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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 5 SOLUTIONS HW 5.1 Solution (a) x ( t ) = cos 3 π 4 t + sin 2 π 3 t If x(t) is periodic with fundamental period p then by definition x ( t + p ) = x ( t ) cos bracketleftbigg 3 π 4 ( t + p ) bracketrightbigg + sin bracketleftbigg 2 π 3 ( t + p ) bracketrightbigg = cos bracketleftbigg 3 π 4 t bracketrightbigg + sin bracketleftbigg 2 π 3 t bracketrightbigg cos bracketleftbigg 3 π 4 t + 3 π 4 p bracketrightbigg + sin bracketleftbigg 2 π 3 t + 2 π 3 p bracketrightbigg = cos bracketleftbigg 3 π 4 t bracketrightbigg + sin bracketleftbigg 2 π 3 t bracketrightbigg See that, both the cos and sin term would have to repeat themselves after p. 3 π 4 p = n 2 π where n = 1, 2, ... (1) 2 π 3 p = m 2 π where m = 1, 2, ... (2) divide (1) and (2) 3 π 4 p 2 π 3 p = n 2 π m 2 π ⇒ n m = 9 8 Plug either n = 9 or m = 8 into (1) or (2) correspondingly, and solve for p. We have p = 24. and ω = 2 π p = π 12 (b) x ( t ) = cos 2 t = 1 2 (1 + cos 2 t ) x ( t + p ) = 1 2 (1 + cos [2( t + p )]) x ( t + p ) = x ( t ) only if 2 p = 2 π ⇒ p = π ⇒ ω = 2 π π = 2 1 (c) x ( t ) = cos( t 2 ) x ( t + p ) = cos[( t + p ) 2 ] = cos( t 2 + 2 tp + p 2 ) See that, no constant p can satisfy 2 tp + p 2 = n 2 π where n = 1, 2, 3, ... Therefore, x(t) is aperiodic. (d) x ( t ) = cos( t ) + sin parenleftbigg 2 π 3 t parenrightbigg If x is periodic with period p , then apply the procedures similiar to part (a). See that, p = 2 πn, n = 1 , 2 , ... 2 π 3 p = 2 πm, m = 1 , 2 , ... n m = p/ 2 π p/ 3 = 3 2 π = irrational Therefore, x(t) is aperiodic. (e) x ( t ) = e i ( 3 π 4 t + 2 π 5 ) = cos( 3 π 4 t + 2 π 5 ) + i sin( 3 π 4 t + 2 π 5 ) x ( t + p ) = cos( 3 π 4 ( t + p ) + 2 π 5 ) + i sin( 3 π 4 ( t + p ) + 2 π 5 ) ⇒ x(t) is periodic with fundamental period p = 8 3 and ω = 3 π 4 2 HW 5.2 Solution Since f is periodic with period T f and g is periodic with period T g , f ( t ) = f ( t + kT f ) , ∀ k ∈ Z g ( t ) = g ( t + lT g ) , ∀ l ∈ Z From the above equations, h ( t ) = f ( t ) + g ( t ) = f ( t + kT f ) + g ( t + lT g ) If ∃ k, l ∈ Z such that kT f = lT g , then T f T g = l k , so l k is a rational number. Thus, in order for the equations to be equivalent T f T g must be a rational number as well and the period is kT f = lT g . HW 5.3 Solution (a) X k = 1 T integraldisplay Δ / 2 − Δ / 2 1 Δ e − jkω t dt = 1 Δ T − 1 jkω ( e − jkω Δ / 2 − e jkω Δ / 2 ) = 1 Δ T − 1 jkω ( − 2 j ) sin( kω Δ / 2) = 1 T Δ 2 kω sin( kω Δ / 2) = 1 T sinc( k 2 π/T Δ / 2) (b) Yes, we can figure out the DC term by noticing that X is the average value over T, which is 1 T (c) X k = 1 T sinc( k 2 π/T ∗ T/ 2) = 1 T sinc( kπ ) = braceleftbigg 1 T k = 0 k negationslash = 0 3 This makes sense because x ( t ) is just a constant....
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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 Berkeley.
 Spring '11
 BABAK
 Electrical Engineering

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