- #1
maverick280857
- 1,774
- 5
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...
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...