Engineering College

Determine (with justification) whether the following systems are (i) memoryless, (ii) causal, (iii) invertible, (iv) stable, and (v) time invariant. For invertibility, either find an inverse system or an example of two inputs that lead to the same output. Note that y[n] denotes the system output and x[n] denotes the system input.

a. y[n] = x[n] x[n-1] + [n+1]
b. y[n] = cos(x[n])

Answers

Answer 1

Answer:

y[n] = x[n] x[n-1] x[n+1]

(i) Memory-less - It is not memory-less because the given system is depend on past or future values.

(ii) Causal - It is non-casual because the present value of output depend on the future value of input.

(iii) Invertible - It is invertible and the inverse of the given system is [tex]\frac{1}{x[n] . x[n-1] x[n+1]}[/tex]

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is not time invariant because the system is multiplying with a time varying function.

y[n] = cos(x[n])

(i) Memory-less - It is memory-less because the given system is not depend on past or future values.

(ii) Causal - It is casual because the present value of output does not depend on the future value of input.

(iii) Invertible - It is not invertible because two or more than two input values can generate same output values .

For example - for x[n] = 0 , y[n] = cos(0) = 1

for x[n] = 2[tex]\pi[/tex] , y[n] = cos(2[tex]\pi[/tex]) = 1

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is time invariant because the system is not multiplying with a time varying function.

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ScenarioUse caseScenariosScenariosUse case

Explanation:

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