Presentation on theme: "Ch.4 Fourier Analysis of Discrete-Time Signals Kamen and Heck."— Presentation transcript:
Ch.4 Fourier Analysis of Discrete-Time Signals Kamen and Heck
4.1 Discrete-Time Fourier Transform X( ) = n=- , x[n] e -j n ( Eq. 4.1) Complex valued function of real variable , the frequency. A sufficient condition for x[n] to have a DTFT in the ordinary sense is that x[n] be absolutely summable.
Example 4.1 Computation of the DTFT Consider x[n] = a n, 0 n q and 0 otherwise. The DTFT is X( ) = n=- , x[n] e -j n = n=0,q a n e -j n = n=0,q (ae -j ) n = [1 – (ae -j ) q+1 ] / [1- (ae -j )] (where the closed form expression for a partial sum exponential is used—(Eq.4.5)
Example 4.2 Rectangular and Polar Forms –Consider x[n] = a n u(n). –This is similar to Ex. 4.1 except we have q . –Consider the DTFT from Ex. 4.1 but let q : X( ) = lim q [1 – (ae -j ) q+1 ] / [1- (ae -j )] This limit exists for |a| < 1. For this case, the DTFT exists in the ordinary sense. X( ) = 1/ [1- (ae -j )] (Eq. 4.16) The rectangular and polar forms are shown on pages
4.1.1 Signals with Even or Odd Symmetry Let x[n] be a real-valued discrete-time signal that is an even function (ie, x[n] = x[- n].) –The DTFT is X( )= x + n=1, 2x[n] cos(n ) Let x[n] be an odd function (ie,x[n]=-x[-n]) –The DTFT is X( )= x - n=1, j2x[n]sin(n )
Example 4.3 DTFT of Rectangular Pulse –Let p[n] = 1 for -q n q and 0 elsewhere. –The signal is even but it is easier to use 4.2. –P( ) = n=-q,q e -j n – =[ e j q – e -j (q+1) ] / [1- e -j ] – = sin[(q + 1/2) ]/[sin( /2)] –This is the discrete-time counterpart to the transform of the rectangular pulse (Ex. 3.9). –Figure 4.3 illustrates the DTFT.
4.1.2 Spectrum of a Discrete-Time Signal For simplification, the discrete-time Fourier series is not discussed. For a discrete time signal that is not a function of sinusoids the spectrum is a continuum of frequency components. The frequency spectrum is made up of the amplitude spectrum and the phase spectrum. The highest value of = .
Example 4.4 Decaying Exponential Assume that x[n] = (.5) n u(n). The signal is plotted in Fig The spectrum is shown in Figure 4.2 Note that most of the spectrum is in the lower frequencies.
Example 4.5 Signal with High- Frequency Components Consider x[n] = (-.5) n u(n). From Figure 4.4 we see that there should be higher frequency components in this signal. From the result of Ex 4.2, the DTFT is: –X( ) = 1/ [1- (-.5e -j )] = 1/ [1 +.5e -j ] –The amplitude and phase spectra are given by equations 4.25 and 4.26 and plotted in Figure 4.5.
4.1.4 Generalized DTFT Example 4.6 DTFT of a Constant Signal Let x[n] =1 for all n. This signal does not have a DTFT in the ordinary sense—(Why?) Figure 4.6 shows the generalized DTFT. Discussion on page 176 illustrates that its inverse is the constant signal.
DTFT Transform Pairs and Properties Transform Pairs—Table 4.1 page 177. Properties—Table 4.2 page 178. –No duality property, but there is a relationship between the inverse of the CTFT and the DTFT. –Result can be used to generate DTFT pairs from CTFT pairs—see Example 4.7.
4.2 Discrete Fourier Transform Let x[n] be a discrete-time signal. Let X( ) is the DTFT of x[n]. Note: the DTFT is a continuous function of . Let N be a positive integer, then the DFT of x[n] is: –X k = n=0,N-1 x[n] e -j2 kn/N, k=0,1,2,…N-1
4.2 The DFT (p.2) In general, X k is a function of the discrete integer k. There are N values in the DFT of x[n]. These values are complex numbers. Polar form: X k = |X k | exp [j X k ] Rectangular form: X k = R k + jI k –See equations 4.36, MATLAB—program on page 180.
4.2 The DFT (p.3) Example 4.8 Computation of the DFT –Finite sequence –page Symmetry –Magnitude of the DFT is symmetric about N/2, for N even. –Phase angle of the DFT has odd symmetry about N/2 when N is even Inverse DFT—see equation 4.40 and MATLAB program and Example 4.9 on page 183.
The DFT (p.4) Sinusoidal Form –The right hand side of the IDFT equation can be written as sinusoids. –See equation 4.45 and Example Relationship to DTFT –If x[n] = 0 for n<0 and n N, the DFT Xk can be viewed as a freqeuency sample version of the DTFT. –X k =X( ) =2 k/N = X(2 k/N ), k = 0,1,2,…,N-1
Example 4.11 DTFT and DFT of a Pulse Consider p[n] from example 4.3. Let x[n] be p[n-q]. Figure 4.10 shows the amplitude spectrum for q=5. Figure 4.11 shows the amplitude of the DFT for q=5 and N= 22. Figure 4.12 shows the amplitude of the DFT for q=5 and N = 88.
4.4 FFT Algorithm Consider the DFT and Inverse DFT: X k = Σ n=0,1,…,N-1 x[n] e -j2 kn/N k=0,1,…,N-1 x[n]= (1/N ) Σ k=0,…,N-1 X k e j2 kn/N, n=0,…,N-1 How many multiplications are needed to compute the DFT? (N 2 ) The FFT algorithm requires N(log 2 N)/2 multiplications.
4.4 FFT Algorithm (p.2) If N = 1024, – DFT requires 1,048,576 multiplications – FFT requires 5,120 multiplications There are different variations of the FFT algorithm. One uses “decimation-in-time”.
4.4 FFT (p.3) Decimation-in-Time –Subdivide the time interval into intervals having a smaller number of points.
4.4 FFT (p.4) X k can be broken up into two parts. –First let exp(-j2 /N) = W N –Then X k = Σ n=0,1,…,N-1 x[n]( W N ) kn k=0,1,…,N-1 –Let N be an even integer: a[n]=x[2n] ; b[n]=x[2n + 1], for n = 0,…,N/2. –Let A k = Σ n=0,…,N/2-1 a[n] (W N/2 ) kn, k=0,1,…N/2-1 –Let B k = Σ n=0,…,N/2-1 b[n] (W N/2 ) kn, k=0,1,…N/2-1 –Then X k = A k + (W N ) k B k, k=0,1,…,N/2 -1 –And X (N/2)+k = A k - (W N ) k B k, k=0,1,…,N/2 -1 –See page 197 for the verification.
4.4 FFT (p.5) Note that the two parts are (N/2) DFTs. This can continue until signals with only one nonzero value are obtained if N is a power of 2. The process is graphically illustrated by Figure To have the outputs in the correct order, a process called bit reversing (see Table 4.3) is used.
4.4.1 Applications of the FFT Algorithm Computation of the Fourier Transform Convolution Data Analysis –Extraction of a Sinusoidal Component Embedded in Noise –Analysis of Sunspot Data –Stock Price Analysis