Review Orthogonality Fourier Series DFT Summary Lecture 5: Fourier Series and Discrete Fourier Transform Mark Hasegawa-Johnson ECE 401: Signal and Image Analysis, Fall 2020
Review Orthogonality Fourier Series DFT Summary Review: Spectrum 1 Orthogonality 2 Fourier Series 3 Discrete Fourier Tranform 4 Summary 5
Review Orthogonality Fourier Series DFT Summary Outline Review: Spectrum 1 Orthogonality 2 Fourier Series 3 Discrete Fourier Tranform 4 Summary 5
Review Orthogonality Fourier Series DFT Summary Two-sided spectrum The spectrum of x ( t ) is the set of frequencies, and their associated phasors, Spectrum ( x ( t )) = { ( f − N , a − N ) , . . . , ( f 0 , a 0 ) , . . . , ( f N , a N ) } such that N � a k e j 2 π f k t x ( t ) = k = − N
Review Orthogonality Fourier Series DFT Summary Fourier’s theorem One reason the spectrum is useful is that any periodic signal can be written as a sum of cosines. Fourier’s theorem says that any x ( t ) that is periodic, i.e., x ( t + T 0 ) = x ( t ) can be written as ∞ � X k e j 2 π kF 0 t x ( t ) = k = −∞ which is a special case of the spectrum for periodic signals: f k = kF 0 , and a k = X k , and F 0 = 1 T 0
Review Orthogonality Fourier Series DFT Summary Analysis and Synthesis Fourier Synthesis is the process of generating the signal, x ( t ), given its spectrum. Last lecture, you learned how to do this, in general. Fourier Analysis is the process of finding the spectrum, X k , given the signal x ( t ). I’ll tell you how to do that today.
Review Orthogonality Fourier Series DFT Summary Outline Review: Spectrum 1 Orthogonality 2 Fourier Series 3 Discrete Fourier Tranform 4 Summary 5
Review Orthogonality Fourier Series DFT Summary Orthogonality Two functions f ( t ) and g ( t ) are said to be orthogonal , over some period of time T , if � T f ( t ) g ( t ) = 0 0
Review Orthogonality Fourier Series DFT Summary Sine and Cosine are Orthogonal For example, sin(2 π t ) and cos(2 π t ) are orthogonal over the period 0 ≤ t ≤ 1:
Review Orthogonality Fourier Series DFT Summary Sinusoids at Different Frequencies are Orthogonal Similarly, sinusoids at different frequencies are orthogonal over any time segment that contains an integer number of periods:
Review Orthogonality Fourier Series DFT Summary How to use orthogonality Suppose we have a signal that is known to be x ( t ) = a cos(2 π 3 t ) + b sin(2 π 3 t ) + c cos(2 π 4 t ) + d sin(2 π 4 t ) + . . . . . . but we don’t know a , b , c , d , etc. Let’s use orthogonality to figure out the value of b : � 1 � 1 x ( t ) sin(2 π 3 t ) dt = a cos(2 π 3 t ) sin(2 π 3 t ) dt 0 0 � 1 + b sin(2 π 3 t ) sin(2 π 3 t ) dt 0 � 1 + c cos(2 π 4 t ) sin(2 π 3 t ) dt 0 � 1 + e sin(2 π 4 t ) sin(2 π 3 t ) dt + . . . 0
Review Orthogonality Fourier Series DFT Summary How to use orthogonality . . . which simplifies to � 1 � 1 sin 2 (2 π 3 t ) dt + 0 + 0 + . . . x ( t ) sin(2 π 3 t ) dt = 0 + b 0 0 The average value of sin 2 ( t ) is 1 / 2, so � 1 x ( t ) sin(2 π 3 t ) dt = b 2 0 If we don’t know the value of b , but we do know how to integrate � x ( t ) sin(2 π 3 t ) dt , then we can find the value of b from the formula above.
Review Orthogonality Fourier Series DFT Summary How to use orthogonality
Review Orthogonality Fourier Series DFT Summary How to use Orthogonality: Fourier Series � We still have one problem. Integrating x ( t ) cos(2 π 4 t ) dt is hard—lots of ugly integration by parts and so on. There are two useful solutions, depending on the situation: 1 Fourier Series: Instead of cosine, use complex exponential: � x ( t ) e − j 2 π ft dt That integral is still hard, but it’s always easier than � x ( t ) cos(2 π 4 t ) dt . It can usually be solved with some combination of integration by parts, variable substitution, etc. 2 Discrete Fourier Transform: Instead of integrating, write it as a sum: � x [ n ] e − j 2 π fn / F s . . . and then write that as a line of python code, and solve it on the computer by typing np.sum() .
Review Orthogonality Fourier Series DFT Summary Outline Review: Spectrum 1 Orthogonality 2 Fourier Series 3 Discrete Fourier Tranform 4 Summary 5
Review Orthogonality Fourier Series DFT Summary Fourier’s Theorem Remember Fourier’s theorem. He said that any periodic signal, with a period of T 0 seconds, can be written ∞ � X k e j 2 π kt / T 0 x ( t ) = k = −∞
Review Orthogonality Fourier Series DFT Summary Fourier’s Theorem and Orthogonality Take Fourier’s theorem, and multiply both sides by e − j 2 πℓ t / T 0 : ∞ x ( t ) e − 2 πℓ t / T 0 = � X k e j 2 π ( k − ℓ ) t / T 0 k = −∞ Now integrate both sides of that equation, over any complete period: � T 0 � T 0 ∞ 1 1 � x ( t ) e − 2 πℓ t / T 0 dt = e j 2 π ( k − ℓ ) t / T 0 dt X k T 0 T 0 0 0 k = −∞
Review Orthogonality Fourier Series DFT Summary Fourier’s Theorem and Orthogonality Now think really hard about what’s inside that integral sign: � T 0 1 e j 2 π ( k − ℓ ) t / T 0 dt T 0 0 � T 0 � 2 π ( k − ℓ ) t � = 1 cos dt T 0 T 0 0 � T 0 + j 1 � 2 π ( k − ℓ ) t � sin dt T 0 T 0 0 If k � = ℓ , then we’re integrating a cosine and a sine over k − ℓ periods. That integral is always zero. If k = ℓ , then we’re integrating � T 0 � T 0 1 cos(0) dt + + j 1 sin(0) dt = 1 T 0 T 0 0 0
Review Orthogonality Fourier Series DFT Summary Fourier Series: Analysis So, because of orthogonality: � T 0 � T 0 ∞ 1 1 x ( t ) e − 2 πℓ t / T 0 dt = � e j 2 π ( k − ℓ ) t / T 0 dt X k T 0 T 0 0 0 k = −∞ = . . . + 0 + 0 + 0 + X ℓ + 0 + 0 + 0 + . . .
Review Orthogonality Fourier Series DFT Summary Fourier Series Analysis (finding the spectrum, given the waveform): � T 0 X k = 1 x ( t ) e − j 2 π kt / T 0 dt T 0 0 Synthesis (finding the waveform, given the spectrum): ∞ � X k e j 2 π kt / T 0 x ( t ) = k = −∞
Review Orthogonality Fourier Series DFT Summary Fourier series: Square wave example
Review Orthogonality Fourier Series DFT Summary Square wave: the X 0 term � T 0 X 0 = 1 x ( t ) e j 2 π 0 t / T 0 dt T 0 0
Review Orthogonality Fourier Series DFT Summary Square wave: the X 1 term � T 0 X 1 = 1 x ( t ) e j 2 π 1 t / T 0 dt T 0 0
Review Orthogonality Fourier Series DFT Summary Square wave: the X 2 term � T 0 X 2 = 1 x ( t ) e j 2 π 2 t / T 0 dt T 0 0
Review Orthogonality Fourier Series DFT Summary Square wave: the X 3 term � T 0 X 3 = 1 x ( t ) e j 2 π 3 t / T 0 dt T 0 0
Review Orthogonality Fourier Series DFT Summary Square wave: the X 5 term � T 0 X 5 = 1 x ( t ) e j 2 π 5 t / T 0 dt T 0 0
Review Orthogonality Fourier Series DFT Summary Fourier Series Analysis (finding the spectrum, given the waveform): � T 0 X k = 1 x ( t ) e − j 2 π kt / T 0 dt T 0 0 Synthesis (finding the waveform, given the spectrum): ∞ � X k e j 2 π kt / T 0 x ( t ) = k = −∞
Review Orthogonality Fourier Series DFT Summary Outline Review: Spectrum 1 Orthogonality 2 Fourier Series 3 Discrete Fourier Tranform 4 Summary 5
Review Orthogonality Fourier Series DFT Summary Discrete-time Fourier Series Suppose you have a signal x [ n ], sampled at a certain number of samples per second. Suppose x [ n ] is periodic with a period of N samples, i.e., x [ n ] = x [ n + N ] Then N − 1 � X k e j 2 π kn / N x [ n ] = k =0
Review Orthogonality Fourier Series DFT Summary Discrete Fourier Transform Suppose you have a signal x [ n ], sampled at a certain number of samples per second. We’ll pretend that x [ n ] is periodic with a period of N samples (even if it’s not really), i.e., we’ll pretend that x [ n ] = x [ n + N ] We’ll define X [ k ] so that N − 1 x [ n ] = 1 � X [ k ] e j 2 π kn / N N k =0 . . . in other words, the DFT is just a scaled-up Fourier series.
Review Orthogonality Fourier Series DFT Summary Fourier’s Theorem and Orthogonality Take Fourier’s theorem, and multiply both sides by e − j 2 πℓ n / N : N − 1 x [ n ] e − 2 πℓ n / N = 1 � X [ k ] e j 2 π ( k − ℓ ) n / N N k =0 Now sum both sides of that equation: N − 1 N − 1 N − 1 x [ n ] e − 2 πℓ n / N = 1 � � � e j 2 π ( k − ℓ ) n / N X [ k ] N n =0 k =0 n =0
Review Orthogonality Fourier Series DFT Summary Fourier’s Theorem and Orthogonality Now think really hard about what’s inside that summation: N − 1 � e j 2 π ( k − ℓ ) n / N dt n =0 N − 1 N − 1 � 2 π ( k − ℓ ) n � � 2 π ( k − ℓ ) n � � � = cos + j sin N N n =0 n =0 If k � = ℓ , then we’re summing a cosine and a sine over k − ℓ periods. That sum is always zero. If k = ℓ , then we’re summing N − 1 N − 1 � � cos(0) + + j sin(0) = N n =0 n =0
Review Orthogonality Fourier Series DFT Summary DFT: Analysis So, because of orthogonality: N − 1 N − 1 N − 1 x [ n ] e − 2 πℓ n / N = 1 � � � e j 2 π ( k − ℓ ) n / N X [ k ] N n =0 k =0 n =0 = 0 + 0 + . . . + 0 + 0 + X [ ℓ ] + 0 + 0 + . . . + 0 + 0
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