ee20-hw05-f09-sol - EECS 20N: Structure and Interpretation...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

This note was uploaded on 12/24/2011 for the course ELECTRICAL 20 taught by Professor Babak during the Spring '11 term at Berkeley.

Page1 / 21

ee20-hw05-f09-sol - EECS 20N: Structure and Interpretation...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online