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A sequence x (n) with the z-transform X (z) = Z4 + Z2 – 2z + 2 – 3Z-4 is applied to an input to a linear time invariant system with the impulse response h (n) = 2δ (n-3). The output at n = 4 will be: 
  • a)
    -6
  • b)
    Zero
  • c)
    2
  • d)
    -4
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A sequence x (n) with the z-transform X (z) = Z4+ Z2– 2z + 2 &nd...
Explanation: H (z) = 2z-3
Then taking the inverse Laplace transform of the equation of Y (z) at n=4 y(n) =0.
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Most Upvoted Answer
A sequence x (n) with the z-transform X (z) = Z4+ Z2– 2z + 2 &nd...
Applying Z-transform to the sequence:
- Given sequence x(n) has the z-transform X(z) = Z^4 + Z^2 - 2z + 2 - 3Z^-4.

Given Impulse Response:
- The system has an impulse response h(n) = 2δ(n-3), where δ(n) is the discrete-time unit impulse function.

Output at n = 4:
- The output of the system at n = 4 can be found by convolving the input sequence x(n) with the impulse response h(n) and evaluating it at n = 4.
- Since the impulse response is non-zero only at n = 3, the output at n = 4 will be 2 * x(4-3) = 2 * x(1).

Calculating x(1) from the z-transform:
- Substituting n = 1 in the z-transform X(z), we get X(z) = z^4 + z^2 - 2z + 2 - 3z^-4.
- Evaluating X(z) at z = 1, we get x(1) = 1^4 + 1^2 - 2(1) + 2 - 3(1)^-4 = 1 + 1 - 2 + 2 - 3 = -1.

Calculating the Output:
- The output at n = 4 will be 2 times x(1), which is 2*(-1) = -2.
Therefore, the correct answer is option B) Zero.
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A sequence x (n) with the z-transform X (z) = Z4+ Z2– 2z + 2 – 3Z-4is applied to an input to a linear time invariant system with the impulse response h (n) = 2δ (n-3). The output at n = 4 will be:a)-6b)Zeroc)2d)-4Correct answer is option 'B'. Can you explain this answer?
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