参数资料
型号: IPR-FFT
厂商: Altera
文件页数: 63/70页
文件大小: 0K
描述: IP FFT/IFFT RENEW
标准包装: 1
系列: *
类型: MegaCore
功能: 快速傅里叶变换处理器
许可证: 续用许可证
Appendix A: Block Floating Point Scaling
A–3
Implementing Scaling
A sample of Verilog HDL code that illustrates the scaling of the output data (for
exponents –11 to –9) with sign extension is shown in the following example:
case (exp)
6'b110101 : //-11 Set data equal to MSBs
begin
full_range_real_out[26:0] <= {real_in[15:0],11'b0};
full_range_imag_out[26:0] <= {imag_in[15:0],11'b0};
end
6'b110110 : //-10 Equals left shift by 10 with sign extension
begin
full_range_real_out[26] <= {real_in[15]};
full_range_real_out[25:0] <= {real_in[15:0],10'b0};
full_range_imag_out[26] <= {imag_in[15]};
full_range_imag_out[25:0] <= {imag_in[15:0],10'b0};
end
6'b110111 : //-9 Equals left shift by 9 with sign extension
begin
full_range_real_out[26:25] <= {real_in[15],real_in[15]};
full_range_real_out[24:0] <= {real_in[15:0],9'b0};
full_range_imag_out[26:25] <= {imag_in[15],imag_in[15]};
full_range_imag_out[24:0] <= {imag_in[15:0],9'b0};
end
.
.
.
endcase
In this example, the output provides a full scale 27-bit word. You must choose how
many and which bits must be carried forward in the processing chain. The choice of
bits determines the absolute gain relative to the input sample level.
Figure A–1 on page A–4 demonstrates the effect of scaling for all possible values for
the 256-point quad output FFT with an input signal level of 0x5000. The output of the
FFT is 0x280 when the exponent = –5. The figure illustrates all cases of valid exponent
values of scaling to the full scale storage register [26..0]. Because the exponent is –5,
you must check the register values for that column. This data is shown in the last two
columns in the figure. Note that the last column represents the gain compensated data
after the scaling (0x0005000), which agrees with the input data as expected. If you
want to keep 16 bits for subsequent processing, you can choose the bottom 16 bits that
result in 0x5000. However, if you choose a different bit range, such as the top 16 bits,
the result is 0x000A. Therefore, the choice of bits affects the relative gain through the
processing chain.
November 2013
Altera Corporation
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