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ee20-hw02-f10-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 3 SOLUTIONS HW 3.1 (i) If δ T ( t )= δ ( t T ) , then ( x δ T )( t )= x ( t T ) . 0 0.5 1 1.5 2 2.5 3 3.5 4 -0.5 0 0.5 1 t Figure 1: 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 2: Plot of 2.2(c)(ii), B = 5W/4 (ii) 1
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-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 3: 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 4: Plot of 2.2(c)(ii), B = 3W HW 3.2 IOU one solution. Check back later. HW 3.3 (a) (i) Fig. 5 shows the sketch of x ( n ) = u ( n ) . From problem 2 , we know that we can write y ( n ) = . . . + x ( 1) f ( n + 1)+ x (0) f ( n )+ 2
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x (1) f ( n 1)+ . . . . Fig. 6 shows this summation and Fig. 7 shows the output for this summation. 0 3 1 2 1 -1 -2 1 1 1 x(n) = u(n) Figure 5: x ( n ) (ii) x ( n ) is shown in Fig. 8. Working out as done in (i), we get y ( n ) as in Fig. 9. (iii) x ( n ) is shown in Fig. 10. Working out as done in (i), we get y ( n ) as in Fig. 11. (iv) x ( n ) is shown in Fig. 12. Working out as done in (i), we get y ( n ) as in Fig. 13. (b) As seen in the figures, we note that the output is nonzero whenever the signal changes. When the input increases with time, the output is positive and vice versa. This gives an indication of the direction of the edge in the input signal. (c) We expect the filter to pass higher frequencies. The magnitude of the output is the difference in the signal and hence when the signal varies slowly, the output is lower, while when the signal varies fast, the output is high. This implies that the filter magnifies higher frequencies more than the lower frequencies. (d) Putting in e iωn into the system, we get e iωn e ( n 1) = (1 e ) e iωn . 3
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0 3 1 2 1 -1 -2 -1 x(0 )f(n) x(1)f(n-1) x(2)f(n-2) 0 3 1 2 1 -1 -2 -1 + + 0 3 1 2 1 -1 -2 -1 + Figure 6: Summation of Terms Hence, the frequency response is F ( ω ) = 1 e = 1 cos( ω )+ i sin( ω ) = 2sin 2 ( ω/ 2)+ i 2sin( ω/ 2)cos( ω/ 2) = 2sin( ω/ 2)(sin( ω/ 2)+ i cos( ω/ 2)) = 2sin( ω/ 2) i (cos( ω/ 2 i sin( ω/ 2)) = 2 i sin( ω/ 2) e iω/ 2 4
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0 3 1 2 1 -1 -2 y(n) Figure 7: y ( n ) 0 3 1 2 1 -1 x( n) 1 1 1 4 Figure 8: x ( n ) (e) Fig. 14 shows the magnitude and phase responses of F ( ω ) . The mag- nitude response shows a high-pass filter, and is thus consistent with re- sults in part (a) and conjecture in part (b). HW 3.4 IOU one solution. Check back later. HW 3.5 (a) For N =3 , y (0)= α y ( 3)+ x (0)=1 5
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0 3 1 2 1 -1 y(n) -1 4 Figure 9: y ( n ) 0 2 1 -1 -2 1 x(n) 1 1 1 1 Figure 10: x ( n ) y (1)= α y ( 2)+ x (1)=0 y (2)= α y ( 1)+ x (2)=0 y (3)= α y (0)+ x (3)= α y (4)= α y (1)+ x (4)=0 y (5)= α y (2)+ x (5)=0 y (6)= α y (3)+ x (6)= α 2 y (7)= α y (4)+ x (7)=0 y (8)= α y (5)+ x (8)=0 6
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0 1 2 -1 y(n) -2 Figure 11: y ( n ) 0 1 2 -1 -2 x(n) 1 -1 1 1 -1 Figure 12: x ( n ) y (9)= α y (6)+ x (9)= α 3 For N =4 , y (0)= α y ( 4)+ x (0)=1 y (1)= α y ( 3)+ x (1)=0 y (2)= α y ( 2)+ x (2)=0 y (3)= α y ( 1)+ x (3)=0 y (4)= α y (0)+ x (4)= α 7
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0 1 2 2 -1 -2 2 y(n) 2 -2 -2 Figure 13: y ( n ) -4 -3 -2 -1 0 1 2 3 4 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 ω Magnitude and Phase of F( ω ) Magnitude Angle(Phase) Figure 14: Magnitude and Phase of F ( ω ) y (5)= α y (1)+ x (5)=0 y (6)= α y (2)+ x (6)=0 y (7)= α y (3)+ x (7)=0 y (8)= α y (4)+ x (8)= α 2 8
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y (9)= α y (5)+ x (9)=0 Generalizing for N , with x ( n )= δ ( n ) f ( n )= y ( n )=
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