Chapter 17-Electrical Circuit Analysis-Problem Solutions, Exercises of Electrical Circuit Analysis

Download Chapter 17-Electrical Circuit Analysis-Problem Solutions and more Exercises Electrical Circuit Analysis in PDF only on Docsity! Chapter 17, Problem 1. Evaluate each of the following functions and see if it is periodic. If periodic, find its period. (a) ttttf 5 cos 3 3 cos 2 cos)( πππ ++= (b) y(t) = sin t + 4 cos 2 π t (c) g(t) = sin 3t cos 4t (d) h(t) = t2cos (e) z(t) = 4.2 sin(0.4 π t + 10º) + 0.8 sin(0.6 π t + 50º) (f) p(t) = 10 (g) q(t) = te π− Chapter 17, Solution 1. (a) This is periodic with ω = π which leads to T = 2π/ω = 2. (b) y(t) is not periodic although sin t and 4 cos 2πt are independently periodic. (c) Since sin A cos B = 0.5[sin(A + B) + sin(A – B)], g(t) = sin 3t cos 4t = 0.5[sin 7t + sin(–t)] = –0.5 sin t + 0.5 sin7t which is harmonic or periodic with the fundamental frequency ω = 1 or T = 2π/ω = 2π. (d) h(t) = cos 2 t = 0.5(1 + cos 2t). Since the sum of a periodic function and a constant is also periodic, h(t) is periodic. ω = 2 or T = 2π/ω = π. (e) The frequency ratio 0.6|0.4 = 1.5 makes z(t) periodic. ω = 0.2π or T = 2π/ω = 10. (f) p(t) = 10 is not periodic. (g) g(t) is not periodic. docsity.com Chapter 17, Problem 2. Using MATLAB, synthesize the periodic waveform for which the Fourier trigonometric Fourier series is ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅⋅⋅+++−= ttttf 5 cos 25 13 cos 9 1 cos4 2 1)( 2π Chapter 17, Solution 2. The function f(t) has a DC offset and is even. We use the following MATLAB code to plot f(t). The plot is shown below. If more terms are taken, the curve is clearly indicating a triangular wave shape which is easily represented with just the DC component and three, cosinusoidal terms of the expansion. for n=1:100 tn(n)=n/10; t=n/10; y1=cos(t); y2=(1/9)*cos(3*t); y3=(1/25)*cos(5*t); factor=4/(pi*pi); y(n)=0.5- factor*(y1+y2+y3); end plot(tn,y) docsity.com Chapter 17, Problem 4. Find the Fourier series expansion of the backward sawtooth waveform of Fig. 17.48. Obtain the amplitude and phase spectra. Figure 17.48 For Probs. 17.4 and 17.66. Chapter 17, Solution 4. f(t) = 10 – 5t, 0 < t < 2, T = 2, ωo = 2π/T = π ao = (1/T) ∫ T 0 dt)t(f = (1/2) ∫ − 2 0 dt)t510( = 2 0 2 )]2/t5(t10[5.0 − = 5 an = (2/T) ∫ ω T 0 o dt)tncos()t(f = (2/2) ∫ π− 2 0 dt)tncos()t510( = ∫ π 2 0 dt)tncos()10( – ∫ π 2 0 dt)tncos()t5( = 2 0 22 tncosn 5 π π − + 2 0 tnsin n t5 π π = [–5/(n2π2)](cos 2nπ – 1) = 0 bn = (2/2) ∫ π− 2 0 dt)tnsin()t510( = ∫ π 2 0 dt)tnsin()10( – ∫ π 2 0 dt)tnsin()t5( = 2 0 22 tnsinn 5 π π − + 2 0 tncos n t5 π π = 0 + [10/(nπ)](cos 2nπ) = 10/(nπ) Hence f(t) = )tnsin( n 1105 1n π π + ∑ ∞ = docsity.com Chapter 17, Problem 5. Obtain the Fourier series expansion for the waveform shown in Fig. 17.49. Figure 17.49 For Prob. 17.5. Chapter 17, Solution 5. 1T/2,2T =π=ωπ= 5.0]x2x1[ 2 1dt)t(z T 1a T 0 o −=π−ππ == ∫ ∫∫∫ π π π π π π = π − π = π − π =ω= 2 2 0 0 T 0 on 0ntsinn 2nt..sin n 1ntdtcos21ntdtcos11dtncos)t(z T 2a ∫∫∫ π π π π π π ⎪⎩ ⎪ ⎨ ⎧ = = π=π + π −= π − π =ω= 2 2 0 0 T 0 on evenn 0, oddn, n 6 ntcos n 2ntcos n 1ntdtsin21ntdtsin11dtncos)t(z T 2b Thus, ntsin n 65.0)t(z oddn 1n ∑ ∞ = = π +−= docsity.com Chapter 17, Problem 6. Find the trigonometric Fourier series for ⎩ ⎨ ⎧ =+ << << = ).()2( and 2,10 0,5 )( tftf t t tf π ππ π Chapter 17, Solution 6. T=2π, ωo=2π/T = 1 2 0 0 1 1 1( ) 5 10 (5 10 ) 7.5 2 2 T oa f t dt dt dtT π π π π π π π ⎡ ⎤ = = + = + =⎢ ⎥ ⎣ ⎦ ∫ ∫ ∫ 2 0 0 2 2( )cos 5cos 10cos 0 2 T n oa f t n tdt ntdt ntdtT π π π ω π ⎡ ⎤ = = + =⎢ ⎥ ⎣ ⎦ ∫ ∫ ∫ 2 0 0 22 2 1 1 1( )sin 5sin 10s cos cos 2 0 T n ob f t n tdt ntdt inntdt nt ntT n n π π π π π ω π π π ⎡ ⎤⎡ ⎤ = = + = − −⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ∫ ∫ ∫ 10 ,5 cos 1 0, n odd n n n n even π π π ⎧− =⎪= − =⎡ ⎤ ⎨⎣ ⎦ ⎪ =⎩ Thus, 10( ) 7.5 sin n odd f t nt nπ ∞ = = − ∑ docsity.com