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Chapter 15-Electrical Circuit Analysis-Problem Solutions, Exercises of Electrical Circuit Analysis

This is solution to problems related Electrical Circuit Analysis course. It was given by Prof. Gurnam Kanth at Punjab Engineering College. Its main points are: Laplace, Transform, Scaling, Property, Signals, Functions, Sinusoidal, Function, Signal

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Download Chapter 15-Electrical Circuit Analysis-Problem Solutions and more Exercises Electrical Circuit Analysis in PDF only on Docsity! Chapter 15, Problem 1. Find the Laplace transform of: (a) cosh at (b) sinh at [Hint: cosh x = ( )xx ee −+ 2 1 , sinh x = ( )xx ee −− 2 1 .] Chapter 15, Solution 1. (a) 2 ee )atcosh( at-at + = [ ] =⎥⎦ ⎤ ⎢⎣ ⎡ + + − = as 1 as 1 2 1 )atcosh(L 22 as s − (b) 2 ee )atsinh( at-at − = [ ] =⎥⎦ ⎤ ⎢⎣ ⎡ + − − = as 1 as 1 2 1 )atsinh(L 22 as a − Chapter 15, Problem 2. Determine the Laplace transform of: (a) cos( θω +t ) (b) sin( θω +t ) Chapter 15, Solution 2. (a) )sin()tsin()cos()tcos()t(f θω−θω= [ ] [ ])tsin()sin()tcos()cos()s(F ωθ−ωθ= LL =)s(F 22s )sin()cos(s ω+ θω−θ (b) )sin()tcos()cos()tsin()t(f θω+θω= [ ] [ ])tsin()cos()tcos()sin()s(F ωθ+ωθ= LL =)s(F 22s )cos()sin(s ω+ θω−θ docsity.com ( )ttue t 4sin2− (c) ( )ttue t 2cosh3− (d) ( )ttue t sinh4− (e) ( )ttute t 2sin− Chapter 15, Solution 3. (a) [ ] =)t(u)t3cos(e-2tL 9)2s( 2s 2 ++ + (b) [ ] =)t(u)t4sin(e-2tL 16)2s( 4 2 ++ (c) Since [ ] 22 as s )atcosh( − =L [ ] =)t(u)t2cosh(e-3tL 4)3s( 3s 2 −+ + (d) Since [ ] 22 as a )atsinh( − =L [ ] =)t(u)tsinh(e-4tL 1)4s( 1 2 −+ (e) [ ] 4)1s( 2 )t2sin(e 2 t- ++ =L If )s(F)t(f ⎯→← )s(F ds d- )t(ft ⎯→← Thus, [ ] ( )[ ]1-2t- 4)1s(2 ds d- )t2sin(et ++=L )1s(2 )4)1s(( 2 22 +⋅++ = [ ] =)t2sin(et -tL 22 )4)1s(( )1s(4 ++ + docsity.com (b) [ ] = + ⋅= 5 2t-4 )2s( !43et3L 5)2s( 72 + (c) =−⋅−=⎥⎦ ⎤ ⎢⎣ ⎡ δ− )01s(4 s 2 )t( dt d 4)t(ut2 2L s4s 2 2 − (d) )t(ue2)t(ue2 -t1)--(t = [ ] =)t(ue2 1)--(tL 1s e2 + (e) Using the scaling property, [ ] =⋅⋅=⋅⋅= s2 1 25 )21(s 1 21 1 5)2t(u5L s 5 (f) [ ] = + = 31s 6 )t(ue6 3t-L 1s3 18 + (g) Let )t()t(f δ= . Then, 1)s(F = . L−′−−=⎥⎦ ⎤ ⎢⎣ ⎡ =⎥⎦ ⎤ ⎢⎣ ⎡ δ −− )0(fs)0(fs)s(Fs)t(f dt d )t( dt d 2n1nn n n n n LL L−⋅−⋅−⋅=⎥⎦ ⎤ ⎢⎣ ⎡ =⎥⎦ ⎤ ⎢⎣ ⎡ δ −− 0s0s1s)t(f dt d )t( dt d 2n1nn n n n n LL =⎥⎦ ⎤ ⎢⎣ ⎡ δ )t( dt d n n L ns docsity.com Chapter 15, Problem 6. Find F(s) given that ( ) ⎪ ⎩ ⎪ ⎨ ⎧ << << = otherwise tt tt tf ,0 21, 10,2 Chapter 15, Solution 6. 1 2 0 0 1 ( ) ( ) 2 2st st stF s f t e dt te dt e dt ∞ − − −= = +∫ ∫ ∫ 22 2 1 2 22 ( 1) 2 (1 ) 0 1 st st s se est e se s s s − − − −− − + = − − − Chapter 15, Problem 7. Find the Laplace transform of the following signals: (a) ( ) ( ) ( )tuttf 42 += (b) ( ) ( ) ( )tuetg t234 −+= (c) ( ) ( ) ( )( ) ( )tuttth 3cos83sin6 += (d) ( ) ( )( ) ( )tutetx t 4cosh2−= Chapter 15, Solution 7. (a) 2 2 4( )F s s s = + (b) 4 3( ) 2 G s s s = + + (c ) 9s 18s8 9s s8 9s 36)s(H 222 + + = + + + = (d) From Problem 15.1, 2 2{cosh } sL at s a = − 2 2 2 2 2( ) ( 2) 4 4 12 s sX s s s s + + = = + − + − 12s4s 2s)d(, 9s 18s8)c(, 2s 3 s 4)b(, s 4 s 2)a( 222 −+ + + + + ++ docsity.com Chapter 15, Problem 8. Find the Laplace transform F(s), given that f(t) is: (a) ( )42 −ttu (b) ( ) ( )2cos5 −tt δ (c) ( )ttue t −− (d) ( ) ( )τ−tut2sin Chapter 15, Solution 8. (a) 2t=2(t-4) + 8 f(t) = 2tu(t-4) = 2(t-4)u(t-4) + 8u(t-4) 4 4 42 2 2 8 2 8( ) s s sF s e e e s s s s − − −⎛ ⎞= + = +⎜ ⎟ ⎝ ⎠ (b) 2 0 0 ( ) ( ) 5cos ( 2) 5cos 5cos 2 2 st st st sF s f t e dt t t e dt te e t δ ∞ ∞ − − − −= = − = = =∫ ∫ (c) ( )t te e eτ τ− − − −= ( )( ) ( )tf t e e u tτ τ τ− − −= − ( 1)1( ) 1 1 s s eF s e e s s τ τ τ − + − −= = + + (d) sin 2 sin[2( ) 2 ] sin 2( )cos 2 cos 2( )sin 2t t t tτ τ τ τ τ τ= − + = − + − ( ) cos 2 sin 2( ) ( ) sin 2 cos 2( ) ( )f t t u t t u tτ τ τ τ τ τ= − − + − − 2 2 2( ) cos 2 sin 2 4 4 s s sF s e e s s τ ττ τ− −= + + + 5cos(2)e–2s docsity.com
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