Fourier Transform E180 Handout Four dierent forms of the Fourier transform Non-periodic, continuous time function x(t), continuous, nonperiodic spectr

Transcrição

1 Fourier Transform E180 Handout Four dierent forms of the Fourier transform Non-periodic, continuous time function xt), continuous, nonperiodic spectrum f) This is the most general form of Fourier transform. f) xt) Z +1,1 xt)e,jft dt Z +1,1 f)e jft df The rst one is the forward transform, and the second one is the inverse transform. Non-periodic, discrete time function xn), continuous, periodic spectrum F f) The discrete time function can be considered as a sequence of samples of continuous time function. The time interval between two consecutive samples xm) and xm +1)ist 0 1F, where F is the sampling rate, which is also the period of the spectrum in the frequency domain. The discrete time function can be written as and its transform is xt) F f) m,1 xm)t, mt 0 ) xm)e,jfmt 0 m,1 xm) 1 F Z +F,F F f)e jfmt 0 df m 0; 1; ; ) 1

2 We can verify that the spectrum is indeed periodic F f + kf) F f + kt 0 ) m,1 xm)e,jf +kt 0)mt 0 F f) for k 1; ; ) because e jmk 1. Periodic, continuous time function x T t), discrete, non-periodic spectrum n) This is the Fourier series expansion of periodic functions. The time period is T, and the interval between two consecutive frequency components is f 0 1T, and its transform is n) 1 T Z +T,T x T t)e,jnf 0t dt x T t) n)e jnf 0t n,1 n 0; 1; ; The discrete spectrum can also be represented as f) n,,1 n)f, nf 0 ) We can verify that the time function is indeed periodic x T t + kt) x T t + kf 0 ) for k 1; ; ) n,1 n)e,jnf 0t+kf 0) x T t)

3 Periodic, discrete time function xm), discrete, periodic spectrum n) This is the discrete Fourier transform DFT). n) 1 T xm)e,jnmf 0t 0 n 0; 1; ;M, 1 xm) 1 F n0 n)e jnmf 0t 0 m 0; 1; ;M, 1 where M is the number of samples in the period T, which is also the number of frequency components in the spectrum M T t 0 1f 0 1F F f 0 We therefore also have TF M and t 0 f 0 1M. The DFT can be redened as n) p 1 xm)e,mn jm M n 0; 1; ;M, 1 xm) p 1 n)e mn jm M n0 where w M e,jm p M. m 0; 1; ;M, 1 n0 w mn M xm) w M,mn n) We can easily verify that the time function and its spectrum are indeed periodic xm + km) xm) and n + km) n).

4 The function The discrete and periodic time function and spectrum can be written as, respectively x T t) F f) m,1 n,1 xm)t, mt 0 ) n)f, nf 0 ) The function used above satises the following 1... t, ) Z +1,1 Z +1,1 0 t 1 t t)dt 1 xt)t, )dt x)

5 Vector form of 1D-DFT The above summation expression for DFT can also be written in more convenient form of matrix-vector multiplication 0) x0) 5 p 1 e,jm ) mn 5 M 5 M, 1) xm, 1) and x0) xm, 1) 5 p 1 M e jm ) mn 5 0) M, 1) It is obvious that the complexity of 1D DFT takes is ON ), which, as we will see later, can be reduced to ONlog N) by Fast Fourier Transform FFT) algorithms. These matrix-vector multiplications can be represented more concisely as and W,1 x x W where both and x are M 1 column vertical) vectors x and W is an M M matrix W 0) M, 1) x0) xm, 1) w mn M1 M1 MM 5

6 where w mn is an element in the mth row and nth column of matrix W dened as 1 w mn p e jm ) mn M whose complex conjugate is w mn 1 p M e,jm ) mn Obviously W is symmetric w mn w nm ) W T W but W is not Hermitian W is a unitary matrix, W T W W W T W W,1 because its rows or columns) are orthogonal W m ;W m 0) M k1 w mkw m 0 k 1 M M k1e,jm ) mk e jm ) m0k 1 M M k1 e jm ) m0,m)k m 0 m * Why?) The DFT pair can be rewritten as W x x W

7 Fast Fourier Transform FFT) Algorithm The M-point DFT of time samples x0);x1); ;xm, 1) is dened as ignoring the coecient 1 p M for now) for n) xm)e,jmnm n 0; 1; ;M, 1 xm)w mn M w M is dened as w M e,jm and it is easy to show that w M has the following properties 1. w km M 1. w k M w k M. w M M,1 Let M N, the above DFT can be written as n) xm)wn mn + xm +1)w m+1)n N The rst summation has all the even terms and the second all the odd ones. Due to the nd property of w M, the above can be rewritten as We dene and n) xm)wn mn + even n) odd n) xm +1)wN mn wn n xm)w mn N xm +1)w mn N They are M-point DFTs. The original M-point DFT becomes n) even n)+ odd n)w n N 1)

8 Here we let the index n cover only the rst half of the original range of the DFT, n 0; 1; ;M, 1N, 1. The second half can be obtained by replacing n in Eq. 1) by n + N n + N) even n + N)+ odd n + N)w n+n N Due to the rst property of w M, we have even n + N) and similarly xm)w mn+n ) N odd n + N) odd n) Also, due to the rd property of w M, we have xm)w mn N even n) w n+n N w n Nw N N,w n N Now the second half of the DFT becomes n + N) even n), odd n)wn n ) The M-point DFT can now be obtained from Eqs. 1), ), once ever n) and odd n) are available. However, since even n) and odd n) are M/- point DFTs, they can be obtained the same way. This process goes on recursively until nally only 1-point DFTs are needed, which are just the time samples themselves. Therefore, the operations of an M-point DFT can be symbolically represented by the following diagram. The complexity is therefore reduced from OM )toomlog M). 8

9 Fourier Transform Real Functions with 1 DFT First we recall the symmetry properties of the DFT. The DFT of xm) x r m)+jx i m) is dened as n) +j[ +j xm)e,jmnm x r m)e,jmnm + j x r m)cosmnm), j x i m)e,jmnm x i m)cosmnm), j x r m)sinmnm) x i m)sinmnm) [x r m)cosmnm)+x i m)sinmnm)] r n)+j i n) [x i m)cosmnm), x r m)sinmnm)] where r n) and i n) are the real and imaginary part of the spectrum respectively. If xm) is real, i.e., x i m) 0, then we have r,n) r n) or i,n), i n),n) r,n)+j i,n) r n), j i n) n) If xm) is imaginary, i.e., x r m) 0, then we have r,n), r n) i,n) i n) or,n) r,n)+j i,n), r n)+j i n), n) 9

10 Next we show how an arbitrary function fx) can be decomposed into the even and odd components f e x) and f o x) fe x) fx)+f,x)) and f o x) fx), f,x)) f e x)+f o x) fx) Now we are ready to show how to Fourier transform two real functions x 1 m) and x m) to get their spectra 1 n) and n) by one DFT. 1. Dene a complex function xm) by the two real functions xm) x 1 m)+jx m) Notice here that we impose j on x m) tomake it imaginary.. Find the DFT of xm) DFT[xm)] n) r n)+j i n). Separate n) into 1 n) and n), the spectra of x 1 m) and x m), using the symmetry properties discussed previously. Since x 1 m) is real, the real part of its spectrum 1 n) is the even component of r n) and the imaginary part of 1 n) is the odd component of i n), i.e., 1 n) 1r n)+j 1i n) rn)+ r,n) +j in), i,n) Since jx m) is imaginary, the real part of its spectrum j n) is the odd component of r n) and the imaginary part of j n) is the even component of i n), i.e., j n) j[ r n)+j i n)] rn), r,n) +j in)+ i,n) Dividing both sides by j, we get n) r n)+j i n) in)+ i,n),j rn), r,n) Note that,n) N, n) because n) is a periodic function. 10

11 Two-Dimensional Fourier Transform D-FT) Similar to 1D-FT, D-FT can also have four dierent forms depending on whether the D signal usually spatial signal) fx; y) is periodic and whether it is discrete. Here we consider only two cases D Fourier transform pair of a Non-periodic, continuous signal fx; y) is F u; v) fx; y) Z Z 1,1 Z Z 1,1 fx; y)e,jux+vy) dx dy F u; v)e jux+vy) du dv where u and v are spatial frequencies in x and y directions, respectively, and F u; v) isthed spectrum of fx; y). D discrete Fourier transform pair of a nite periodic) and discrete signal xm; n); 0 m M, 1; 0 n N, 1) is k; l) p 1 MN n0 xm; n) p 1 MN l0 k0 mk,j xm; n)e M + nl N ) j mk k; l)e M + nl N ) 0 m; k M, 1; 0 n; l N, 1) where M and N are the numbers of samples in x and y directions, respectively, and k; l) is the D discrete spectrum of xm; n). Both k; l) and xm; n) can be considered as elements in two M by N matrices [x] and [], respectively. Example 1 fx; y) 1 if, a <x< a ;, b <y< b ) 0 else 11

12 F u; v) Z Z 1 Z a,a sinua) u,1 fx; y)e,jux+vy) dx dy e,jux dx Z b,b sinvb) v See Fig.. on page 85 of the text book. Example 1 x + y fx; y) <R 0 else e,jvy dy It is more convenient to use polar coordinate system in both spatial and frequency domains. Let x r cos; y r sin r p x + y ; tan,1 yx) and dx dy rdr d u cos; v sin p u + v ; tan,1 vu) we have du dv d d F u; v) Z Z 1,1 Z R Z [ 0 0 Z R Z [ 0 0 Z R Z 0 [ 0 fx; y)e,jux+vy) dx dy e,jrcoscos+sinsin) d]rdr e,jrcos,) d]rdr e,jrcos d]rdr 1

13 To continue, we need to use 0th order Bessel function J 0 x) dened as J 0 x) 1 Z 0 e,jx cos d which is related to the 1st order Bessel function J 1 x) by i.e. d dx xj 1x)) xj 0 x) Z x Substituting r for x, we have 0 F u; v) F ; ) xj 0 x)dx xj 1 x) Z R 1 RJ 1R) 0 r J 0 r)dr We see that the spectrum F u; v) F ; ) is independent of angle and therefore is central symmetric. See the top example in Fig.. on page 8 of the text book. 1

14 Matrix Form of D DFT Reconsider the D DFT where k; l) 1 p MN 1 p N n0 n0 [,j mk xm; n)e M ]e,j nl N 0,j nl k; n)e N 0 m; k M, 1; 0 n; l N, 1) 0 k; n) p 1 M,j mk xm; n)e M As the summation is with respect to the row index m and the column index n can be treated as a xed parameter, this expression can be considered as the Fourier transform of the nth column of [x], which can be written in column vector vertical) form as for all columns n 0; ;N, 1. 0 n W x n Putting all these N columns together, we can write h i 0 0 ; ; 0 W [x 0 ; ; x ] or more concisely [ 0 ]W [x] where W is a N by N Fourier transform matrix. We then notice that the summation expression for k; l) is with respective to the column index n and the row index number k can be treated as a xed parameter, the expression is the Fourier transform of the kth row, which can be written in row vector horizontal) form as T k W 0 k) T 0T k W T ; k 0; ;M, 1) 1

15 Putting all these M rows together, we can write T 0 5 0T 0 5 W T 0T W is symmetric W T W ), or more concisely [] [ 0 ] W Substituting [ 0 ] by W [x], we have [] W [x] W This transform expression indicates that D DFT can be implemented by transforming all the rows of [x] and then transforming all the columns of the resulting matrix. The order of the row and column transforms is not important. Similarly, the inverse D DFT can be written as [x] W [] W Again note that W is a symmetric Unitary matrix W,1 W T W It is obvious that the complexity of D DFT is OM ) assuming M N), which can be reduced to OM log M) if FFT is used. 15