Question

Problem 31: (34 points) 1. (10 points) A pulse width modulated (PWM) signal fPwM(t) in Figure 2. The symbol D represents a duty cycle, a number between zero and one. Determine the compact trigonometric Fourier series coefficients (Co C,11 %) of the signal f(t). 2. (10 points) One use of PWM is to generate variable DC voltages. While the PWM signal is not DC, you should be able to see from your results in part 1 that it hss a DC component that depends on the duty cyle, D. If the voltage is applied to a system that is low pass in nature, it will respond to the DC component of the signal PWM DT, TO Figure 2: A pulse width modulated signal. while being largely unaffected by the other frequency components Assume the PWM signal is applied to the low-pass system represented by the frequency response function where the time constant T. Show that the sinusoidal steady-state output y(t) can be expressed as y(t)-G+Σ cos n uo n tan ( nT) 3. (14 points) Use MATLAB to generate and plot the signal g(t) for three values of duty cycle, 0.25, 0.58 and 0.8 using the following parameters and guideline: Set A 5 V, To 10 ms, and 21 ms. Use a time vector that spans the interval -TO,4701 with 10,001 uniformly spaced points. .Approximate the output of the filter network as 100 y(t) ~G+Σ . Plot all three results for y(t) as a function of time in ms. -For D 0.58, plot y(t) using a dashed blue line. TL T -For D 0.25, plot y(t) using a solid red line. For D0.8, plot y(t) using a dotted black line. . Use the MATLAB function legend to distinguish the plots. . Appropriately label the x and y axes and title the graph; no credit is given for MATLAB plots whose axes are unlabeled! . Use the MATLAB command gtext to place your name and section name within the figure

Answer 1

The given square wave is We have to determine compact tri geometric fourier series coefficient C_0 = a_0 c_n = squareroot a^2_n + b^2_n theta_n = tan^-1 (-b_n/a_n) and a_n = 2/T_0 integral^T_0_0 f(t) cos n w_0 t dt a_n = 2/T_0 integral^T_0_0 A cos n w_0 t dt a_n = 2A/T_0 [sin n w_0 t/n w_0[^DT_0_0 a_n = 2A/T_0 [sin n w_0 DT_0/nw_0] \r\n and w_0 2pi/T_D implies w_0 T_D = 2 pi a_n = 2A/2 pi n [sin n T_D D times pi.T_D] a_n = 2A/2 pi n sin (n D 2 pi) a_n = A/pi n sin (2 pi n D) a_n = A/pi_n sin (2pi n D) and b_n = 2/T_0 integral^T_0_0 A sin n w_0 t dt = 2/T_0 integral^DT_0_- A sin n w_0 t dt = 2A/T_D [cos n w_0 t)/n w_0]^0_DT_0 = 2A/T_0 [1 - cos (n w_0 D T_0/n w_0] = 2A/2 pi n [1 - cos (2 pi n D)/1] b_n = A/pi n [1 - cos (2 pi n D)]