Digital Signal Processing: Signal Transforms

Transcription

1 Digital Signal Processing: Signal Transforms Aishy Amer, Mohammed Ghazal January 19, 1 Instructions: 1. This tutorial introduces frequency analysis in Matlab using the Fourier and z transforms.. More Matlab resources are under 3. Use the help xyz command to get more information on command xyz(). 1 Signal Transforms Figure 1: Relationships between signal transforms. Discrete-time Signal Transforms DTFT: Discrete-time Fourier tarnsform. DFS : Discrete Fourier series. DFT : Discrete Fourier transform 1

2 ELEC Digital Signal Processing: Signal Transforms /9 Transform Name Forward Transform Inverse Transform Notes Partial fractions, z-transform X(z) = x[n]z n Power series, has ROC n= z = re jω Inspection. DTFT X(e jω ) = π x[n]e jωn x[n] = 1 X(e jω )e jωn dω * continuous in freq. DFS n= N 1 X[k] = n= x[n]w nk N π π x[n] = 1 N N 1 k= Periodic (π) X[k]W nk N W N = e j π N = e jω o * periodic signal x[n] = x[n + N] * discrete in freq. N is the period DFT X[k] = N 1 x[n]wn nk x[n] = 1 N 1 N X[k]W nk N W N = e j π N = e jω o * discrete in freq. * samples from DTFT. 3 DTFT, DFT, and DFS n= k= Suppose that x[n] = { cos ( ) 3πn, if n 31, otherwise. (1) The discrete-time Fourier transform (DTFT) of x[n] is X(e jω ) = e j 31 (ω+3π ) sin((ω + 3π )) e j sin( 1 (ω + 3π + )) 31 (ω 3π ) sin((ω 3π )) sin( 1 (ω 3π )) () As can be seen from (), X(e jω ) is a continuous function of ω. For this particular x[n], we are able to obtain a closed-form expression for the DTFT and can easily plot the signal and the magnitude response X(e jω ) with Matlab as follows: Example 1: Defining and plotting the DTFT magnitude response 1 N = 3; n = [ :N 1]; 3 x = cos(3 pi/ n ) ; w = :.1: pi ; % step s i ze i s.1 5 X = exp( i (31/) (w+3 pi /))/. ( sin ( w+ pi ) ). / ( sin (w/+3 pi /))+... exp( i (31/) (w 3 pi /))/. ( sin ( w pi ) ). / ( sin (w/ 3 pi /)); 7 subplot (1,,1); 9 stem(n, x ) ; 1 xlabel ( n ) ; ylabel ( x [ n ] ) ; grid on ; 11 1 subplot (1,,); 13 plot (w, abs(x) ) ; 1 xlabel ( \omega ) ; ylabel ( X( eˆ{ j \omega }) ) ; grid on ;

3 ELEC Digital Signal Processing: Signal Transforms 3/9 and the output is Fig.. x[n] n X(e jω ) ω Figure : Output of example 1. Left: x[n]. Right: DTFT of x[n]. It is not always this easy to obtain a closed-form expression of X(e jω ). Moreover, to plot X(e jω ), we uniformly sampled it from to π with the very small step size of.1 (the step size is small to simulate continuous frequency). This means that we used a discrete approximation of the DTFT (i.e., both w and X are vectors), since we cannot easily deal with continuous-time signals using computers. To work around these issues, we use a transform that will take the discrete values of our signal and conveniently give us samples of X(e jω ) without having to analytically find X(e jω ). This transform is known as the discrete Fourier transform (DFT). An added benefit of using the DFT comes from a fast implementation of how to calculate it called the fast Fourier transform (FFT). In Matlab, this implementation is in the function fft(x,l). It takes as arguments our discrete-time singal x and how many samples we want to take L and returns the DFT coefficients X[k]. The default value for L is the size of our signal x[n], i.e., N. We demonstrate the use of fft() in the following example: Example : Calculating the DFT coefficients 1 X = fft (x ) ; % Notes : % 1. This c a l l i s the same as f f t (x,l), where L=length (x)=n. 3 %. The values returned are direct samples of the DTFT % 3. The values returned are the DFT c o e f f i c i e n t s X[ k ] These DFT coefficients are complex numbers. One way to visualize them is to plot their magnitude X[k] against the coefficient number (or index) k. Example 3: Plotting the DFT against coefficient number 1 k = [ : length(x) 1]; stem(k, abs(x) ) ; 3 grid on ; xlabel ( k ) ; 5 ylabel ( X[ k ] ) ; and the output is in Fig. 3. There are two important observations we can make about Fig. 3. First, the discrete approximation we

4 ELEC Digital Signal Processing: Signal Transforms / X[k] k Figure 3: Output of example 3 obtained for X(e jω ) is not a particularly good one, because we lost a significant amount of information by choosing only 3 samples for L. In other words, our choice for resolution could be better. The second important observation is that by examining Fig. 3, we cannot determine the fundamental frequency of our sinusoidal signal (i.e., 3π ) directly. We can observe the spike (two actually), but with the x-axis being the index number, we cannot determine the frequency (or frequencies) at which this spike occurs. To obtain a better approximation of X(e jω ), we increase L to and 1 and plot the magnitude of the DFT coefficients as follows: Example : The effect of increasing L 1 L = ; L1= 1; 3 k = : L 1; k1= : L1 1; 5 X = fft (x, L ) ; X1= fft (x, L1 ) ; 7 subplot (1,,1); stem(k, abs(x ) ) ; 9 grid on ; 1 xlabel ( k ) ; 11 ylabel ( X {}[k ] ) ; 1 subplot (1,,); 13 stem( k1, abs(x1 ) ) ; 1 grid on ; 15 xlabel ( k ) ; ylabel ( X {1}[k ] ) ; and the output is in Fig.. To make the x-axis more meaningful, we wish to plot every coefficient against its corresponding angular frequency ω (i.e., get an accurate x-axis). To do this, realizing that fft calculates DFT at equally-spaced angles from to π, we can do the following

5 ELEC Digital Signal Processing: Signal Transforms 5/ X [k] X 1 [k] k k Figure : Output of example Example 5: Obtaining the frequency axis 1 w = [ :N 1]/N ( pi ) ; stem(w, abs(x) ) ; 3 xlabel ( \omega k ( rad/s ) ) ; ylabel ( X[ k ] ) ; grid on ; and get Fig X[k] ω k (rad/s) Figure 5: Output of example 5 Now we can easily observe the spike at ω = 3π = The other (right most) spike is located at ω = π 3π = This is because Matlab s fft() uses frequencies from to π. Frequencies from π to π are the negative frequencies from the next repetition of the spectrum. A better way to plot the DFT coefficients is to sample from π to π. This way, the spikes will be located at ω = +/ 3π = +/ as per what we are familiar with. There is a command in Matlab just for this purpose called fftshift() that is used as follows:

6 ELEC Digital Signal Processing: Signal Transforms /9 Example : Shifting the spectrum 1 w = [ N/:N/ 1]/N ( pi ) ; X = fft (x ) ; 3 X = fftshift (X) ; stem(w, abs(x) ) ; 5 xlabel ( \omega k ( rad/s ) ) ; ylabel ( X[ k ] ) ; grid on ; and the output is Fig X[k] ω k (rad/s) Figure : Output of example Often in practice, we wish to plot the magnitude response against the frequency in Hertz (Hz) as it is more meaningful to us. First of all, we do not care about redundant frequencies from a practical point of view. This means, we only need to plot the first half of the frequencies and disregard the rest. Moreover, we must convert ω to f using f = ω π f s (by dividing over π and multiplying with the sampling frequency f s ). We can do all this as follows: Example 7: Changing the frequency axis 1 N = 3; % length of our signal n = [ :N 1]; 3 x = cos ((3 pi /) n ) ; % x [ n ] 5 L = 1; % Size of DFT ( i. e., L point DFT) f s = 1; % Sampling frequency 7 w = [ : L 1]/L ( pi ) ; % angular freq. 9 f = w/( pi ) fs ; % freq. in Hz 1 11 X k = fft (x,l ) ; 1 subplot (1,3,1); 13 stem(w(1:l/),abs(x k (1:L/))); 1 xlabel ( w k ( rad/s ) ) ; ylabel ( X[ k ] ) ; grid on ;

7 ELEC Digital Signal Processing: Signal Transforms 7/ X ejw = fft (x,l ) ; 1 subplot (1,3,); 19 plot (w(1:l/),abs( X ejw (1:L/))); xlabel ( w ( rad/s ) ) ; ylabel ( X( eˆ{ j \omega }) ) ; grid on ; 1 X c = fft (x,l)/n; % normalize by length of signal 3 subplot (1,3,3); plot ( f (1:L/),abs( X c (1:L/))); 5 xlabel ( f (Hz) ) ; ylabel ( X( f ) ) ; grid on ; and the output is Fig X[k] X(e jω ) X(f) w (rad/s) k w (rad/s) f (Hz) Figure 7: Output of example 7 Note that the left plot in Fig. 7 shows ω k in the frequency axis and the DFT coefficients X[k] in the y-axis. The middle plot shows an approximation of the DTFT. Note how we now have a value of the DTFT at every ω from to π (i.e., continuous function). This happened by interpolating the plot to the left. In the right plot in Fig. 7, the frequency values (x-axis) in Hz are calculated from ω using f = ω π f s. Moreover, the magnitude response is normalized (by dividing over the signal length N). Note that the maximum value stands at.5 (half our signal amplitude of 1) as we expect from the spectrum of a cosine signal. The discrete Fourier series (for discrete-time periodic signals) is the same as the discrete Fourier transform. Therefore, there is no special function to calculate it. The assumption with the discrete Fourier series is that the signal you provide to fft() corresponds to one period of your periodic signal. z-transform As for the z-transform, Matlab provides a number of useful commands. We explore them in this example. Suppose that two discrete-time LTI systems have the following transfer functions H 1 (z) = 1 1z 1 z + z 3 z 1 z H (z) =.9(1 + z 1 + z + z 3 + z ) 1 +.z +.177z We can use Matlab s residuez() command to find the partial fraction expansion of H 1 (z) as follows: Example : Partial fraction expansion

8 ELEC Digital Signal Processing: Signal Transforms /9 1 Num = [1 1 ] ; %numerator polynomial c o e f f i c i e n t s Den = [ ]; % denominator polynomial c o e f f i c i e n t s 3 [ r, p, k ] = residuez (Num, Den) and the output is r =[1.5.5] p =[ -1] k =[1.5-1.] To interpret the answer, check the following equation B(z) A(z) = r(1) 1 p(1)z r(n) 1 p(n)z 1 + k(1) + k()z 1.. (3) which basically means that the partial fraction expansion of H 1 (z) is H 1 (z) = z z z 1 () We can use Matlab s zplane() command to generate the pole-zero plot of H (z) as follows: Example 9: Pole-zero plot 1 b =.9 [1 1 ] ; a = [1..177]; 3 zplane (b, a ) ; and the output is Fig Imaginary Part Real Part Figure : Output of example 9 We can use the command freqz() to evaluate H (e jw ) directly from the transfer function of the system at frequencies we choose as follows: Example 1: Evaluate H(e jw ) from the transfer function 1 N = 5; % The same as the s i z e of the DFT [H,w] = freqz (b, a,n) ; % This command returns the frequency axis also

9 ELEC Digital Signal Processing: Signal Transforms 9/9 3 Habs = abs(h) ; plot (w, Habs ) ; grid on ; 5 xlabel ( \omega ) ; ylabel ( H( eˆ{ j \omega }) ) ; and the output is Fig H(e jω ) ω Figure 9: Output of example 1 Finally, we can use Matlab to find the z-transform analytically as follows: Example 11: Finding the z-transform analytically 1 syms a t n % These now are symbols or variables H = ztrans ( h ) and the output is %H_ = -z/(a - z) To get an expression for the DTFT, one can substitute e jω in the transfer function.