一次在线字典的学习算法代码

说明：  实现bpsk在AWGN下的调制与解调，误码率计算。(psk in Rayleigh fading channel simulation model. Simulink achieve. Simulation can see Rayleigh fading channels for BPSK performance impact. And to be compared Rayleigh Channel and AWGN under BPSK performance)

read MIT ECG for the students(read ECG for the MIT students)

此程序给出了换热网络旁路设计中的相关增益分析，可计算出相对增益矩阵。(This program gives the bypass heat exchanger network design-related gain analysis to calculate the relative gain matrix.)

二维FDTD程序，采用Mur吸收边界，实现了较好的吸收效果，可以用来与UPML吸收边界比较。(2D FDTD program, the Mur absorbing boundary to achieve a better absorption, can be used with UPML absorbing boundary.)

针对转子系统非线性动力学图谱分析，绘制转子的轨迹图，相图，频谱图以及庞加莱图。(analysis about nonlinear dynamics of the rotor bearing system)

说明：  手写体识别 lenet5 LeNet5由7层CNN（不包含输入层）组成，上图中输入的原始图像大小是32×32像素，卷积层用Ci表示，子采样层（pooling，池化）用Si表示，全连接层用Fi表示。下面逐层介绍其作用和示意图上方的数字含义。(Lenet5 is composed of seven layers of CNN)

用Wilson方法计算结构动力反应(无阻尼)(Wilson calculated using the structure of dynamic response (without damping))

个人整理的，最齐全的Bursa_Wolf，布尔莎坐标转换资料，含matlab实例，说明，以及相应的引文资料。 纯属个人原创，请注明引用出处！

说明：  curvelet 工具包，功能强大，版本最新，是图像处理，多尺度分析的主要工具。(curvelet toolkits, powerful, version of the latest is the image processing, multi-scale analysis of the major tools.)

产生右图所示图像f1(m,n)，其中图像大小为256×256，中间亮条为128 ×32，暗处=0，亮处=100。对其进行FFT： ① 同屏显示原图f1(m,n)和FFT(f1)的幅度谱图； ② 若令f2(m,n)=（-1）m+n f1(m,n)，重复以上过程，比较二者幅度 谱的异同，简述理由； ③ 若将f2(m,n)顺时针旋转90 度得到f3(m,n)，试显示FFT(f3)的幅 度谱，并与FFT(f2)的幅度谱进行比较； ④ 若将f1(m,n) 顺时针旋转90 度得到f4(m,n)，令f5(m,n)=f1(m,n)+f4(m,n)，试显 示FFT(f5)的幅度谱，并指出其与FFT(f1)和FFT(f4)的关系； ⑤ 若令f6(m,n)=f2(m,n)+f3(m,n)，试显示FFT(f6)的幅度谱，并指出其与FFT(f2)和 FFT(f3)的关系，比较FFT(f6)和FFT(f5)的幅度谱。(Generating an image f1 (m, n) shown in the figure, wherein the image size is 256 256, the intermediate light bar 128 32 0 = dark, bright Department 100. Its FFT: ① screen display picture f1 (m, n) and the FFT (f1) of the amplitude spectrum ② If so f2 (m, n) = (-1) m+n f1 (m, n), repeat The above process, comparing the amplitude spectrum of the similarities and differences between the two, brief reasons ③ If f2 (m, n) 90 degrees clockwise to get f3 (m, n), try to display FFT (f3) the amplitude spectrum and with the FFT (f2) comparing the amplitude spectrum ④ If f1 (m, n) obtained by 90 degrees clockwise f4 (m, n), so f5 (m, n) = f1 (m, n)+f4 (m, n ), try to display FFT (f5) amplitude spectrum, and pointed out its relationship with the FFT (f1) and FFT (f4) of ⑤ If so f6 (m, n) = f2 (m, n)+f3 (m, n) and try to display the FFT (f6) amplitude spectrum, and pointed out its relationship with the FFT (f2) and FFT (f3), comparing FFT (f6) and FFT (f5) amplitude spectrum.)