DeCarlo Ch22 Solutions

Linear Circuit Analysis: The Time Domain and Phasor Approach

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CHAPTER 22 PROBLEM SOLUTIONS SOLUTION TO PROBLEM 22.1 (a) For figure P22.1a, T 0 = 2 and ϖ 0 = π . Let t 0 = -1 in equation 22.5b. Then f(t) = δ (t) and c n = 0.5 ( t ) e - jn π t dt - 1 1 = 0.5 for all n From equation 22.6, a n = 1 and b n = 0 for all n. Finally from equation 22.2 f ( t ) = 0.5 + cos( n π t ) n = 1 (b) For figure P22.1b, T 0 = 2 and ϖ 0 = π . Let t 0 = 0 in equation 22.5b. Then f(t) = - δ (t -1) and c n = - 0.5 ( t - 1) e - jn π t dt - 1 1 = - 0.5 e - jn π From equation 22.6, b n = 0 for all n, and a n = -1 for n even a n = 1 for n odd Finally from equation 22.2 f ( t ) = - 0.5 + cos( π t ) - cos(2 π t ) + cos(3 π t ) - cos(4 π t ) + K SOLUTION TO PROBLEM 22.2 (a) T 0 = 1 and ϖ 0 = 2 π . Let t 0 = 0 in equation 22.5b. Then f ( t ) = e - l n (2) ( 29 t and from equation 22.5b c n = e - l n (2) ( 29 t e - j 2 n π t dt 0 1 = e - l n (2) + j 2 n π ( 29 t dt 0 1 = - 1 l n (2) + j 2 n π ( 29 e - l n (2) + j 2 n π ( 29 - 1 [ ] = 0.5 l n (2) + j 2 n π ( 29 (b) Using the above result for c n c 0 = 0.5 ln2 = 0.7213 , c 1 = 0.5 ln2+ j2 π = 0.7213e -j1.4609 c 2 = 0.5 ln2+ j4 π = 0.397e -j1.516
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From equation 22.6 d 0 = c 0 = 0.7213 d 1 =2c 1 = 0.158, θ 1 =-1.461x180/ π =-83.7 o d 2 =2c 2 = 0.0795 θ 2 =-1.516x180/ π =-86.84 o Thus, f(t) in the form of equation 22.3 is f(t) = 0.7213 + 0.158cos(2 π t - 83.7 o ) + 0.0795os(2 π t - 86.84 o ) SOLUTION TO PROBLEM 22.3. (a) T 0 = 1 and ϖ 0 = 2 π . Let t 0 = 0 in equation 22.5b. Then f ( t ) = e - l n (2) ( 29 t u ( t ) - u ( t - 0.5) [ ] and from equation 22.5b c n = e - l n (2) ( 29 t e - j 2 n π t dt 0 0.5 = e - l n (2) + j 2 n π ( 29 t dt 0 0.5 = - 1 l n (2) + j 2 n π ( 29 e - 0.5 l n (2) + j 2 n π ( 29 - 1 [ ] = 1 - ( - 1) n 2 l n (2) + j 2 n π ( 29 (b) Using the above result for c n , and MATLAB to evaluate the numerical result, »n= 0; »c0= (1- (-1)^n/sqrt(2))/(log(2) + j*2*n*pi c0 4.2256e-01 »n=1; »c1= (1- (-1)^n/sqrt(2))/(log(2) + j*2*n*pi) c1 = 2.9612e-02- 2.6843e-01i »abs(c1) ans = 2.7006e-01 »degreec1=angle(c1)*180/pi degreec1 = -8.3705e+01 »n=2; »c2= (1- (-1)^n/sqrt(2))/(log(2) + j*2*n*pi) c2 = 1.2817e-03- 2.3237e-02i
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»abs(c2) ans = 2.3272e-02 degreec2= angle(c2)*180/pi degreec2 = -8.6843e+01 From equation 22.6 and equation 22.3 f(t) = 0.4226 + 0.54cos(2 π t - 83.7 o ) + 0.04654os(2 π t - 86.84 o ) SOLUTION PROBLEM 22.4. (a) f(t) = cos(4t) sin(2t) = 0.5[ sin(6t) - sin (2t)] . The fundamental angular frequency of f(t) is ϖ 0 = 2 rad/s. The given f(t) can be expressed as f(t) = -0.5 sin( ϖ 0 t) + 0.5sin(3 ϖ 0 t) . Observe that b 1 = -0.5, b 3 = 0.5 and all other a i and b i are zero. From equation 22.4 , d 1 = 0.5 /-90 o and d 3 = 0.5 /90 o. From equations 22.6a and 22.6b. c 1 = 0.25j and c 3 = -0.25j. All other c n are zero for n positive (b) f(t) = sin 2 (4t) cos 2 (8t)= 0.5[1 - cos(8t)]x0.5 1+ cos(16t) = 0.25 [ 1 - cos(8t) + cos(16t) - cos(8t) cos(16t)] = 0.25 - 0.375 cos(8t) + 0.25cos(16t) - 0.125cos(24t) The fundamental angular frequency of f(t) is ϖ 0 = 8 rad/s.
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DeCarlo Ch22 Solutions - CHAPTER 22 PROBLEM SOLUTIONS...

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