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.
a = RG and Z0 = RG ?
.
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
?
?
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.
.
There is negative number present in first column. Thus network is unstable.
i(t) = Vr + (ii - ij)e-1/p = 0.5 - 0.125e-1000t .
Apply KVL to mesh 2
-2 + Vy + 1 = 0
? Vy = 1
Apply KVL to mesh 1
? -8 + Vx + 2 = 0
? Vx = 6 V
Apply KVL to mesh 3
-1 + Vz -2 = 0
? Vz = 3 V.
s = x2 - ex
x(t) = cos t.
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