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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 closedform expression for x ( t ) . EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 6 Solution: Start with fact #2: F 1 { (1 + j ) X ( j ) } = Ae 2 t u ( t ) Take the Fourier transform of both sides: F n F 1 { (1 + j ) X ( j ) } o = F n Ae 2 t u ( t ) o (1 + j ) X ( j ) = A 2 + j X ( j ) = A ( 2 + j )( 1 + j ) We would like to find an expression for x ( t ) . Currently X ( j ) is not in the form whose inverse FT we know right away. EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 7 Use partial fractions to convert X ( j ) into a more familiar form. X ( j ) = A ( 2 + j )( 1 + j ) = a 2 + j + b 1 + j with unknown constants a and b . X ( j ) = A ( 2 + j )( 1 + j ) = a (1 + j ) + b (2 + j ) (2 + j )(1 + j ) Then A = a (1 + j ) + b (2 + j ) Since A is a constant, it must be that a + 2 b = A and a + b = 0 . Then a = A and b = A . Then we can write: X ( j ) = A ( 1 1 + j + 1 2 + j ) EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 8 In time domain: x ( t ) = A ( e t e 2 t ) u ( t ) To completely specify x ( t ) , we need to find the value of A . Recall fact #3: Z   X ( j )  2 d = 2 By Parsevals relation: 1 2 Z   X ( j )  2 d = Z   x ( t )  2 dt which means that Z   x ( t )  2 dt = 1 A 2 Z  e t e 2 t  2 dt = A 2 Z ( e 2 t 2 e 3 t + e 4 t ) dt = 1 A 2 ( 1 2 2 3 + 1 4 ) = A 2 12 = 1 EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 9 A = 12 Then the closed form expression for...
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