Основы радиоэлектроники. Невежин Ю.В - 12 стр.

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38
Выполнить задания, сформулированные в примере, для следующих типов
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1.
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11.
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                                                                     38
           В ы п ол ни ть зад ани я, сформ у л и рованны е в п ри м ере, д л я сл ед у ющ и х ти п ов
           си гнал ов:



           1. s(t ) = s0 exp  −(t t 0) ,T = 0.08, τ j = 5t 0 j , t 01 = 6.5 ⋅ 10 −3 , t 0 2 = 7 ⋅ 10 −3 ,
                                       4
                                       
           t 03 = 8.5 ⋅ 10 −3 , t 0 4 = 9 ⋅ 10 −3 , t 05 = 9.6 ⋅ 10 −3 , t 0 6 = 10 −3 , t 0 7 = 2.2 ⋅ 10 −2 .
           2. s(t ) = s0 exp[ −| t | t 0] , T = 0.22, τ j = 6t 0 j , t 01 = 6 ⋅ 10 −3 , t 0 2 = 8.7 ⋅ 10 −3 ,
           t 03 = 0.011, t 0 4 = 0.014, t 05 = 0.017, t 06 = 0.019, t 07 = 0.022.
                            s0
           3. s(t ) =                  , T = 0.044, τ j = 6t 0 j , t 01 = 6 ⋅ 10 −4 , t 0 2 = 10 −3 ,
                      1 + ch(t / t 0 )
           t 03 = 1.4 ⋅ 10 −3 , t 0 4 = 18
                                         . ⋅ 10 −3 , t 05 = 2.2 ⋅ 10 −3 , t 0 6 = 2.6 ⋅ 10 −3 , t 0 7 = 3 ⋅ 10 −2 .
                              s0
           4. s(t ) =                     , T = 0.24, τ j = 10t 0 j , t 01 = 4 ⋅ 10 −3 , t 0 2 = 6.3 ⋅ 10 −3 ,
                         1 + (t / t 0 )2
           t 03 = 8.7 ⋅ 10 −3 , t 0 4 = 0.011, t 05 = 0.013, t 0 6 = 0.016, t 0 7 = 0.018.
                                     2 πt  
           5. s(t ) = s0 1 + cos            , T = 0.044, τ j = 0.04t 0 j , t 01 = 0.1, t 0 2 = 0.167,
                                     T t 0 
           t 03 = 0.233, t 04 = 0.3, t 05 = 0.367, t 06 = 0.433, t 07 = 0.5.
                               s0
           6. s(t ) =                                                          . ⋅ 10 −3 , t 0 2 = 2.58 ⋅ 10 −3 ,
                                           ,T = 0.052, τ j = 10t 0 j , t 01 = 13
                        1 + exp(t / t 0 )2
           t 03 = 3.87 ⋅ 10 −3 , t 0 4 = 515
                                          . ⋅ 10 −3 , t 05 = 6.43 ⋅ 10 −3 , t 06 = 7.72 ⋅ 10 −3 , t 0 7 = 9 ⋅ 10 −3 .
                            s0
           7. s(t ) =                   ,T = 0.038, τ j = 15t 0 j , t 01 = 13  . ⋅ 10 −3 , t 0 2 = 2.08 ⋅ 10 −3 ,
                      1 + (t / t 0 )2
           t 03 = 2.87 ⋅10 −3 , t 0 4 = 3.65 ⋅10 −3 , t 05 = 4.43 ⋅ 10 −3 , t 06 = 5.22 ⋅10 −3 , t 07 = 6 ⋅ 10 −3 .
                                               2
                                 2 πt  
           8. s(t ) = s0 1 + cos        , T = 0.04, τ j = 0.04t 0 j , t 01 = 0.15, t 0 2 = 0.225,
                                 T t 0  
           t 03 = 0.3, t 0 4 = 0.375, t 05 = 0.45, t 06 = 0.525, t 07 = 0.6.
                                   t 
           9. s(t ) = s0 1 − tanh    ,T = 0.048, τ j = 6t 0 j , t 01 = 3 ⋅ 10 −3 , t 0 2 = 4 ⋅ 10 −3 ,
                                   t 0 
           t 03 = 5 ⋅ 10 −3 , t 0 4 = 6 ⋅ 10 −3 , t 05 = 7 ⋅ 10 −3 , t 0 6 = 8 ⋅ 10 −3 , t 07 = 9 ⋅ 10 −3 .
                                   s0
           10. s(t ) =                         , T = 0.04, τ j = 6t 0 j , t 01 = 10 −3 , t 0 2 = 15   . ⋅ 10 −3 ,
                        1 + sinh(t / t 0 )
           t 03 = 2 ⋅ 10 −3 , t 0 4 = 2.5 ⋅ 10 −3 , t 05 = 3 ⋅ 10 −3 , t 06 = 35
                                                                               . ⋅ 10 −3 , t 07 = 4 ⋅ 10 −3 .
                                   s0
           11. s(t ) =                          ,T = 0.038, τ j = 6t 0 j , t 01 = 8 ⋅ 10 −4 , t 0 2 = 117 . ⋅ 10 −3 ,
                        [1 + ch(t / t 0 )]   2




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