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Two discrete-time signals x[n] and h[n] are both non-zero for n = 0, 1, 2 and are zero otherwise. It is given that x [0] = 1, x [1] = 2, x [2] = 1, h [0] = 1.
Let y[n] be the linear convolution of x[n] and h[n]. Given that y[1] = 3 and y[2] = 4, the value of the expression (10y[3] + y[4]) is _________.
Correct answer is '30.00 to 31.00'. Can you explain this answer?
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Verified Answer
Two discrete-time signals x[n] and h[n] are both non-zero for n = 0, 1...
y[n] = {1, 2 + a, 2a + b +1, 2b + a, b}
It is given that y [1] = 3
∴ 2+a =3 ⇒ a =1
Similarly 2a + b + 1 = 4 Þ b = 3 - 2 (1) = 1
b = 1
∴ y [3] = 2 (1) + 1 = 3
y [4] = b = 1
∴10y [3] + y [4] = 30 + 1 = 31
Most Upvoted Answer
Two discrete-time signals x[n] and h[n] are both non-zero for n = 0, 1...
Given Information:
- The signals x[n] and h[n] are both non-zero for n = 0, 1, 2 and zero otherwise.
- x[0] = 1, x[1] = 2, x[2] = 1, and h[0] = 1.
- y[n] is the linear convolution of x[n] and h[n].
- y[1] = 3 and y[2] = 4.
Calculating y[n]:
To calculate y[n], we need to perform the linear convolution of x[n] and h[n]. The linear convolution can be obtained by taking the sum of the products of corresponding elements of x[n] and h[n] for all possible values of n.
The linear convolution of x[n] and h[n] can be written as:
y[n] = x[0] * h[n] + x[1] * h[n-1] + x[2] * h[n-2]
Substituting the given values, we get:
y[n] = 1 * h[n] + 2 * h[n-1] + 1 * h[n-2]
Since h[0] = 1, h[1] = h[2] = 0, the equation simplifies to:
y[n] = h[n] + 2 * h[n-1] + h[n-2]
Calculating y[3] and y[4]:
Using the equation derived above, we can calculate y[3] and y[4] as follows:
y[3] = h[3] + 2 * h[2] + h[1]
Since h[3] is zero, y[3] = 0 + 2 * 0 + 1 = 1
y[4] = h[4] + 2 * h[3] + h[2]
Since h[4] is zero, y[4] = 0 + 2 * 0 + 0 = 0
Calculating the expression (10y[3] - y[4]):
Substituting the calculated values, we can evaluate the expression (10y[3] - y[4]) as follows:
(10y[3] - y[4]) = 10 * 1 - 0 = 10
Therefore, the value of the expression (10y[3] - y[4]) is 10.
However, the correct answer is given as '30.00 to 31.00', which is not consistent with the calculations. It's possible that there is an error in the given answer, or there may be additional information or calculations not provided in the question.
- The signals x[n] and h[n] are both non-zero for n = 0, 1, 2 and zero otherwise.
- x[0] = 1, x[1] = 2, x[2] = 1, and h[0] = 1.
- y[n] is the linear convolution of x[n] and h[n].
- y[1] = 3 and y[2] = 4.
Calculating y[n]:
To calculate y[n], we need to perform the linear convolution of x[n] and h[n]. The linear convolution can be obtained by taking the sum of the products of corresponding elements of x[n] and h[n] for all possible values of n.
The linear convolution of x[n] and h[n] can be written as:
y[n] = x[0] * h[n] + x[1] * h[n-1] + x[2] * h[n-2]
Substituting the given values, we get:
y[n] = 1 * h[n] + 2 * h[n-1] + 1 * h[n-2]
Since h[0] = 1, h[1] = h[2] = 0, the equation simplifies to:
y[n] = h[n] + 2 * h[n-1] + h[n-2]
Calculating y[3] and y[4]:
Using the equation derived above, we can calculate y[3] and y[4] as follows:
y[3] = h[3] + 2 * h[2] + h[1]
Since h[3] is zero, y[3] = 0 + 2 * 0 + 1 = 1
y[4] = h[4] + 2 * h[3] + h[2]
Since h[4] is zero, y[4] = 0 + 2 * 0 + 0 = 0
Calculating the expression (10y[3] - y[4]):
Substituting the calculated values, we can evaluate the expression (10y[3] - y[4]) as follows:
(10y[3] - y[4]) = 10 * 1 - 0 = 10
Therefore, the value of the expression (10y[3] - y[4]) is 10.
However, the correct answer is given as '30.00 to 31.00', which is not consistent with the calculations. It's possible that there is an error in the given answer, or there may be additional information or calculations not provided in the question.
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