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.

thanks in advance.

 

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.

 

thx, I will have a look at that

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

 

Is a simple LCR circuit considered a cascaded circuit ?

 

Hello,

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

Greetings,
Bertus

 

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

 

Hello,

Then take a look at this page with links on resonant circuits:
http://www.educypedia.be/electronics/electricityrlc.htm

Greetings,
Bertus

 

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.

 

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.

thanks in advance.

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.

 

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.

 

Start by considering the bandwidth: BW = fc/Q

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

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

 

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

Your right. I appologize for my sloppy math.