Can't understand Verilog arithmetics

Ubix2014 Guest 2016-03-04 10:01 #4493336

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Hi,

I am trying to understand some Verilog source code, but I am confused 
about the following

reg   [29-PSR-1: 0] kp_reg        ;
wire  [    29-1: 0] kp_mult       ;
always @(posedge clk_i) begin
   if (rstn_i == 1'b0) begin
      kp_reg  <= {29-PSR{1'b0}};
   else begin
      kp_reg <= kp_mult[29-1:PSR] ;
assign kp_mult = $signed(error) * $signed(set_kp_i);

 where PSR is a paramter equal = 12.

So, kp_mult keeps the value of the product. But then it is truncated and 
only MSB are assigned to kp_reg. I am wondering what this is happened? 
My guess is that it is related to the fixed-point arithmetics and PSR 
defines the position of the binary point but I am not sure.

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Andi Meerbann Guest 2016-03-04 13:53 #4493790

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You are right with your assumption!

Multiplications tends to blow up the bit widths of your signals because 
the resulting bit width is always the sum of both input bit widths. 
That's why you often have to truncate the result after a multiplication. 
And that's exactly what happens here.

When you consider a number as fixed point you can easily cut off some of 
the fractional bits when the required accuracy is not so high.
Of course rounding would be better than truncating, but this is more 
complicated and requires more logic resources.

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Ubix2014 Guest 2016-03-04 14:24 #4493839

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Hi Andi,

I would agree with you if kp_mult is used for the accumulation. However, 
it just a single multiplication. Consider for example that kp_mult 
equals to 3 in decimal, kp_mult = 0000 0000 0000 0000 0000 0000 0000 11. 
If we move only 17 MSB to kp_reg it will equal to zero. And it will be 
zero until really big number (2^11). This is why, I am not sure that 
just truncation is a reasonable explanation.

Apart from it, error and set_kp_i are integer values.