ee20-hw02-f09-sol - EECS 20N: Structure and Interpretation...

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Unformatted text preview: EECS 20N: Structure and Interpretation of Signals and Systems Department of Electrical Engineering and Computer Sciences UNIVERSITY OF CALIFORNIA BERKELEY Problem Set 2 SOLUTIONS HW 2.1 (a) (i) Let x ( t ) = a + bi and = c + di then F ( x ) = F (( a + bi )( c + di )) = F ( ac bd + ( ad + bc ) i ) = Re [ ac bd + ( ad + bc ) i ] = ac bd But F ( x ) = ( c + di ) Re [ a + bi ] = ac + adi negationslash = F ( x ) (ii) Let x ( t ) = a + bi and = c + di then f ( x ) = F (( a + bi )( c + di )) = F ( ac bd + ( ad + bc ) i ) = ( ac bd + ( ad + bc ) i ) = ac bd ( ad + bc ) i But F ( x ) = ( c + di )( a + bi ) = ac + bd + ( ad bc ) i negationslash = f ( x ) (b) Let X 1 = bracketleftbigg 1 2 bracketrightbigg and X 2 = bracketleftbigg 2 5 bracketrightbigg . Then F ( X 1 ) + F ( X 2 ) = 0 + 7 = 7 . But F ( X 1 + X 2 ) = F ( bracketleftbigg 1 7 bracketrightbigg ) = 8 negationslash = F ( X 1 ) + F ( X 2 ) . HW 2.2 (a) (i) ( t )-1-0.8-0.6-0.4-0.2 0.2 0.4 0.6 0.8 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 t Figure 1: Magnitude response of 2.2(a)(i) (ii) e a ( t 1) For the plot, a is set to -2. (iii) cos ( 2 + 2 ) ( t 1 2 ) = cos ( ) ( t 1 2 ) = ( t 1 2 ) (iv) ( w ) (b) (i) e a (ii) sin ( 2 ) = 1 (iii) X ( 2 W 3 ) = 1 3 (iv) 1 + 2 k =1 e k 1 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0.02 0.04 0.06 0.08 0.1 0.12 0.14 t Figure 2: Magnitude response of 2.2(a)(ii)-1-0.8-0.6-0.4-0.2 0.2 0.4 0.6 0.8 1-1-0.8-0.6-0.4-0.2 t Figure 3: Magnitude response of 2.2(a)(iii) (c) (i) 2-1-0.8-0.6-0.4-0.2 0.2 0.4 0.6 0.8 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 w Figure 4: Magnitude response of 2.2(a)(iv) 3 0.5 1 1.5 2 2.5 3 3.5 4-0.5 0.5 1 t Figure 5: Plot of 2.2(c)(i)-2-1.5-1-0.5 0.5 1 1.5 2 0.2 0.4 0.6 0.8 1 1.2 /W magnitude Figure 6: Plot of 2.2(c)(ii), B = 5W/4 (ii) 4-5-4-3-2-1 1 2 3 4 5 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 /W magnitude Figure 7: Plot of 2.2(c)(ii), B = 2W-5-4-3-2-1 1 2 3 4 5 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 /W magnitude Figure 8: Plot of 2.2(c)(ii), B = 3W (d) (i) u ( t ) = integraldisplay ( ) u ( t ) d = integraldisplay t ( ) u ( t ) d + integraldisplay t + ( ) u ( t ) d 5 = integraldisplay t ( )1 d + integraldisplay t + ( )0 d = integraldisplay t ( ) d Differentiating of both sides of equation (2) leads to equation (1)....
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This note was uploaded on 12/24/2011 for the course ELECTRICAL 20 taught by Professor Babak during the Spring '11 term at University of California, Berkeley.

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ee20-hw02-f09-sol - EECS 20N: Structure and Interpretation...

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