某二阶LTI系统,y"(t)+a0y'(t)+a1y(t)=b0f'(t)+b1f(t),...
某二阶LTI系统,y"(t)+a0y'(t)+a1y(t)=b0f'(t)+b1f(t),在激励e-2tu(t)作用下的全响应为(-e-t+4e-2t-e-3t)u(t);而在激励δ(t)-2e-2tu(t)作用下的全响应为(3e-t+e-2t-5e-3t)u(t),设初始状态同定,求:
某二阶LTI系统,y"(t)+a0y'(t)+a1y(t)=b0f'(t)+b1f(t),在激励e-2tu(t)作用下的全响应为(-e-t+4e-2t-e-3t)u(t);而在激励δ(t)-2e-2tu(t)作用下的全响应为(3e-t+e-2t-5e-3t)u(t),设初始状态同定,求:
A.y(t)=t+τe-t/τB.y(t)=-t+τe-t/τ
C.y(t)=t-τ+τe-t/τD.y(t)=2t-τ+τe-t/τ
若f(u)可导,且y=f(ex),则有( ).
A.dy=f'(ex)dx B.dy=f'(ex)dex
C.dy=[f(ex)]'dexD.dy=f'(ex)exdx
void f(int y,int *x)
{ y=y+*x; *x=*x+y; }
main()
{int x=2,y=4;
f(y,&x);
ptintf("%d %d\n",x,y);
}
执行后输出结果是【 】。
A、Compared to x(t), y(t) is advanced
B、Compared to x(t), y(t) is delayed
C、Compared to x(t), y(t) is stretched
D、Compared to x(t), y(t) is compress
A.8,2,3,4,5,6,7,1,
B.5,6,7,8,1,2,3,4,
C.1,2,3,4,5,6,7,8,
D.8,7,6,5,4,3,2,1
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