DeCarlo Ch16a Solutions

Linear Circuit Analysis: The Time Domain and Phasor Approach

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1/25/02 P16-1 © R. A. DeCarlo, P. M. Lin P ROBLEM S OLUTIONS Solution 16.1. (a) By the definition of the convolution integral f 2 ( t ) f 2 ( t ) = f 2 ( t - ) f 2 ( ) d = -∞ 2 u ( t - )2 u ( ) d = 4 u ( t - ) d 0 -∞ The integrand, u ( t - ) , is nonzero only when t . This suggests that there are two regions of consideration: t < 0 and t 0. Case 1: t < 0. Here u ( t - ) = 0 since is restricted to the interval [0, ). Hence f 2 ( t ) f 2 ( t ) = 0 , for t < 0. Case 2: t 0. f 2 ( t ) f 2 ( t ) = 4 u ( t - ) d = 4 d 0 t -∞ = 4 t , for t 0. In sum, f 2 ( t ) f 2 ( t ) = 0, t < 0 4 t , t 0 (b) By the definition of the convolution integral f 2 ( t ) f 3 ( t ) = f 2 ( t - ) f 3 ( ) d = -∞ 2 u ( t - )4 e - 2 u ( ) d = 8 e - 2 u ( t - ) d 0 -∞ The integrand, u ( t - ) , is nonzero only when t . This suggests that there are two cases to consider: t < 0 and t 0. Case 1: t < 0. Here u ( t - ) = 0 since is restricted to the interval [0, ). Hence f 2 ( t ) f 3 ( t ) = 0, for t < 0. Case 2: t 0. f 2 ( t ) f 3 ( t ) = 8 e - 2 d = - 4 e - 2 ] 0 t 0 t = 4(1 - e - 2 t ) In sum, f 2 ( t ) f 2 ( t ) = 0, t < 0 4(1 - e - 2 t ), t 0 (c) By the definition of the convolution integral and the sifting property of the delta function f 1 ( t ) f 3 ( t ) = f 1 ( t - ) f 3 ( ) d = -∞ 5 ( t - )4 e - 2 u ( ) d = -∞ = 20 e - 2 u ( ) ] = t = 20 e - 2 t u ( t ) (d) By the definition of the convolution integral
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1/25/02 P16-2 © R. A. DeCarlo, P. M. Lin f 3 ( t ) f 3 ( t ) = f 3 ( t - ) f 3 ( ) d = -∞ 4 e - 2( t - ) u ( t - )4 e - 2 u ( ) d = 16 e - 2 t u ( t - ) d 0 -∞ The integrand, u ( t - ) , is nonzero only when t . This suggests that there are two cases to consider: t < 0 and t 0. Case 1: t < 0. Here u ( t - ) = 0 since is restricted to the interval [0, ). Hence f 3 ( t ) f 3 ( t ) = 0 , for t < 0. Case 2: t 0. f 3 ( t ) f 3 ( t ) = 16 e - 2 t d = 16 te - 2 t 0 t In sum, f 3 ( t ) f 3 ( t ) = 0, t < 0 16 te - 2 t , t 0 (e) By the definition of the convolution integral and the sifting property of the delta function f 1 ( t + 2) f 2 ( t + 4) = f 1 ( t + 2 - ) f 2 ( + 4) d = 5 ( t + 2 - )2 u ( + 4) d = -∞ -∞ = 10 u ( + 4) ] = t + 2 = 10 u ( t + 6) (f) By the distributive property of convolution f 2 ( t ) f 2 ( t ) + f 3 ( t ) [ ] = f 2 ( t ) f 2 ( t ) + f 2 ( t ) f 3 ( t ) Using the results of parts (a) and (b) the result follows immediately f 2 ( t ) f 2 ( t ) + f 3 ( t ) [ ] = 0, t < 0 4(1 + t - e - 2 t ), t 0 Solution 16.2. (a) By definition f 3 ( t ) = f 1 ( ) f 2 ( t - ) d = -∞ K 1 u ( - T 1 ) K 2 u ( t - - T 2 ) d = K 1 K 2 u ( t - - T 2 ) d T 1 -∞ Here observe that u ( t - - T 2 ) = 0 for t - T 2 . Hence there are two cases to consider: t - T 2 < T 1 and t - T 2 T 1 . Case 1: t - T 2 < T 1 . Here u ( t - - T 2 ) = 0 , since is restricted to the interval [ T 1 , ) . f 3 ( t ) = 0 Case 2: t - T 2 T 1 .
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1/25/02 P16-3 © R. A. DeCarlo, P. M. Lin f 3 ( t ) = K 1 K 2 d = K 1 K 2 ( t - T 2 - T 1 ) T 1 t - T 2 In sum, f 3 ( t ) = 0, t < T 1 + T 2 K 1 K 2 ( t - T 2 - T 1 ), t T 1 + T 2 (b) By definition f 3 ( t ) = f 1 ( t - ) f 2 ( ) d = -∞ K 1 u ( t - + T 1 ) K 2 u ( + T 2 ) d = K 1 K 2 u ( t - + T 1 ) d - T 2 -∞ Here observe that u ( t - + T 1 ) = 0 for t + T 1 . Hence there are two cases to consider: t + T 1 < - T 2 and t + T 1 ≥ - T 2 .
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