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102_1_lecture17

102_1_lecture17 - UCLA Fall 2011 Systems and Signals...

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UCLA Fall 2011 Systems and Signals Lecture 17: Sampling Theorem I November 28, 2011 EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 1
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Agenda Today’s topics Sampling of continuous-time signals Interpolation of band-limited signals (Sampling theorem) Processing of continuous-time signals using discrete-time systems EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 2
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Laplace Transform Review Example: Find the Laplace transform of the following functions: 1) f 1 ( t ) = e - at u ( t ) 2) f 2 ( t ) = - e - at u ( - t ) EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 3
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Solution: L [ e - at u ( t )] = Z 0 e - at e - st dt = - e - ( s + a ) t s + a 0 = 1 s + a , ROC: <{ s } > - a L [ - e - at u ( - t )] = Z 0 -∞ - e - at e - st dt = e - ( s + a ) t s + a 0 -∞ = 1 s + a , ROC: <{ s } < - a f 1 ( t ) and f 2 ( t ) have the same Laplace transform functions, but different ROCs. In other words, given a Laplace transform function, with different ROCs, we obtain different inverse Laplace transforms. EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 4
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Example: Consider a continuous-time LTI system with below system function: H ( s ) = 1 s 2 - s - 2 Determine the impulse response function h ( t ) for each of the following cases: 1) The system is stable 2) The system is causal 3) The system is neither stable nor causal EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 5
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Solution: H ( s ) = 1 s 2 - s - 2 = 1 / 3 s - 2 - 1 / 3 s + 1 This system has two poles at s = - 1 and s = 2 . 1) An LTI system is stable if and only if the ROC of its system function H ( s ) includes the entire -axis. Therefore, ROC of H ( s ) has to be - 1 < <{ s } < 2 . We obtain: h ( t ) = - 1 3 e 2 t u ( - t ) - 1 3 e - t u ( t ) 2) The ROC associated with the system function H ( s ) for a causal system is a right-half plane. For a system with a rational system function, causality of the system is equivalent to the ROC being the right-half plane to the right of the rightmost pole. Therefore, the ROC for H ( s ) has to be <{ s } > 2 . As a result, h ( t ) = 1 3 e 2 t u ( t ) - 1 3 e - t u ( t ) EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 6
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3) If the system is neither stable nor causal, the ROC for H ( s ) is <{ s } < - 1 . Therefore, h ( t ) = - 1 3 e 2 t u ( - t ) + 1 3 e - t u ( - t ) EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 7
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The Initial- and Final-Value Theorems Initial-Value Theorem: It has been shown in previous lectures that L [ f 0 ( t )] = sF ( s ) - f (0) . If f ( t ) doesn’t have discontinuity at t = 0 , then f (0) = f (0 - ) = f (0 + ) . We have: Z 0 e - st f 0 ( t ) dt = sF ( s ) - f (0 + ) As s → ∞ the left hand side of the above equation goes to zero, then f (0 + ) = lim s →∞ sF ( s ) EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 8
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Final-Value Theorem: If f ( ) exists, then we have: Z 0 f 0 ( t ) dt = f ( ) - f (0) We also have: Z 0 f 0 ( t ) dt = lim s 0 Z 0 e - st f 0 ( t ) dt = lim s 0 sF ( s ) - f (0) Compare the above equations, we obtain: f ( ) = lim t →∞ f ( t ) = lim s 0 sF ( s ) EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 9
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The Initial- and Final-Value Theorems Initial-Value Theorem: If x ( t ) = 0 for t < 0 and x ( t ) contains no impulses or higher-order singularities at t = 0 , then x (0 + ) = lim s →∞ sX ( s ) Final-Value Theorem: If x ( t ) = 0 for t < 0 and x ( t ) has a finite limit as t → ∞ , then lim t →∞ x ( t ) = lim s 0 sX ( s ) EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 10
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Introduction Many signals around us are not continuous
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