Solutions to Assignment 4

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1 EE35 Spectrum Analysis and Discrete Time Systems (Fall 5) Solutions to Assignment. Consider the continuous-time periodic signal: x(t) = sin(t 3) + sin(6t) (8) [] (a) Obviously, the fundamental frequency is =. Using Euler s relation one has: sin(t 3) = sin( 3) = ( e ( ) e ) ( ) = e e t e3 e t sin(6t) = sin(3 t) = ( e 3 t e ) 3 t = e3 t e 3 t x(t) = e 3 t e3 e t + e e t + e3 t From Equation (9), the nonzero Fourier series coefficients a k are identified to be: a = = = e( ) a = e3 = e3 = e ( +3) = e (3 3 ) = e.7 a = e = e = e [ ( +3)] = e ( 3 ) = e.7 a 3 = = = e( ) [3] (b) With a a k e θ k, then ak and θ k are the magnitude and phase of a k, respectively. Also, as a convention, the phase angles are always converted to the range [, ] by adding (or subtracting) with multiples of. The magnitude and phase spectrum of x(t) are sketched in Figure and. Note that the magnitude spectrum is even and the phase spectrum is odd. This is expected since x(t) is a real-valued signal. Remark: This question is accidentally given to be the same as the example discussed in class as well as in Lecture Notes (page ). (9) [5]. A continuous-time periodic signal x(t) is real-valued and has a fundamental period T = 8 the fundamental frequency is =. The nonzero Fourier series coefficients for x(t) are specified as: a = a =, a 5 = a 5 = x(t) = k= a k e k t = (e 5 t + e 5 t ) + (e t e t ) = cos(5 t) sin( t) = cos ( t + ) + cos(5 t) Electrical Engineering, University of Saskatchewan Page 6

2 EE35 Spectrum Analysis and Discrete Time Systems (Fall 5) Magnitude Spectrum Phase Spectrum 5 5 Normalized Frequency (/ ) Figure : Magnitude spectrum of x(t). Comparing the terms gives:, α, 5 ; φ, otherwise 3. (Properties of Fourier Series Coefficients) {,, otherwise ; k = k,,,... [] (a) Signals (i) and (ii) are real-valued since their magnitude spectra are even and phase spectra are odd. Signal (iii) is a complex-valued since its phase spectrum is not odd. [] (b) Since e ± = and e =, the FS coefficients of signal (ii) are real-valued. Thus signal (ii) is both real-valued and even function. None of the signal is both real-valued and odd since an real-valued and odd signal must have an odd phase spectrum and the phase values can only be, / and / only. [] (c) Because signal (i) is a real-valued signal, any form of FS can be used to write x(t). For example, x(t) = e e t + e / e t + + e / e t + e e t [ ( = + cos t + ) ] + cos ( t + ) Obviously, the dc component is a = volt. The average power is simply found by applying Parseval s relation: k= a k = = watts. Electrical Engineering, University of Saskatchewan Page 7

3 EE35 Spectrum Analysis and Discrete Time Systems (Fall 5) (i) 3 3 (ii) 3 3 (iii) 3 3. (Fourier series of the sawtooth waveform) [] (a) The fundamental period is T = and the fundamental frequency is = T =. [3] (b) To compute the trigonometric Fourier series coefficients B k and C k, consider x(t) in one period, from t. Then { t, t < x(t) =, < t Electrical Engineering, University of Saskatchewan Page 8

4 EE35 Spectrum Analysis and Discrete Time Systems (Fall 5) Similarly B T x(t) cos(k t)dt = T = [ cos(kt) + t ] (k) k sin(kt) C T x(t) sin(k t)dt = T = [ sin(kt) t ] (k) k cos(kt) t cos(kt)dt = ( )k (k) t sin(kt)dt = ( )k k The DC component of x(t) is simply a = T T x(t)dt = t d t = t = [3] (c) The magnitude and phase spectrum of x(t) are plotted in Figure 5. Recall that the magnitude spectrum is the plot of a k = Bk + C k, while the phase spectrum is the plot of a tan (C k /B k ). Again, notice that since x(t) is a real function, it has even amplitude and odd phase spectrum..3 Magnitude Spectrum Phase Spectrum 5 5 Normalized Frequency (/ ) Figure 5: Magnitude spectrum of x(t). Electrical Engineering, University of Saskatchewan Page 9

5 EE35 Spectrum Analysis and Discrete Time Systems (Fall 5) [3] (d) Consider the following approximation for x(t): N ˆx(t) = a + [B k cos(k t) C k sin(k t)] k= Note the Gibb s phe- Plots of ˆx(t) for N = 5,, 5 are shown in Figure 6. nomenon at the points of discontinuity..5 N=5 3.5 N= 3.5 N=5 3 t (sec) Figure 6: Partial Fourier series approximations of x(t) with N = 5,, 5. [3] (e) The average power of x(t) can be computed as P T = x (t)dt = t dt = t3 T T 6 = 6 Using Matlab, the smallest values of N in the above approximation such that ˆx(t) captures 95% and 99% power in x(t) are found to be N = 6 and N = 3, respectively. The Matlab program is listed below. P_x=/6;a=/;w=pi; percentage=.95;p_xhat=a^;k=; while P_xhat<percentage*P_x k=k+; ak=((-).^k-)./(*(k*pi).^)+*(-).^k./(*pi*k); P_xhat=P_xhat+*(abs(ak))^; end N=k; Electrical Engineering, University of Saskatchewan Page