NOTE: For both a) and b) assume x(t) has finite length.
a) Is the following system time invariant?
\(y(t) = \int_{-a}^{\infty}x(\tau)d\tau\)
b) Is the following system time invariant?
\(y(t) = \int_{-\infty}^{\infty}x(\tau)d\tau\)
My attempt at the solution,
a)
Shifting the output gives,
\(y(t-T) = \int_{-a}^{\infty}x(\tau-T)d\tau\)
If the input is delayed,
\(x_{d}(t) = x(t-T) \rightarrow y_{d}(t) = \int_{-a}^{\infty}x_{d}(\tau)d\tau =\int_{-a}^{\infty}x(\tau-T)d\tau = y(t-T) \)
Therefor it is time invariant.
b) I can't see how anything changes from case a) to case b). Does anything change? Why?
Thanks again!
a) Is the following system time invariant?
\(y(t) = \int_{-a}^{\infty}x(\tau)d\tau\)
b) Is the following system time invariant?
\(y(t) = \int_{-\infty}^{\infty}x(\tau)d\tau\)
My attempt at the solution,
a)
Shifting the output gives,
\(y(t-T) = \int_{-a}^{\infty}x(\tau-T)d\tau\)
If the input is delayed,
\(x_{d}(t) = x(t-T) \rightarrow y_{d}(t) = \int_{-a}^{\infty}x_{d}(\tau)d\tau =\int_{-a}^{\infty}x(\tau-T)d\tau = y(t-T) \)
Therefor it is time invariant.
b) I can't see how anything changes from case a) to case b). Does anything change? Why?
Thanks again!