order bandpass filter.

I started my analysis with the normalized portion of the transfer function since it appeared to me that the constant portion of the transfer function had no impact on the bandwidth calculation.

$H(s)\ =\ \frac{3s}{s^2\ +\ 3s\ +\ 1}$

My first step was to form the expression for the magnitude of the transfer function as seen below.

$\left|H(s)\left|\ =\ \frac{\sqrt{(3s)^2}}{\sqrt{(1\ -\ s^2)^2\ +\ (3s)^2}}$

My next step was to set the magnitude expression equal to 1/√2.

$\frac{\sqrt{(3s)^2}}{\sqrt{(1\ -\ s^2)^2\ +\ (3s)^2}}\ =\ \frac{1}{sqrt{2}}$

My next step was to simplify the above expression by squaring the numerator and the denominator on both sides of the equal sign.

$\frac{(3s)^2}{{(1\ -\ s^2)^2\ +\ (3s)^2}}\ =\Large \ \frac{1}{{2}}$

My next step was to expand the squared terms.

$\frac{9\omega^2}{{(\omega^4\ -\ 2\omega^2\ +\ 1)\ +\ (9\omega^2)}}\ =\Large \ \frac{1}{{2}}$

Next I carried out the cross-multiplication to begin to get the expression into a more manageable form.

$(\omega^4\ +\ 7\omega^2\ +\ 1) \ =\ 18\omega^2$

Futher simplification involved collecting terms over on the left-hand side of the expression so that I would have the more familiar form of the polynomial expression set equal to zero. Notice that the order of the expression is four.

$(\omega^4\ -\ 11\omega^2\ +\ 1) \ =\ 0$

I then observed that I could use variable substitution to get the equation into a more manageable second order expression that I knew how to tackle.

Let $\omega^'\ =\ \omega^2$

then the equation becomes:

$(\omega^'^2\ -\ 11\omega^'\ +\ 1)\ =\ 0$

Using the time-honored quadratic formula it is possible to calculate the two solutions to the above equation. I will let the reader figure out the rest of the manipulations since it is nothing more than applying the quadratic equation and turning the algebraic crank.

$\omega^'_h\ =\ \frac{11\ +\ sqrt{11^2\ -\ 4}}{2}\ =\ 10.908326913$

$\omega^'_l\ =\ \frac{11\ -\ sqrt{11^2\ -\ 4}}{2}\ =\ 0.091673087$

$\omega^'\ =\ \omega^2\$ therefore $\ \omega\ =\ sqrt{\omega^'}$

$Bandwidth\ =\ sqrt{\omega^'_h}\ -\ sqrt{\omega^'_l}\ =\ sqrt{10.908326913}\ -\ sqrt{0.091673087}\ =\ 3.302775638\ -\ 0.302775638\ = \ 3\ rads/sec$

Thankfully, all of that effort yielded the answer that was consistent with the one mentioned in the initial post in this thread. I will not pretend that this was easy and lie and tell you that I sorted this all out on the back of a napkin while eating at a local restaurant. I went through several attempts and used a lot of paper, ink, and pencil lead to get to what I felt was accurate enough to present here.

phew!!!
hgmjr

Last edited: Jun 9, 2011
18. ### steveb Senior Member

Jul 3, 2008
2,433
469
Well, thank you for making that good point! Once again I managed to miss the forest for the trees.

So, the bandwidth is exactly equal to Wb. The thing that is not exact is the claim that lower limit is exactly Wo-Wb/2 and upper limit Wo+Wb/2. Instead it is asymmetrical around Wo. It becomes nearly symmetrical in the case of high Q.

OK, so that leaves me with a little identity to prove tonight, so that I can add it to my derivation. That complicated formula for $\beta$ should be equal to Wb if I continue the algebra. I don't blame myself for not seeing it in the formula, because it's so complex looking. But, my God, I should have seen it from the numbers I quoted ... 3.303 and 0.303. Clearly the difference is 3 !

19. ### steveb Senior Member

Jul 3, 2008
2,433
469
OK, so please add the attached page to the previous derivation. This shows symbolically exactly hgmjr's point. The answer of 3 rad/s in this case, and more generally $\omega_B$ is the correct and exact bandwidth.

The only thing that is an approximation for high Q is the idea that the 3 dB points are ${{\omega_B}\over{2}}$ away from the center. In general, it's not symmetrical. For a low Q system, the exact cutoff frequencies can be found with the formula.

$\omega_{H,L}=\sqrt{\gamma\pm \sqrt{\gamma^2-\omega_o^4}}$

where

$\gamma=\omega_o^2+{{\omega_B^2}\over{2}}$

• ###### Bandpass3.pdf
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