M 0 M 2 K

Author: h | 2025-04-24

To convert from m / s 2 to k m / h 2, multiply the acceleration value by . k m / h 2 = ( m / s 2 ) To convert from k m / h 2 to m / s 2, multiply the acceleration value by 0. .

$ sum_{k=1}^m k(k-1){m choose k} = m(m-1) 2^{m-2}$

Problem 681For a square matrix $M$, its matrix exponential is defined by\[e^M = \sum_{i=0}^\infty \frac{M^k}{k!}.\]Suppose that $M$ is a diagonal matrix\[ M = \begin{bmatrix} m_{1 1} & 0 & 0 & \cdots & 0 \\ 0 & m_{2 2} & 0 & \cdots & 0 \\ 0 & 0 & m_{3 3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & m_{n n} \end{bmatrix}.\]Find the matrix exponential $e^M$. Add to solve laterSponsored Links Solution.First, we find $M^k$ for each integer $k \geq 0$. The first couple powers can be calculated directly,\[M^0 = \begin{bmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix} , \quad M = \begin{bmatrix} m_{1 1} & 0 & 0 & \cdots & 0 \\ 0 & m_{2 2} & 0 & \cdots & 0 \\ 0 & 0 & m_{3 3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & m_{n n} \end{bmatrix},\]\[M^2 = \begin{bmatrix} m^2_{1 1} & 0 & 0 & \cdots & 0 \\ 0 & m^2_{2 2} & 0 & \cdots & 0 \\ 0 & 0 & m^2_{3 3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & m^2_{n n} \end{bmatrix} , \quad M^3 = \begin{bmatrix} m^3_{1 1} & 0 & 0 & \cdots & 0 \\ 0 & m^3_{2 2} & 0 & \cdots & 0 \\ 0 & 0 & m^3_{3 3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & m^3_{n n} \end{bmatrix}.\]The general pattern can now be seen:\[M^k = \begin{bmatrix} m^k_{1 1} & 0 & 0 & \cdots & 0 \\ 0 & m^k_{2 2} & 0 & \cdots & 0 \\ 0 & 0 & m^k_{3 3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & m^k_{n n} \end{bmatrix}.\]Now, we can calculate the infinite series $e^M$:\begin{align*}e^M &= \sum_{k=0}^{\infty} \frac{ M^k }{k!} \\&= \sum_{k=0}^\infty \frac{1}{k!} \begin{bmatrix} m^k_{1, 1} & 0 & 0 & \cdots & 0 \\ 0 & m^k_{2, 2} & 0 & \cdots & 0 \\ 0 & 0 & m^k_{3, 3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & m^k_{n, n} \end{bmatrix} \\&= \begin{bmatrix} \sum_{k=0}^\infty \frac{ m^k_{1 1} }{k!} & 0 & 0 & \cdots & 0 \\ 0 & \sum_{k=0}^\infty \frac{ m^k_{2 2} }{k!} & 0 & \cdots & 0 \\ 0 & 0 & \sum_{k=0}^\infty \frac{ m^k_{3 ] [(k[m:0] (1m) ? (k[3:m] (m+1)) + k[m:0] : (k[3:m] (m+1)) + (k[m:0]-(1m))) + (1m) ]), .factor_real(factor_real[k[m:0](1m)? k[m:0] : k[m:0]-(1m) ]), .factor_imag(factor_imag[k[m:0](1m)? k[m:0] : k[m:0]-(1m) ]), //output data .valid (en_connect[ (m+1)*4 + k ] ), .yp_real (xm_real[ m+1 ][k[m:0] (1m) ? (k[3:m] (m+1)) + k[m:0] : (k[3:m] (m+1)) + (k[m:0]-(1m))] ), .yp_imag (xm_imag[ m+1 ][(k[m:0]) (1m) ? (k[3:m] (m+1)) + k[m:0] : (k[3:m] (m+1)) + (k[m:0]-(1m))] ), .yq_real (xm_real[ m+1 ][(k[m:0] (1m) ? (k[3:m] (m+1)) + k[m:0] : (k[3:m] (m+1)) + (k[m:0]-(1m))) + (1m) ]), .yq_imag (xm_imag[ m+1 ][((k[m:0]) (1m) ? (k[3:m] (m+1)) + k[m:0] : (k[3:m] (m+1)) + (k[m:0]-(1m))) + (1m) ]) ); end end endgenerate assign valid = en_connect[12]; assign y0_real = xm_real[3][0] ; assign y0_imag = xm_imag[3][0] ; assign y1_real = xm_real[3][1] ; assign y1_imag = xm_imag[3][1] ; assign y2_real = xm_real[3][2] ; assign y2_imag = xm_imag[3][2] ; assign y3_real = xm_real[3][3] ; assign y3_imag = xm_imag[3][3] ; assign y4_real = xm_real[3][4] ; assign y4_imag = xm_imag[3][4] ; assign y5_real = xm_real[3][5] ; assign y5_imag = xm_imag[3][5] ; assign y6_real = xm_real[3][6] ; assign y6_imag = xm_imag[3][6] ; assign y7_real = xm_real[3][7] ; assign y7_imag = xm_imag[3][7] ;endmoduletestbenchtestbench 编写如下，主要用于 16 路实、复数据的连续输入。因为每次 FFT 只有 8 点数据，所以送入的数据比较随意，并不是正弦波等规则的数据。实例 `timescale 1ns/100psmodule test ; reg clk; reg rstn; reg en ; reg signed [23:0] x0_real; reg signed [23:0] x0_imag; reg signed [23:0] x1_real; reg signed [23:0] x1_imag; reg signed [23:0] x2_real; reg signed [23:0] x2_imag; reg signed [23:0] x3_real; reg signed [23:0] x3_imag; reg signed [23:0] x4_real; reg signed [23:0] x4_imag; reg signed [23:0] x5_real; reg signed [23:0] x5_imag; reg signed [23:0] x6_real; reg signed [23:0] x6_imag; reg signed [23:0] x7_real; reg signed [23:0] x7_imag; wire valid; wire signed [23:0] y0_real; wire signed [23:0] y0_imag; wire signed [23:0] y1_real; wire signed [23:0] y1_imag; wire signed [23:0] y2_real; wire signed [23:0] y2_imag; wire signed [23:0] y3_real; wire signed [23:0] y3_imag; wire signed [23:0] y4_real; wire signed [23:0] y4_imag; wire signed [23:0] y5_real; wire signed [23:0] y5_imag; wire signed [23:0] y6_real; wire signed [23:0] y6_imag; wire signed [23:0] y7_real; wire signed [23:0] y7_imag; initial begin clk = 0; //50MHz rstn = 0 ; #10 rstn = 1; forever begin #10 clk = ~clk; //50MHz end end fft8 u_fft ( .clk (clk ), .rstn (rstn ), .en (en ), .x0_real (x0_real), .x0_imag (x0_imag), .x1_real (x1_real), .x1_imag (x1_imag), .x2_real (x2_real), .x2_imag (x2_imag), .x3_real (x3_real), .x3_imag (x3_imag), .x4_real (x4_real), .x4_imag (x4_imag), .x5_real (x5_real), .x5_imag (x5_imag), .x6_real (x6_real), .x6_imag (x6_imag), .x7_real (x7_real), .x7_imag (x7_imag), .valid (valid), .y0_real (y0_real), .y0_imag (y0_imag), .y1_real (y1_real), .y1_imag (y1_imag), .y2_real (y2_real), .y2_imag (y2_imag), .y3_real (y3_real), .y3_imag (y3_imag), .y4_real (y4_real), .y4_imag (y4_imag), .y5_real (y5_real), .y5_imag (y5_imag), .y6_real (y6_real), .y6_imag (y6_imag), .y7_real (y7_real), .y7_imag (y7_imag)); //data input initial begin en = 0 ; x0_real =

Show $ sum_{k=0}^{m} binom{n-k}{m-k}= binom{n1}{m}$

13bits because of Wnr assign yp_real = {yp_real_r[39], yp_real_r[13+23:13]}; assign yp_imag = {yp_imag_r[39], yp_imag_r[13+23:13]}; assign yq_real = {yq_real_r[39], yq_real_r[13+23:13]}; assign yq_imag = {yq_imag_r[39], yq_imag_r[13+23:13]}; assign valid = en_r[2];endmodule顶层例化根据 FFT 算法结构示意图，将蝶形单元例化，完成最后的 FFT 功能。可根据每一级蝶形单元的输入输出对应关系，依次手动例化 12 次，也可利用 generate 进行例化，此时就需要非常熟悉 FFT 中"组"和"级"的特点：(1) 8 点 FFT 设计，需要 3 级运算，每一级有 4 个蝶形单元，每一级的组数目分别是 4、2、1。(2) 每一级的组内一个蝶形单元中两个输入端口的距离恒为 （m 为级标号，对应左移运算 1(3) 每一级相邻组间的第一个蝶形单元的第一个输入端口的距离为 （对应左移运算 2例化代码如下。其中，矩阵信号 xm_real（xm_imag）的一维、二维地址是代表级和组的标识。在判断信号端口之间的连接关系时，使用了看似复杂的判断逻辑，而且还带有乘号，其实最终生成的电路和手动编写代码例化 12 个蝶形单元的方式是完全相同的。因为 generate 中的变量只是辅助生成实际的电路，相关值的计算判断都已经在编译时完成。这些变量更不会生成实际的电路，只是为更快速的模块例化提供了一种方法。实例 timescale 1ns/100psmodule fft8 ( input clk, input rstn, input en, input signed [23:0] x0_real, input signed [23:0] x0_imag, input signed [23:0] x1_real, input signed [23:0] x1_imag, input signed [23:0] x2_real, input signed [23:0] x2_imag, input signed [23:0] x3_real, input signed [23:0] x3_imag, input signed [23:0] x4_real, input signed [23:0] x4_imag, input signed [23:0] x5_real, input signed [23:0] x5_imag, input signed [23:0] x6_real, input signed [23:0] x6_imag, input signed [23:0] x7_real, input signed [23:0] x7_imag, output valid, output signed [23:0] y0_real, output signed [23:0] y0_imag, output signed [23:0] y1_real, output signed [23:0] y1_imag, output signed [23:0] y2_real, output signed [23:0] y2_imag, output signed [23:0] y3_real, output signed [23:0] y3_imag, output signed [23:0] y4_real, output signed [23:0] y4_imag, output signed [23:0] y5_real, output signed [23:0] y5_imag, output signed [23:0] y6_real, output signed [23:0] y6_imag, output signed [23:0] y7_real, output signed [23:0] y7_imag ); //operating data wire signed [23:0] xm_real [3:0] [7:0]; wire signed [23:0] xm_imag [3:0] [7:0]; wire en_connect [15:0] ; assign en_connect[0] = en; assign en_connect[1] = en; assign en_connect[2] = en; assign en_connect[3] = en; //factor, multiplied by 0x2000 wire signed [15:0] factor_real [3:0] ; wire signed [15:0] factor_imag [3:0]; assign factor_real[0] = 16'h2000; //1 assign factor_imag[0] = 16'h0000; //0 assign factor_real[1] = 16'h16a0; //sqrt(2)/2 assign factor_imag[1] = 16'he95f; //-sqrt(2)/2 assign factor_real[2] = 16'h0000; //0 assign factor_imag[2] = 16'he000; //-1 assign factor_real[3] = 16'he95f; //-sqrt(2)/2 assign factor_imag[3] = 16'he95f; //-sqrt(2)/2 //输入初始化，和码位有关倒置 assign xm_real[0][0] = x0_real; assign xm_real[0][1] = x4_real; assign xm_real[0][2] = x2_real; assign xm_real[0][3] = x6_real; assign xm_real[0][4] = x1_real; assign xm_real[0][5] = x5_real; assign xm_real[0][6] = x3_real; assign xm_real[0][7] = x7_real; assign xm_imag[0][0] = x0_imag; assign xm_imag[0][1] = x4_imag; assign xm_imag[0][2] = x2_imag; assign xm_imag[0][3] = x6_imag; assign xm_imag[0][4] = x1_imag; assign xm_imag[0][5] = x5_imag; assign xm_imag[0][6] = x3_imag; assign xm_imag[0][7] = x7_imag; //butter instantiaiton //integer index[11:0] ; genvar m, k; generate //3 stage for(m=0; m2; m=m+1) begin: stage for (k=0; k3; k=k+1) begin: unit butterfly u_butter( .clk (clk ) , .rstn (rstn ) , .en (en_connect[m*4 + k] ) , //是否再组内？组编号+组内编号：下组编号+新组内编号 .xp_real (xm_real[ m ] [k[m:0] (1m) ? (k[3:m] (m+1)) + k[m:0] : (k[3:m] (m+1)) + (k[m:0]-(1m))] ), .xp_imag (xm_imag[ m ] [k[m:0] (1m) ? (k[3:m] (m+1)) + k[m:0] : (k[3:m] (m+1)) + (k[m:0]-(1m))] ), .xq_real (xm_real[ m ] [(k[m:0] (1m) ? (k[3:m] (m+1)) + k[m:0] : (k[3:m] (m+1)) + (k[m:0]-(1m))) + (1m) ]), //增加蝶形单元两个输入端口间距离 .xq_imag (xm_imag[ m. To convert from m / s 2 to k m / h 2, multiply the acceleration value by . k m / h 2 = ( m / s 2 ) To convert from k m / h 2 to m / s 2, multiply the acceleration value by 0. .

Combinatoric proof that $ sum_ {k=0}^m binom {m} {k} binom {nk} {m

N are processed by an AHC algorithm [47]. This algorithm groups the input TDLPs into a multilevel binary cluster tree (or dendrogram). The linkage criterion adopted in this paper for grouping is the so-called Ward’s minimum-variance method, which minimizes the within-cluster variance, namely, the sums of squares of the distances between the TDLPs in a cluster and its centroid. This choice is due to the fact that the Ward’s method was already successfully adopted in studies on load profiles modelling and it is recognized as one of the best linkage methods [34], as it prevents the formation of large clusters [29]. Denoting with N i the set of clusters at the i-th level of the dendrogram, the steps of the AHC algorithm implemented in this work are summarized below. Initially, the elements of N 0 (i.e., at level 0 for a sequence number j = 0 ) are exactly N clusters consisting of 1 TDLP each. As a consequence, the initial NOC value M is M = | N 0 | = N and the M × M dissimilarity matrix D j is computed as follows: D j = 0 d 12 2 ⋯ d 1 ( M − 1 ) 2 d 1 M 2 d 21 2 0 ⋯ d 2 ( M − 1 ) 2 d 2 M 2 ⋮ ⋮ ⋮ ⋮ ⋮ d M 1 2 d M 2 2 ⋯ d M ( M − 1 ) 2 0 (1) where d r k is the distance between each pair of clusters r and k. If the Ward’s method is used, initially d r k coincides with the squared Euclidean distance between TDLPs p ¯ r ( t ) and p ¯ k ( t ) , i.e., d r k = | | p ¯ r ( t ) − p ¯ k ( t ) | | 2 .Starting from the current matrix D j , the clusters with the least dissimilarity, i.e., those with indexes ( r * , k * ) = arg min r , k ∈ N i d r k (2) are merged into a new cluster and the sequence number j is incremented by 1. Furthermore, matrix D j is updated by deleting the rows and columns associated with clusters r * and k * , and by adding a new row and column including T ⋯ x 2 + ( m − 1 ) t ⋮ ⋮ ⋯ ⋮ x M x M + t ⋯ x N ] . (1) At this time, the correlation integral is Formula (2), where θ ( x ) = { 0 x 0 1 x ≥ 0 . According to the statistical conclusion of BDS, the range of M and r k can be obtained when N > 3000 ; m ∈ { 2 , 3 , 4 , 5 } , and r k = k × 0.5 σ which is a real number representing a given range of distances. σ is the standard deviation of time series, and k ∈ { 1 , 2 , 3 , 4 } . Correlation integral indicates the probability that the distance between any two points in the phase space is less than r k . C ( m , N , r k , t ) = 2 M ( M − 1 ) ∑ 1 ≤ i j ≤ M θ ( r k − ‖ X i − X j ‖ ) . (2) We define the test statistic S and Δ S , and use the block averaging strategy, as shown in Formula (3). { S ( m , N , r k , t ) = 1 t ∑ i = 1 M C i ( m , N t , r k , t ) − C i m ( m , N t , r k , t ) Δ S ( m , N , t ) = max [ S ( m , N , r k , t ) ] − min [ S ( m , N , r k , t ) ] . (3) Formula (4) is the average of S and Δ S . Rounding the t value of the first zero of S ¯ or the first minimum of Δ S is the optimal delay topt. { S ¯ ( t ) = 1 4 × 4 ∑ m = 2 5 ∑ k = 1 4 S ( m , N , r k , t ) Δ S ¯ ( t ) = 1 4 ∑ m = 2 5 Δ S ( m , N , t ) . (4) Formula (5) is the test statistic. The global minimum of S c o r ( t ) is the optimal embedded window t ω . S c o r ( t ) = Δ S ¯ ( t ) + | S ¯ ( t ) | . (5) Then: t ω = ( m o p t − 1 ) t o p t . (6) Therefore, the optimal delay topt determined by Equation (4) and the optimal embedded window t ω determined by Equation (5) can be substituted into Formula (6) and rounded to obtain the optimal embedding dimension mopt. 2.2. Deep Neural NetworkA neural network generally consists of

Is k m ? (1) m = 4(k 1) (2) m = k 2 : Data Sufficiency (DS)

+ K i K f ) ] G 3 = L m m ema + m el L m + J m L m η ema k ema 2 R G 2 = η ema k ema 2 R ( J m R m + J m K i K f + B fm L m ) + B fr L m + ( m ema + m el ) ( R m + K i K f ) G 1 = η ema k ema 2 R ( C T C E + B fm R m + K i K f B fm ) + B fr ( R m + K i K f ) + S t L m G 0 = S t ( R m + K i K f ) By dividing both the numerator and the denominator on the right side of Equation (11) by the coefficient of u, we obtain: F ema = u − ( F 3 s 3 + F 2 s 2 + F 1 s ) x a ( s 2 m el + S t ) R k ema η ema K i K v C T [ ( s 2 m el + S t ) R k ema η ema K i K v C T ] ( G 3 s 3 + G 2 s 2 + G 1 s + G 0 ) (12) It can be seen from Figure 13 that to suppress the disturbance force, the feedforward correction element should be used to counteract the part following u in the numerator of Equation (12): G T = F 3 s 3 + F 2 s 2 + F 1 s ( s 2 m el + S t ) R k ema η ema K i K v C T (13) The simulation model of the DEMA force servo system introduced with a PID controller and a feedforward correction element is shown in Figure 14 [48,49,50].As shown in Figure 14, we added several modules in the model to simulate the sampling and data processing of the processor, making the simulation model more realistic. In order to make the simulation model easier to understand and operate, we use PMSM and planetary-roller screw pairs as super components, and integrate feedforward and its sampling and data processing into the super components.The disturbance of

Find the Value of K If M = `[(1,2),(2,3)]` and `M^2 - Km - I_2 = 0

D 12 d C 2 D 12 cD 11 + D 12 d D 21 32 H  Fixed-Order Controller Design Choose matrices a, b, c, d to minimize the H  norm of the transfer function The dimension of a is k by k (order, between 0 and n=dim(A)) The dimension of b is k by p (number of system inputs) The dimension of c is m (number of system outputs) by k The dimension of d is m by p Total number of variables is (k+m)(k+p) The case k=0 is static output feedback A + B 2 d C 2 B 2 c b C 2 a B 1 + B 2 d D 21 b D 21 C 1 + D 12 d C 2 D 12 cD 11 + D 12 d D 21 33 H  Fixed-Order Controller Design Choose matrices a, b, c, d to minimize the H  norm of the transfer function The dimension of a is k by k (order, between 0 and n=dim(A)) The dimension of b is k by p (number of system inputs) The dimension of c is m (number of system outputs) by k The dimension of d is m by p Total number of variables is (k+m)(k+p) The case k=0 is static output feedback When B1,C1 are empty, all I/O channels are in performance measure A + B 2 d C 2 B 2 c b C 2 a B 1 + B 2 d D 21 b D 21 C 1 + D 12 d C 2 D 12 cD 11 + D 12 d D 21 34 HIFOO: H  Fixed-Order Optimization Aims to find a, b, c, d for which H  norm is locally optimal Begins by minimizing the spectral abscissa max(real(eig(A-block))). To convert from m / s 2 to k m / h 2, multiply the acceleration value by . k m / h 2 = ( m / s 2 ) To convert from k m / h 2 to m / s 2, multiply the acceleration value by 0. . SI unit of M is k g, that of L is m, T is s and temperature (K) is K. So, SI unit of [ M 0 L 2 T − 2 K − 1 ] is m 2 s − 2 K − 1 . Was this answer helpful?

proving combinatorics identity - $ sum_{k=0}^m{n-k choose m-k

System is based on twenty-two. The twenty two symbols used are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, G, H, I, J, K and L.Base-23Base 23 or triovigesimal numeral system is based on twenty-three. The twenty three symbols used are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, G, H, I, J, K, L and M.Base-24The base-24 system is a numeral system with 24 as its base. The symbols used in this system are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, G, H, I, J, K, L, M, and N.Base-25The base-25 system is a numeral system with 25 as its base. The symbols used in this system are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, G, H, I, J, K, L, M, N and O.Base-26A hexavigesimal numeral system has a base of twenty-six. The symbols used in this system are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, G, H, I, J, K, L, M, N, O and P.Base-27A septemvigesimal numeral system has a base of twenty-seven. The symbols used in this system are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P and Q.Base-28The base 28 numberal system is based on twenty-eight and uses 28 different symbols ( 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q and R.)Base-29The base 29 numberal system is based on twenty-nine and uses 29 different symbols ( 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R and S.)Base-3Ternay or trinary is the base-3 numeral system. Ternary requireds only three symbols: 0, 1 and 2.Base-30Trigesimal or base