如何在Matlab中產生具有波動的正弦信號？

Question

我需要一個頻率為50 Hz，振幅為50的正弦波。此外，我還需要一個正弦波的（+10到-10）三角形波紋和一個頻率為1000 Hz的正弦波，如下所示

平均值是正弦波，但不是實際的波形是帶有波紋（即以三角形形式脈動）。 原始波形是正弦波形式的三角傳播（移動）

`

f_inv_ac = 50; 

fs_inv = 1e3;

p_inv_ac = 150e3;  % power in Watt

u_inv_ac_rms = 400; % output voltage (RMS) in V

i_inv_ac_rms = p_inv_ac/u_inv_ac_rms/sqrt(3);

t_inv_ac = 1/f_inv_ac; % Inverter output period in s

ts_inv = 1/fs_inv; % Inverter switching period in s

vector_calc = zeros(1, num_inv_switch);

t_inv = vector_calc;

angle_inv_rn  = vector_calc;


for i = 1:num_inv_switch

    t_inv(i) = i*ts_inv;

    angle_inv_rn(i) = 2*pi*t_inv(i)/t_inv_ac;

end


u_inv_ac_rn = u_inv_ac_rms*sqrt(2/3)*sin(angle_inv_rn); % R-N voltage

i_inv_ac_r = i_inv_ac_rms*sqrt(2)*sin(angle_inv_rn); % R current

figure 

plot(t_inv*1e3,i_inv_ac_r)


%% for ripple calulation

 fs_inv = 1e3; % frequency ripple

% current ripple

delta_i_inv_r = zeros(1, num_inv_switch);

% current peak-value

i_peak_inv_r = zeros(1, num_inv_switch);

% current bottom-value

i_bottom_inv_r = zeros(1, num_inv_switch);

 % Current ripple (peak-to-peak) (in version 1.00 no ripple considered)

    delta_i_inv_r(i) = i_inv_ac_r(i)*0.30;


 % Current-peak value

 i_peak_inv_r(i) = i_inv_ac_r(i) + delta_i_inv_r(i)/2;

 i_bottom_inv_r(i) = i_inv_ac_r(i) - delta_i_inv_r(i)/2;`

我不知道下一步如何進行。

Answer 1

我將不使用您的代碼就做這個問題，希望您不會受到冒犯，但是很難理解您所有的變量，並且您的某些變量丟失了，我無法運行您的代碼：

首先讓我們創建一個范圍：

t = 0:0.00001:0.1; %test range - change 0.00001 to get more samples.
y=50*sin(2*pi*50*t); %generated simple Sine wave
% generate Triangle wave
period=1/2000; % same as 1/frequency in your case, 2000Hz
phase=0; % 0 unless you want it phase shifted
ratio=0.5; % 50:50 ratio, if you want sawtooth use 0.99
amp=20*1.25; %generates the 0 to 20 range for triangle wave
u1=rem(t+phase*period/(2*pi)+period,period);
s=(u1<=period*ratio).*u1*amp/(period*ratio)+(u1>period*ratio).*(period-u1)*amp/(period*(1-ratio))-10; %NOTE this is actually Simulink's triangle wave generator

%combine them
z = s+y;

figure
plot(t,y,'r')
hold on
plot(t,z)

Answer 2

我認為在您的情況下，您想嘗試sawtooth的頻率和幅度。

amplitude=1;
amplitude2=0.1;
amplitude3=0.3;
t=1:0.01:10;
frequency=0.1;
frequency2=1;
frequency3=4;
sinusoid = amplitude * (sin(frequency * 2 * pi * t));

sinusoid2 = sinusoid + amplitude2 * (sin(frequency2 * 2 * pi * t));

square_wave_1 = sinusoid + amplitude2 * square(frequency2*2*pi*t,50);

square_wave_2 = sinusoid + amplitude2 * square(frequency2*2*pi*t,33);

sawtooth = sinusoid + amplitude3 * sawtooth(frequency3*pi*t);


figure;hold on
plot(t,sinusoid)
plot(t,sinusoid2,'k')
plot(t,square_wave_1,'r')
plot(t,square_wave_2,'g')
plot(t,sawtooth,'c')

請注意，我的鋸齒代碼在某些頻率和振幅下看起來很糟糕。