发布时间:2023-11-17 15:00
clc,clear,close all;
x = linspace(-2,2,10);
y = exp(-x.^2);
figure(1)
stem(x,y,"LineWidth",1.5)
grid on
title('f(x)')
%interp1线性插值
xq = linspace(-2,2,20);
vq1 = interp1(x,y,xq);
figure(2)
plot(x,y,'o',xq,vq1,':.',"LineWidth",1.5);
xlim([-2 2]);
grid on
title('interp1线性插值');
%polyfit多项式拟合插值
p = polyfit(x,y,7);
y1 = polyval(p,x);
figure(3)
plot(x,y,'g-o',"LineWidth",1.5)
hold on
plot(x,y1,"LineWidth",1.5)
hold off
xlim([-2 2]);
grid on
title('polyfit多项式拟合插值');
%spline三次样条插值
yy = spline(x,y,xq);
figure(4)
plot(x,y,'o',xq,yy,"LineWidth",1.5)
xlim([-2 2]);
grid on
title('spline三次样条插值');
问:已知输入信号为混有噪声的信号
x(t)=sin(250πt)+cos(500πt)+cos(700πt)
①画出输入信号时域波形和频谱图,并指出其包含的频率成分(以Hz为单位);
②假定输入信号的中间频率成分为噪声信号,请问应如何设计数字滤波器处理该混合信号,给出合理的设计指标并说明理由,画出滤波器的频响特性:要求两种以上的实现方案。
③用(2)中设计的滤波器完成滤波并验证设计方案,画出滤波器输出信号时域波形和频谱图。
答:
clc,clear,close all;
t = -0.08:0.0001:0.08;
n = -100:100;
L = length(n);
fs = 1000;
x0 = sin(2*pi*125.*n/fs); %时域采样后的信号t=nT=n/fs
x1 = cos(2*pi*250.*n/fs);
x2 = cos(2*pi*350.*n/fs);
x3 = x0 + x1 + x2;
%时域信号
figure(1)
subplot(411)
plot(t,sin(2*pi*125.*t),"LineWidth",1.5)
grid on
subplot(412)
plot(t,cos(2*pi*250.*t),"LineWidth",1.5)
grid on
subplot(413)
plot(t,cos(2*pi*350.*t),"LineWidth",1.5)
grid on
subplot(414)
plot(t,sin(2*pi*125.*t)+cos(2*pi*250.*t)+cos(2*pi*350.*t),"LineWidth",1.5)
grid on
%滤波器设计
wn = 3/5; %截止频率wn为3/5pi(300Hz),wn = 2pi*f/fs
N = 60; %阶数选择
hn = fir1(N-1,wn,boxcar(N)); %10阶FIR低通滤波器
figure(2)
freqz(hn,1);
figure(3)
y = fftfilt(hn,x3); %经过FIR滤波器后得到的信号
plot(n,y,"LineWidth",1.5)
grid on
%频谱分析
X = fft(x3); %未滤波前的频谱
p2 = abs(X/L);
p1 = p2(1:L/2+1);
p1(2:end-1) = 2*p1(2:end-1);
Y = fft(y); %输出信号的fft
P2 = abs(Y/L);
P1 = P2(1:L/2+1);
P1(2:end-1) = 2*P1(2:end-1);
f = fs*(0:(L/2))/L;
figure(4)
subplot(211)
plot(f,p1,"LineWidth",1.5)
title('Single-Sided Amplitude Spectrum of X(t)')
xlabel('f (Hz)')
ylabel('|p1(f)|')
grid on
subplot(212)
plot(f,P1,"LineWidth",1.5)
title('Single-Sided Amplitude Spectrum of Y(t)')
xlabel('f (Hz)')
ylabel('|P1(f)|')
grid on
时域波形:分别为125(2pi*125=250pi)、250和350Hz的余弦信号
截止频率为300Hz的FIR低通滤波器:
滤波前和滤波后的频谱:
滤波后的时域信号:
问题相关代码:
clear; clc; close all
[,,raw ] = xlsread('体育设施热力图数据.xlsx');
mat = cell2mat(raw(2:end, 2:end));
imagesc(mat); %生成热图
c=colorbar;
colormap hot;
ylabel(c,'Population Heat');
caxis([1 11]) %更改右侧颜色条最大最小值
xticks(1:13) %x轴分成13等分
xticklabels(raw(1,2:end))
yticks(1:15)
yticklabels(raw(2:end,1))
预期效果:
答:能是数据量较少的原因,预期效果应该用了插值;
而不仅仅是用的imagesc,更像是颜色插值后的伪彩图,建议尝试pcolor()
clc,clear,close all;
data=round(rand(1,900)*100);
data=reshape(data,10,90);
h=pcolor(data);
h.FaceColor = 'interp';
set(h,'LineStyle','none');
clc,clear,close all;
data=round(rand(1,900)*100);
data=reshape(data,10,90);
h=pcolor(data);
h.FaceColor = 'interp';
colormap jet;
colorbar;
set(h,'LineStyle','none');