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Consider a random process X(t)= 3V(t) - 8, where V(t) is a zero-mean stationary random process with autocorrelation Ru(T)= 4E-5|T| . The power in P(1&X(t)) is;
- a)120
- b)100
- c)80
- d)45
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
Consider a random process X(t)= 3V(t) - 8, where V(t) is a zero-mea...
Calculation of Power in P(1&X(t)):
Given Parameters:
- X(t) = 3V(t) - 8
- V(t) is a zero-mean stationary random process
- Autocorrelation Ru(T) = 4E-5|T|
Power Calculation:
- Power P(1&X(t)) is defined as the expectation of the square of X(t) at time t=1.
- P(1&X(t)) = E[(X(1))^2]
- P(1&X(t)) = E[(3V(1) - 8)^2]
- P(1&X(t)) = E[9(V(1))^2 - 48V(1) + 64] (Expanding the square)
Using the Autocorrelation Property:
- E[V(1)V(1)] = Ru(0) (since V(t) is a zero-mean stationary random process)
- E[(V(1))^2] = Ru(0) (by definition of autocorrelation)
- E[(V(1))^2] = 4E-5 * 0 = 0 (at T=0)
Substituting in the Power Calculation:
- P(1&X(t)) = 9 * 0 - 48 * 0 + 64
- P(1&X(t)) = 64
Therefore, the power in P(1&X(t)) is 64, which corresponds to option 'B' (100).
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Community Answer
Consider a random process X(t)= 3V(t) - 8, where V(t) is a zero-mea...
X(t) = 3V(t)- 8
Rx (t) = E [x(t) × (t+τ)]
= E(3v(t) - 8)(3v(t + τ) - 8)]
= E[(9v(t) v (t+τ) - 24v(t) - 24v(t+τ) + 64]
= 9 Rv (τ) - 48 E[v[t] + 64
Px (τ):-∘
= Power in X (t) = 9Rv (0) + 64
= 36 + 64 = 100w
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