DSP Laboratory (EELE 4110) Lab#5 DTFS & DTFT

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1 Islamic University of Gaza Faculty of Engineering Electrical Engineering Department EG.MOHAMMED ELASMER Spring-22 DSP Laboratory (EELE 4) Lab#5 DTFS & DTFT Discrete-Time Fourier Series (DTFS) The discrete-time Fourier series (DTFS) applies only for periodic signals whereas most realistic signals are aperiodic. Furthemore it does not apply to systems. These are the two reasons why the DTFS has limited use and we will go through it quickly. A periodic signal of period (in figure 5., = 8) can be expressed mathematically as x n = x n +, all n Such a signal can be expanded into a series of components: k 2 j kn x ( n) c e, n,2,..., k Figure 5. : Period signal with period of 8 samples The coefficients C k are the frequency components or spectral components (coefficients) of the signal x(n). They are given by 2 j kn ck x ( n) e, k =,,2,..., n otice that a signal of period sampled is expanded into the same number of spectral components. The series as defined by C k is periodic with the period, this is quite different from the continuoustime series which is aperiodic,see figure 5.2.

2 Figure 5.2 : Fourier series and transform for CT and DT EXAMPLE 5. Consider a periodic sequence of 64 samples of which the first sample is the unit sample δ(t) and the next 63 samples are zero (Figure 5.3a ),Find its magnitude and phase spectrum. Solution The spectral coefficients C k are 2 2 j kn j kn ck x ( n) e = ( n) e = n n =, k =,,2,...,63 64 Thus the magnitude spectrum equals /64 at every value of k (Figure 5.3b) and the phase spectrum equals to zero at every value of k (Figure 5.3c).

3 Figure 5.3 : periodic unit sample sequence with = 64 and its magnitude and phase spectrum Power spectral density of periodic signal The average power of a discrete-time periodic signal with period is defined as: P x = x n 2 = C k 2 n = k= EXAMPLE 5.2 Consider a periodic sequence x n = cos 2π 3 n + sin 2π 5 n, Find the DTFS coefficients C k. Solution We first must check if this sequence is a periodic sequence and find its period. From the first term =3, and from the second term 2=5, so the total period=lcm(3,5)=5 =5; for n=:4; x(n+)=cos(2*pi*n/3)+sin(2*pi*n/5); for k=:; SUM=; for n=:; C(k)=(/)*x(n)*exp(-j*2*pi*(k-)*(n-)/); SUM=SUM+C(k); C(k)=SUM; C =,,, -.5i,,.5,,,,,.5,,.5i,,

4 The power of x(n) can be evaluated either from x(n) or from C k.the power of the signal x(n) in example 7.2 can be evaluated as following: for n=:; PSD_(n)=(abs(x(n))).^2; PSD_=sum(PSD_)/ for k=:; PSD_2(k)=(abs(C(k))).^2; PSD_2=sum(PSD_2) PSD_ = PSD_2 = Figure 5.4 show the amplitude and phase spectrum of the signal in example 5.2 n_2=:4; stem(n_2,abs(c)), title('abs(ck)') stem(n_2,angle(c)),title('phase(ck)').7.6 Abs(Ck) 2 Phase(Ck) (a) (b) Figure 5.4 : (a) Amplitude spectrum (b) Phase spectrum Exercise 5. Consider a sequence x n = 2 + 2cos π 4 n + cos π 2 n + 2 cos 3π 4 n Using Matlab (a) Find the DTFS coefficients C k. (b) Find the power of the signal over one period. (c) Sketch its amplitude and phase spectrum for one period. Check for your solution using formulas.

5 Discrete-Time Fourier Transform (DTFT) DTFT transforms a discrete signal x(n) into a complex-valued continuous function X(e jω ) of real variable ω, called digital frequency, which is measured in radians. If x(n) is absolutely summable,that is given by x n <, then its discrete-time Fourier transform is The inverse discrete-time Fourier transform (IDTFT) of X(e jω ) is given by The DTFT X(e jω ) is periodic in ω with period 2π. j jn ( DTFT ) X ( e ) X ( z ) x ( n) e If x(n) is of infinite duration, then MATLAB cannot be used directly to compute X(e jω ) form x(n). However, we can use it to evaluate the expression X e jω over, π frequencies. EXAMPLE 5.3 Determine the discrete-time Fourier transform of x n =.5 n u(n) The sequence x(n) is absolutely summable; therefore its discrete-time Fourier transform exists. j z e n j jn ( IDTFT ) x ( n) X ( e ) e d 2 2 X e jω = x(n)e jωn n = =.5 n e jωn n = =.5e jω n =.5e jω n = = e jω e jω.5,.5e jω < EXAMPLE 5.4 Determine the discrete-time Fourier transform of x n = {, 2, 3,4,5} The sequence x(n) is absolutely summable; therefore its discrete-time Fourier transform exists. X e jω = x(n)e jωn n= = e jω e jω + 4e j2ω + 5e j3ω EXAMPLE 5.5 Determine the discrete-time Fourier transform of h n = {,,} The sequence x(n) is absolutely summable; therefore its discrete-time Fourier transform exists. H e jω = h(n)e jωn n= = + e jω + e j2ω = e jω e jω + + e jω = e jω + cos(ω)

6 Since X e jω is a complex-valued function, we will have to plot its magnitude and its phase with respect to ω. ow ω is a real variable between and, which would mean that we can plot only a part of X e jω function using MATLAB. Some useful properties of DTFT: Periodicity: the DTFT X e jω is periodic in ω with period 2π, X e jω j [ω +2π ] = X e Symmetry: for real-valued x(n), X e jω is conjugate symmetric. X e jω = X e jω (even symmetry) phase X e jω = phase X e jω (odd symmetry) EXAMPLE 5.6 Sketch the phase and magnitude spectrum of H e jω in example 5.5 n = :2; h = ones(size(n)); subplot(3) stem(n,h) grid axis([- 3-2]) xlabel('n') ylabel('h[n]') title('h[n] = \delta(n) + \delta(n-) + \delta(n-2)') % DTFT of h[n] subplot(32) W=-pi:.:pi; H = ( + 2*cos(W)).*exp(-j*W); plot(w,abs(h),'r') grid xlabel('w '); ylabel(' H(w) '); title('dtft of h[n] = \delta(n) + \delta(n-) + \delta(n-2)') subplot(33) plot(w,angle(h),'r') grid xlabel('w'); ylabel('phase'); title('dtft of h[n] = \delta(n) + \delta(n-) + \delta(n-2)')

7 Phase (degrees) Magnitude (db) phase H(w) h[n] 2 h[n] = (n) + (n-) + (n-2) n DTFT of h[n] = (n) + (n-) + (n-2) w DTFT of h[n] = (n) + (n-) + (n-2) w Figure 7.5 : Plots in example 7.6 EXAMPLE 5.7 Determine the discrete-time Fourier transform of h n = {,,} from its z-transform. H z = h n z n n= = + z + z 2 = z2 + z + z 2 H ω = H z z=e jω = + e jω + e j2ω num = [ ]; den = [ ]; freqz(num, den) ormalized Frequency ( rad/sample) ormalized Frequency ( rad/sample)

8 Figure 5.6 : Plots in example 5. 7 Exercise 5.2 Determine and sketch the magnitude and phase response of DTFT for following system: y n = 2 x n + x(n ) Frequency response function from difference equations The discrete-time Fourier transform of an impulse response is called the Frequency Response of an LTI system. EXAMPLE 5.8 An LTI system is specified by the difference equation y n =.8y n + x n a- Determine H ω b- Calculate and plot the frequency response to x n = cos.5πn u(n) Solution: a- H ω =.8e jω b- The input with frequency ω o =.5π and θ o =. The response of the system is Therefore, H ω =.5π = = 4.928e j e j.5π y n = cos.5πn.5377 u(n) This means that at the output the sinusoid is scaled by and shifted by 3.42 samples. This can be verified by Matlab. b= ; a=[,-.8] n=: x=cos(.5*pi*n) y=filter(b,a,x) subplot(2) stem(n,x) title('input sequence') subplot(22) stem(n,y) title('output sequence')

9 Figure 5.7 : Plots in example 5. 8 Exercise 5.3 An LTI system is specified by the difference equation y n = x n x n 2 a- Determine H ω b- Plot the magnitude and phase spectrum of H ω c- Calculate and plot the frequency response to : x n = 4 + sin πn

Overview of Discrete-Time Fourier Transform Topics Handy Equations Handy Limits Orthogonality Defined orthogonal

Overview of Discrete-Time Fourier Transform Topics Handy equations and its Definition Low- and high- discrete-time frequencies Convergence issues DTFT of complex and real sinusoids Relationship to LTI