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Hi,
Can any of you please help me with the question below?
Some paper said that CIC filter has a high gain and needed to be
compensated?
From Mattew P. Donadio:
 "For a CIC decimator, the normalized gain at the output of the last
comb lies in the interval of 1/2 and 1.  When R (decimation ratio) is
a power of two, the gain is unity"
So which one is correct high gain or unity gain?
Thanks,
Eng

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On 27 Sep 2004 10:32:50 -0700, e...@yahoo.com (Eng Gan) wrote:

>Hi,

Hi Eng,

>Can any of you please help me with the question below?
>Some paper said that CIC filter has a high gain and needed to be
>compensated?

Humm, I wonder where you read that(?). 

>From Mattew P. Donadio:
> "For a CIC decimator, the normalized gain at the output of the last
>comb lies in the interval of 1/2 and 1.  When R (decimation ratio) is
>a power of two, the gain is unity"
>So which one is correct high gain or unity gain?
>Thanks,
>Eng

The CIC filter's gain at DC (zero Hz) is equal to the 
sum of the filter's impulse response samples.
And, as it turns out, the sum of the CIC filter's 
impulse response samples is equal to the length of the 
comb filter.

[-Rick-]

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Rick,
Thanks for replying.

> Humm, I wonder where you read that(?). 

From this link,
http://www.techonline.com/community/ed_resource/tech_paper/31819
a paper "Practical FIR Filter Design in Matlab"  written by Ricardo A.
Losada has the following statement:
"The filter exhibits a (sin(x)/x)^5 shape.  It also has a large DC
gain (more that 180 dB), that has to be compensated for. To compensate
for this large gain, the "4016" provides a power of two scaling prior
to data enterning the filter, in order to avoid overflow."

Question 1:What is power of two scaling?
Question 2: Did it mean if there's no overlow that high gain is
compensated?
 
> The CIC filter's gain at DC (zero Hz) is equal to the 
> sum of the filter's impulse response samples.
> And, as it turns out, the sum of the CIC filter's 
> impulse response samples is equal to the length of the 
> comb filter.

If the length of the CIC filter is 3 and the oversampling ratio is 16,
with 16^3 it has the gain of 4096.  The data output will have have a
range of -4096 to 4096
Question 3:  How do you present this output to the next stage of
filter?  If 14 bit is used, with MSB being the sign bit, 2^13 should
be closed to 8192, if 13 bit is used, 4096 is overflow.  We just have
the chip back that I minus 1 from my output and so the output is in
the range of 4095 to -4096 which is a full rail signal. Functionality
it works fine, but I only get a 8 bit resolution on my last stage of
filter where the output bit is 16.

Thanks,
Eng
> [-Rick-]

l

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