Multiplication of 2 8 bit numbers

[gold_kiss]

Hi,
say i have a [7:0] and b [7:0] as two inputs to a multiplication block.
what should be the output z [7:0] or z [15:0] or z [31:0]?

can any one also tell me what should in boolean equation?
z[0] :=
z[1] :=
z[2] :=

....and so on

z[7] :=

Thanks,
Gold_kiss

 

[aravind]

if u multiply m*m u will get 2m-1 always
.if u r design any dsp /processor accurual chip.u can change the datapath path size.it should be m
u much truncate or roundoff the 2m-1 to m
u much scale the design
search scaling in vlsi at google
sorry if i say anything wrong
this r the information i known

 

[moorhuhn]

You have to be more specific. If your 8-bit numbers are in 2's complement, the multiplier will
differ from those for multiplication of positive numbers. Multiplier and their different implementations were often published in IEEE Journal of Solid State Circuits. If you do not
targetting the high speed you may be satisfied with a product of any synthesizer, where
you just need to write:
z = a * b;

It is self evident, that the result of the 8 x 8bit multiplication is 16bit number. For the DSP purposes you should use the accumulator, which has to be wider.

 

[lordwolf]

If you're looking for the hardware structural description, well there are two ways you can implement a multiplier: (1) Serial multiplier (2) Parallel Multiplier.

Serial multiplier takes much less space compared to a parallel multiplier, but it takes more time and parallel multiplier vice-versa. It's a matter of trade-off between size & speed. Multiplication is actually a sequence of add-shift process.

So, if you're building an nxn bit multiplier, for a serial multiplier you need only 1 n-bit adder with 1 2*n-bit shift register and 1 n-bit register. You need at least 8 clock cycles for this circuit. For a parallel multiplier, you need n n-bit adder but you circuit becomes a single combinational logic block.

-lordwolf-

 

[zyphor1]

pls refer the attachment file