Continuous and Discrete Time Signals and Systems (Mandal &amp; Asif) solutions - chap03

# Continuous and Discrete Time Signals and Systems (Mandal & Asif) solutions - chap03

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Chapter 3: Time Domain Analysis of LTIC Systems Problem 3.1 Linearity: For x 3 ( t ) = α x 1 ( t ) + β x 2 ( t ) applied as the input, the output y 3 ( t ) is given by )) ( ) ( ( )) ( ) ( ( )) ( ) ( ( )) ( ) ( ( ) ( 2 1 0 2 1 1 1 2 1 1 1 2 1 3 0 3 1 1 3 1 1 3 t x t x b dt t x t x d b dt t x t x d b dt t x t x d b t y a dt dy a dt y d a dt y d m m m m m m n n n n n β α + β + α + + β + α + β + α = + + + + " " . Rearranging the terms on the right hand side of the equation, we get + + + + β + + + + + α = + + + + ) ( ) ( ) ( 2 0 2 1 1 2 1 1 2 1 0 1 1 1 1 1 1 1 3 0 3 1 1 3 1 1 3 t x b dt dx b dt x d b dt x d b t x b dt dx b dt x d b dt x d b t y a dt dy a dt y d a dt y d m m m m m m m m m m m m n n n n n " " " Expressing the higher order derivatives of x 1 ( t ) and x 2 ( t ) in terms of y 1 ( t ) and y 2 ( t ), we get + + + + β + + + + + α = + + + + ) ( ) ( ) ( 1 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 3 0 3 1 1 3 1 1 3 t y a dt dy a dt y d a dt y d t y a dt dy a dt y d a dt y d t y a dt dy a dt y d a dt y d n n n n n n n n n n n n n n n " " " or, () ) ( ) ( ) ( ) ( ) ( ) ( 2 1 0 2 1 1 1 2 1 1 1 2 1 3 0 3 1 3 1 3 t y t y a dt y y d a dt y y d a dt y y d t y a dt dy a dt y d a dt y d n n n n n n n n n n β + α + β + α + + β + α + β + α = + + + + " " which implies that ) ( ) ( ) ( 2 1 3 t y t y t y β + α = . The system is therefore linear. Time-invariance: For x ( t t 0 ) applied as the input, the output y 1 ( t ) is given by ) ( ) ( ) ( ) ( ) ( 0 0 0 1 1 0 1 1 0 1 0 1 1 1 1 1 1 1 t t x b dt t t dx b dt t t x d b dt t t x d b t y a dt dy a dt y d a dt y d m m m m m m n n n n n + + + + = + + + + " " Substituting τ = t t 0 (which implies that dt = d τ ), we get ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( 0 1 1 1 1 0 1 0 0 1 1 1 0 1 1 1 0 1 τ + τ τ + + τ τ + τ τ = + τ + τ + τ + + τ + τ + τ + τ x b d dx b d x d b d x d b t y a d t dy a d t y d a d t y d m m m m m m n n n n n " " Comparing with the original differential equation representation of the system, we get

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78 Chapter 3 ) ( ) ( or, ) ( ) ( 0 1 0 1 t y y t y y τ = τ + τ = τ , proving that the system is time-invariant. Note that the time invariance property is only valid if the coefficients a r ’s and b r ’s are constants. If a r ’s and b r ’s are functions of time, then the substitution ( τ = t t 0 ) will also affect them. Clearly, y ( τ ) y 1 ( τ + t 0 ) in such a case and the system will NOT be time- invariant. Problem 3.2 (i) 4 () 4 () 8 () () () w i th , (0 ) 0 , and (0 . t yt xt e ut y y ++= + = = = ±± ± ± ± (a) Zero-input response of the system: The characteristic equation of the LTIC system (i) is 0 8 4 2 = + + s s , which has roots at s = 2 ± j 2. The zero-input response is given by ) 2 sin( ) 2 cos( ) ( 2 2 t Be t Ae t y t t zi + = for t 0, with A and B being constants. To calculate their values, we substitute the initial conditions 0 ) 0 ( = y and 0 ) 0 ( = y ± in the above equation. The resulting simultaneous equations are 0 2 2 0 = + = B A A that has the solution, A = 0 and B = 0. The zero-input response is therefore given by . 0 ) ( = t y zi Because of the zero initial conditions, the zero-input response is also zero.
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## This note was uploaded on 01/19/2012 for the course EE 421 taught by Professor Zhicheng during the Spring '11 term at Ohio State.

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Continuous and Discrete Time Signals and Systems (Mandal & Asif) solutions - chap03

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