= 6.903 x 10-8
?
?
fB = - 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.
?
? = - 8VP - 16
? + 8VP + 16 = 0
? (VP + 4)2 = 0
? VP = -4 V.
Emax - Emin = 0.5 x 1 kV = 0.5 kV
2Emax = 1.5 kVi, Emax = 0.75 kV
Emin = 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.
W = B = Bandwidth
C = 6 x 106 x 12 = 72 Mbps.
Then ROC for x1(n) + x2(n)
R1 ? R2 (R1 < R2)
Then ROC for x1(n) + x2(n)
R1 ? R2(R1 < R2)
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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