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Hi Elnaz,
If a is of type INT (int8, int16 or int32) in the 30-60 range, the result of 1/a will always give zero (e.g. 1/35 = 0,0285714, but INT(0,0285714)=0 ) because of implicit conversion (operation with two INT will give the result of type INT).
If you need to store the result into the...
Hi Dalibor,
Thank you for the correction. I made a mistake, it should be convolution, not correlation. (P.S. you made a small typo - "multiplication in the frequency domain is equivalent to convolution, not correlation, in the frequency domain." - it should be "time domain")
Regarding the...
Some facts:
1. Filter transfer function H(f) (in frequency domain) and filter impulse response h(t) (in time domain) are Fourier transform pair, i.e. F{h(t)} = H(f)
2. If you have the serie of N different filters with corresponding transfer functions H1(f), H2(f), ..., Hn(f), then overall...
Well, it depends on the application (what is throughput needed). This is the Xilinx Application Note on Xilinx Spartan-3 family interface to DDR2 :
https://www.xilinx.com/support/documentation/application_notes/xapp454.pdf
Well, since Visual Studio 2008 and now in 2010 there is a MS Chart control that gives you a huge possibilities for drawing a certain type of chart. More about this on: **broken link removed**
I guess that calculating FFT for the block of data under C# is not a problem (just google a little bit)
Of course it is possible. But, you need some fast logic to interface the memory modules. One way is FPGA connected to DDR2 module. You can visit the Xilinx.com and Altera.com and look under the Evaluation Kits and Boards. Practically, every board has DDR2 module, and you can look in the...
Lets say that P1 is PSD of signal #1, and P2 is PSD of signal #2. The arrays have the same length.
I guess in Matlab it would be:
P1_max = max(P1); %find the max value of P1
P2_max = max(P2); %find the max value of P2
P1_norm = P1/P1_max; % P1 normalized within [0, 1] range
P2_norm = P2/P2_max...
I've tried your program with Matlab R2008a.
For
g=conv(im,b)
it works fine.
For
g=conv2(im,b)
I get the same error.
But if I do this
g=conv2(double(im),b)
then it gives me no error, and the result is
g =
Columns 1 through 18
0 0 0 0 0 0 0 0 0 0 0...
Hi bluespec9,
You say that the signal in the baseband occupies [-20MHz, +20MHz] spectrum, and then you say that low pass filtering has passband up to 16MHZ and stopband from 20MHz. This means that the transition part of the filter attenuation is from 16MHz to 20MHz, and as I understand all...
I suggest you to ask this question on Analog Device's Engineer Zone website:
EngineerZone: Community: Blackfin Processors
The registration is free, and they can really help you with advices.
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