msdryxon
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Hi, Now I'm using verilog-a to do some research on modeling kinematic structures. We all know that ddt or idt is accepted in verilog-a, such like the following sentences: Pos(v)<+ddt(x);
But if Fdrv is from a source outside my module and i want to output the x in the following differential equation, how can i?
a*x''+b*x'+cx=Fdrv
i've written these codes, .......(includings) module test(Fdrv,x); inout Fdrv,x; kinematic Fdrv,x; kinematic v; //temp. var. to hold ddt(x) parameter real a=xxx, b=xxx, c=xxx; // assume a,b,c are properly defined analog begin Pos(v)<+ddt(x); F(Fdrv)<+a*ddt(Pos(v)) +b*Pos(v) +c*Pos(x); end endmodule
if i input x to module test, i will get proper output of Fdrv, but if i want to input Fdrv and get the proper value of x, which means i want to solve this differential equation, it never works. How can i solve this problem? thank you very very much....
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