Stopband Rejection and Attenuation

The adjacent curves are used to determine the out-of-band or stopband attenuation for ClearComm's Discrete Element Bandpass Filters. These curves show the attenuation as multiples of the normalized stopband frequency to the 3 dB bandwidth for filters up to 12 sections. For requirements with greater rejections needed, please contact the factory.

These curves below show the stopband frequencies normalized to the 3 dB bandwidth for filter designs of 2 to 8 sections. This ratio of stopband frequency to 3 dB bandwidth is used to calculate the number of sections required.

The curves show a slightly asymmetric frequency response resulting from the circuit used. Other schematics may be utilized to yield different attenuation characteristics (i.e. steeper on the high frequency side of the passband and shallower on the low side). Please note that the frequency response curves shown are based on a low ripple Chebyshev design. Exact performance is related directly to the unloaded Q, component selection and package size. If you have a critical parameter please contact the factory so full compliance may be assured. Example: A filter has a center frequency of 300 MHz and a 3 dB bandwidth of 20 MHz. There is a 35 dB Rejection needed at 250 MHz and 60 dB Rejection needed at 380 MHz. The percentage bandwidth is calculated as follows: 3 dB Bandwidth / Center Frequency So %BW = 20 / 300 = 6.67% For the first stopband requirement, calculate the number of 3 dB Bandwidths away from Center Frequency. We have: (300-250) / 20 = 2.5

For the second stopband requirement, We have:

(380-300) / 20 = 4.0

Since our percentage bandwidth is >7% use the curve labeled >7% From the 7-50% bandwidth attenuation curve, we find that a minimum of 3 sections is required.

The greater number of sections must be used to insure full specification compliance; therefore, a 5 section should be used.

Example:

Cutoff Frequency (f0) = 230 MHz
40 dB attenuation required at 315 MHz

We must first calculate the ration of the two frequencies:

315 / 230 = 1.37

We can see from the table above that to achieve 40 dB attenuation at 1.37 away that we need a minimum of 5 sections. Always use the higher number of sections.

Insertion Loss

LOSS CONSTANT VS. FREQUENCY
Center Frequency (MHz )	Loss Constant
0.1 - 10	9.8
10 - 159	8.8
160 – 199	7.8
200 – 299	6.8
300 – 399	5.7
400 – 499	4.9
500 - 599	4.4
600 – 699	4.0
700 – 799	3.7
800 – 899	3.5
900 – 999	3.25
1000 – 6000	3.0

Knowing the number of sections, center frequency and bandwidth of the filter, insertion loss may be calculated
using the following formula:

I.L. =	Low Constant x (N-1.5)	+ 0.2
	%3dB BW

Example:

1. Percentage BW = 145 / 725 x 100 = 20%
2. LC from table = 6.8
3. Number of Section = 7

I.L. =	6.8 x (7 - 1.5)	+ 0.2 = 1.87 dB
	20