# building a bandpass filter with ...

Discussion in 'Homework Help' started by 1233, Dec 26, 2009.

1. ### 1233 Thread Starter New Member

Dec 26, 2009
10
0
I am asked to design a bandpass filter give the selectivity, the center frequency, and the passband gain.

Q= 10
fo = 5MHz
Passband gain Ko=1

Edit: using cascaded 1st order stages.

I don't know where to start from.

Last edited: Dec 26, 2009
2. ### ABoul Member

Mar 30, 2009
15
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i'm in a bit of a hurry so this is the best i can do: http://en.wikipedia.org/wiki/Q_factor

go under the section titled "Physical interpretation of Q" (more specifically, the response equation, noting that s = jw) and the first part of the next section. if you play about with the equations, you should get values for R, L and C.

EDIT: by inspection, LC = (2 * pi * 5000000)^2.

3. ### 1233 Thread Starter New Member

Dec 26, 2009
10
0
thx, I will have a look at that

edit : forgot to mention that I should use cascaded 1st order stages.

Last edited: Dec 26, 2009
4. ### 1233 Thread Starter New Member

Dec 26, 2009
10
0
Is a simple LCR circuit considered a cascaded circuit ?

Apr 5, 2008
15,534
2,301
Hello,

With what parts may you make the filter ?
Take a look at this PDF for filter design.
sloa96_filter_design.pdf

Greetings,
Bertus

6. ### 1233 Thread Starter New Member

Dec 26, 2009
10
0
I think I should build the filter using inductor, capacitor and resistors only. I am sure that it is not so difficult, but I don't know where to start from

Apr 5, 2008
15,534
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8. ### Nanophotonics Active Member

Apr 2, 2009
365
3
It's OK to design filters using RLC, but I think it's not practical to build it. The range of values available for L is limited and there are various other disadvantages. I would probably go for opamps.

9. ### PRS Well-Known Member

Aug 24, 2008
989
35
Start by considering the bandwidth: BW = fc/Q

This, obviously is 500kHz
So your f(low is 4.5MHz) and f(high)=5.5MHz.

With the instructions to use "stages" I'd guess you're allowed to use transistors for amplification in order to offset the attenuation of the filter.

A single common emmiter stage should do.
Use the input capacitor to obtain f(low) and at the output put a capacitor in parallel with the load to get your f(high). To calculate the C use the input resistance in series with the source resistancefor fL and and the output resistance in parallel with the load resistance for fH. Put these in the equation C = 1/(2*Pi*R*f).

At least, this is how I understand the problem. Better ask your instructor for more details.

10. ### Wendy Moderator

Mar 24, 2008
20,735
2,499
I'm betting not, it just means you have to predict the 1.5db points, and adjust accordingly.

Generally filters use T attenuators to offset differences in terminations. A 6db attenuator will have a 12 db return loss.

11. ### 1233 Thread Starter New Member

Dec 26, 2009
10
0

if B.W is 500 kHz, then f(low) is 4.75MHz and f(high)=5.25MHz, am I right ?

12. ### PRS Well-Known Member

Aug 24, 2008
989
35
Your right. I appologize for my sloppy math.