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# ch06 - Solutions to Additional Problems 6.26 A signal x(t...

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Solutions to Additional Problems 6.26. A signal x ( t ) has Laplace transform X ( s ) as given below. Plot the poles and zeros in the s -plane and determine the Fourier transform of x ( t ) without inverting X ( s ). (a) X ( s ) = s 2 +1 s 2 +5 s +6 X ( s ) = ( s + j )( s j ) ( s + 3)( s + 2) zeros at: ± j poles at: 3 , 2 X ( ) = X ( s ) | s = = ω 2 + 1 ω 2 + 5 + 6 Pole-Zero Map Real Axis Imag Axis -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 Figure P6.26. (a) Pole-Zero Plot of X ( s ) (b) X ( s ) = s 2 1 s 2 + s +1 X ( s ) = ( s + 1)( s 1) ( s + 0 . 5 j 3 4 )( s + 0 . 5 + j 3 4 ) zeros at: ± 1 1

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poles at: 1 ± j 3 2 X ( ) = X ( s ) | s = = ω 2 1 ω 2 + + 1 Pole-Zero Map Real Axis Imag Axis -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 Figure P6.26. (b) Pole-Zero Plot of X ( s ) (c) X ( s ) = 1 s 4 + 2 s 2 X ( s ) = 3( s 10 3 ) ( s 4)( s 2) zero at: 10 3 poles at: 4 , 2 X ( ) = X ( s ) | s = = 1 4 + 2 2 2
Pole-Zero Map Real Axis Imag Axis 0 0.5 1 1.5 2 2.5 3 3.5 4 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 Figure P6.26. (b) Pole-Zero Plot of X ( s ) 6.27. Determine the bilateral Laplace transform and ROC for the following signals: (a) x ( t ) = e t u ( t + 2) X ( s ) = −∞ x ( t ) e st dt = −∞ e t u ( t + 2) e st dt = 2 e t (1+ s ) dt = e 2(1+ s ) 1 + s ROC: Re(s) > -1 (b) x ( t ) = u ( t + 3) X ( s ) = 3 −∞ e st dt = e 3 s s ROC: Re(s) < 0 3

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(c) x ( t ) = δ ( t + 1) X ( s ) = −∞ δ ( t + 1) e st dt = e s ROC: all s (d) x ( t ) = sin( t ) u ( t ) X ( s ) = 0 1 2 j ( e jt e jt ) e st dt = 0 1 2 j e t ( j s ) dt 0 1 2 j e t ( j + s ) dt = 1 2 j 1 j s 1 j + s = 1 (1 + s 2 ) ROC: Re(s) > 0 6.28. Determine the unilateral Laplace transform of the following signals using the defining equation: (a) x ( t ) = u ( t 2) X ( s ) = 0 x ( t ) e st dt = 0 u ( t 2) e st dt = 2 e st dt = e 2 s s (b) x ( t ) = u ( t + 2) X ( s ) = 0 u ( t + 2) e st dt = 0 e st dt = 1 s (c) x ( t ) = e 2 t u ( t + 1) 4
X ( s ) = 0 e 2 t u ( t + 1) e st dt = 0 e t ( s +2) dt = 1 s + 2 (d) x ( t ) = e 2 t u ( t + 2) X ( s ) = 0 e 2 t u ( t + 2) e st dt = 2 0 e t (2 s ) dt = e 2(2 s ) 1 2 s (e) x ( t ) = sin( ω o t ) X ( s ) = 0 1 2 j ( e o t e o t ) e st dt = 1 2 j 0 e t ( o s ) dt 0 e t ( o + s ) dt = 1 2 j 1 o s 1 o + s = ω o s 2 + ω 2 o (f) x ( t ) = u ( t ) u ( t 2) X ( s ) = 2 0 e st dt = 1 e 2 s s (g) x ( t ) = sin( πt ) , 0 < t < 1 0 , otherwise X ( s ) = 1 0 1 2 j ( e jπt e jπt ) e st dt = π (1 + e s ) s 2 + π 2 5

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6.29. Use the basic Laplace transforms and the Laplace transform properties given in Tables D.1 and D.2 to determine the unilateral Laplace transform of the following signals: (a) x ( t ) = d dt { te t u ( t ) } a ( t ) = te t u ( t ) L u ←−−−→ A ( s ) = 1 ( s + 1) 2 x ( t ) = d dt a ( t ) L u ←−−−→ X ( s ) = s ( s + 1) 2 (b) x ( t ) = tu ( t ) cos(2 πt ) u ( t ) a ( t ) = tu ( t ) L u ←−−−→ A ( s ) = 1 s 2 b ( t ) = cos(2 πt ) u ( t ) L u ←−−−→ s s 2 + 4 π 2 x ( t ) = a ( t ) b ( t ) L u ←−−−→ X ( s ) = A ( s ) B ( s ) X ( s ) = 1 s 2 ( s 2 + 4 π 2 ) (c) x ( t ) = t 3 u ( t ) a ( t ) = tu ( t ) L u ←−−−→ A ( s ) = 1 s 2 b ( t ) = ta ( t ) L u ←−−−→ B ( s ) = d ds A ( s ) = 2 s 3 x ( t ) = tb ( t ) L u ←−−−→ X ( s ) = d ds B ( s ) = 6 s 4 (d) x ( t ) = u ( t 1) e 2 t u ( t 1) a ( t ) = u ( t ) L u ←−−−→ A ( s ) = 1 s b ( t ) = a ( t 1) L u ←−−−→ B (
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