# Center Frequency and Bandwith of Bandpass Filter?

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1. May 15, 2015

### Gwozdzilla

1. The problem statement, all variables and given/known data
Determine the center frequency and bandwidth of the bandpass filter.

2. Relevant equations
center frequency of bandpass filter = ωc occurs when the magnitude of H = 1

H = V0/Vs

Center frequency occurs when Im(Z) = 0
Zc = 1/(jωC) = 1/s
ZL = jωL
ZR = R

I don't know a formula for finding bandwidth.
3. The attempt at a solution
Zseries = 1 + 1/s
Zll = ((1/s)(1 + 1/s))/(2/s + 1)
Zeq = 1 + Zll
Simplifying gives...
Zeq = (1 + 3s + s2)/(2s + s2)

Vs = IZeq
Vll = IZll
V0 = I2(1Ω)
Vseries = I2Zseries = Vll

V0Zseries = Vll = IZll

V0Zseries = (VsZll)/Zeq

V0/Vseries = Zll/(ZeqZseries)

Plugging in values from above and simplifying gives...

V0/Vseries = s/(s + 3s + 1) = H

s2 = -ω2

How do I find the magnitude of H? What is a bandwidth?

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2. May 15, 2015

### rude man

First thing you should do is label the components R1, C1, C2, R2 left to right. Then work with these symbols.

Then: your H(s) can't be right since it has gain = 1/4 at high frequencies whereas the network H(s) must have zero gain there, thanks to C1.

3. May 15, 2015

### BvU

Hello GZ,

Took me a while too to untangle this. In the mean time Rudy chipped in, but here's my 2 cents as well:

It would have been easier for me (and probably for others too) if you would have told us what you mean with the various symbols.

Up to Vseries= ... I agree. Then V0 Zseries=Vll x (1Ω) otherwise the dimensions don't fit. But numerically you're still good.

V0/Vseries = Zll x (1Ω)/(ZeqZseries) is something you could have gotten more straightforward as

V0/V|| = 1Ω / Zseries multiplied with

V||/Vs = Zll / Zeq

The typo that distracted Rudy demonstrates your derivation is a bit complex (too many Z, too many inversions and then inverting again). Never mind.

Magnitude of H follows from $\sqrt{ H^{\displaystyle ^*} H }$ as with all complex numbers.

(At first, LaTeX let me down here, with $H^{\displaystyle ^*}$ I mean the complex conjugate of $H$ )

Bandwith is $| \omega_1 - \omega_2 |$ where $H^2(\omega_{1,2}) = H^2_{max}/2$

Last edited: May 15, 2015
4. May 15, 2015

### Gwozdzilla

So from my value for H = s/(s2 + 3s + 1), I can convert back to using jω:

H = (jω)/(-ω2 + 3jω + 1)
H = (jω)/(3jω + (1-ω2))

The complex conjugate of the denominator is then 3jω - (1-ω2). I can multiply the numerator and denominator by this and simplify to get:
H = ((ω2-1)(jω) - 3ω2)/(-9ω2 -1)

√(H*H) = 1 → H*H = 1

((ω2-1)(jω) - 3ω2)/(-9ω2 -1) = 1
2-1)(jω) - 3ω2 = -9ω2 -1
(6 + j)ω2 - jω + 1 = 0
The quadratic equation gives the result:
ω = -.0219 -.331j

I don't think I did that right. Did I determine the complex conjugate incorrectly?

So for the bandwidth, am I supposed to find dH/dω = 0, and plug that ω back in to find Hmax, then solve Hmax2/2 or am I misunderstanding?

5. May 15, 2015

### rude man

You never performed HH*. When you multiply numerator & denominator of H(jw) by the cc of the denominator you wind up with the original H(jw), not the cc
You compute |H| = √|H(jw) H*(jw)|, which is the magnitude of H(jw) and is a real number.
You can then set d|H|/dw = 0 to solve for w of |H|. Plug into |H| and you get |Hmax|. Then set |H| = 1/√2 Hmax and solve for w again. This time you get two w. Rest is as BvU stated.

Last edited: May 15, 2015
6. May 15, 2015

### Gwozdzilla

How do I find the complex conjugate of something in the form of (jω)/(3jω + (1-ω2))? I don't see how to simplify it into an x + iy form.

Is the complex conjugate (-jω)/(3jω - (1-ω2))?
If this is the complex conjugate then I get...

√[(jω)/(3jω + (1-ω2))][(-jω)/(3jω - (1-ω2))] = 1

[(jω)/(3jω + (1-ω2))][(-jω)/(3jω - (1-ω2))] = 1
Simplifying gives...
ω4 + 8ω2 + 1 = 0 which only has imaginary roots, so it's probably also wrong.

7. May 15, 2015

### BvU

No need to simplify. Just replace every instance of j by $-$j

8. May 15, 2015

### BvU

That's the idea. ($d|H|\over d\omega$ you mean, of course ... ).

9. May 15, 2015

### Gwozdzilla

Okay, so when I replace every instance of j by -j, the complex conjugate is (-jω)/(-3jω + (1-ω2)).

[(jω)/(3jω + (1-ω2))][(-jω)/(-3jω + (1-ω2))] = 12

This simplifies to 0 = ω4 + 6ω2 + 1
so ω = ±j(.414) and ω = ±j(2.414)

Is this part correct?

10. May 15, 2015

### rude man

Why are you setting |H|2 = 1?

11. May 15, 2015

### Gwozdzilla

I thought that a bandpass filter was defined as having its |H(ωc)| = 1. Is this not the case?

12. May 15, 2015

### BvU

Check it out ! You already have an expression for $|H(\omega)|$

13. May 15, 2015

### rude man

No it isn't. You need to go d|H|/dw = 0, solve for w, replace in |H|.

14. May 15, 2015

### Gwozdzilla

|H(ω)| = [(jω)/(3jw + (1-ω2))][(-jω)/(-3jw + (1-ω2))]
|H| = ω2/(7ω2 + ω4 + 1)

d|H|/dω = 2ω(7ω2 + ω4 + 1)-12(7ω2 + ω4 + 1)-2(14ω + 4ω3) = 0
Simplifying gives ω = 1. Is this the center frequency?

|H(1)| = 1/9 Is this the bandwidth?

15. May 15, 2015

### rude man

Yes, if you did the math correctly. I did not check it.
No. |H|(1) = max. gain.
I did not check your math but your approach is correct up to then.
Follow what I said in post 5 near the end to get the bandwidth.

16. May 15, 2015

### Gwozdzilla

|H(ω)| = ω2/(7ω2 + ω4 + 1) = (1/√2)Hmax = 1/(9√2)

Solving for ω gives ω = ±2.355 and ±.425

Bandwidth = B = ω2 - ω1 = 2.78 or 1.93

Is this correct?

17. May 15, 2015

### rude man

ω cannot be negative, at least not for you now.
Otherwise, looks right if you did the math right.
Check #1: does |H| → 0 as ω → 0? Yes.
Check #2. does |H| → 0 as ω → ∞? Yes.
Check #3: does √(ω1ω2) = ω0 = 1? √[(2.355)(0.425)] = 1. Yes. (This test rests on what I state below in my last sentence. So you most probably got the mid-band and corner frequencies right.

EDIT: But it's not sure. What you can say is that if √(ω1ω2) is not = ωmax then you got something wrong. In other words, it's a necessity but not a sufficiency test.)

So, H is a bandpass, since gain is zero at DC (ω=0) and at arbitrarily high frequencies, and gain > 0 somewhere inbetween.
But you don't have the right answer for bandwidth. Bandwidth is the width of the passband around the peak, with rolloff frequencies at gain = |Hmax|/√2 on either side of the max. frequency ω0. If you're using log paper (for the x axis), the two points w1 and w2 are equidistant from w0. ω0 is the geometric mean between ω1 and ω2. So take another shot at bandwidth.

Last edited: May 15, 2015
18. May 15, 2015

### Gwozdzilla

If I know that ω1 and ω2 are correct by calculating that √ω1ω2 = 1, and I know that I should take the positive values of ω1,2, then why isn't the bandwidth = |ω12| = |.425-2.355| = |-1.93| = 1.93?

19. May 15, 2015

### rude man

I said ω is always > 0 but you still got the same right answer: bandwidth = 2.355 - 0.425 = 1.93 rad/s.

BTW look again at my post #17. I edited it a bit.

Last edited: May 15, 2015