我有一臺處于靜止模式的機器。當它開始運行時,它的頻率線性變化(從 0 到完全作業頻率),然后變得恒定,最后開始線性下降(從完全作業頻率到 0)。
我為此撰寫了以下代碼:
te = 300; % seconds - time to get fully operative
% (to arrive from 0 to fully operational frequency)
% and to stop (from fully operational to 0)
me = 10; % eccentric mass
e = 0.5; % eccentricity of the eccentric mass
omega=0.2; % fully operational force frequency
% i need three time vectors
tt0=[0:0.1:te]; % uphill frequency part (from 0 to fully operational frequency)
tt1=[te 0.1:0.1:1.5*te]; % permanent fully operational frequency part
tt2=[1.5*te 0.1:0.1:2.5*te]; % downward frequency part (from fully operational frequency to 0)
tt_1=[tt0,tt1,tt2]; % whole time vector
n7=numel(tt_1);
n5=numel(tt0);
n6=numel(tt1);
n8=numel(tt2);
% create a 3 force frequency vectors
delta_omega1=omega/(n5-1);
delta_omega2=omega/(n8-1);
omega000=[0:delta_omega1:omega]; % from 0 to fully operational frequency
omega001=omega*ones(1,n6); % permanent fully operational frequency part
omega002=[omega:-delta_omega2:0]; % from fully operational frequency to 0
omega00=[omega000,omega001,omega002]; % whole force frequency vector
p00=zeros(n7,1);
p_0010=zeros(n5,1);
p_0011=zeros(n6,1);
p_0012=zeros(n8,1);
p_001=zeros(n7,1);
% force amplitude calculation
for j=1:n7
p00(j,1) = (me*e*omega00(j)^2);
end
% then i create 3 sin force vectors (for 3 different force frequency laws)
for j=1:n5
p_0010(j,1) = p00(j,1)*sin(omega/(2*te)*(tt0(j))^2);
end
for j=1:n6
p_0011(j,1) = p00(j n5,1)*sin(omega*te omega*(tt1(j)-te));
end
for j=1:n8
t2=tt2(1);
p_0012(j,1) = p00(j n6 n5,1)*sin((-1/2*omega*te omega*tt2(j)-...
1/2*omega/te*((t2-te)^2 (tt2(j)-t2)^2)));
end
p_001=[p_0010;p_0011;p_0012]; % whole force vector
figure (1)
plot (tt_1,omega00)
xlabel('time')
ylabel('force frequency')
grid on;
hold on
figure (2)
plot (tt_1,p_001)
xlabel('time')
ylabel('applied force')
grid on;
hold on
這導致
這是對的。
但力圖:
是不正確的。
我怎么解決這個問題?我不需要力圖上的任何不連續性(因為它會產生我系統的奇怪行為,例如這個奇異點中的“寄生共振”)。
uj5u.com熱心網友回復:
你對力量的表達是錯誤的。對于不平衡力,它的運算式me*omega^2*sin(omega*t)
只有在轉速 ω 恒定的情況下。如果轉速不是恒定的(如它線性增加或減少的情況),它的完整運算式是
對于旋轉機器,您應該在水平和垂直方向上具有不平衡力(上述力矢量的第二項和第三項)。δ 是相位角, φ_dot 是旋轉速度, φ_ddot 是旋轉加速度。請注意,對于線性增加或減少旋轉速度, φ_ddot = α = 常數,因此您的力運算式應該包含一個項
α*cos(phi)
對于恒定加速度,角位置就像您定義的那樣。只需正確定義每個時間間隔的初始角位置phi_0
(它們不為零,它們是前一個時間間隔的最后位置)。
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標籤:matlab
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