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#1
04-24-2013, 06:21 PM
 rayraysayshi New Member Join Date: Oct 2012 Location: Golden Posts: 4
How to attenuate an undesired signal

Hello, I am working on some homework and wish to learn how to attenuate a strong signal so it doesn't interfere with another desired frequency. My question is; a strong signal at 2.45 MHz is interfering with an AM signal at 980 Hz. Design a filter that will attenuate the undesired signal by at least 60 dB.

First I extract given data;
2.45 MHz (signal to be attenuated)
980 kHz (desired signal)
60 dB (factor which undesired frequency needs to be attenuated)

I know I wish to implement a low-pass filter but I dont know how to go about creating a circuit from the 3 given values. I tried to use the voltage gain relationship, Gain [dB] = 20log(Vout/Vin) but, this just gives me Vout/Vin which I fail to see useful when I am only given an undesired frequency and a desired one.

I'm confused what order the filter needs/can be. I'm confused about how to get the transfer function from the given information. I believe once I have the transfer function I can draw the circuit.

Any ideas, obvious misunderstanding or suggestions? Thanks for reading.

Last edited by rayraysayshi; 04-24-2013 at 06:27 PM. Reason: Hz -> kHz
#2
04-24-2013, 06:26 PM
 bertus Administrator Join Date: Apr 2008 Location: Amsterdam,Holland (GMT + 1) Posts: 12,097

Hello,

I assume you are talking about a 980 kHz AM signal and not a 980 Hz signal, wich would be audio.

You could try to make a notch filter tuned to the 2.45 Mhz.
http://en.wikipedia.org/wiki/Band-stop_filter

Bertus
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Last edited by bertus; 04-24-2013 at 06:37 PM.
#3
04-24-2013, 06:39 PM
 rayraysayshi New Member Join Date: Oct 2012 Location: Golden Posts: 4

I'm reading about notch circuits right now and this seems like the right approach; via wikipedia;

"...that passes most frequencies unaltered, but attenuates those in a specific range to very low levels. It is the opposite of a band-pass filter. A notch filter is a band-stop filter with a narrow stopband (high Q factor)."
#4
04-24-2013, 06:44 PM
 #12 Senior Member Join Date: Nov 2010 Location: 15 miles west of Tampa, Florida Posts: 9,054 Blog Entries: 9

Filters are rated at db/decade. I forgot how many. How many decades of frequency do you have to work with? A little over 2. Will a single pole filter do 30 db per decade? I don't think so.

See me trying to arrive at what order of filter will be required?
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#5
04-24-2013, 06:54 PM
 rayraysayshi New Member Join Date: Oct 2012 Location: Golden Posts: 4

Quote:
 Originally Posted by #12 How many decades of frequency do you have to work with? A little over 2.
I'm reading why it should be a little over 2 now, I don't fully understand dB/decade.

Okay...I'm starting to see some equations appear; first I converted the frequency to radians/s:

and now log(15.39E6/6.157E6) = 2.5 decades which means I will need more than one pole in the denominator.

Last edited by rayraysayshi; 04-24-2013 at 07:12 PM. Reason: progress on understanding how dB relates to decades, conversion of Hz to radians/s
#6
04-24-2013, 07:09 PM
 #12 Senior Member Join Date: Nov 2010 Location: 15 miles west of Tampa, Florida Posts: 9,054 Blog Entries: 9

Oops. It's not 2. It's a bit over 1 decade. 1Mhz is 10 times 100KHz.

Let me google that for you:

http://www.soundfirst.com/Filter_Slopes.html
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 The Following User Says Thank You to #12 For This Useful Post: rayraysayshi (04-24-2013)
#7
04-24-2013, 07:17 PM
 bertus Administrator Join Date: Apr 2008 Location: Amsterdam,Holland (GMT + 1) Posts: 12,097

Hello,

If the Q of the LC filter is high enough, it would be no problem to eliminate the 2.45 Mhz signal.
A series LC circuit would already be a simple solution:

How would it otherwise be possible to recieve a single station on the radio?

Bertus
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It would be nice to have the Timezone ( GMT +/- x ) in the location field in the profile.
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 The Following 2 Users Say Thank You to bertus For This Useful Post: #12 (04-24-2013), rayraysayshi (04-24-2013)
#8
04-24-2013, 07:34 PM
 #12 Senior Member Join Date: Nov 2010 Location: 15 miles west of Tampa, Florida Posts: 9,054 Blog Entries: 9

Darn. I wasn't even THINKING in RF terms.
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#9
04-24-2013, 07:48 PM
 rayraysayshi New Member Join Date: Oct 2012 Location: Golden Posts: 4

So a second order circuit is now in my crosshairs...There are second-order bandstop* filter equations;

http://imgur.com/u5rAqju

I wouldn't be able to tune to a different station if I've adjusted my Q factor very high--it would only tune into that frequency. But for now I am assuming that I don't care about receiving any other frequencies and only care to receive the AM.

Last edited by rayraysayshi; 04-24-2013 at 07:55 PM. Reason: Bandpass -> Bandstop
#10
04-24-2013, 08:37 PM
 crutschow Senior Member Join Date: Mar 2008 Location: L.A. USA Posts: 6,323 Blog Entries: 1

If you wanted to use a low-pass filter you would select a corner frequency somewhat above the desired frequency (so it isn't appreciably attenuated) say 1MHz. From that to the 2.45MHz frequency is a little over an octave so you want a rolloff of near 60dB/octave. This can be accomplished with an 8th-order Butterworth filter or a 6th-order Chebyshev with 0.5dB passband ripple, which would require 4 or 3 high-speed op amp 2nd-order Sallen-Key filters respectively.

That becomes fairly complex so perhaps Bertus's design to use a resonant notch filter for the desired attenuation would be preferable. But you may need some way to tune the notch frequency to compensate for component tolerances.
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