If x₁[n] x₁(z) with ROC = R₁ and x₂[n] x₂(z) with ROC = R₂ then ROC for x₁[n] + x₂[n] will be __________ (where R_i ? Region of convergence and x₁[n] and x₂[n] are causal).

V_T(new) = V_T(odd) +

= 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.

(s³ + 2s² + 5s + 6)y(s) = (s² + 4) x (s)

s³y(s) + 2s²y(s) + 5sy(s) + 6y(s) = s² x (s) + 4 x (s)

Replacing s by and y(s) by y(t) and x(s) by x(t) we get

y(t) + 2y(t) + 5y(t) + 6y(t)

= x(t) + 4x(t)

+ 2 + 5 + 6 = + 4x

This is required differential equation.

Since autocorrelation function and power spectral density bears a Fourier transform relation, then since required in frequency domain will five rectangular convolutions in time domain thus it is triangular function.

Distortion less

a = RG and Z₀ = RG ?

.

Statement 1 is false, since f{h(t)e^3t} corresponds to the value of the Laplace transform of h(t) at s = 3.

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.

Probability of no-error + probability one error

.

Routh array

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