Tuto9_sol - The ROC of the second expression is Re{s}...

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EE3210 Tutorial 9 (Solution) 1. 5 } Re{ , 5 ) 1 ( ) ( ) 5 ( ) 5 ( 1 5 > + = = = + + s s e dt e dt e t u e s X s t s st t 2. a) No. From Property 3 of ROC for Laplace transform, for a finite-length signal, the ROC is the entire s -plane. b) Yes. Since the signal is absolutely integrable or stable, the ROC must include the j ω -axis. Furthermore, X ( s ) has a pole at s = 2. Therefore, one valid ROC is Re{ s } < 2, which corresponds to a left-sided signal. c) No. Since the signal is absolutely integrable or stable, the ROC must include the j -axis. Furthermore, X ( s ) has a pole at s = 2. Therefore, we can never have an ROC of the form Re{ s } > α where is a negative number. d) Yes. Since the signal is absolutely integrable or stable, the ROC must include the j -axis. Furthermore, X ( s ) has a pole at s = 2. Therefore, one valid ROC is < Re{ s } < 2, where is a negative number. 3. Let 3 4 12 7 ) 2 ( 2 ) ( 2 + + + = + + + = s B s A s s s s X We have A = 4 and B = 2. The ROC of the first expression is Re{s} > -4.
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Unformatted text preview: The ROC of the second expression is Re{s} &gt; -3. (Note: the intersection of the above two regions should be the same as the ROC of the original expression.) Taking the inverse Laplace transform, we obtain ) ( 2 ) ( 4 ) ( 3 4 t u e t u e t x t t − − − = . 4. The Laplace transforms of x 1 ( t ) and x 2 ( t ) are 2 } Re{ , 2 1 ) ( 1 − &gt; + = s s s X 3 } Re{ , 3 1 ) ( 2 − &gt; + = s s s X Using the time-shifting and time-scaling properties, we obtain 2 } Re{ , 2 ) ( ) 2 ( 2 1 2 1 − &gt; + = ↔ − − − s s e s X e t x s s L 3 } Re{ , 3 ) ( ) 3 ( 3 2 3 2 &lt; + − = − ↔ + − − − s s e s X e t x s s L . Using the convolution property yields 3 } Re{ 2 , 3 2 ) ( 3 2 &lt; &lt; − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + − ⋅ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + = − − s s e s e s Y s s...
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Tuto9_sol - The ROC of the second expression is Re{s}...

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