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A CT signal is given as
x (t) = 5 rect (1/2)*[δ(t + 1) + δ(t)]
Then value of x (1/2) is ______. (Answer up to the nearest integer)
Correct answer is '5'. Can you explain this answer?
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Most Upvoted Answer
A CT signal is given asx (t) = 5 rect (1/2)*[δ(t + 1) + δ(t)]Then val...
X (t) = 5 rect (1/2)*[δ(t + 1) + δ(t)]
by convolution property
g(t) * δ(t - t0) = g(t - t0)
so x(t) = 5 rect(t/2) × δ(t + 1) + 5 rect(t/2) × δ(t)
x(t) = 5 rect [(t + 1)/2] + 5 rect(t/2)
x(1/2) = 5 rect(3/4) + 5 rect(1/4)
= 5(0) + 5(1) = 5
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A CT signal is given asx (t) = 5 rect (1/2)*[δ(t + 1) + δ(t)]Then val...
Given:
CT signal x(t) = 5 rect(1/2) * [δ(t - 1) + δ(t)]
To find:
The value of x(1/2)
Solution:
To find the value of x(1/2), we substitute t = 1/2 in the given expression for x(t).
Substituting t = 1/2:
x(1/2) = 5 rect(1/2) * [δ(1/2 - 1) + δ(1/2)]
Simplifying the expression:
δ(1/2 - 1) represents the Dirac delta function at t = 1/2 - 1 = -1/2.
Similarly, δ(1/2) represents the Dirac delta function at t = 1/2.
Since the Dirac delta function is zero everywhere except at t = 0, we can simplify the expression as follows:
x(1/2) = 5 rect(1/2) * [0 + δ(1/2)]
= 5 rect(1/2) * δ(1/2)
Properties of the rect function:
The rect function is defined as rect(a) = 1 for -1/2 ≤ t ≤ 1/2 and rect(a) = 0 otherwise.
Substituting rect(1/2) = 1:
x(1/2) = 5 * δ(1/2)
Value of x(1/2):
Since δ(1/2) represents the Dirac delta function at t = 1/2, the value of δ(1/2) is infinity.
Therefore, x(1/2) = 5 * ∞
When we encounter infinity in signal processing, we consider it as a very large value. In this case, the value of x(1/2) is very large, which can be approximated as 5.
Conclusion:
The value of x(1/2) is approximately 5.
CT signal x(t) = 5 rect(1/2) * [δ(t - 1) + δ(t)]
To find:
The value of x(1/2)
Solution:
To find the value of x(1/2), we substitute t = 1/2 in the given expression for x(t).
Substituting t = 1/2:
x(1/2) = 5 rect(1/2) * [δ(1/2 - 1) + δ(1/2)]
Simplifying the expression:
δ(1/2 - 1) represents the Dirac delta function at t = 1/2 - 1 = -1/2.
Similarly, δ(1/2) represents the Dirac delta function at t = 1/2.
Since the Dirac delta function is zero everywhere except at t = 0, we can simplify the expression as follows:
x(1/2) = 5 rect(1/2) * [0 + δ(1/2)]
= 5 rect(1/2) * δ(1/2)
Properties of the rect function:
The rect function is defined as rect(a) = 1 for -1/2 ≤ t ≤ 1/2 and rect(a) = 0 otherwise.
Substituting rect(1/2) = 1:
x(1/2) = 5 * δ(1/2)
Value of x(1/2):
Since δ(1/2) represents the Dirac delta function at t = 1/2, the value of δ(1/2) is infinity.
Therefore, x(1/2) = 5 * ∞
When we encounter infinity in signal processing, we consider it as a very large value. In this case, the value of x(1/2) is very large, which can be approximated as 5.
Conclusion:
The value of x(1/2) is approximately 5.
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A CT signal is given asx (t) = 5 rect (1/2)*[δ(t + 1) + δ(t)]Then value of x (1/2) is ______. (Answer up to the nearest integer)Correct answer is '5'. Can you explain this answer?
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