Laplace Transform is widely used to convert time domain differential equation into frequency domain equations.This is the basic idea that you should be concerned with.Now let's get started with the codes:-
>> ft= t;
>> laplace (ft) {after this press enter}
output=1/s^2
>> ft=t^2;
>> laplace (ft)
output= 2/s^3
>> ft=exp(-5*t);
>> laplace (ft)
output=1/s+5
>> ft=(t^2)*[exp(-5*t)];
>> laplace (ft)
output= 1/(s+5)^3
>> ft=sin(5*t);
>> laplace (ft)
output=5/(s^2+25)
>>ft= cos(3*t);
>> laplace (ft)
output=s/(s^2+9)
>> ft=[exp(-5*t)]*sin(3*t);
>> laplace (ft)
output= 3/(s+5)^2+9
If you experience any problem using this codes,do write it in the comment box!!That's all for today.
- Find the laplace transform of f(t)=t
>> ft= t;
>> laplace (ft) {after this press enter}
output=1/s^2
- Find the laplace transform of t^2.
>> ft=t^2;
>> laplace (ft)
output= 2/s^3
- Find the laplace transform of e^-5t.
>> ft=exp(-5*t);
>> laplace (ft)
output=1/s+5
- Find the laplace transform of t^2.e^-5t
>> ft=(t^2)*[exp(-5*t)];
>> laplace (ft)
output= 1/(s+5)^3
- Find the laplace transform of sin5t.
>> ft=sin(5*t);
>> laplace (ft)
output=5/(s^2+25)
- Find the laplace transform of cos3t.
>>ft= cos(3*t);
>> laplace (ft)
output=s/(s^2+9)
- Find the laplace transform of e^-5t(sin3t).
>> ft=[exp(-5*t)]*sin(3*t);
>> laplace (ft)
output= 3/(s+5)^2+9
If you experience any problem using this codes,do write it in the comment box!!That's all for today.









