102_1_lecture8

102_1_lecture8 - UCLA Fall 2011 Systems and Signals Lecture...

Info iconThis preview shows pages 1–10. Sign up to view the full content.

View Full Document Right Arrow Icon
UCLA Fall 2011 Systems and Signals Lecture 8: Continuous Time Fourier Series II October 19, 2011 EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Introduction Today’s topics: Review: Fourier series Properties of continuous time Fourier series Introduction to filtering EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 2
Background image of page 2
Review . Last time we’ve seen that complex exponentials are useful for studying LTI systems. If a system has a transfer function H ( s ) and an input can be written as: x ( t ) = X k a k e s k t The system output will be: y ( t ) = X k a k H ( s k ) e s k t LTI system scales complex exponentials by its transfer function H ( s ) This allows us to avoid calculating the convolution integral! EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 3
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
We’ve also seen how to decompose continuous, periodic signals into sums of complex exponentials. Continuous Time Fourier Series x ( t ) = X k = -∞ a k e jkω 0 t (Synthesis) a k = 1 T Z T x ( t ) e - 0 t dt (Analysis) The two equations allow us to alternate between time domain and frequency domain. The transformation is invertible: you can first compute a k from x ( t ) (go from time to frequency domain), and then compute x ( t ) from a k (return to time domain). Both representations uniquely specify the signal. EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 4
Background image of page 4
We will sometimes use a shorthand notation for this transformation (your book also uses this): x ( t ) FS --→ a k a k - 1 ---→ x ( t ) Notice some facts about the representations. Signals in time domain are continuous and periodic (for now!. .). Frequency domain is discrete (by construction, signals we deal with only contain frequencies at ω = 0 , where ω 0 = 2 π T ). Fourier coefficients are, in general, complex (even if the signal is real). We can gain insight into frequency content of a signal by considering magnitude and phase of Fourier coefficients ( | a k | and 6 a k ). EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 5
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Example . Given Fourier coefficients a k , find associated signal x ( t ) that’s periodic in T = 8 . a k = 1 if k = 0 2 j if k = 1 - 2 j if k = - 1 0 otherwise . In this case, it’s enough to look at the synthesis equation: x ( t ) = X k = -∞ a k e jkω 0 t (Synthesis) x ( t ) = 1 e j 0 ω 0 t + 2 j e j 1 ω 0 t - 2 j e j ( - 1) ω 0 t EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 6
Background image of page 6
x ( t ) = 1 + 4 ± e 0 t - e - 0 t 2 j ² You immediately see this as a sin function. Since ω 0 = 2 π/ 8 , this can be rewritten as: x ( t ) = 1 + 4 sin( π 4 t ) Example. Find the Fourier series coefficients of the periodic (period T , ω 0 = 2 π T ) square wave defined as: x ( t ) = ³ 1 if | k | < T 1 0 if T 1 < | t | < T . t T 1 T x(t) -T 1 EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 7
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Use the analysis equation. Integrate from - T/ 2 to 2 . a k = 1 T Z 2 - 2 x ( t ) e - jkω 0 t dt a k = 1 T Z T 1 - T 1 e - 0 t dt a k = - 1 0 T e - 0 t ± ± ± ± ± T 1 - T 1 a k = - 1 0 T ² e - 0 T 1 - e 0 T 1 ³ Difference of complex exponentials is can be rewritten a sin function EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 8
Background image of page 8
a k = 2 0 T ± e jkω 0 T 1 - e - 0 T 1 ² 2 j a k = 2 0 T sin( 0 T 1 ) Recalling that ω 0 = 2 π T , 2 0 T = 2 T k 2 πT = 1 .
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 10
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 41

102_1_lecture8 - UCLA Fall 2011 Systems and Signals Lecture...

This preview shows document pages 1 - 10. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online