2.1. A number of systems are specified below in terms of...

2.1. A number of systems are specified below in terms of their input-output relationships. For each case, determine if the system is linear and/or time-invariant. a. $y(t) = |x(t)| + x(t)$ b. $y(t) = tx(t)$ c. $y(t) = e^{-t}x(t)$ d. $y(t) = int_{-infty}^{t} x(lambda) dlambda$ e. $y(t) = int_{t-1}^{t} x(lambda) dlambda$ f. $y(t) = (t+1)int_{-infty}^{t} x(lambda) dlambda$

Transcript

00:01 In this question we have to find the time variant graph.

00:02 So here first part we have y of t is equal to x of t.

00:06 So magnitude of x of t plus x of t.

00:09 So here for x greater than zero we have y of t is equal to two times the x of t and for x of t less than zero we have y of t is equal to zero.

00:23 Now we can have the delay in input.

00:25 So delay in input is given by magnitude of x into t minus t naught minus so plus x of t minus t naught.

00:35 So here we can have direct relay.

00:38 So direct delay is equal to it is y of t minus t naught which is equal to so it is magnitude of x into t minus t naught plus x into t minus t naught.

00:53 Therefore this is time variant.

00:55 So time invariant.

00:57 Then for the second case we have y of t.

01:02 So second one is y of t is equal to t into x of t.

01:07 So here we have a into y1 of t is equal to t into a x1 of t and we have b y2 of t is equal to t into b y2 so x2 of t.

01:21 So here we have output delay is given by a x1 of t plus b x2 of t is equal to so is equal to it is y of t which is equal to a into x1 of t plus b into x2 of t and we have y of t again can be written as so y of t can be written as so y of t can be written as a into y1 of t for linear system.

01:54 So in case of linear system plus b into y2 of t.

02:00 So then we have y of t at t is equal to t naught we have this is t into x minus x naught.

02:08 So t can be written as so it can be written and then we have y of t at t is equal to t naught is not equal to y of t minus t naught.

02:18 Therefore this is time variant system.

02:21 Then we have for third question we have y of t is equal to e power minus t x of t.

02:31 So we have a y1 of t is equal to e power minus t a x of t and we have a y2 of t is equal to e power minus t b x2 of t.

02:45 So then we have y3 of t will be equal to it is e power minus t into a x1 of t plus b x2 of t.

02:56 So for the linear system we have y3 of t is equal to e power minus t into so a y1 of t plus b y2 of t...

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