{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ee20-hw05-f09-sol

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

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

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

View Full Document

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

View Full Document
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

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern