已知x(n)={1, 2, 3},y(n)={1, 2, 1},则x(n)*y(n)=________。(下划线表示n=0) A: {1, 4, 8, 8, 3} B: {1, 4, 8, 8, 3} C: {1, 4, 8, 8, 3} D: {1, 4, 8, 8, 3}
已知x(n)={1, 2, 3},y(n)={1, 2, 1},则x(n)*y(n)=________。(下划线表示n=0) A: {1, 4, 8, 8, 3} B: {1, 4, 8, 8, 3} C: {1, 4, 8, 8, 3} D: {1, 4, 8, 8, 3}
设 X ~ N(3, 12),Y ~ N(2, 4),且 X,Y 独立,则 X − Y ~ N(1, 8) .
设 X ~ N(3, 12),Y ~ N(2, 4),且 X,Y 独立,则 X − Y ~ N(1, 8) .
已知()y()=()ln()x(),则()y()(()n())()=()。A.()(()−()1())()n()n()!()x()−()n()"()role="presentation">()(()−()1())()n()n()!()x()−()n();()B.()(()−()1())()n()(()n()−()1())()!()x()−()2()n()"()role="presentation">()(()−()1())()n()(()n()−()1())()!()x()−()2()n();()C.()(()−()1())()n()−()1()(()n()−()1())()!()x()n()"()role="presentation">()(()−()1())()n()−()1()(()n()−()1())()!()x()-n();()D.()(()−()1())()n()−()1()n()!()x()−()n()+()1()"()role="presentation">()(()−()1())()n()−()1()n()!()x()−()n()+()1().
已知()y()=()ln()x(),则()y()(()n())()=()。A.()(()−()1())()n()n()!()x()−()n()"()role="presentation">()(()−()1())()n()n()!()x()−()n();()B.()(()−()1())()n()(()n()−()1())()!()x()−()2()n()"()role="presentation">()(()−()1())()n()(()n()−()1())()!()x()−()2()n();()C.()(()−()1())()n()−()1()(()n()−()1())()!()x()n()"()role="presentation">()(()−()1())()n()−()1()(()n()−()1())()!()x()-n();()D.()(()−()1())()n()−()1()n()!()x()−()n()+()1()"()role="presentation">()(()−()1())()n()−()1()n()!()x()−()n()+()1().
计算下列序列的N点DFT。(1)x(n)=1(2)x(n)=δ(n)(3)x(n)=δ(n一n0),0<n0<N(4)x(n)=Rm(n),0<m<N(7)x(n)=ejω0nRN(n)(8)x(n)=sin(ω0n)RN(n)(9)x(n)=cos(ω0n)RN(n)(10)x(n)=nRN(n)
计算下列序列的N点DFT。(1)x(n)=1(2)x(n)=δ(n)(3)x(n)=δ(n一n0),0<n0<N(4)x(n)=Rm(n),0<m<N(7)x(n)=ejω0nRN(n)(8)x(n)=sin(ω0n)RN(n)(9)x(n)=cos(ω0n)RN(n)(10)x(n)=nRN(n)
设随机变量X的分布律为P{X=k}=a/N,k=1,2,…,N,则a=()。 A: 1/8 B: 1/4 C: 1/2 D: 1
设随机变量X的分布律为P{X=k}=a/N,k=1,2,…,N,则a=()。 A: 1/8 B: 1/4 C: 1/2 D: 1
已知随机变量X~B(n,1/2),且P{X=5}=1/32,则n=___________ A: 5 B: 6 C: 7 D: 8
已知随机变量X~B(n,1/2),且P{X=5}=1/32,则n=___________ A: 5 B: 6 C: 7 D: 8
设X~N(1, 2), Y~N(-1, 3),且X与Y相互独立,则2X-Y~( ) A: N(3, 8) B: N(3, 5) C: N(3, 11) D: N(3,25)
设X~N(1, 2), Y~N(-1, 3),且X与Y相互独立,则2X-Y~( ) A: N(3, 8) B: N(3, 5) C: N(3, 11) D: N(3,25)
设有一个递归算法如下:int X(int n){ if(n<=3) return 1;else return X(n-2)+X(n-4)+1}则计算X(X(8))时需要计算X函数()次.
设有一个递归算法如下:int X(int n){ if(n<=3) return 1;else return X(n-2)+X(n-4)+1}则计算X(X(8))时需要计算X函数()次.
\( {1 \over {1 + x}} \)的麦克劳林公式为( )。 A: \( {1 \over {1 + x}} = 1 + x + { { {x^2}} \over 2} + \cdots + { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) \) B: \( {1 \over {1 + x}} = 1 + x + {x^2} + \cdots + {x^n} + o\left( { { x^n}} \right) \) C: \( {1 \over {1 + x}} = 1 - x + {x^2} - \cdots + {( - 1)^n}{x^n} + o\left( { { x^n}} \right) \) D: \( {1 \over {1 + x}} = 1 - x - { { {x^2}} \over 2}- \cdots - { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) \)
\( {1 \over {1 + x}} \)的麦克劳林公式为( )。 A: \( {1 \over {1 + x}} = 1 + x + { { {x^2}} \over 2} + \cdots + { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) \) B: \( {1 \over {1 + x}} = 1 + x + {x^2} + \cdots + {x^n} + o\left( { { x^n}} \right) \) C: \( {1 \over {1 + x}} = 1 - x + {x^2} - \cdots + {( - 1)^n}{x^n} + o\left( { { x^n}} \right) \) D: \( {1 \over {1 + x}} = 1 - x - { { {x^2}} \over 2}- \cdots - { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) \)
\( {1 \over {1 + x}} \)的麦克劳林公式为( ). A: \( {1 \over {1 + x}} = 1 + x + { { {x^2}} \over 2} + \cdots + { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) \) B: \( {1 \over {1 + x}} = 1 + x + {x^2} + \cdots + {x^n} + o\left( { { x^n}} \right) \) C: \( {1 \over {1 + x}} = 1 - x + {x^2} - \cdots + {( - 1)^n}{x^n} + o\left( { { x^n}} \right) \)
\( {1 \over {1 + x}} \)的麦克劳林公式为( ). A: \( {1 \over {1 + x}} = 1 + x + { { {x^2}} \over 2} + \cdots + { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) \) B: \( {1 \over {1 + x}} = 1 + x + {x^2} + \cdots + {x^n} + o\left( { { x^n}} \right) \) C: \( {1 \over {1 + x}} = 1 - x + {x^2} - \cdots + {( - 1)^n}{x^n} + o\left( { { x^n}} \right) \)