# Signals & Systems questions

#### Georacer

Joined Nov 25, 2009
5,182
I don't like doing this, for some reason, but it seems I gotta get all the help I have available to get those courses out of my back.

I 'll be asking some S&S related questions here, have a look at them when you get the time:

Q1: Prove that δ'(-t)=-δ'(t)

As hinted, I attempted this by using $$\int \phi (t) f_1(t) dt =\int \phi (t) f_2(t) dt \Leftrightarrow f_1(t)=f_2(t)$$

So,
$$\int_{-\infty}^{\infty} \phi(t) (- \delta ' (t)) dt=\int_{-\infty}^{\infty} \phi '(t) \delta (t) dt=\phi'(0)$$

and

$$\int_{-\infty}^{\infty} \phi(t) \delta ' (-t) dt=-\int_{\infty}^{-\infty} \phi(-\tau) \delta ' (\tau) d\tau$$
but I can't see the last step, to get to the same result. Any hints?

#### Zazoo

Joined Jul 27, 2011
114
Make the substitution $$t=-\tau$$ on the RHS

#### Georacer

Joined Nov 25, 2009
5,182
In which equation of all? If you are talking about the last one, that's what I did to get it.

#### steveb

Joined Jul 3, 2008
2,436
Consider what happens if $$\phi (t)$$ is chosen to be an odd function. Does this give you anything useful?

#### Zazoo

Joined Jul 27, 2011
114
In which equation of all? If you are talking about the last one, that's what I did to get it.
Ah, I misunderstood, sorry. I thought you were asking how to get the RHS to match the LHS in that step.

#### Georacer

Joined Nov 25, 2009
5,182
I thought about it, but shouldn't φ(t) be a completely arbitrary test function for my hint equivalence to be valid?

#### steveb

Joined Jul 3, 2008
2,436
I thought about it, but shouldn't φ(t) be a completely arbitrary test function for my hint equivalence to be valid?
I don't think it is required to be arbitrary. The thing you are trying to prove has nothing to do with the test function. Hence, if you need to restrict the function to have odd symmetry in order to extract a proof, I don't see a problem with that. If the "hint" relation is valid for any arbitrary function, then it is valid for any arbitrary odd function, as well.

Just looking visually, an odd function will allow you to make the next step in your proof, and then it's almost done from there.

#### Georacer

Joined Nov 25, 2009
5,182
Yes, if I can use an odd function for φ(t), consider the problem solved. It's just that I 'd like some confirmation on whether I can use it or not.

For example, if I choose φ(t)=u(t), then, any two functions that are equal for x>=0, but not equal for x<0, would appear as equal, under that scope.

#### steveb

Joined Jul 3, 2008
2,436
Yes, if I can use an odd function for φ(t), consider the problem solved. It's just that I 'd like some confirmation on whether I can use it or not.

For example, if I choose φ(t)=u(t), then, any two functions that are equal for x>=0, but not equal for x<0, would appear as equal, under that scope.
Yes, I see your point, and I think it is a valid one. The original relation is not fully stated, but it says that that relation is valid provided that it is true for all arbitrary function (or at least a very broad class of functions), and not just for one function (or a highly restricted class of funtions). You would like to develop a corollary of this which says the class of functions can be odd, and that is sufficient.

I would take the approach that any function can be represented as a sum of an odd function and an even function. This is a known theorem that isn't hard to prove. I think it will not be hard to use this fact on the original function and show that it is sufficient to restrict the class of functions to odd only. Look at the following relation.

$$\int \phi_e(t) f_1(t) dt+\int \phi_o(t) f_1(t) dt=\int \phi_e(t) f_2(t) dt+\int \phi_o(t) f_2(t) dt$$

If this relation is true for arbitrary odd and even test functions, then the corresponding terms must also be equal, it seems. Can we draw the necessary conclusion from this? If not, you will be forced to prove your relation for both odd and even functions.

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#### steveb

Joined Jul 3, 2008
2,436
Can we draw the necessary conclusion from this? If not, you will be forced to prove your relation for both odd and even functions.
Actually, this is probably the way to go. If you let phi be a sum of odd and even functions, then you can show that the integral of the odd parts are equal. Hence the even parts must be zero (otherwise they would add a value to one side and subtract it from the other side and destroy the equality). The integral of the even part of phi with the derivative of the delta function is zero if the latter is an odd function. But, that is exactly what you are trying to prove.

EDIT: Scratch all that. I tried these out on paper and they don't seem to work.

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#### steveb

Joined Jul 3, 2008
2,436
Question?

Have you incorporated all the important properties of the delta function into that final relation you have? It seems to me that the properties of delta are not factored in yet. It looks like integration by parts is used, and the fact that delta is zero at the limits is included. But, the fact that delta is an even function is not included, as far as I can tell. Have I missed something?

I think if the even property of delta is included (possibly using the integral definition of delta), then the odd property of delta prime will be embedded, and then breaking phi and delta prime into even and odd parts will allow you to prove that delta prime is odd.

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#### Georacer

Joined Nov 25, 2009
5,182
I think I got it. The part that troubled me was that $$-\int_{\infty}^{-\infty} \phi(-\tau) \delta ' (\tau) d\tau$$ wasn't of the form of $$\int_{-\infty}^{\infty} \phi(t) \delta ' (t) dt$$, because of the negative sign on τ, inside φ.

But letting g(t)=φ(-τ), we have:
$$\int_{-\infty}^{\infty} \phi(t) \delta ' (-t) dt= -\int_{\infty}^{-\infty} \phi(-\tau) \delta ' (\tau) d\tau= \int_{-\infty}^{\infty} g(t) \delta ' (t) dt= g'(0)= \left\ {\frac {dg(t)}{dt}} \right| _{t=0}=\\ \left\ {\frac {-d \phi(t)}{dt}} \right| _{t=0}=\\ -\phi ' (0)$$

It can also be proven though Wolframalpha's equation 17: $$x \delta(x)=-\delta(x)$$, using the fact that δ(t) is even.

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#### t_n_k

Joined Mar 6, 2009
5,455

#### Georacer

Joined Nov 25, 2009
5,182
I don't know about any contribution of the Hermitian tables to any turnover rule, and I don't need another chapter of maths invading in my brain right now.

But thanks, anyway!

#### t_n_k

Joined Mar 6, 2009
5,455
Perhaps in the end it boils down to the finer points of what a mathematician and an engineer consider an adequate proof in this matter.

I suspect the proof while intuitively obvious (for some) is somewhat impenetrable for the majority of us mere mortals.

#### Georacer

Joined Nov 25, 2009
5,182
I 'm content with the proof of post #13. I was given theorems and proven an identity. That's enough for me.

#### t_n_k

Joined Mar 6, 2009
5,455
One interesting point the material in that link makes is that the test function would satisfy the condition of being .....

"a differentiable function f(x) that vanishes at plus and minus infinity"