Final answer:
To sketch the output of this linear system for an impulse input, the output is a single spike at t = 0 with a height of -252. To sketch the input signal w(t) = 3rect(1/3), the input signal is a rectangular pulse of height 3 centered at t = 0 with a width of 6 units. To sketch the output of the system for the input w(t), we need to convolve the input signal with the impulse response function.
Step-by-step explanation:
To sketch the output of this linear system for an impulse input, we need to evaluate the response function. For an impulse input, $\delta(t)$ is defined as a function that is 0 for all values of $t$ except when $t = 0$, where it is infinitely large. When $t = 0$, the response function can be simplified to $h(0) = 38(0-6)+28(0-7) = -252$. Therefore, the output of the system for an impulse input is a single spike at $t = 0$ with a height of -252.
To sketch the input signal $w(t) = 3rect(1/3)$, we need to evaluate the rectangular function $rect(1/3)$. The rectangular function is a function that is 1 for values of $t$ within a certain range, and 0 for values outside that range. In this case, $rect(1/3)$ is 1 when $|t| < 3$ and 0 when $|t| > 3$. Therefore, the input signal is a rectangular pulse of height 3 centered at $t = 0$ with a width of 6 units.
To sketch the output of the system for the input $w(t)$, we need to convolve the input signal with the impulse response function. The convolution is defined as the integral of the product of the input signal and the shifted and scaled impulse response function. In this case, the convolution is given by:
$w(t) * h(t) = 3rect(1/3) * (38(t-6)+28(t-7))$