Convolution and Impulse Response

  • Thread starter mym786
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  • #1
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If one has input x(t), then convolving x(t) with impulse response of the system would give the zero-state of the system.

For example, we have a system described as :
(D[tex]^{2}[/tex] + 4D + 3)y(t) = (D+5)f(t).

I computed system impulse response which is :

h(t) = 2e[tex]^{-t}[/tex] - e[tex]^{-3t}[/tex]

Now if say f(t) = e[tex]^{t}[/tex]

f(t) (convolved) h(t) does not equal the particular solution of D.E. (Particular solution is zero-state) .

Why ??
 

Answers and Replies

  • #2
My DSP is a little dusty, but I'll try to work this out with you.

First the impulse response:
Taking Laplacian and assuming y(0)=0 and y'(0)=0

[tex]h(s)=\frac{(s+5)}{(s+3)(s+1)}[/tex] which can be rewritten as

[tex]h(s)=\frac{s+\frac{3}{2}+\frac{7}{2}}{(s+\frac{3}{2})^2 +\frac{3}{4}}[/tex]

[tex]h(s)=\frac{s+\frac{3}{2}}{(s+\frac{3}{2})^2 +\frac{3}{4}} + \frac{7}{2}\frac{2}{\sqrt{3}}\frac{\frac{\sqrt{3}}{2}}{(s+\frac{3}{2})^2 +\frac{3}{4}} [/tex]

These two expressions are in standard form. Taking the inverse Laplace Transform we get:

[tex]h(t)=e^{-\frac{\sqrt{3}}{2}t} (sin\frac{\sqrt{3}}{2}t + \frac{7}{\sqrt{3}}cos\frac{\sqrt{3}}{2}t)[/tex]

Can you show us how you came to your expression for the impulse response?
 
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  • #3
uart
Science Advisor
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My DSP is a little dusty, but I'll try to work this out with you.

First the impulse response:
Taking Laplacian and assuming y(0)=0 and y'(0)=0

[tex]h(s)=\frac{(s+5)}{(s+3)(s+1)}[/tex] which can be rewritten as

[tex]h(s)=\frac{s+\frac{3}{2}+\frac{7}{2}}{(s+\frac{3}{2})^2 +\frac{3}{4}}[/tex]

No that step is wrong. You don't want to express it in "completed square" form unless the quadratic has no real roots. Obviously from the previous step it has real roots.

BTW. Your completed square form is for the wrong polynomial => s^2 + 3s + 3
 
  • #4
Sorry for that. Messed up the polynomial working it out in tex. Not to hijack this thread but I've forgotten how to work out the solution to the DE. The characteristic solution comes from putting f(t)=0 and working out the roots etc. How would you work out the particular integral in this case? And how would you work out the transfer function from the laplacians.
 
  • #5
uart
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Method 1. Homogeneous and particular solutions.

For a linear DE like this one start with the solution of the homogeneous equation (RHS replaced with zero).

The homogeneous solution is :

[tex]y_h = A e^{-t} + B e^{-3t}[/tex]

Then find the particular solution to the actual DE, which is typically of the same form as the excitation function, x(t). In this case,

[tex]y_p = \frac{3}{4} e^t[/tex]

Then select constants A and B to satisfy the initial conditions, in this case the zero initial conditions y(0)=0 and y'(0)=0. Here we get [itex]A = -\frac{3}{2}[/itex] and [itex]B = \frac{3}{4}[/itex].

So

[tex] y(t) = y_h + y_p = \frac{3\,e^t}{4} - \frac{3\,e^{-t}}{2} + \frac{3\,e^{-3t}}{4}[/itex]
 
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  • #6
uart
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Method 2. Using the convolution integral.

The transfer function is :

[tex] H(s) = \frac{s+5}{s^2 + 4s + 3} [/tex]

And it's inverse Laplace transform h(t) is the impulse response. So in this case the impulse response is,

[itex]h(t) = \left( 2\,e^{-t} - e^{-3t} \right) U(t)[/itex]

By the convolution integral we get,

[tex]y(t) = (x \star h)(t) = \int_{-\infty}^{+\infty} h(\lambda) \, x(t - \lambda) \, d\lambda [/tex]

Applying the convolution integral (see *note) with [itex]x = e^t \, U(t)[/itex] we get,

[tex] y(t) = \frac{3\,e^t}{4} -e^{-t} + \frac{e^{-3t}}{4}[/tex]

We see that though this answer is definitely of the correct form it does not fully satisfy the zero initial conditions.
That is, y(0) = 0 but y'(0) [itex]\neq[/itex] 0. This is the concern expressed by the OP.
We'll get some insight into why this is the case when I look at method three (the method of Laplace transforms) in one moment.

*Note : It's important to include the step function, U(t), in the definition of h(t) and x(t) so that everything starts at zero. This reduces the convolution integral to finite limits.
 
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  • #7
uart
Science Advisor
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Method Three. Laplace Transforms.

Here we transform the DE using the following relations,

[itex] L\{f'\} = s L\{f\} - f(0) [/itex] and [itex] L\{f''\} = s^2 L\{f\} - s f(0) - f'(0)[/itex]

So for the given DE and zero initial conditions (on y) we get,

[tex] (s^2 + 4s + 3) Y = (s + 5) X - x(0) [/tex]

Substituting in the transform for x(t) and the initial condition x(0)=1 (and factorizing) gives,

[tex] Y = \frac{s + 5}{(s+1)(s+3)(s-1)} - \frac{1}{(s+1)(s+3)} [/tex]

Taking inverse Laplace transforms you get precisely the the same results as per method one (post #5 above). Interestingly if you fail to include the second term in the above expression, the one that accounts for the non zero initial value of x(t), then you get the same results as the convolution integral of method two above.

So I guess the answer to the OP question is that the convolution integral doesn't cope with the non zero initial conditions on [itex]x(t)= e^t U(t)[/itex], which abruptly jumps from x=0 to x=1 at t=0.

You can verify that if you repeat the problem but with [itex]x(t) = e^t - 1[/itex] then in this case all three of the above methods will yield identical solutions.
 
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