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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 Today’s 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 ( jω ) e jωt • Scaling x ( at ) ↔ 1  a  X ( j ω a ) • Derivative dx ( t ) dt ↔ jωX ( jω ) • Modulation x ( t ) e jω t ↔ X ( j ( ω ω )) • Duality X ( t ) ↔ 2 πx ( ω ) EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 3 • Parseval R ∞∞  x ( t )  2 dt = 1 2 π R ∞∞  X ( jω )  2 dω • ? The convolution theorem ? x ( t ) * y ( t ) ↔ X ( jω ) Y ( jω ) 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 ( jω ) = X ( jω ) H ( jω ) G ( jω ) 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 ( jω ) . Suppose we are given the following facts. • x ( t ) is real and nonnegative. • F 1 { (1 + jω ) X ( jω ) } = Ae 2 t u ( t ) , where A is independent of t . • R ∞∞  X ( jω )  2 dω = 2 π 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 + jω ) X ( jω ) } = Ae 2 t u ( t ) Take the Fourier transform of both sides: F n F 1 { (1 + jω ) X ( jω ) } o = F n Ae 2 t u ( t ) o (1 + jω ) X ( jω ) = A 2 + jω X ( jω ) = A ( 2 + jω )( 1 + jω ) We would like to find an expression for x ( t ) . Currently X ( jω ) 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 ( jω ) into a more familiar form. X ( jω ) = A ( 2 + jω )( 1 + jω ) = a 2 + jω + b 1 + jω with unknown constants a and b . X ( jω ) = A ( 2 + jω )( 1 + jω ) = a (1 + jω ) + b (2 + jω ) (2 + jω )(1 + jω ) Then A = a (1 + jω ) + b (2 + jω ) 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 ( jω ) = A ( 1 1 + jω + 1 2 + jω ) 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 ( jω )  2 dω = 2 π By Parseval’s relation: 1 2 π Z ∞∞  X ( jω )  2 dω = 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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This note was uploaded on 12/13/2011 for the course EE 102 taught by Professor Levan during the Fall '08 term at UCLA.
 Fall '08
 Levan
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