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Question

Five facts about real signal x(t) with Laplace transform X(s) are given below : 1. X(s) has exactly two poles. 2.X(s) has no zeros in the finite s-plane. 3. X(s) has a pole at s = -1 + j. 4. e2tx(t) is not absolutely integrable. 5. X(0) = 8.

A
8s2+2s+1,Re[s]>1
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B
16s2+2s+2,Re[s]>1
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C
16s2+2s+2,Re[s]<1
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D
8s2+2s+1,Re[s]<1
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Solution

The correct option is B 16s2+2s+2,Re[s]>−1X(s) is of the form, X(s)=A(s+a)(s+b) Given that one pole is at s = -1 + j. Since x(t) is real that poles of X(s) must occure in conjugate reciprocal pairs. s=−1−j is another pole of X(s). ∴X(s)=A(s+1−j)(s+1+j)=A(s+1)2+1 =As2+2s+2 Given, X(0)=8 X(0)=A2 8=A2 A=16 ∴X(s)=16s2+2s+2 Let R denote the ROC of X(s). From the pole locations we know that there are two possible choices of R R may either be Re[s] < -1 or Re[s]> -1 Given that, e2tx(t)L.T⟶X(s−2) The ROC of Y(s) is R shifted by 2 to the right. Since it is given that y(t) is not absolutely integrable. The ROC of Y(s) should not include the jω axis. This is possible only if R is Re[s]>-1.

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