Signals & Systems. Lecture 5 Continuous-Time Fourier Transform. Alp Ertürk

Transcription

1 Signals & Systems Lecture 5 Continuous-Time Fourier Transform Alp Ertürk alp.erturk@kocaeli.edu.tr

2 Fourier Series Representation of Continuous-Time Periodic Signals Synthesis equation: x t = a k e jkω 0t = a k e jk 2π/T 0 t Analysis equation: a k = 1 T T x(t)e jkω 0t dt = 1 T T x(t)e jk 2π/T 0 t dt

3 Limit of the Fourier Series Consider the finite-duration signal x(t) that is not periodic We cannot represent x(t) with a Fourier series However, we can use x(t) to define one period of a periodic function as long as the period is longer than the duration of x(t) An aperiodic signal can be viewed as a periodic signal with an infinite period

4 Limit of the Fourier Series Note that as the period increases, fundamental frequency decreases and the harmonically related frequencies become closer in frequency As the period becomes infinite, the frequency components form a continuum and the Fourier series sum becomes an integral

5 Limit of the Fourier Series

6 Limit of the Fourier Series Consider the finite-duration rectangular pulse given by: x t = 1 t < T 1 0 T 1 < t < T 2 = u t + T 1 u t T 1

7 Limit of the Fourier Series The Fourier coefficients of the periodic form of this square wave are: a k = 2 sin kω 0T 1 kω 0 T = 2 sin k 2π T T 1 k 2π T T An envelope can be obtained by multiplying by T and denoting ω = kω 0 : Ta k = 2 sin ωt 1 ω

8 Limit of the Fourier Series T = 4T 1 T = 8T 1 T = 16T 1

9 Consider a signal x(t) that has finite duration. That is x(t) = 0 if t > T 1 We can construct a signal x(t) for which x(t) is one period

10 a k = 1 T T x(t)e jkω 0t dt T/2 = 1 x(t)e jkω0t dt T T/2 = 1 T/2 x(t)e jkω0t dt = 1 x(t)e jkω0t dt T T/2 T X jω = Ta k = x(t)e jkω 0t dt

11 x t = a k e jkω 0t = 1 T X jkω 0 e jkω 0t Since ω 0 = 2π T = 1 2π X jkω 0 e jkω 0t ω 0

12 X jω = x(t)e jkω 0t dt x t = 1 2π X jkω 0 e jkω 0t ω 0 As T, ω 0 0, x t approaches x(t) and the right-hand side of the equation above becomes an integral Therefore,

13 Inverse Continuous-Time Fourier Transform: x(t) = 1 2π X jω e jωt dω Forward Continuous-Time Fourier Transform: X jω = x(t)e jωt dt

14 These equations were derived for an arbitrary signal with finite-duration But the equations remain valid for an extremely broad class of signals of infinite duration The set of conditions for these equations to be valid are similar to the convergence rules we covered in Fourier series

15 Dirichlet conditions: 1) x(t) must be absolutely integrable, that is: x(t) dt < 2) x(t) must have a finite number of maxima and minima within any finite interval 3) x(t) must have a finite number of discontinuities within any finite interval. Furthermore, each of these discontinuities must be finite

16 Example: Find the Fourier Transform of x t = e 7t u(t) X jω = = 0 e 7t u(t)e jωt dt e 7t e jωt dt = e 7t e jωt (7 + jω) 0 = 1 (7 + jω)

17 Example: Fourier Transform of x t = e at u t, a > 0 X jω = e at u(t)e jωt dt = 1 (a + jω), a > 0 X jω = 1 a 2 + ω 2, X jω = tan 1 ω a

18 X jω = 1 a 2 + ω 2, X jω = tan 1 ω a

19 X jω = 1 a 2 + ω 2, X jω = tan 1 ω a

20 Example: Fourier Transform of x t = e a t, a > 0 X jω = e a t e jωt dt = 0 e at e jωt dt + 0 e at e jωt dt = 1 (a jω) + 1 (a + jω) = 2a a 2 + ω 2

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23 Example: Fourier Transform of x t = δ t X jω = δ t e jωt dt = 1

24 Example: Inverse Fourier Transform of X jω = 2πδ ω x(t) = 1 2π 2πδ ω e jωt dω = 1

25 Example: Fourier Transform of x t = 1, t < T 1 0, t > T 1 X jω = x(t) e jωt dt = T 1 e jωt dt T 1 = 2 sin ωt 1 ω

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28 Example: Find the signal whose Fourier transform is: X jω = 1, ω < W 0, ω > W x(t) = 1 2π = 1 W 2π W X jω e jωt dω e jωt dω = sin(wt) πt

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31 For a periodic signal, the Fourier transform can be directly constructed from its Fourier series representation The resulting transform consists of a train of impulses in the frequency domain The areas of the impulses proportional to the Fourier series coefficients

32 Consider a Fourier transform that is a single impulse of area 2π Then, X jω = 2πδ(ω ω 0 ) x(t) = 1 2π 2πδ(ω ω 0 )e jωt dω = e jω 0t

33 If X jω is of the form of a linear combination of impulses equally spaced in frequency, that is, Then, X jω = 2πa k δ(ω kω 0 ) k= x(t) = a k e jkω 0t k= Hence, the Fourier transform of a periodic signal with Fourier series coefficients a k can be interpreted as a train of impulses occurring at harmonically related frequencies

34 Example: Consider a periodic square wave, one period of which is given by x t = 1 t < T 1 0 T 1 < t < T 2

35 The Fourier series coefficients for this signals are: a k = 2 sin kω 0T 1 kω 0 T = sin kω 0T 1 πk Then, X jω = k= 2π sin kω 0T 1 πk δ(ω kω 0 )

36 X jω = 2 sin kω0 T 1 k= k δ(ω kω 0 )

37 Example: x t = sin ω 0 t a 1 = 1 2j, a 1 = 1 2j a k = 0, if k 1 or 1 X jω = 2π 1 2j δ ω ω 0 1 2j δ(ω + ω 0)

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39 Example: x t = k= δ(t kt) T/2 a k = 1 δ t e jkω0t dt = 1 T T/2 T X jω = k= 2π 1 T δ(ω kω 0) = 2π T k= δ ω 2πk T

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41 Inverse Continuous-Time Fourier Transform: x(t) = 1 2π X jω e jωt dω (Forward) Continuous-Time Fourier Transform: X jω = x(t)e jωt dt