How Does the Derivative of the Delta Function Affect y(t) in Signal Processing?

[ace1719]

I have an evil TA (who makes the assignments) who likes to give us torturously difficult assignments on stuff we haven't been taught (and in many cases don't even understand conceptually).

Homework Statement

The input signal, x(t) is a real-valued bandlimited signal with bandwidth W. Find y(t).

Homework Equations

I'm using mathematica notation here, so I'm not sure whether it will come out properly or not.

a. y(t)=x(t)p(t)

b. p(t)=$\sum\delta'(t-kT)$ where k goes from -∞ to ∞
and $\delta'(t)=\frac{d}{dt}\delta(t)$

The Attempt at a Solution

x(t) is not explicitly given in the question, but it's spectrum is, however the real issue here is finding p(t). I know the integral of the delta function is 1, so does that mean the derivative of the delta function is 0, therefore making the summation (essentially an integral) a constant in discrete time?

 

[micromass]

The delta function is a distribution and so is the derivative of the delta function. The defining property of the delta function is that

$$\int_{-\infty}^{+\infty}f(x)\delta(x)dx = f(0)$$

To find what the derivative does, you can just do integration by parts formally (note that you can assume that $f$ vanishes outsides a closed interval). Thus

$$\int_{-\infty}^{+\infty} f(x)\delta^\prime(x)dx = - \int_{-\infty}^{+\infty} f^\prime(x)\delta(x) = -f^\prime(0)$$

There are other formulas for the derivative of the delta function for example

$$\delta^\prime(x) = -\frac{\delta(x)}{x}$$

but these should be interpreted properly.