# Question A system may or may not be Memoryless Time invariant Linear Causal Stable Determine which of these properties hold and which do not hold for each of the following discrete-time system. Explain your answers. (In each example, y is the system output and x is the system input.) (Each of these properties are 15 points for each question.) y[n] = x + 2] = (75 points) If the system is time-invariant, make it time-varying, if it is time-varying, make it time-invariant by changing the system. (25 points)

IKHYUJ The Asker · Computer Science

Transcribed Image Text: A system may or may not be Memoryless Time invariant Linear Causal Stable Determine which of these properties hold and which do not hold for each of the following discrete-time system. Explain your answers. (In each example, y is the system output and x is the system input.) (Each of these properties are 15 points for each question.) y[n] = x + 2] = (75 points) If the system is time-invariant, make it time-varying, if it is time-varying, make it time-invariant by changing the system. (25 points)
More
Transcribed Image Text: A system may or may not be Memoryless Time invariant Linear Causal Stable Determine which of these properties hold and which do not hold for each of the following discrete-time system. Explain your answers. (In each example, y is the system output and x is the system input.) (Each of these properties are 15 points for each question.) y[n] = x + 2] = (75 points) If the system is time-invariant, make it time-varying, if it is time-varying, make it time-invariant by changing the system. (25 points)
Community Answer
U08TYM

SOLUTION:- SOME EXAMPLES- A) (a) y(t)=x(t-2)+x(2-t)Let us check for linearity.{:[x_(1)(t) rarry_(1)(t)=x_(1)(t-2)+x_(1)(2-t)],[x_(2)(t) rarry_(2)(t)=x_(2)(t-2)+x_(2)(2-t)],[ax_(1)(t)+bx_(2)(t)=x_(3)(t) rarry_(3)(t)=x_(3)(t-2)+x_(3)(2-t)],[=ax_(1)(t-2)+bx_(2)(t-2)+ax_(1)(2-t)+bx_(2)(2-t)],[=a(x_(1)(t-2)+x_(1)(2-t))+b(x_(2)(t-2)+x_(2)(2-t))],[=ay_(1)(t)+by_(2)(t)]:}Hence linear.Let us check for time-invariance.{:[x_(1)(t) rarry_(1)(t)=x_(1)(t-2)+x_(1)(2-t)],[x_(1)(t-t_(o))=x_(2)(t) rarry_(2)(t)=x_(2)(t-2)+x_(2)(2-t)],[=x_(1)(t-t_(o)-2)+x_(2)(2-t-t_(o))],[!=y_(1)(t-t_(o))]:}Note that y_(1)(t-t_(o))=x_(1)(t-t_(o)-2)+x_(1)(2-t+t_(o)). Hence time-variant.Suppose |x(t)| < B. Then y(t) < B+B=2B (because |x(t-2)| < B and |x(2-t)| < B). Hence stable.Not memoryless as the present output at time t depends on t-2.Non-Causal because y(-1)=x(-3)+x(3). So depends on future inputs. B) Let us check for linearity.{:[x_(1)(t) rarry_(1)(t)=[cos(3t)]x_(1)(t)],[x_(2)(t) rarry_(2)(t)=[cos(3t)]x_(2)(t)],[ax_(1)(t)+bx_(2)(t)=x_(3)(t) rarry_(3)(t)=[cos(3t)]x_(3)(t)],[=[cos(3t)](ax_(1)(t)+bx_(2)(t))],[=ay_(1)(t)+by_(2)(t)]:}Hence linear.Let us check for time-invariance.{:[x_(1)(t) rarry_(1)(t)=[cos(3t)]x_(1)(t)],[x_(1)(t-t_(o))=x_(2)(t) rarry_(2)(t)=[cos(3t)]x_(2)(t)],[=[cos(3t)]x_(1)(t-t_(o))],[!=y_(1)(t-t_(o))]:}Note that y_(1)(t-t_(o))=[cos(3(t-t_(o)))]x_(1)(t-t_(o)). Hence time-variant.Stable as |y(t)| < oo, when |x(t)| < B.Memoryless as the output at time t depends only on inputs at time t.Clearly causal. C)(c) y(t)=int_(-oo)^(2t)x(tau)d tauLet us check for linearity.{:[x_(1)(t)rarry_(1)(t)=int_(-oo)^(2t)x_(1)(tau)d tau],[x_(2)(t)rarry_(2)(t)=int_(-oo)^(2t)x_(2)(tau)d tau],[ax_(1)(t)+bx_(2)(t)=x_(3)(t)rarry_(3)(t)=int_(-oo)^(2t)x_(3)(tau)d tau],[=int_(-oo)^(2t)(ax_(1)(tau)+bx_(2)(tau))d tau],[=aint_(-oo)^(2t)x_(1)(tau)d tau+bint_(-oo)^(2t)x_(2)(tau)d tau],[=ay_(1)(t)+by_(2)(t)]:}Hence linear. (d) y(t)={[0,t < 0],[x(t)+x(t-2),t >= 0]:}.By using the same method as we used for the above parts, it is linear, causal and stable and not memoryless. Now let us check for time-invariance.{:[x_(1)(t) rarry_(1)(t)={[0,t < 0],[x_(1)(t)+x_(1)(t-2),t >= 0], ... See the full answer