Location:
Search - fc n
Search list
Description: 经典的模拟器游戏，是网络上游戏爱好者提供的！我也帮在这个供大家取用-classic simulator games that network game lovers! I also helped in the access for all
Platform: |
Size: 5703680 |
Author: hou |
Hits:
Description: 通信信号相干调制解调源码 Fc=10 ％载频
Fs=40 ％系统采样频率
Fd=1 ％码速率
N=Fs/Fd
df=10
numSymb=25 ％进行仿真的信息代码个数
M=2 ％进制数-Coherent modulation and demodulation of communication signal source Fc = 10 carrier frequency Fs = 40 systematic sampling frequency Fd = 1 code rate N = Fs/Fd df = 10 numSymb = 25 of the information code to simulate the number of M = 2 hexadecimal number
Platform: |
Size: 1024 |
Author: 包耐峰 |
Hits:
Description: FIR滤波器设计与实现，Fc=2000Hz，N=65，FIR_LPF.-FIR Filter Design and Implementation ,Fc=2000Hz，N=65，FIR_LPF.
Platform: |
Size: 5120 |
Author: 闵晓东 |
Hits:
Description: fm=input( Enter the message signal frequency fm= )
fc=input( Enter the carrier frequency fc= )
Am=input( Enter the message signal amplitude Am= )
Ac=input( Enter the carrier signal frequency Ac= )
n=input( Enter the no of DFT points: )
tm=(1/fm)
tc=(1/fc)
nfc=40
t=0:(tc/nfc):4*tm
msg=Am*cos(2*pi*fm*t)
car=Ac*cos(2*pi*fc*t)
Modulation
dsb=msg.*car
AM=(dsb/Ac)+car
figure(1)
subplot(2,1,1)
plot(t,msg)
subplot(2,1,2)
plot(t,car)
figure(2)
subplot(2,1,1)
plot(t,dsb)
subplot(2,1,2)
plot(t,AM) -fm=input( Enter the message signal frequency fm= )
fc=input( Enter the carrier frequency fc= )
Am=input( Enter the message signal amplitude Am= )
Ac=input( Enter the carrier signal frequency Ac= )
n=input( Enter the no of DFT points: )
tm=(1/fm)
tc=(1/fc)
nfc=40
t=0:(tc/nfc):4*tm
msg=Am*cos(2*pi*fm*t)
car=Ac*cos(2*pi*fc*t)
Modulation
dsb=msg.*car
AM=(dsb/Ac)+car
figure(1)
subplot(2,1,1)
plot(t,msg)
subplot(2,1,2)
plot(t,car)
figure(2)
subplot(2,1,1)
plot(t,dsb)
subplot(2,1,2)
plot(t,AM)
Platform: |
Size: 1024 |
Author: dasu |
Hits:
Description: virtualNES模拟器代码，供学习编写FC模拟器的同学参考。-virtualNES simulator code, and for people learning FC simulator programming.
Platform: |
Size: 631808 |
Author: Jazzy_K |
Hits:
Description: 在电脑上玩FC/NES红白机游戏少不了模拟器，网上下个NES格式的模拟游戏文件就可以玩了。VC6工程，需要DirectX SDK的支持，如编译出错请将DIRECTX目录下的DDraw.h和DInput.h两个文件拷到工程根目录下，还有个IDC_什么的未定义，直接将该语句删除就可以了。一般是一次编译通过。编译好的.exe文件需要.vlp文件，也压到压缩包里了-In the computer playing FC/NES NES simulator games and ultimately, the next online simulation game NES format files can be played. VC6 project requires the support of DirectX SDK, such as compiler error please DIRECTX directory by copying the two files DDraw.h and DInput.h project root directory, there are not defined what a IDC_ directly delete the statement on it. Generally is a compile. Compiled. Exe files need to be. Vlp file, but also pressure to compress the bag was
Platform: |
Size: 959488 |
Author: wanghan |
Hits:
Description: Dado x(t) = 3[1+0.2(cos100πt+cos200πt)]cos10,000πt + n(t), donde n(t) es Ruido Blanco (en realidad coloreado) Gausiano de banda limitada con Sn(f) = 2 µ w/Hz hasta una frecuencia de 10,000 Hz. Asumiendo fc=5,000 Hz y B=600 Hz, determine la (S/N) a la entrada del BPF, a la entrada del LPF y a la salida del LPF de la siguiente figura. Asuma la existencia de un capacitor de bloqueo de DC a la salida del LPF en caso de requerirlo.-Dado x(t) = 3[1+0.2(cos100πt+cos200πt)]cos10,000πt + n(t), donde n(t) es Ruido Blanco (en realidad coloreado) Gausiano de banda limitada con Sn(f) = 2 µ w/Hz hasta una frecuencia de 10,000 Hz. Asumiendo fc=5,000 Hz y B=600 Hz, determine la (S/N) a la entrada del BPF, a la entrada del LPF y a la salida del LPF de la siguiente figura. Asuma la existencia de un capacitor de bloqueo de DC a la salida del LPF en caso de requerirlo.
Platform: |
Size: 56320 |
Author: erierieri |
Hits:
Description: Dado x(t) = 3[1+0.2(cos100πt+cos200πt)]cos10,000πt + n(t), donde n(t) es Ruido Blanco (en realidad coloreado) Gausiano de banda limitada con Sn(f) = 2 µ w/Hz hasta una frecuencia de 10,000 Hz. Asumiendo fc=5,000 Hz y B=600 Hz, determine la (S/N) a la entrada del BPF, a la entrada del LPF y a la salida del LPF de la siguiente figura. Asuma la existencia de un capacitor de bloqueo de DC a la salida del LPF en caso de requerirlo.-Dado x(t) = 3[1+0.2(cos100πt+cos200πt)]cos10,000πt + n(t), donde n(t) es Ruido Blanco (en realidad coloreado) Gausiano de banda limitada con Sn(f) = 2 µ w/Hz hasta una frecuencia de 10,000 Hz. Asumiendo fc=5,000 Hz y B=600 Hz, determine la (S/N) a la entrada del BPF, a la entrada del LPF y a la salida del LPF de la siguiente figura. Asuma la existencia de un capacitor de bloqueo de DC a la salida del LPF en caso de requerirlo.
Platform: |
Size: 129024 |
Author: erierieri |
Hits:
Description: 存储区域网中N端口的设计和实现
存储区域网中N端口的设计和实现-N-port storage area network design and implementation of storage area network design and implementation of N ports
Platform: |
Size: 2378752 |
Author: 秦清全 |
Hits:
Description: 当前，矢量控制是一种优越的交流电机控制方式，它模拟直流电机的控制方式使得交流电机也能取得与直流电机相媲美的控制
效果。依据矢量控制的基本原理和方法，用
456752 8 9:4 7:<=
模块构建了一个基于转子磁场定向的旋转坐标系下的交流异步
电机矢量控制系统仿真模型，对其速度控制器提出并设计一种模糊滑模变结构控制算法。该控制算法具备模糊逻辑控制和滑模
变结构控制两者优点，并且较好地解决了滑动模态的抖动问题。仿真结果表明了该设计的可行性以及具有良好的动静态性能、
较强的鲁棒性-6.C+D& EFB@.A B.*@A.G ! H’ , )?+ ?1IFA).A B.*@A.G J.CF .K )*C1B@).* J.@.AL :@ B+* .>@+)* @(F ?+JF B.*@A.G FKKFB@ .K M’ J.@.A >D J.CFG)*/ @(F
B.*@A.G J.CF .K M’ J.@.AL 5BB.AC)*/ @. EFB@.A B.*@A.G N ? IA)*B)IGF+*C JF@(.C& 456752 8 9:4 7:<= J.C1GF O+? 1?FC @. >1)GC+ ?)J1P
G+@).* J.CFG .K 5’ J.@.A EFB@.A B.*@A.G ?D?@FJ >+?FC .* @(F A.@.A KG1Q .A)F*@FC B..AC)*+@F )* @()? I+IFAL R1SSD ?G)C)*/ E+A)+>GF ?@A1B@1AF
B.*@A.G O+? IA.I.?FC+*C FJIG.DFC )* ?IFFC B.*@A.GGFAL 6(F B.*@A.GGFA I.??F??FC @(F JFA)@? .K K1SSD G./)B B.*@A.G+*C ?G)C)*/ J.CF E+A)+>GF
?@A1B@1AF B.*@A.G
，
+*C )@ B+* ?.GEF @(F IA.>GFJ .K @(F B(+@@FA)*/ .K ?G)C)*/ J.CFL 6(F AF?1G@? .K ?)J1G+@).* )*C)B+@F @(+@ @(F B.*@A.GGFA O+?
O.AT+>GF O)@( /..C A.>1?@*F??+*C /..C CD*+J)B+*C ?@+@)B IFAK.AJ+*BF F?IFB)+GGD )* @(F K)FGC .K F*/)*FFA)*/
Platform: |
Size: 380928 |
Author: teamm |
Hits:
Description: 用verilog实现8为计数器频率范围20-80kHz，根据DDS原理来一个时钟计数器记一下，n=n+1，根据公式fout=（fc÷x）÷2，fout=80 fc=320，所以n≥2时，再取反，又由公式 fout=(k.fc)÷2^n，k=50hz，fout=80khz，fc=320，所以数据的位宽n≥7。
设计要求两路方波信号的相位差在0-360゜可调，可以根据延时来实现。具体的-8 is realized with verilog counter frequency range 20-80kHz, according to DDS principle to remember what a clock counter, n = n+1, according to the formula fout = (fcx) 2, fout = 80 fc = 320, so n ≥ 2:00 then negated, but also by the formula fout = (k.fc) 2 ^ n, k = 50hz, fout = 80khz, fc = 320, so the data bit width n ≥ 7. Design requirements for two square wave signal phase at 0-360 ゜ adjustable delay can be achieved. Specific
Platform: |
Size: 24576 |
Author: 倪飞 |
Hits:
Description: 1. 设计一个 N = 10 的低通 fir 滤波器，对 fs = 8kHz 的信号 fc = 2.5kHz
2. 设计等同性能的 IIR 滤波器
3. 定点实现，不能使用浮点
4. 测试多正弦波合成文件 MultiToneTest.wav 分析输出结果
5. 测试语音文件 sc03.wav 得到输出结果-1. Design a lowpass fir filter N = 10, for fs = fc 8kHz signal design equivalent of = 2.5kHz 2. 3. IIR filter performance fixed-point implementation, you can not use floating-point multi-sine wave synthesis document 4. Testing 5. Test MultiToneTest.wav analysis output file sc03.wav get voice output
Platform: |
Size: 901120 |
Author: 王艳 |
Hits:
Description: empirical formula with kaiser
clc
clear all
fs=1000
fc=250
df=50
r=0.001
f=fc/fs
dw=2*pi*(df/fs)
a=-20*log(r)
n=floor(((a-8)/(2.285*dw))+1)
if a>50
b=0.1102*(a-8.7)
elseif a>=21 && a<=50
b=0.5842*((a-21)^0.4)+0.07886*(a-21)
elseif a<21
b=0.0
end
w=kaiser(n,b)
for i=1:n
if i~=(n-1)/2
hd(i)= (2*f*sin((i-((n-1)/2))*2*pi*f))/((i-((n-1)/2))*2*pi*f)
elseif i==(n-1)/2
hd(i)=2*f
end
end
for j=1:n
h(j)=w(j)*hd(j)
end
subplot(3,1,1), plot(w)
subplot(3,1,2), plot(h)
subplot(3,1,3), plot(h,n)
- empirical formula with kaiser
clc
clear all
fs=1000
fc=250
df=50
r=0.001
f=fc/fs
dw=2*pi*(df/fs)
a=-20*log(r)
n=floor(((a-8)/(2.285*dw))+1)
if a>50
b=0.1102*(a-8.7)
elseif a>=21 && a<=50
b=0.5842*((a-21)^0.4)+0.07886*(a-21)
elseif a<21
b=0.0
end
w=kaiser(n,b)
for i=1:n
if i~=(n-1)/2
hd(i)= (2*f*sin((i-((n-1)/2))*2*pi*f))/((i-((n-1)/2))*2*pi*f)
elseif i==(n-1)/2
hd(i)=2*f
end
end
for j=1:n
h(j)=w(j)*hd(j)
end
subplot(3,1,1), plot(w)
subplot(3,1,2), plot(h)
subplot(3,1,3), plot(h,n)
Platform: |
Size: 81920 |
Author: rezwan |
Hits:
Description: empirical formula with kaiser
clc
clear all
fs=1000
fc=250
df=50
r=0.001
f=fc/fs
dw=2*pi*(df/fs)
a=-20*log(r)
n=floor(((a-8)/(2.285*dw))+1)
if a>50
b=0.1102*(a-8.7)
elseif a>=21 && a<=50
b=0.5842*((a-21)^0.4)+0.07886*(a-21)
elseif a<21
b=0.0
end
w=kaiser(n,b)
for i=1:n
if i~=(n-1)/2
hd(i)= (2*f*sin((i-((n-1)/2))*2*pi*f))/((i-((n-1)/2))*2*pi*f)
elseif i==(n-1)/2
hd(i)=2*f
end
end
for j=1:n
h(j)=w(j)*hd(j)
end
subplot(3,1,1), plot(w)
subplot(3,1,2), plot(h)
subplot(3,1,3), plot(h,n)
- empirical formula with kaiser
clc
clear all
fs=1000
fc=250
df=50
r=0.001
f=fc/fs
dw=2*pi*(df/fs)
a=-20*log(r)
n=floor(((a-8)/(2.285*dw))+1)
if a>50
b=0.1102*(a-8.7)
elseif a>=21 && a<=50
b=0.5842*((a-21)^0.4)+0.07886*(a-21)
elseif a<21
b=0.0
end
w=kaiser(n,b)
for i=1:n
if i~=(n-1)/2
hd(i)= (2*f*sin((i-((n-1)/2))*2*pi*f))/((i-((n-1)/2))*2*pi*f)
elseif i==(n-1)/2
hd(i)=2*f
end
end
for j=1:n
h(j)=w(j)*hd(j)
end
subplot(3,1,1), plot(w)
subplot(3,1,2), plot(h)
subplot(3,1,3), plot(h,n)
Platform: |
Size: 541696 |
Author: rezwan |
Hits:
Description: clc
fm1=3000000
fm2=5000000
fc=500000000
fs=100000000
Am1=2
Am2=0.3
A=2
N=512
K=N-1
n=0:N-1
t=(0:1/fs:K/fs)
m1=Am1*sin(2*pi*fm1*t)+Am2*sin(2*pi*fm2*t)
figure(1)
subplot(1,1,1),plot(t,m1),title( m2(t) )-clc
fm1=3000000
fm2=5000000
fc=500000000
fs=100000000
Am1=2
Am2=0.3
A=2
N=512
K=N-1
n=0:N-1
t=(0:1/fs:K/fs)
m1=Am1*sin(2*pi*fm1*t)+Am2*sin(2*pi*fm2*t)
figure(1)
subplot(1,1,1),plot(t,m1),title( m2(t) )
Platform: |
Size: 2048 |
Author: yahe |
Hits: