How to attenuate an undesired signal

Thread Starter

rayraysayshi

Joined Oct 9, 2012
4
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.
 
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Thread Starter

rayraysayshi

Joined Oct 9, 2012
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)."
 

#12

Joined Nov 30, 2010
18,224
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?
 

Thread Starter

rayraysayshi

Joined Oct 9, 2012
4
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:
2.45 MHz = 15.39E6 rad/s
980 kHz = 6.157E6 rad/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:

bertus

Joined Apr 5, 2008
22,270
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
 

Thread Starter

rayraysayshi

Joined Oct 9, 2012
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:

crutschow

Joined Mar 14, 2008
34,285
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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