HW_07 - ‫ﺻﻔﺤﻪ 1‬ ‫ﺗﺒﺪﻳﻞ...

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Unformatted text preview: ‫ﺻﻔﺤﻪ 1‬ ‫ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ‬ ‫‪π‬‬ ‫4‬ ‫ﻣﺴﺎﺋﻞ ﻓﺼﻞ ﺷﺸﻢ‬ ‫1‐ ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ ﺗﻮﺍﺑﻊ ﺯﻳﺮ ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﻧﻤﺎﺋﻴﺪ .‬ ‫+ ‪sin(ω t‬‬ ‫)‬ ‫,‬ ‫− ‪e −t cos(ω t‬‬ ‫‪π‬‬ ‫4‬ ‫)‬ ‫,‬ ‫− ‪cos(t‬‬ ‫‪π‬‬ ‫3‬ ‫− ‪)u (t‬‬ ‫‪π‬‬ ‫3‬ ‫)‬ ‫،‬ ‫− ‪cos(t‬‬ ‫‪π‬‬ ‫4‬ ‫− ‪)u (t‬‬ ‫‪π‬‬ ‫2‬ ‫)‬ ‫، 3 ‪δ (t ) + 2δ ′′(t ) + 2t + 6t‬‬ ‫‪e − 3t + 3te − t + 4t 3 e − 3t‬‬ ‫،‬ ‫+ ‪e −2t + 3 . cos(4t‬‬ ‫‪π‬‬ ‫4‬ ‫52 + )‬ ‫2 – ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ ﺷﮑﻞ ﻣﻮﺝ ﻫﺎﯼ ﺯﻳﺮ ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﻧﻤﺎﺋﻴﺪ .‬ ‫3 – ﺗﺒﺪﻳﻞ ﻣﻌﮑﻮﺱ ﻻﭘﻼﺱ ﺗﻮﺍﺑﻊ ﺯﻳﺮ ﺭﺍ ﻣﺸﺨﺺ ﻭ ﺭﺳﻢ ﻧﻤﺎﺋﻴﺪ .‬ ‫1 + ‪s + 3s‬‬ ‫3 )2 + ‪s( s + 1)( s‬‬ ‫2‬ ‫،‬ ‫1‬ ‫، ) ‪(1 − e −2s + 2e − 4s‬‬ ‫‪s‬‬ ‫‪( s + 4 )e‬‬ ‫) 8 + ‪( s + 1)( s 2 + 4s‬‬ ‫‪− 5s‬‬ ‫،‬ ‫‪1 1 −s‬‬ ‫‪−e‬‬ ‫‪ss‬‬ ‫‪1 + e −2 s‬‬ ‫,‬ ‫‪1 − e−s‬‬ ‫‪(1 + e − s ) s‬‬ ‫4 – ﺑﻪ ﮐﻤﮏ ﻗﻀﻴﻪ ﮐﺎﻧﻮﻟﻮﺷﻦ، ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ ﻣﻌﮑﻮﺱ ﺗﻮﺍﺑﻊ ﺯﻳﺮ ﺭﺍ ﻣﺸﺨﺺ ﻧﻤﺎﺋﻴﺪ ﻭ ﺑﺎﺭ ﺩﻳﮕﺮ ﺑﻪ ﮐﻤﮏ ﮔﺴﺘﺮﺵ ﺑﻪ‬ ‫ﮐﺴﺮﻫﺎﯼ ﺟﺰﺋﯽ ﻣﺴﺌﻠﻪ ﺭﺍ ﺣﻞ ﮐﻨﻴﺪ .‬ ‫4 + ‪s 2 + 4s‬‬ ‫)31 + ‪( s + 1)( s 2 + 6 s‬‬ ‫،‬ ‫) ‪s 4 (e − s − e −2 s‬‬ ‫)4 + 2 ‪( s 2 + 1)( s‬‬ ‫،‬ ‫3 + ‪2s‬‬ ‫2 )1 + ‪( s + 2) 2 ( s‬‬ ‫،‬ ‫1+ ‪s‬‬ ‫) ‪s ( s + 1)(1 − e − s‬‬ ‫2‬ ‫2‬ ‫5 – ﺍﻧﺘﮕﺮﺍﻝ ﻫﺎﯼ ﻣﻘﺎﺑﻞ ﺭﺍ ﺑﻪ ﮐﻤﮏ ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ ﻣﺤﺎﺳﺒﻪ ﻧﻤﺎﺋﻴﺪ .‬ ‫∫‬ ‫∞‬ ‫‪o‬‬ ‫‪e −2t (sin t + 2 cos 2t + 3 )dt‬‬ ‫،‬ ‫∫‬ ‫∞‬ ‫‪o‬‬ ‫‪(2t 3 + 3t 2 + t − 4)e − 3 t dt‬‬ ‫6 – ﻣﻌﺎﺩﻻﺕ ﻣﻘﺎﺑﻞ ﺭﺍ ﺑﻪ ﮐﻤﮏ ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ ﺣﻞ ﮐﻨﻴﺪ .‬ ‫) ‪x ′′(t ) + 5 x ′(t ) + 4 x(t ) = (e + e )u (t‬‬ ‫‪−t‬‬ ‫‪−4 t‬‬ ‫‪e t f (t ) + 2 ∫ f (τ )eτ dτ + ∫ f (τ )e 2τ e −2 t dτ = 2t‬‬ ‫‪o‬‬ ‫‪o‬‬ ‫‪t‬‬ ‫‪t‬‬ ‫7 – ﺑﻪ ﺩﻭ ﺭﻭﺵ ﻣﺨﺘﻠﻒ، ﻣﻘﺎﺩﻳﺮ ) + ‪ f (∞), f (o‬ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﻧﻤﺎﺋﻴﺪ .‬ ‫= )‪F ( s‬‬ ‫‪2s‬‬ ‫)2 − ‪( s + 1)( s‬‬ ‫،‬ ‫= )‪F (s‬‬ ‫2 + ‪s 2 + 3s‬‬ ‫1 + ‪s 3 + 3s2 + 3s‬‬ ‫،‬ ‫= )‪F (s‬‬ ‫)1 + ‪s( s‬‬ ‫) 3 + ‪( s + 2)( s‬‬ ‫ﺻﻔﺤﻪ 2‬ ‫1‬ ‫ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ‬ ‫ﻣﺴﺎﺋﻞ ﻓﺼﻞ ﺷﺸﻢ‬ ‫‪is‬‬ ‫‪i‬‬ ‫1‬ ‫‪vC‬‬ ‫1‬ ‫1‬ ‫4‬ ‫‪iL‬‬ ‫1‬ ‫3‬ ‫8 –ﻣﻌﺎﺩﻻﺕ ﻣﺪﺍﺭ ﻣﻘﺎﺑﻞ ﺭﺍ ﺑﻪ ﺭﻭﺵ ﮔﺮﻩ ﺩﺭ ﺣﻮﺯﻩ ﺯﻣﺎﻥ ﺑﻨﻮﻳﺴﻴﺪ‬ ‫ﻭ ﺑﻪ ﮐﻤﮏ ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ ) ‪ vc (t‬ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﻧﻤﺎﻳﻴﺪ .‬ ‫ﺣﺎﻟﺖ ﺍﻭﻟﻴﻪ ‪ iL (o − ) = a‬ﻭ ‪ vc (o − ) = b‬ﺍﺳﺖ .‬ ‫ﻭﺭﻭﺩﯼ ﺭﺍ ) ‪ is = 4e −4 t u (t‬ﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳﺪ .‬ ‫9 – ﻣﻌﺎﺩﻻﺕ ﻣﺶ ﻣﺪﺍﺭ ﺯﻳﺮ ﺭﺍ ﺩﺭ ﺣﻮﺯﻩ ﺯﻣﺎﻥ ﺑﻨﻮﻳﺴﻴﺪ‬ ‫ﻭ ﺑﮑﻤﮏ ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ ﭘﺎﺳﺦ ﻭﺭﻭﺩﯼ ﺻﻔﺮ ﻭ ﻫﻤﭽﻨﻴﻦ‬ ‫ﭘﺎﺳﺦ ﮐﺎﻣﻞ ) ‪ v0 (t‬ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﻧﻤﺎﺋﻴﺪ .‬ ‫ﺣﺎﻟﺖ ﺍﻭﻟﻴﻪ 1 = ) − ‪ i1 (o‬ﻭ ‪ i2 (o − ) = o‬ﻭ 2 = ) − ‪ vc (o‬ﻭ‬ ‫ﻭﺭﻭﺩﯼ ) ‪ v s = 2δ (t ) + 3u (t‬ﺍﺳﺖ .‬ ‫01 – ﺍﻟﻒ– ﻣﻌﺎﺩﻻﺕ ﻣﺶ ﻣﺪﺍﺭ ﺯﻳﺮ ﺭﺍ ﺩﺭ ﺣﻮﺯﻩ ﺯﻣﺎﻥ‬ ‫ﺑﻨﻮﻳﺴﻴﺪﻭﺍﺯ ﺁﻥ ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ ﺑﮕﻴﺮﻳﺪ .‬ ‫‪ i L1 ( o − ) = i L 2 ( o − ) = I o‬و ‪v c ( o − ) = o‬‬ ‫ﺏ– ﻭﺭﻭﺩﯼ ﭼﮕﻮﻧﻪ ﺍﻧﺘﺨﺎﺏ ﺷﻮﺩ ﺗﺎ ﭘﺎﺳﺦ ﺣﺎﻟﺖ‬ ‫ﺻﻔﺮﻭ ﺣﺎﻟﺖ ﻭﺭﻭﺩﯼ ﺻﻔﺮ ﻳﮑﺴﺎﻥ ﺑﺎﺷﺪ .‬ ‫2‬ ‫1‪i‬‬ ‫2‪i‬‬ ‫1‬ ‫2‬ ‫‪vs‬‬ ‫1‬ ‫2‬ ‫1‬ ‫‪vC‬‬ ‫2‬ ‫‪vo‬‬ ‫2‬ ‫1‪iL‬‬ ‫1‬ ‫‪vs‬‬ ‫1‬ ‫2 ‪iL‬‬ ‫‪vC‬‬ ‫1‬ ‫‪vo‬‬ ‫11‐ ﭘﺎﺳﺦ ﺿﺮﺑﻪ ﻳﮏ ﻣﺪﺍﺭ ﺧﻄﯽ ﻭ ﺗﻐﻴﻴﺮﻧﺎﭘﺬﻳﺮ ﺑﺎ ﺯﻣﺎﻥ ﺑﺮﺍﺑﺮ ) ‪ h(t ) = (e −4 t − e −2t + 2te −2t‬ﻣﯽ ﺑﺎﺷﺪ، ﭘﺎﺳﺦ ﺣﺎﻟﺖ ﺻﻔﺮ ﻣﺪﺍﺭ‬ ‫ﺭﺍ ﺑﻪ ﻭﺭﻭﺩﯼ ) ‪ is (t ) = 3δ (t ) + 2e −2t u (t‬ﺑﻪ ﺩﻭ ﺭﻭﺵ ﻣﺨﺘﻠﻒ، ﻳﮑﺒﺎﺭ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺍﻧﺘﮕﺮﺍﻝ ﮐﺎﻧﻮﻟﻮﺷﻦ )ﺭﻭﺵ ﺣﻮﺯﻩ ﺯﻣﺎﻥ(‬ ‫ﻭ ﺑﺎﺭ ﺩﻳﮕﺮ ﺑﻪ ﻳﺎﺭﯼ ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ )ﺭﻭﺵ ﺣﻮﺯﻩ ﻓﺮﮐﺎﻧﺲ( ﺑﻴﺎﺑﻴﺪ .‬ ‫21‐ﺍﻟﻒ‐ ﻣﻌﺎﺩﻻﺕ ﮔﺮﻩ ﺭﺍ ﺩﺭ ﺣﻮﺯﻩ ﺯﻣﺎﻥ ﺑﻪ ﻓﺮﻡ ﻣﺎﺗﺮﻳﺴﯽ ‪Yn ( D)e = i‬‬ ‫1 1‪iL‬‬ ‫2 ‪iL‬‬ ‫2‬ ‫1‬ ‫‪i‬‬ ‫3 ‪iL‬‬ ‫1‬ ‫2‬ ‫ﺑﺮﺍﯼ ﻣﺪﺍﺭ ﻣﻘﺎﺑﻞ ﺑﻨﻮﻳﺴﻴﺪ ﻭ ﺍﺯ ﺁﻥ ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ ﺑﮕﻴﺮﻳﺪ ﻭ ﺑﺎ‬ ‫ﻣﺤﺎﺳﺒﻪ ﻭﻟﺘﺎﮊ ﮔﺮﻩ ، ﺟﺮﻳﺎﻥ ) ‪ i (t‬ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﻧﻤﺎﻳﻴﺪ .‬ ‫ﺏ ‐ ﺟﺮﻳﺎﻥ ) ‪ i (t‬ﺭﺍ ﺑﺎﺭ ﺩﻳﮕﺮ ﺑﻪ ﮐﻤﮏ ﺭﻭﺵ ﺗﺤﻠﻴﻞ ﻣﺶ ﻭ ﺗﺒﺪﻳﻞ‬ ‫ﻻﭘﻼﺱ ﻣﺤﺎﺳﺒﻪ ﻧﻤﺎﻳﻴﺪ . ﺣﺎﻟﺖ ﺍﻭﻟﻴﻪ ﻣﺪﺍﺭ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮﻣﯽ ﺑﺎﺷﺪ:‬ ‫‪⎧i L1 (o) = 1A‬‬ ‫⎪‬ ‫‪⎨i L 2 (o) = 2 A‬‬ ‫‪⎪i (o) = 3 A‬‬ ‫3‪⎩ L‬‬ ‫ﺻﻔﺤﻪ 3‬ ‫1‬ ‫ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ‬ ‫‪io‬‬ ‫ﻣﺴﺎﺋﻞ ﻓﺼﻞ ﺷﺸﻢ‬ ‫‪is‬‬ ‫31‐ ﺍﻟﻒ‐ ﻣﻌﺎﺩﻻﺕ ﮔﺮﻩ ﺭﺍ ﺩﺭ ﺣﻮﺯﻩ ﺯﻣﺎﻥ ﺑﺮﺍﯼ ﻣﺪﺍﺭ ﻣﻘﺎﺑﻞ‬ ‫ﺑﻨﻮﻳﺴﻴﺪ ﻭ ﺳﭙﺲ ﺍﺯ ﺁﻥ ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ ﺑﮕﻴﺮﻳﺪ ﻭ ﭘﺎﺳﺦ‬ ‫ﮐﺎﻣﻞ ) ‪ V2 ( s‬ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﻧﻤﺎﻳﻴﺪ ﻭ ﺍﺯ ﺁﻥ ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ‬ ‫1‬ ‫1‬ ‫1‪v‬‬ ‫1‬ ‫2‪v‬‬ ‫1‬ ‫‪2i‬‬ ‫ﭘﺎﺳﺦ ﺣﺎﻟﺖ ﺻﻔﺮ، ﺗﺎﺑﻊ ﺷﺒﮑﻪ ﻭ ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ ﭘﺎﺳﺦ‬ ‫ﻭﺭﻭﺩﯼ ﺻﻔﺮ ﺭﺍ ﻣﺸﺨﺺ ﻧﻤﺎﻳﻴﺪ. ﺣﺎﻟﺖ ﺍﻭﻟﻴﻪ‬ ‫ﺭﺍ ‪ v1 (o − ) = a‬ﻭ ‪ v 2 (o − ) = b‬ﻣﻨﻈﻮﺭ ﻧﻤﺎﻳﻴﺪ .‬ ‫ﺏ‐ ﭘﺎﺳﺦ ﮐﺎﻣﻞ ) ‪ v2 (t‬ﻭ ﭘﺎﺳﺦ ﮐﺎﻣﻞ ) ‪ io (t‬ﭼﻨﺎﻧﭽﻪ، 2 = ‪ a = b‬ﻭ ﻭﺭﻭﺩﯼ ) ‪ is (t ) = 2e −t u (t‬ﺑﺎﺷﺪ ﻣﺤﺎﺳﺒﻪ ﺷﻮﺩ .‬ ‫‪o‬‬ ‫1‬ ‫2‪Kv‬‬ ‫1 1‪v‬‬ ‫2‬ ‫1‬ ‫1‪is‬‬ ‫‪i‬‬ ‫1‪4v‬‬ ‫1‬ ‫2‬ ‫1‬ ‫2‪v‬‬ ‫‪3i‬‬ ‫2 ‪is‬‬ ‫41‐ﺍﻟﻒ‐ ﻣﻌﺎﺩﻻﺕ ﻣﺪﺍﺭ ﻣﻘﺎﺑﻞ ﺭﺍ ﺑﻪ ﻫﺮ ﺭﻭﺷﯽ ﮐﻪ ﻣﻨﺎﺳﺒﺘﺮ‬ ‫ﻣﯽ ﺩﺍﻧﻴﺪ ﺑﻨﻮﻳﺴﻴﺪ ﻭ ﺍﺯ ﺁﻥ ﺩﺭ ﻓﺮﻡ ﻣﺎﺗﺮﻳﺴﯽ‬ ‫ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ ﺑﮕﻴﺮﻳﺪ ﻭ ﭼﻨﺎﻧﭽﻪ ﺣﺎﻟﺖ ﺍﻭﻟﻴﻪ‬ ‫ﺻﻔﺮ ﺑﺎﺷﺪ ؛ ﭘﺎﺳﺦ )‪ V2 ( s‬ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﻧﻤﺎﻳﻴﺪ .‬ ‫ﺏ‐ ﻣﻘﺪﺍﺭ ‪ k‬ﭼﮕﻮﻧﻪ ﺍﺧﺘﻴﺎﺭ ﺷﻮﺩ ﺗﺎ ﻣﺪﺍﺭ‬ ‫ﺟﻮﺍﺏ ﻳﮑﺘﺎ ﺩﺍﺷﺘﻪ ﺑﺎﺷﺪ؟‬ ‫ﺑﺮﺍﯼ ﺣﺎﻟﺘﯽ ﮐﻪ 2 = ‪ k‬ﻭ ) ‪ is1 (t ) = 2e −t u (t‬ﻭ ) ‪ is 2 (t ) = sin(t )u (t‬ﺍﺳﺖ، ) ‪ v2 (t‬ﺭﺍ ﺑﻴﺎﺑﻴﺪ .‬ ‫‪1H‬‬ ‫‪io‬‬ ‫‪2Ω‬‬ ‫‪1H‬‬ ‫‪1H‬‬ ‫‪2Ω‬‬ ‫2 ‪is‬‬ ‫1‪is‬‬ ‫51‐ﺍﻟﻒ‐ﺑﻪ ﮐﻤﮏ ﺟﻤﻊ ﺁﺛﺎﺭ ﻭ ﺗﺠﺰﻳﻪ ﻭ ﺗﺤﻠﻴﻞ ﺣﺎﻟﺖ ﺩﺍﺋﻤﯽ‬ ‫ﺳﻴﻨﻮﺳﯽ، ﭘﺎﺳﺦ ﺣﺎﻟﺖ ﺩﺍﺋﻤﯽ ) ‪ io (t‬ﺩﺭ ﻣﺪﺍﺭ ﻣﻘﺎﺑﻞ ﺭﺍ‬ ‫ﻣﺤﺎﺳﺒﻪ ﻧﻤﺎﻳﻴﺪ .‬ ‫ﺏ‐ﭘﺎﺳﺦ ﺣﺎﻟﺖ ﺩﺍﺋﻤﯽ ) ‪ io (t‬ﺭﺍ ﺑﻪ ﮐﻤﮏ ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ‬ ‫ﺑﻴﺎﺑﻴﺪ .‬ ‫⎞ 1 ‪⎛v‬‬ ‫‪is‬‬ ‫61‐ﺍﻟﻒ‐ ﻣﻌﺎﺩﻻﺕ ﺣﺎﻟﺖ ﺭﺍ ﺑﺮﺍﯼ ﻣﺪﺍﺭ ﻣﻘﺎﺑﻞ ﺑﺮﺍﯼ ﺑﺮﺩﺍﺭ ⎟ ‪ x = ⎜ c‬ﺑﻔﺮﻡ‬ ‫⎟ ‪⎜i‬‬ ‫⎠‪⎝L‬‬ ‫ﻣﺎﺗﺮﻳﺴﯽ ﺑﻨﻮﻳﺴﻴﺪ ﻭ ﺍﺯ ﺁﻥ ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ ﺑﮕﻴﺮﻳﺪ ﺣﺎﻟﺖ ﺍﻭﻟﻴﻪ‬ ‫‪iL‬‬ ‫‪i‬‬ ‫1‬ ‫3‬ ‫1‪vC‬‬ ‫1 2 ‪vC‬‬ ‫‪ i L (o − ) = 1A‬ﻭ ‪ vC 1 (o − ) = 3V‬ﺍﺳﺖ .‬ ‫1‬ ‫4‬ ‫ﺏ‐ ﺍﺯﻣﻌﺎﺩﻻﺕ ﺑﺎﻻ ﺗﺎﺑﻊ ﺷﺒﮑﻪ ﺍﺭﺗﺒﺎﻁ ﺩﻫﻨﺪﻩ ﺧﺮﻭﺟﯽ ‪ i‬ﻭ‬ ‫ﻭﺭﻭﺩﯼ ‪ i s‬ﺭﺍ ﻣﺸﺨﺺ ﻧﻤﺎﻳﻴﺪ .‬ ‫ﭖ‐ ‪ e At‬ﺭﺍ ﺑﺮﺍﯼ ﻣﺪﺍﺭ ﻣﺸﺨﺺ ﻧﻤﺎﻳﻴﺪ ﻭ ﭘﺎﺳﺦ ﮐﺎﻣﻞ ﺭﺍ ﻭﻗﺘﻴﮑﻪ ) ‪ is (t ) = 2u (t‬ﻣﯽ ﺑﺎﺷﺪ ، ﺑﺮﺍﯼ ) ‪ i (t‬ﻣﺤﺎﺳﺒﻪ ﻧﻤﺎﻳﻴﺪ .‬ ‫71‐ ﻣﺴﺎﻟﻪ 31 ﺭﺍ ﺑﺎ ﻧﻮﺷﺘﻦ ﻣﻌﺎﺩﻻﺕ ﺣﺎﻟﺖ ﻭ ﻣﺤﺎﺳﺒﻪ ‪ e At‬ﺁﻥ ﺑﮑﻤﮏ ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ ﺣﻞ ﻭ ) ‪ v2 (t‬ﺭﺍ ﺑﻴﺎﺑﻴﺪ .‬ ‫5‬ ‫‪i‬‬ ‫4‬ ‫ﺻﻔﺤﻪ 4‬ ‫‪vk‬‬ ‫1‬ ‫2‬ ‫ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ‬ ‫ﻣﺴﺎﺋﻞ ﻓﺼﻞ ﺷﺸﻢ‬ ‫81‐ﺍﻟﻒ‐ ﺩﺭ ﻣﺪﺍﺭ ﺑﺎﻻ ‪ v k‬ﺭﺍ ﺑﺮﺍﯼ ﺗﻤﺎﻡ ﺯﻣﺎﻧﻬﺎ ﺑﻴﺎﺑﻴﺪ .‬ ‫ﺏ‐ ﺩﺭ ﻣﺪﺍﺭ ﭘﺎﻳﻴﻦ ‪ ik‬ﺭﺍ ﺑﺮﺍﯼ ﺗﻤﺎﻡ ﺯﻣﺎﻧﻬﺎ ﺑﻴﺎﺑﻴﺪ .‬ ‫ﮐﻠﻴﺪﻫﺎ ﺩﺭ ﻟﺤﻈﻪ ‪ t = o‬ﺗﻐﻴﻴﺮ ﻭﺿﻌﻴﺖ ﻣﯽ ﺩﻫﻨﺪ .‬ ‫1‬ ‫3‬ ‫3‬ ‫1‬ ‫4‬ ‫1‬ ‫3‬ ‫‪ik‬‬ ‫1‬ ‫2‬ ‫2‬ ‫3‬ ‫1‬ ‫3‬ ‫1‬ ‫2‬ ‫3‬ ‫1‬ ‫91‐ﺍﻟﻒ‐ﻣﻌﺎﺩﻻﺕ ﺍﻧﺘﮕﺮﺍﻝ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﮔﺮﻩ ﺭﺍ ﺑﺮﺍﯼ ﻣﺪﺍﺭ ﻣﻘﺎﺑﻞ ﺑﻪ‬ ‫ﻓﺮﻡ ﻣﺎﺗﺮﻳﺴﯽ ﺑﻨﻮﻳﺴﻴﺪ ﻭ ﺷﺮﺍﻳﻂ ﺍﻭﻟﻴﻪ ﻻﺯﻡ ﺭﺍ ﺑﺮ ﺣﺴﺐ‬ ‫1‪v‬‬ ‫ﻭﻟﺘﺎﮊ ﺍﻭﻟﻴﻪ ﺧﺎﺯﻥ ﻫﺎ ﻣﺸﺨﺺ ﻧﻤﺎﻳﻴﺪ .‬ ‫1‬ ‫1‬ ‫‪is‬‬ ‫1‬ ‫2‪v‬‬ ‫3‪v‬‬ ‫2‬ ‫ﺏ‐ ﺍﺯ ﻣﻌﺎﺩﻻﺕ ﻣﺎﺗﺮﻳﺴﯽ ﺑﺎﻻ ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ ﺑﮕﻴﺮﻳﺪ ﻭ ﺍﺯ‬ ‫2‬ ‫ﻣﻌﺎﺩﻻﺕ ﺟﺒﺮﻱ ﺣﺎﺻﻞ )‪ E 3 ( s‬ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ ﭘﺎﺳﺦ‬ ‫ﮐﺎﻣﻞ ﻭﻟﺘﺎﮊ ﮔﺮﻩ 3‪ e‬ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﻧﻤﺎﻳﻴﺪ ﻭ ﭘﺎﺳﺦ ﻫﺎﯼ‬ ‫ﺣﺎﻟﺖ ﺻﻔﺮ ﻭ ﻭﺭﻭﺩﯼ ﺻﻔﺮ ﺭﺍ ﺗﻔﮑﻴﮏ ﻧﻤﺎﻳﻴﺪ ﻭ ﺗﺎﺑ ﺷﺒﮑﻪ ﺍﺭﺗﺒﺎﻁ‬ ‫ﺩﻫﻨﺪﻩ ﭘﺎﺳﺦ ﺣﺎﻟﺖ ﺻﻔﺮ ﺑﻪ ﻭﺭﻭﺩﯼ ‪ i s‬ﺭﺍ ﻣﺸﺨﺺ‬ ‫ﻧﻤﺎﻳﻴﺪ ﻭﭼﻨﺎﻧﭽﻪ ‪ vc (o − ) = 1V , vc (o − ) = 2V , vc (o − ) = 3V‬ﻭ ﻭﺭﻭﺩﯼ ) ‪ i s (t ) = δ (t ) + 3u (t‬ﺑﺎﺷﺪ ﭘﺎﺳﺦ ﮐﺎﻣﻞ‬ ‫) ‪ e3 (t‬ﺭﺍ ﺑﺮﺍﯼ ‪ t ≥ o‬ﺑﻴﺎﺑﻴﺪ .‬ ‫1‪e‬‬ ‫2‪e‬‬ ‫1‬ ‫2‬ ‫1‬ ‫3‪e‬‬ ‫1‬ ‫2‬ ‫3‬ ‫02‐ ﺍﻟﻒ‐ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ ) ‪ F ( s‬ﺗﺎﺑﻊ ) ‪ f (t‬ﺷﮑﻞ ﻣﻘﺎﺑﻞ ﺭﺍ ﺑﻴﺎﺑﻴﺪ:‬ ‫= ) ‪ G ( s‬ﺑﺎﺷﺪ , ) ‪ g (t‬ﺭﺍ ﺑﺮﺍﯼ ‪ t > o‬ﺑﻴﺎﺑﻴﺪ .‬ ‫‪(1 − e −3 s )e −4 s‬‬ ‫ﺏ‐ﺍﮔﺮ‬ ‫) ‪( s + 3 )(1 + e − s‬‬ ...
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This note was uploaded on 02/06/2011 for the course ECE 423 taught by Professor Dolatabadi during the Spring '11 term at Amirkabir University of Technology.

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