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ee20-hw02-f09-sol

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

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EECS 20N: Structure and Interpretation of Signals and Systems Department of Electrical Engineering and Computer Sciences U NIVERSITY OF C ALIFORNIA B ERKELEY Problem Set 2 SOLUTIONS HW 2.1 (a) (i) Let x ( t ) = a + bi and α = c + di then F ( αx ) = F (( a + bi )( c + di )) = F ( ac bd + ( ad + bc ) i ) = Re [ ac bd + ( ad + bc ) i ] = ac bd But αF ( x ) = ( c + di ) Re [ a + bi ] = ac + adi negationslash = F ( αx ) (ii) Let x ( t ) = a + bi and α = c + di then f ( αx ) = F (( a + bi )( c + di )) = F ( ac bd +( ad + bc ) i ) = ( ac bd +( ad + bc ) i ) = ac bd ( ad + bc ) i But αF ( x ) = ( c + di )( a + bi ) = ac + bd + ( ad bc ) i negationslash = f ( αx ) (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) δ ( t ) -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 0 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 δ ( t 1) For the plot, a is set to -2. (iii) cos ( π 2 + π 2 ) δ ( t 1 2 ) = cos ( π ) δ ( t 1 2 ) = δ ( t 1 2 ) (iv) δ ( w ) (b) (i) e a (ii) sin ( π 2 ) = 1 (iii) X ( 2 W 3 ) = 1 3 (iv) 1 + 2 k =1 e k 1

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0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 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 0.2 0.4 0.6 0.8 1 -1 -0.8 -0.6 -0.4 -0.2 0 t Figure 3: Magnitude response of 2.2(a)(iii) (c) (i) 2
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 0 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

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0 0.5 1 1.5 2 2.5 3 3.5 4 -0.5 0 0.5 1 t Figure 5: Plot of 2.2(c)(i) -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 0 0.2 0.4 0.6 0.8 1 1.2 ω /W magnitude Figure 6: Plot of 2.2(c)(ii), B = 5W/4 (ii) 4
-5 -4 -3 -2 -1 0 1 2 3 4 5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 ω /W magnitude Figure 7: Plot of 2.2(c)(ii), B = 2W -5 -4 -3 -2 -1 0 1 2 3 4 5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 ω /W magnitude Figure 8: Plot of 2.2(c)(ii), B = 3W (d) (i) u ( t ) = integraldisplay −∞ δ ( τ ) u ( t τ ) = integraldisplay t −∞ δ ( τ ) u ( t τ ) + integraldisplay t + δ ( τ ) u ( t τ ) 5

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= integraldisplay t −∞ δ ( τ )1 + integraldisplay t + δ ( τ )0 = integraldisplay t −∞ δ ( τ ) Differentiating of both sides of equation (2) leads to equation (1). (ii) integraldisplay −∞ u ( τ ) g ( t τ ) = u ( τ ) g ( t τ ) | −∞ + integraldisplay −∞ u ( τ )( g ( t τ )) = g ( t − ∞ ) + integraldisplay 0 g ( t τ ) = g ( t − ∞ ) g ( t − ∞ ) + g ( t ) = g ( t ) (e) noting that τ = at t = τ a Also, note that δ ( t ) = braceleftbigg 1 , x = 0 0 , otherwise δ ( t ) = braceleftbigg 1 , x = 0 0 , otherwise
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