DeCarlo Ch16b 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 SOLUTION 16.31. In order to compute the area beneath v ( t - ) h ( ) seven regions will be considered: t < 0, 0 t < 1, 1 t < 2, 2 t < 3, 3 t < 4, 4 t < 5 and 5 t . Step 1: t < 0. For t in this region v ( t - ) h ( ) = 0 for all . Hence y ( t ) = v ( t ) h ( t ) = 0 for t < 0. Step 2: 0 t < 1. In this case v ( t - ) h ( ) = v 0 × h 0 for 0 t and is zero otherwise. Therefore the area beneath v ( t - ) h ( ) equals y ( t ) = v ( t ) h ( t ) = v 0 × h 0 × t for 0 t < 1. Step 3: 1 t < 2. For t in this region we have v ( t - ) h ( ) = v 1 × h 0, 0 t - 1 v 0 × h 0, t - 1 < < 1 v 0 × h 1, 1 < t 0, otherwise Therefore the area beneath v ( t - ) h ( ) equals y ( t ) = v ( t ) h ( t ) = v 1 × h 0 × ( t - 1) - 0 [ ] + v 0 × h 0 × 1 - ( t - 1) [ ] + v 0 × h 1 × ( t - 1) = = t × ( v 1 × h 0 - v 0 × h 0 + v 0 × h 1) - v 1 × h 0 + 2 × v 0 × h 0 - v 0 × h 1, for 1 t < 2 Step 4: 2 t < 3. In this case v ( t - ) h ( ) = v 1 × h 0, t - 2 < < 1 v 1 × h 1, 1 t - 1 v 0 × h 1, t - 1 < < 2 v 0 × h 2, 2 < t 0, otherwise Hence, for 2 t < 3, y ( t ) = v ( t ) h ( t ) = = v 1 × h 0 × 1 - ( t - 2) [ ] + v 1 × h 1 × ( t - 1) - 1 [ ] + v 0 × h 1 × 2 - ( t - 1) [ ] + v 0 × h 2 × ( t - 2) = = t × ( - v 1 × h 0 + v 1 × h 1 - v 0 × h 1 + v 0 × h 2) + 3 × v 1 × h 0 - 2 × v 1 × h 1 + 3 × v 0 × h 1 - 2 × v 0 × h 2 Step 5: 3 t < 4. In this case v ( t - ) h ( ) = v 1 × h 1, t - 2 < < 2 v 1 × h 2, 2 t - 1 v 0 × h 2, t - 1 < < 3 0, otherwise Hence, for 3 t < 4, y ( t ) = v ( t ) h ( t ) = = v 1 × h 1 × 2 - ( t - 2) [ ] + v 1 × h 2 × ( t - 1) - 2 [ ] + v 0 × h 2 × 3 - ( t - 1) [ ] = = t × ( - v 1 × h 1 + v 1 × h 2 - v 0 × h 2) + 4 × v 1 × h 1 - 3 × v 1 × h 2 + 4 × v 0 × h 2 Step 6: 4 t < 5. In this case v ( t - ) h ( ) = v 1 × h 2 for t - 2 < < 3 and is zero otherwise. Therefore y ( t ) = v ( t ) h ( t ) = v 1 × h 2 × 3 - ( t - 2) [ ] = v 1 × h 2 × (5 - t ) for 4 t < 5.

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1/25/02 P16-2 © R. A. DeCarlo, P. M. Lin Step 7: 5 t . For t in this region v ( t - ) h ( ) = 0 for all . Hence y ( t ) = v ( t ) h ( t ) = 0 for 5 t . In sum, y ( t ) = v 0 × h 0 × t , 0 t < 1 t × ( v 1 × h 0 - v 0 × h 0 + v 0 × h 1) - v 1 × h 0 + 2 × v 0 × h 0 - v 0 × h 1, 1 t < 2 t × ( - v 1 × h 0 + v 1 × h 1 - v 0 × h 1 + v 0 × h 2) + 3 × v 1 × h 0 - 2 × v 1 × h 1 + 3 × v 0 × h 1 - 2 × v 0 × h 2, 2 t < 3 t × ( - v 1 × h 1 + v 1 × h 2 - v 0 × h 2) + 4 × v 1 × h 1 - 3 × v 1 × h 2 + 4 × v 0 × h 2, 3 t < 4 v 1 × h 2 × (5 - t ), 4 t < 5 0, otherwise Hence, y 1 = y (1) = v 0 × h 0 = 6 y 2 = y (2) = v 0 × h 1 + v 1 × h 0 = 8 y 3 = y (3) = v 0 × h 2 + v 1 × h 1 = - 6 y 4 = y (4) = v 1 × h 2 = 4 (b) Using the expressions of p ( x ) and q ( x ) it follows that p ( x ) × q ( x ) = x 3 × ( v 0 × h 0) + x 2 × ( v 0 × h 1 + v 1 × h 0) + x × ( v 0 × h 0 + v 1 × h 1) + v 1 × h 2 We observe that the coefficients of p ( x ) × q ( x ) are exactly y 1 , y 2 , y 3 and y 4 , respectively. Therefore r ( x ) = p ( x ) × q ( x ) . SOLUTION 16.32. (a) This part will be solved using the techniques of convolution algebra. Therefore we can write f 3 ( t ) as f 3 ( t ) = f 1 ( - 1) ( t ) f 2 (1) ( t ) Where the superscript (-1) means integration and the superscript (1) means differentiation. From figure P16.32 we observe that f 1 ( t ) = 4 u ( t ) - u ( t - 4) [ ] Hence f 1 ( - 1) ( t ) = 4 tu ( t ) - ( t - 4) u ( t - 4) [ ] = = 4 r ( t ) - r ( t - 4) [ ] By inspection, from the same figure, we have f 2 (1) ( t ) = 4 ( t ) - 2 ( t - 2) + 2 ( t - 4) - 2 ( t - 6) + ( t - 8) [ ] Using the sifting property of the delta function f 3 ( t ) can be computed as follows f 3 ( t ) = 4 r ( t ) - r ( t - 4) [ ] { } 4 ( t ) - 2 ( t - 2) + 2 ( t - 4) - 2 ( t - 6) + ( t - 8) [ ] { } = = 16 r ( t ) - 2 r ( t - 2) + r ( t - 4) - r ( t - 8) + 2 r ( t - 10) - r ( t - 12) [ ]
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