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ps7sol

# ps7sol - 18.03 Problem Set 7 Solutions 26[Part I problems...

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18.03 Problem Set 7: Solutions 26. [Part I problems: 4] (a) [10] G ( s ) = 0 f ( at ) e - st dt . To make this look more like F ( s ) = 0 f ( t ) e - st dt , make the substitution u = at . Then du = a dt and G ( s ) = 0 f ( u ) e - su/a du a = 1 a 0 f ( u ) e - ( s/a ) u du = 1 a F ( s/a ) . For example, take f ( t ) = sin t , so F ( s ) = 1 s 2 + 1 , g ( t ) = sin( at ), G ( s ) = a s 2 + a 2 . Now compute 1 a F s a = 1 a 1 ( s/a ) 2 + 1 = a s 2 + a 2 = G ( s ). (b) [10] Compute F ( s ) G ( s ) = 0 0 f ( x ) e - sx g ( y ) e - sy dxdy = R f ( x ) g ( y ) e - s ( x + y ) dxdy , where R is the first quadrant. The suggested substitution is x = t - τ , y = τ . To convert to these coordinates, note that the Jacobian is det ∂x ∂t ∂x ∂τ ∂y ∂t ∂y ∂τ = det 1 - 1 0 1 = 1 For fixed τ , t ranges over numbers greater than τ , and τ ranges over positive numbers; or, what is the same, t ranges over positive numbers and τ ranges over numbers between 0 and t . So, continuing, and noticing that x + y = t , F ( s ) G ( s ) = 0 t 0 f ( t - τ ) g ( τ ) e - st dτdt = 0 t 0 f ( t - τ ) g ( τ ) e - st dt = 0 ( f ( t ) * g ( t )) e - st dt = 0 h ( t ) e - st dt = H ( s ). 27. [Part I problems: 5] (a) [10] Unit impulse response is the solution w ( t ) to a ˙ w + bw = δ ( t ) with rest initial

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ps7sol - 18.03 Problem Set 7 Solutions 26[Part I problems...

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