Now, suppose that the signal is of finite duration (length) NN. Specifically, the signal is zero outside the range 0 nn < NN. Then

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1 EE 464 Discrete Fourier Transform Fall 2018 Read Text, Chapter 4. Recall that for a complex-valued discrete-time signal, xx(nn), we can compute the Z-transform, XX(zz) = nn= xx(nn)zz nn. Evaluating on the unit circle at zz = ee jj2ππππππ yields the discretetime Fourier transform, XX(ee jj2ππππππ ). Now, suppose that the signal is of finite duration (length) NN. Specifically, the signal is zero outside the range 0 nn < NN. Then NN 1 XX(zz) = xx(nn)zz nn. nn=0 Next, evaluate z on the unit circle, yielding the DTFT NN 1 XX ee jj2ππππππ = xx(nn)ee jj2ππππππππ. nn=0 Finally, discretize frequency, ff, by sampling ee jj2ππππππ at NN values uniformly around the unit circle, so at the frequency samples ff = kk, kk = 0,1,, NN 1. This then yields the DFT NNNN NN 1 XX(kk) XX ee jj2ππππ NN = xx(nn)ee jj2ππππππ NN nn=0, kk = 0,1,, NN 1 1

2 Note that we may think of the DFT (values XX(kk)) as samples of the Z-transform, XX(zz), spread evenly around the unit circle, with the caveat that the signal is of finite duration over the samples nn = 0,, NN 1. (There is a related transform, called the chirp ZZ-transform, that evaluates XX(zz) at locations in the zz-plane not on the unit circle.) The DFT has an inverse, called the inverse DFT (IDFT) xx(nn) = 1 NN NN 1 XX(kk)eejj2ππππππ NN kk=0, nn = 0,1,, NN 1 Show that the IDFT is the inverse of the DFT. (Substitute the DFT equation into the IDFT equation, and simplify.) 2

3 The DFT is defined as NN 1 XX(kk) = xx(nn)ee jj2ππππππ NN nn=0, kk = 0,1,, NN 1 For each kk, the sequence ee jj2ππππππ NN, nn = 0,1,, NN 1, is one of the basis signals of the transform. The DFT can be expressed as a matrix multiplication. XX(0) XX(1) 1 ee jj2ππ NN = XX(NN 1) 1 ee jj2ππ(nn 1) NN ee jj2ππ(nn 1) NN ee jj2ππ(nn 1)2 NN xx(0) xx(1) xx(nn 1) written more compactly as XX = AAxx, which is a linear transformation. The rows of the matrix A are referred to as the basis vectors of the DFT. Observations. 1. The DFT basis signals are orthogonal. Proof. Compute the inner product between ee jj2ππππππ NN and ee jj2ππππππ NN. 3

4 ee jj2ππππππ NN NN 1, ee jj2ππππππ NN = ee jj2ππππππ NN NN 1 nn=0 = ee jj2ππ(kk mm)nn NN nn=0 ee jj2ππππππ NN NN, kk = mm + NNNN = 0, kk mm + NNNN where rr = integer. Hence, the basis vectors of the DFT are orthogonal, but not normalized. 2. The value λλ = ee jj2ππ NN is an NNth root of unity, satisfying λλ NN = 1, and λλ nn 1 for 1 nn < NN. So, the DFT is of the form NN 1 XX(kk) = xx(nn)λλ nnnn. nn=0 NN Such a value, λλ = 1 exists for every NN 1 in the complex field of numbers, so a length-nn DFT exists for any length signal. 3. The DFT can be used to diagonalize any cyclic (circulant) matrix. If BB is a circulant matrix, then AAAAAA HH = DD, where DD is a diagonal matrix. (This property is used, for example, in orthogonal frequency division multiplexing, OFDM.) 4

5 To get familiar with the DFT, let s explicitly compute the 8- point DFT of the signal xx(nn) = (1,1,1,0,0,0,1,1). Then NN 1 XX(kk) = xx(nn)ee jj2ππππππ 8 nn=0, kk = 0,1,,7 Using the signal values, and noting ee jj2ππππ7 8 = ee jj2ππππ1 8 and ee jj2ππππ6 8 = ee jj2ππππ2 8, XX(kk) = 1 + ee jj2ππππ1 8 +ee jj2ππππ2 8 +ee jj2ππππ6 8 +ee jj2ππππ7 8 = 1 + ee jj2ππππ1 8 + ee jj2ππππ ee jj2ππππ2 8 + ee jj2ππππ2 8 = cos 2ππππ + 2 cos 4ππππ 8 8. So, evaluating we get XX(0) = 5 XX(1) = = XX(7) XX(2) = 1 = XX(6) XX(3) = 1 2 = XX(5) XX(4) = 1 5

6 We can view this simply as a 1-1 mapping from N (complexvalued) numbers to N (complex-valued) numbers: (1,1,1,0,0,0,1,1) NN pppp DDDDDD (5, 1 + 2, 1, 1 2, 1, 1 2, 1, 1 + 2) Note that in this example both xx(nn) and XX(kk) are real-valued. 6

7 Properties 1. Periodic extension. If we evaluate the N-point DFT for values of k outside the range {0,1,,N-1}, then we get XX(kk) = XX(kk mod NN). So, if N = 8, then XX(10) = XX(2), and XX( 3) = XX(5), etc. Similarly, evaluating the inverse DFT at values of nn outside the range {0,, NN 1} yields a periodic extension of the signal. Proof: 7

8 8 2. Linearity

9 9 3. If xx(nn) is a real-valued sequence, then XX(kk) = XX (NN kk), kk = 0,1,, NN 1. Proof.

10 4. Any length-n sequence can be decomposed into its conjugate symmetric and conjugate antisymmetric components: xx ee (nn) = 1 2 [xx(nn) + xx (NN nn)] xx oo (nn) = 1 2 [xx(nn) xx (NN nn)] Note that xx(nn) = xx ee (nn) + xx oo (nn). A real-valued signal satisfying xx(nn) = xx(nn nn), nn = 0,1,, NN 1 is called DFT even, and if it satisfies xx(nn) = xx(nn nn), nn = 0,1,, NN 1 then it is called DFT odd. Example. Which of the following are DFT-even? Which are DFT-odd? (1,1,0,0,0,1) (0,-1,1,0,-1,1) (0,1,1,-1,-1) (1,-1,1,0,0,1,-1) 10

11 a) If xx(nn) is real and DFT-even, then XX(kk) is real and DFT-even. Proof. Property 3 shows xx(nn) real-valued implies XX(kk) = XX (NN kk). Now, just show XX(kk) must be realvalued. XX(kk) = xx(0) + xx(1)ee jj2ππππ NN + + xx(nn 1)ee jj2ππ(nn 1)kk NN. Next, use xx(nn) = xx(nn nn), nn = 0,1,, NN 1, so XX(kk) = xx(0) + xx(1) ee jj2ππππ NN + ee jj2ππ(nn 1)kk NN + Then, use ee jj2ππ(nn 1)kk NN = ee jj2ππππ NN, so XX(kk) = xx(0) + xx(1) ee jj2ππππ NN + ee jj2ππππ NN + And finally, XX(kk) = xx(0) + xx(1) 2cos 2ππππ NN + which is real-valued. b) If xx(nn) is real and DFT-odd, then XX(kk) is imaginary and the imaginary part is DFT-odd. 11

12 Example. A DFT-odd signal and its transform are shown below. 12

13 Problem 1. Suppose that the signals xx 1 (nn) and xx 2 (nn) are realvalued and of length NN, and that xx 1 (nn) is DFT-even and xx 2 (nn) is DFT-odd. How can one DFT computation be used to compute the DFT of both signals? 13

14 Problem 2. Suppose that we have available a computer program to compute the NN-pt DFT. Show how, with some pre/post-processing, we can use the program to compute the NN-pt inverse DFT. 14

15 Problem 3. Suppose that xx 1 (nn) and xx 2 (nn) are two real-valued signals of length NN. Show how, with some pre/postprocessing, the NN-point DFT of both signals can be computed using a single DFT computation. 15

16 5. Frequency resolution. Let xx(nn) be samples of an analog signal, xx aa (tt), with sampling rate FF ss = 1 samples/sec. The TT NN-point DFT of xx(0),, xx(nn 1) is computed (as XX(kk)). What is the frequency resolution of the DFT samples, XX(kk)? To compute the DFT, we begin with the DTFT, XX(ee jj2ππππππ ), with ff in units of Hz. The samples of ff are taken at the frequencies ff = kk Hz, kk = 0,1,, NN 1. The frequency resolution is then the spacing between frequency samples, namely ff = 1 NNNN Hz. Example. Suppose an analog signal is sampled at the rate of 20 MHz. How many samples are required (what value of NN is required) to get a frequency resolution of 200 Hz? Solution. Using ff = 1 Hz, NNNN NN = 1 = = 10 5 samples. ffff 200 NNTT 16

17 Example. Suppose we are interested in studying the spectrum of a signal of 10 khz bandwidth that is located in the frequency range [8.200, 8.300] MHz. If we sample at the Nyquist rate (or faster) what length DFT is necessary to obtain a frequency resolution of 200 Hz? If bandpass sampling is used, what length DFT is needed to obtain the same 200 Hz frequency resolution? 17

18 6. Cyclic shifting. In using the DFT, by delaying, or shifting, a length-nn sequence is meant cyclical shifting, as illustrated below: xx(nn) = (xx(0), xx(1),, xx(nn 2), xx(nn 1)) xx(nn 2) = (xx(nn 2), xx(nn 1), xx(0),, xx(nn 3)) Fact: If xx(nn) DDDDDD XX(kk) then DDDDDD xx(nn nn 0 ) ee jj2ππnn 0kk NN XX(kk) Dual: xx(nn)ee jj2ππππkk 0 NN DDDDDD XX(kk kk 0 ) Application: Let NN be even, so that NN/2 is an integer. Case 1. Then, from the shifting property above xx nn NN 2 DDDDDD ee jj2ππππππ/2 NN XX(kk) = ( 1) kk XX(kk) So, circular shifting by NN/2 causes the NN-point DFT values to be weighted by ( 1) kk. 18

19 Case 2. Using the dual expression, xx(nn)ee jj2ππππππ/2 NN = ( 1) nn xx(nn) DDDDDD XX(kk NN 2 ) This shifts the kk = 0 term of the DFT to the center of the DFT sequence, as illustrated below. Note that the indices for the bottom plot are corrected to kk 4, and so range from -4 to 3. Hence, the X(0) coefficient is in the center of the plot, and the frequency range is from ππ rad/sec, assuming a sampling rate of 1/TT samples/sec. TT to ππ TT 19

20 7. Zero padding. (Computing additional values of XX(ee jj2ππππππ )) Given the data values xx(0),, xx(nn 1), the NN-point DFT provides NN samples of the DTFT, XX(ee jj2ππππππ ), spaced uniformly over the frequency range [0, 1 TT ] Hz. Specifically, the samples are at the frequencies kk NNNN, for kk = 0,1,, NN 1. The frequency resolution is ff = 1 Hz. Zero-padding yields additional samples of the same XX(ee jj2ππππππ ), but with less spacing between samples. Typically, this is done as follows. a. For integer mm > 1, append (mm 1)NN zeros to the xx(nn) data, so that the total length is now mmmm samples. Call this new data sequence xx 1 (nn). b. Compute XX 1 (kk), the mmmm-point DFT of xx 1 (nn). Then XX 1 (kk) = XX kk, kk = rrrr, rr an integer mm "in between", otherwise That is, every mmth DFT value is the same as from the NNpoint DFT, with the in between mm 1 values the additional samples of the underlying XX(ee jj2ππππππ ). Proof. Directly evaluate the mmmm-point DFT. NNNN 20

21 XX 1 (kk) = mmmm 1 xx 1 (nn)ee jj2ππππππ mmmm nn=0 NN 1 = xx(nn)ee jj2ππππππ mmmm = nn=0 NN 1 xx(nn)ee jj2ππ( nn=0 = XX kk, kk = rrrr, rr an integer mm "in between", otherwise. kk mm )nn NN Example. Let xx aa (tt) = sin(2ππ3tt) + sin (2ππ6tt). The signal is sampled at rate 1 = 100 Hz, and 32 samples TT taken to form xx(nn) = xx aa (nnnn), nn = 0,1,,31. The 32- point DFT is computed, and its magnitude is shown below. 21

22 Next, try zero-padding with m=3 (so a 128-point DFT). Note that with the use of zero-padding the two spectral peaks (due to the sinusoids at 3 and 6 Hz, respectively) are plainly visible, whereas they were merged and indistinguishable using the 32-point DFT. Note also that the DTFT, XX(ee jj2ππππππ ), is unchanged. Using zero-padding just provides more samples (in frequency) of the fixed DTFT. So, in the sense of the underlying DTFT, there is no increase in the frequency resolution. But in the sense of the plot of DFT values, there is increased resolution in the plot. 22

23 Example continued frequency resolution in the two plots above. The sampling rate is 1 = 100 Hz. If N = 32 TT (first figure) then the frequency resolution is ff = 1 = 3.125Hz. If the horizontal frequency axis is NNNN expressed in Hz, the ffff 0, 1 = [0, 100) Hz, with TT spacing ff between DFT values. So, the first figure above then can be plotted using a frequency axis variable ff 32 = ff [0: 1: 31], using the Matlab notation for a vector of integers. For the zero-padding case, a length NN 1 = 4 NN = 128 point DFT is computed, so the frequency resolution is 1 NN 1 TT = Hz, and the corresponding frequency axis variable for plotting is ff 128 = 1 [0: 1: 127]. The two DFT magnitudes are NN 1 TT shown below, with relevant Matlab code. Note that the frequency axis ranges over [0, 100) Hz in both DFT plots. The two spectral peaks (at 3 and 6 Hz) are evident using zero padding. 23

24 24 >> n=[0:1:31]; >> T=0.01; >> x=sin(2*pi*3*n*t)+sin(2*pi*6*n*t); >> X=fft(x,32); >> X1=fft(x,128); >> f=n/(32*t); >> f1=[0:1:127]/(128*t); >> figure(1) >> subplot(3,1,1) >> subplot(3,1,1) >> plot(n,x,'.-') >> ylabel('x(n)') >> xlabel('time Sample Index, n') >> title('signal, N=32 Point DFT, 128-point DFT with Zero Padding') >> subplot(3,1,2) >> plot(f,abs(x),'._') >> plot(f,abs(x),'.-') >> subplot(3,1,3) >> plot(f1,abs(x1),'.-')

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