If the power spectral density of stationary random process is a sine-squared function of frequency, the shape of its autocorrelation is

Then ROC for x₁(n) + x₂(n)

R₁ ? R₂ (R₁ < R₂)

Then ROC for x₁(n) + x₂(n)

R₁ ? R₂(R₁ < R₂)

= 6.903 x 10^-8

?

?

f_B = - 8.6 x 10¹¹

The threshold voltage is always negative for p-channel and hence implant is of p-type.

(C) is without over modulation.

?

? = - 8V_P - 16

? + 8V_P + 16 = 0

? (V_P + 4)² = 0

? V_P = -4 V.

E_max - E_min = 0.5 x 1 kV = 0.5 kV

2E_max = 1.5 kVi, E_max = 0.75 kV

E_min = 0.25 kV.

a = RG and Z₀ = RG ?

.

If this converges, it implies that s = - 3 is in the ROC.

A casual and stable system must always have its ROC to the right of all its poles. However, s = - 3 is not to the right of the pole at s = - 2.

Statement 2 is false, because it is equivalent to stating that H(0) = 0. This contradicts the fact that H(s) does not have a zero at the origin.

Statement 3 is false. If h(t) is of finite duration, then if its Laplace transform has any points in its ROC, ROC must be the entire s-plane.

However, this is not consistent with H(s) having a pole at s = - 2.

Statement 4 is false. If it were true, then H(s) has a pole at s = - 2, it must also have a pole at s = 2.

This is inconsistent with the fact that all the poles of a causal and stable system must be in the left half of the s-plane.

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There is negative number present in first column. Thus network is unstable.

i(t) = V_r + (i_i - i_j)e^-1/p = 0.5 - 0.125e^-1000t .