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How to Solve with subs

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Kyle Langford
Kyle Langford 2022-2-18
评论： Kyle Langford 2022-2-25
I am trying to first solve for Tau first, and then plug the known value back in and solve for y(t) @ 1.5 seconds.
it is given that @ y(1.2)=80.
clear;clc;
syms T %Tau
syms yt
t=1.2;
y0=0;
KA=100; %steady response
U_t=(y0-KA)*exp(-t/T); %transient response
y_t=80; %given that y(t), if y(1.2)=80
eq3_5a=y_t==KA+U_t
Tau=vpasolve(eq3_5a,T)
subs(U_t,T,Tau)
eq3_5b=yt==KA+U_t
solve(eq3_5b,yt)

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采纳的回答

VBBV
VBBV 2022-2-18
Ran in:
clear;clc;
syms T yt
t=1.2;
y0=0;
KA=100; %steady response
U_t=(y0-KA)*exp(-t/T) %transient response
U_t = 
y_t=80; %given that y(t), if y(1.2)=80
eq3_5a=y_t==KA+U_t
eq3_5a = 
Tau=vpasolve(eq3_5a,T)
Tau = 
0.74560192147153417284863926581669
subs(U_t,T,Tau)
ans = 
eq3_5b=yt==KA+U_t
eq3_5b = 
Yt = solve(eq3_5b,yt)
Yt = 
vpa(subs(Yt,T,1.5),4)
ans = 
55.07
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VBBV
VBBV 2022-2-18
编辑：VBBV 2022-2-18
Ran in:
clear;clc;
syms T yt t
% t=1.2;
y0=0;
KA=100; %steady response
U_t= (y0-KA)*exp(-t/T) %transient response
U_t = 
y_t=80; %given that y(t), if y(1.2)=80
eq3_5a=y_t==KA+U_t;
Tau=vpasolve(subs(eq3_5a,t,1.2),T)
Tau = 
0.74560192147153417284863926581669
subs(U_t,T,Tau);
eq3_5b=yt==KA+U_t
eq3_5b = 
Yt = solve(eq3_5b,yt);
vpa(subs(Yt,[T t],[Tau 1.5]),4)
ans = 
86.63
Ok, It is possible to Subs 2 variables like above
Kyle Langford
Kyle Langford 2022-2-25
Sweet. Thanks! I tried to do something similar using {} but it didn't work. I see i was using the wrong characters.

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