How Do Impulse Trains Relate to the Derivative of a Periodic Signal?

[maverick280857]

Hi

The following question is from Oppenheim/Wilsky/Nawab chapter 1.

Consider a periodic signal

$$x(t) = 1$$ for $$0 \leq t \leq 1$$
$$x(t) = -2$$ for $$1 < t <2$$

with period $T = 2$. The derivative of this signal is related to the impulse train

$$g(t) = \sum_{k = -\infty}^{\infty}\delte(t-2k)$$

with period T = 2. It can be shown that

$$\frac{dx(t)}{dt} = A_{1}g(t-t_{1}) + A_{2}g(t-t_{2})$$

Determine the values of $A_{1}$, $t_{1}$, $A_{2}$, and $t_{2}$.

I got stuck with this one. Anyway here's my solution. Would appreciate any help in solving the problem.

$$x(t) = \sum_{k = -\infty}^{\infty}(u(t-2k) - u(t-2k-1)) + (-2)(u(t-2k-1) - u(t-2k-2))$$

so

$$x(t) = \sum_{k = -\infty}^{\infty}u(t-2k) - 3\sum_{k = -\infty}^{\infty}u(t-2k-1)) -2\sum_{k = -\infty}^{\infty}u(t-2k-2)$$

so

$$\frac{dx}{dt} = g(t) - 3g(t-1) - g(t-2)$$

which is wrong...

 

[maverick280857]

Okay I got it graphically, but what if I want to do it algebraically?

 

[ravenprp]

Hello,

Thank you for sharing your solution. Your approach is correct, but there is a small error in your final equation for the derivative. The correct equation should be:

\frac{dx}{dt} = g(t) - 3g(t-1) + 2g(t-2)

This is because the signal x(t) is equal to 1 from 0 to 1, and then -2 from 1 to 2. Therefore, when taking the derivative, the -2 will become +2.

From this, we can determine the values of A1, t1, A2, and t2:

A1 = 1, t1 = 0
A2 = -3, t2 = 1
A3 = 2, t3 = 2

I hope this helps and clarifies any confusion. Let me know if you have any further questions. Good luck with your studies!