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102_1_lecture12_student - UCLA Fall 2011 Systems and...

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Unformatted text preview: UCLA Fall 2011 Systems and Signals Lecture 12: Fourier Transform and Frequency Response of Systems November 7, 2011 EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 1 Todays topics : Review Fourier Transforms of periodic signals Limiting transforms EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 2 Review Properties of Fourier Transform Shift x ( t- t ) X ( j ) e- jt Scaling x ( at ) 1 | a | X ( j a ) Derivative dx ( t ) dt jX ( j ) Modulation x ( t ) e j t X ( j ( - )) Duality X ( t ) 2 x (- ) EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 3 Parseval R - | x ( t ) | 2 dt = 1 2 R - | X ( j ) | 2 d ? The convolution theorem ? x ( t ) * y ( t ) X ( j ) Y ( j ) EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 4 Example. x(t) y(t) System H System G *h(t) H(jw) *g(t) G(jw) Output is given by: y ( t ) = x ( t ) * h ( t ) * g ( t ) This calculation is tedious. In frequency domain, output is given by: Y ( j ) = X ( j ) H ( j ) G ( j ) Analysis is often simpler EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 5 Example (Oppenheim & Willsky 4.14). Consider a signal x ( t ) with Fourier transform X ( j ) . Suppose we are given the following facts. x ( t ) is real and nonnegative. F- 1 { (1 + j ) X ( j ) } = Ae- 2 t u ( t ) , where A is independent of t . R - | X ( j ) | 2 d = 2 Determine a closed-form expression for x ( t ) . EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 6 Solution: EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 7 Solution cont.: EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 8 Another version of convolution theorem Weve studied: x ( t ) * y ( t ) X ( j ) Y ( j ) Convolution in time is equivalent to multiplication in frequency domain Another version: 1 2 X ( j ) * Y ( j ) x ( t ) y ( t ) Convolution in frequency domain is equivalent to multiplication in time domain EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 9 Suppose Z ( j ) = X ( j ) * Y ( j ) : z ( t ) = 1 2 Z - Z ( j ) e jt d Z ( j ) = X ( j ) * Y ( j ) = Z - X ( j ) Y ( j ( - )) d z ( t ) = 1 2 Z - Z - X ( j ) Y ( j ( - )) d e jt d Rearrange the integrals: z ( t ) = 1 2 Z - X ( j ) Z - Y ( j ( - )) e jt d d Do u-substitution. LetDo u-substitution....
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102_1_lecture12_student - UCLA Fall 2011 Systems and...

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