Impulse Response using Dirac(t)
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Consider the following code segment used to generate the step response
syms y t
y=dsolve('D2y+5*Dy+6*y=heaviside(t)','y(0)=0','Dy(0)=0','t');
Now, the impulse response can be determined (and plotted) using
ir = diff(y);
subplot(211)
ezplot(ir,[0,5]);
title('Impulse response from Step reponse');
However, when I try to find (and plot) the impulse response using the following code, it gives the same shape as before, but with (almost) half the magnitude.
x=dsolve('D2y+5*Dy+6*y=dirac(t)','y(0)=0','Dy(0)=0','t');
subplot(212)
ezplot(x,[0,5]);
title('Impulse response From dirac')
Although I personally think that the answer must be the same, can anyone explain why it isn't the case??
采纳的回答
Teja Muppirala
2011-10-16
I think this is a very good question. Here is my explanation, but maybe someone can give a better one.
In this equation:
dsolve('D2y+5*Dy+6*y=dirac(t)','y(0)=0','Dy(0)=0','t');
Dy is actually discontinuous at the origin. The way the initial conditions are interpreted is that Dy(0) means the average of Dy(0+) and Dy(0-).
Here (0+) means the limit from the positive side, (0-) means the limit from the negative side. If you look at x around the origin, you will indeed see that the derivative is actually discontinuous at the origin. At (0-) it is negative, and at (0+) it is positive with the same magnitude so that the AVERAGE value is zero, It's a little different than how it is defined for Laplace transforms, where initial conditions are taken at (0-).
Try this for example:
dsolve('Dy = dirac(t)', 'y(0) = 0')
And you will see that it is not exactly the step function, but rather a step function that is shifted down by one half to make the average value at y(0) = 0.
ans = heaviside(t) - 1/2
In fact these two evaluate to be the same:
dsolve('D2y+5*Dy+6*y=dirac(t)','y(0)=0','Dy(0)=0','t');
diff(dsolve('D2y+5*Dy+6*y=heaviside(t)-1/2','y(0)=0','Dy(0)=0','t'));
Note, also, that the following two expressions where I simply added a delay will also give you the same answers, because in this case there is no discontinuity at the t = 0 and the initial conditions are in agreement regardless of which definition you use:
dsolve('D2y+5*Dy+6*y=dirac(t-1)','y(0)=0','Dy(0)=0','t');
diff(dsolve('D2y+5*Dy+6*y=heaviside(t-1)','y(0)=0','Dy(0)=0','t');
2 个评论
Ultrazord
2011-10-17
Teja Muppirala
2011-10-17
There is a very easy workaround. Use limits to enforce your boundary conditions:
x = dsolve('D2y+5*Dy+6*y=dirac(t)','y(e)=0','Dy(e)=0','t');
x = limit(x,'e',0,'left')
ezplot(x,[-1 5])
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