LABORATORY 1 DISCRETE-TIME SIGNALS

Transcription

1 LABORATORY DISCRETE-TIME SIGNALS.. Introduction A discrete-time signal is represented as a sequence of numbers, called samples. A sample value of a typical discrete-time signal or sequence is denoted as: x[, N n N (.) In MATLAB these sequences can be defined as a row or column vectors, with real or complex elements. A first limitation arises from the fact that these vectors are of finite length while the digital signal processing problems can work with infinite length sequences... Generation of Sequences Two basic discrete-time sequences are the unit impulse sequence and the unit step sequence. The unit impulse sequence, n 0 [ (.) 0, n 0

2 Delayed unit impulse, n n0 [ n n0 ] (.3) 0, n n0 Since MATLAB cannot define sequences of infinite length must be stated range of values for n. To facilitate the definition of sequences of this kind will create a MATLAB function function [y, = impuls(li,ls,k) %IMPULS the unit (Dirac) impulse in discrete time, %defined on a finite temporal support % Output parameters: % y = delta(n-k) (line vector) on the support li:ls % n = temporal support li:ls (line vector) % Input parameters: % li = the inferior limit of the temporal support; % ls = the superior limit of the temporal support; % k = the index from delta(n-k) if (nargin~=3) (nargout>) error(check the input/output arguments!') end if li>=ls error('invalid temporal support!') end if (k<li) (k>ls) error('index outside the temporal support!') end L = ls-li+; y = zeros(,l); y(k-li+) = ; n = li:ls; Observations: The rules for proper function definition and usage are listed here:

3 The function s name should also be the name of the file containing that function. Do NOT use names that are already used for Matlab toolbox functions! The functions will be called from other Matlab scripts saved in the same current directory! A function will NOT be executed as a normal script by pushing the Run button! Examples: Define and plot the following discrete time sequences:. x[ [. x [ n ] 0,5 [ n 3], for 0 n 0. [y, = impuls(-0,0,0) figure, stem(n,y), grid [y, = impuls(-0,0,3); figure, stem(n,0.5*y), grid The unit step sequence, n 0 u [ (.4) 0, n 0 Delayed unit step sequence:, n n0 u [ n n0] (.5) 0, n n0 A MATLAB function can be also created to define the unit step sequences: function [y, = treapta(li,ls,k) %TREAPTA unit step sequence in discrete time, %defined on a finite temporal support % Output parameters: % y = u(n-k) (line vector) for the support li:ls % n = the temporal support li:ls % Input parameters: % li = the inferior limit of the temporal support; % ls = the superior limit of the temporal support; % k = the index from u(n-k) if (nargin~=3) (nargout>) 3

4 error(' Check the input/output arguments!!') end if li>=ls error(' Invalid temporal support! ') end if (k<li) (k>ls) error('index outside the temporal support!') end L=ls-li+; y=zeros(,l); y(k-li+:l)=; n=li:ls; Examples: Define and plot the following discrete time sequences:. x[ u[. x[ 0,7( un [ 3] un [ 3]), for 5 n 0. [y, = treapta(-5,0,0); figure, stem(n,y), grid [y, = treapta(-5,0,-3); [y, = treapta(-5,0,3); y=0.7*(y-y); figure, stem(n,y), grid E. Exercises: Define and plot the following discrete time sequences:. x n δnδn for 0 n 0. x n δn0.5δn 0.3δn δn for 0 n 0 3. x n un0.5un 4 0.5un 4 for 0 n 0 4. x n δn un 5 δn δn 9, for 0 n 0 5. x n for 0n0 6. x n lncos sin for 0 n 0 7. x n cos for 0n0 4

5 A discrete-time signal results from the sampling of a continuous signal using a sampling period of time denoted T. The samples stored in the vector that represents the discrete signal are the amplitude values of the continuous signal taken at multiples of the sampling period n T, n N A x () a t sn s nt t T s T s nt s -A A -A x( n) n For example, for a sinusoidal signal of frequency F the sampled signal is: s t A sinπ F t sn A sinπ F n T A sin π F F n By denoting the normalized frequency f, respectively the normalized angular frequency ω πf π results in: sn A sinπ f n A sinω n 5

6 Example:. Represent graphically using stem function a discrete signal obtained by sampling a sine wave of 300 Hz, total duration 0 ms and amplitude. The sampling frequency is Fs = 4kHz. How many samples does the resulting discrete signal has? F0 = 300; Fs = 4000; Tmax = 0^(-); % signal duration is 0 ms N = Tmax*Fs; % number of samples N=0ms*4kHz w0 = *pi*f0/fs; % normalized angular frequency n = 0:N-; x = *sin(w0*n); figure(), stem(n,x), grid. Represent graphically, using the plot function, the continuous signal obtained from the previous discrete one, through digital to analogue conversion (Fs = 4kHz). The representation will be done in absolute time (ms). Determine the period and non-normalized frequency of the analogue signal. F0 = 300; Fs = 4000; Tmax = 0^(-); % signal duration is 0 ms Ts = /Fs; t = 0:Ts:Tmax-Ts; xa = *sin(*pi*f0*t); figure(), plot(t,xa,'-o'), xlabel('t[ms]'), grid.3. Linear convolution of discrete-time signals Linear convolution of two discrete-time sequences x[ and x[ (assumed as infinite length sequences) is defined as: [ x [ x[ x[ k] x k x [ n k] (.6) In MATLAB, the sequences are arrays of finite length N starting at index. Therefore the above formula is changed to: N k0 x [ n ] x [ k ] x[ n k] (.7) 6

7 MATLAB function to implement convolution is: y=conv(x,x) returns in the vector y the result of linear convolution of the vectors x and x. The resulting vector length is equal to length of vector x plus length of vector x minus. Example: Compute and plot the linear convolution of the sequences x [ u[ u[ n 5] ( 0 n 0 ) and n [ (0,9 ( 0 0 x ) n ). x=treapta(0,0,0)-treapta(0,0,5); n=0:0; x=0.9.^n; x=conv(x,x); subplot(,,),stem(0:0,x),title( x ),grid subplot(,,),stem(n,x),title( x ),grid subplot(,,),stem(0:length(x)-,x),title( x ),grid x x x E. Exercises: Compute and plot the linear convolution of the following sequences:. x [ u[ n 5] u[ n 5] x [ 0 n for 0 n 0. x [ n ] sin0. n x [ [ n 5] for 0 n 0 3. x [ cos(0,5 n) n x [ ( ) for 0 n 0 7

8 .4. Discrete Fourier Transform Discrete Time Fourier Transform (DTFT) of a sequence x [ is defined by: X j jn ( e ) x[ e (.8) n F where is the normalized angular frequency:. FS F is the un-normalized frequency (measured in Hz) and FS is the sampling frequency. F Also, the normalized frequency is: f. FS j The function X ( e ) is periodical of period, so it is sufficient to know the behavior in the interval [, ) (base interval). Because this function is defined over a continuous variable which can take an infinite number of values, it is not possible implementation on a computing machine. In order to achieve, however, a frequency analysis, it is used the Discrete Fourier Transform (DFT), computed by replacing over interval [ 0, ) into N uniformly distributed points: k k, with k 0,,, N. N Therefore, the Discrete Fourier Transform of a sequence x [ is defined by relation: j kn N Xk [ ] xne [ ] with k 0,,, N (.9) n The figure below shows the spectrum of a discrete-time signal representation based on the normalized angular frequency or normalized frequency and the correspondence with analog frequency. We also notice the correspondence between the DFT spectral components of index k and the spectrum represented in normalized angular frequency. 8

9 X a j S S S max Ω[rad/s] -Ωmax 0 Ωmax -Fmax 0 Fmax FS/ FS F[Hz] j X e -π -ωmax 0 ωmax π π fmax 0 fmax 0.5 F S F f F S f X k k 0,..., N k k N k 0 3 N/ N- In MATLAB the function used to compute the discrete Fourier transform is FFT function (Fast Fourier Transform) and it uses for DFT calculating a fast algorithm. Syntax: y = fft(x) for a vector x the result is the discrete Fourier transform (DFT) of vector x returned in a vector y having the same length to x. for a matrix x the result is a matrix y of the same size to x, the fft operation is applied to each column of matrix x. 9

10 y = fft(x,n) is the N-point fft, padded with zeros if x has less than N points and truncated if it has more. Examples: Compute the discrete Fourier transform of the following sequence: x [ u[ u[ n 0], for 0 n 0 Plot the magnitude, phase, real and imaginary parts for the computed Fourier transform. x=treapta(0,0,0)-treapta(0,0,0); X=fft(x); figure(),plot(x) // results a wrong graphical representation since X has complex values // the magnitude of Fourier transform can be represented using absolute value figure(),plot(abs(x)) // the length of computed Fourier transform is the same to length of vector x // for a better representation we will compute the Fourier transform in a large number of points (N = 5): X=fft(x,5); figure(3),plot(abs(x)),grid This representation corresponds to the frequency interval [ 0, ). Usually we want a symmetrical representation of the spectrum in the interval [, ). Since the discrete Fourier transform is periodical of period then the representation in interval [,0) corresponds to representation in interval [, ). It must therefore be done a reversal of the two halves of the vector X. This is done using the MATLAB function fftshift which swaps the left and right halves of vector X. This is useful for visualizing the Fourier transform with the zero-frequency component in the middle of the spectrum. 0

11 In addition, to have representation over the abscissa within [, ) interval, a linear step vector must be generated having a total number of elements equal to the length of the represented spectrum vector: w=-pi:*pi/5:pi-*pi/5; figure(4),plot(w,fftshift(abs(x))),grid // Figure 4 plots the amplitude characteristics and figure 5 plots the phase characteristics using the function angle(). figure(5),plot(w,fftshift(angle(x))),grid // Finally we represent real and imaginary parts of DFT figure(6) subplot(,,),plot(w,fftshift(real(x))),grid subplot(,,),plot(w,fftshift(imag(x))),grid

12 Comments: The representation of spectrums was performed according to the normalized angular frequency [, ). The values computed by X=fft(x,5) represent the spectrum k computed in k, with k 0,,, N. The number of points N to N calculate the FFT determines the spectral resolution. As N is bigger the j spectrum approximation of X ( e ) is better. If the x [ sequence length is less than N, the sequence is filled with zeros, the calculated spectrum representing the convolution between the spectrum of infinite length sequence function and the spectrum of the square gate sequence. The spectrum computed by fft can be represented also depending of: o DFT index: k 0,,, N. F o Normalized frequency: f, f [ 0.5,0.5). FS f = -0.5:/N:0.5-/N; o Un-normalized frequency F f F [Hz]. F[ F /, F /). S Es Es E3. Example:. a) Generate the discrete sequence s[ obtained by sampling with the sampling frequency FS 8kHz of the signal s() t sinπft 0 with frequency F0 500Hz and duration tmax 40ms. How many samples does the discrete sequence have? Fs = 8000; F0 = 500; t_max = 0.04; t = 0:/Fs:t_max-/Fs; s = sin(*pi*f0*t); No_samples = length(s); % or: No_samples = t_max*fs // we got the sequence sn by stepping with a stride T F for as long as t seconds. b) Compute the DFT of the sequence in N=56 points using fft and plot the amplitude spectrum (modulus of the DFT) as a function of k=0:n-.

13 N_fft = 56; S = fft(s,n_fft); k = 0:N_fft-; figure(),plot(k,abs(s)) k k // in figure() there are clearly visible Dirac impulses, witch corresponds theoretically to the Fourier transform of a sine wave: Fsinω tω i δω ω δω ω The values of normalized angular frequency for the Dirac impulses will be determined, knowing the representation corresponds to a period of 0, π. Subsequently, the indexes k and k for the DFT samples are: ω π k N ω πf π F F k N F F ω π k πω N π f π F F F k N F F 40 F 6 c) Plot the amplitude spectrum as function of the normalized angular frequency [, ). w = -pi: *pi/n_fft :pi-*pi/n_fft; figure(),plot(w,fftshift(abs(s))),grid // figure() represents the amplitude spectrum for the base period π, π. It can be verified (using Zoom in) that the normalized angular frequency ω πf /F 0.39 has the same value in the spectrum. 3

14 d) Plot the amplitude spectrum as function of the normalized frequency [ 0.5,0.5) f. f = -0.5: /N_fft :0.5-/N_fft; % or: f = w/(*pi) figure(3),plot(f,fftshift(abs(s))),grid // in figure(3), the amplitude spectrum is represented as a function of the normalized frequency f, and so the Dirac pulses correspond to the value f e) Plot the amplitude spectrum as function of the un-normalized frequency [Hz]. 4

15 F = -Fs/: Fs/N_fft :Fs/-Fs/N_fft; %sau:f=w/(*pi)*fs figure(4),plot(f,fftshift(abs(s))),grid // in figure(4), the amplitude spectrum of the sine wave is represented directly as a function of un-normalized frequency and so, the unit impulses stand straight on the value F, the sinusoidal signal frequency. f) Plot the amplitude and the phase spectrum as function of the normalized frequency on the same figure, using subplot. figure(5),subplot(),plot(f,fftshift(abs(s))),grid subplot(),plot(f,fftshift(angle(s))),grid 5

16 g) Plot the real and the imaginary part of the DFT as function of the normalized frequency on the same figure, using subplot. figure(6),subplot(),plot(f,fftshift(real(s))),grid subplot(),plot(f,fftshift(imag(s))),grid Each spectrum will be represented on a separate figure (use the figure() function). The plot will contain the functions: grid, title and xlabel.. For the program from the previous exercise, modify the following parameters and explain the changes that occur in the spectrum representation. a) The signal s frequency F0 00Hz. b) The number of points of the DFT N=04 (for the initial signal with F0 500Hz ). 6