# Discussion #12 (2020-04-10).pdf - EECS 216 –...

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EECS 216 – Discussion #12 (2020-04-10) – [email protected] Inverse Laplace Transform o −1 { 𝐻𝐻 ( 𝑠𝑠 )} ≡ ℎ ( 𝑡𝑡 ) = 1 𝑗𝑗2𝜋𝜋 𝐻𝐻 ( 𝑠𝑠 ) 𝑒𝑒 𝑠𝑠𝑠𝑠 𝑑𝑑𝑠𝑠 𝑐𝑐+𝑗𝑗∞ 𝑐𝑐−𝑗𝑗∞ for 𝑡𝑡 ≥ 0 and real constant 𝑐𝑐 ∈ ROC o This is a “contour integral” in the complex plane; you’d need to use techniques from the calculus of complex variables, which has similarities to multivariable calculus (after all, 𝐻𝐻 ( 𝑠𝑠 ) is in a way a function of two variables: Re{ 𝑠𝑠 } and Im{ 𝑠𝑠 } ) o In practice, the approach we mostly use to find Inverse Laplace Transforms is to pattern-match with LT pairs in the table and utilize the LT properties o Caution: Due to the fact that the Unilateral Laplace Transform disregards the values of the time domain signal for 𝑡𝑡 < 0 , the associated Inverse Laplace Transform, e.g. −1 { 𝐻𝐻 ( 𝑠𝑠 )} , will only be able to “recover” the values of the time domain signal, e.g. ( 𝑡𝑡 ) , for 𝑡𝑡 ≥ 0 Consequently, the equivalence of convolving in time and multiplying in s-domain for the Unilateral LT works only when dealing with right- sided input/output signals and causal (right-sided) impulse responses For other situations, e.g. pure sinusoids/complex exponentials, first ensure that the 𝑗𝑗𝑗𝑗 axis is in the ROC, and if so, then evaluate 𝐻𝐻 ( 𝑠𝑠 ) , the transfer function, along the imaginary axis, e.g. 𝑒𝑒 𝑗𝑗𝜔𝜔 0 𝑠𝑠 → 𝐻𝐻 ( 𝑠𝑠 ) → 𝐻𝐻 ( 𝑗𝑗𝑗𝑗 0 ) 𝑒𝑒 𝑗𝑗𝜔𝜔 0 𝑠𝑠 (important note: 𝑒𝑒 𝑗𝑗𝜔𝜔 0 𝑠𝑠 ≠ 𝑒𝑒 𝑗𝑗𝜔𝜔 0 𝑠𝑠 𝑢𝑢 ( 𝑡𝑡 ) , cos( 𝑗𝑗 0 𝑡𝑡 ) cos( 𝑗𝑗 0 𝑡𝑡 ) 𝑢𝑢 ( 𝑡𝑡 ) , etc.) Partial Fractions o For many of the problems we will encounter, we will need to use the technique of Partial Fraction Decomposition in order to rewrite the 𝐻𝐻 ( 𝑠𝑠 ) (or 𝑋𝑋 ( 𝑠𝑠 ) or 𝑌𝑌 ( 𝑠𝑠 ) ) in a form so as to be able to utilize the LT pairs/properties tables o Here are a few of the scenarios The general form of a rational function is 𝑋𝑋 ( 𝑠𝑠 ) = 𝑁𝑁 ( 𝑠𝑠 ) 𝐷𝐷 ( 𝑠𝑠 ) = ( 𝑠𝑠−𝑏𝑏 1 ) ⋅⋅⋅ ( 𝑠𝑠−𝑏𝑏 𝑚𝑚 ) ( 𝑠𝑠−𝑎𝑎 1 ) ⋅⋅⋅ ( 𝑠𝑠−𝑎𝑎 𝑛𝑛 ) Note: 𝑎𝑎 1 , … , 𝑎𝑎 𝑛𝑛 and 𝑏𝑏 1 , … , 𝑏𝑏 𝑚𝑚 could be complex-valued If 𝑚𝑚 < 𝑛𝑛 Here 𝑋𝑋 ( 𝑠𝑠 ) is called a “proper rational function” , i.e. the degree of the numerator is strictly less than the degree of the denominator If 𝒂𝒂 𝟏𝟏 , … , 𝒂𝒂 𝒏𝒏 are all distinct… o 𝑋𝑋 ( 𝑠𝑠 ) = 𝐴𝐴 1 𝑠𝑠−𝑎𝑎 1 + + 𝐴𝐴 𝑛𝑛 𝑠𝑠−𝑎𝑎 𝑛𝑛 o Here, constants 𝐴𝐴 𝑖𝑖 = lim 𝑠𝑠→𝑎𝑎 𝑖𝑖 ( 𝑠𝑠 − 𝑎𝑎 𝑖𝑖 ) 𝑋𝑋 ( 𝑠𝑠 ) for 𝑖𝑖 = 1, … , 𝑛𝑛 If you have repeated roots o For example, if 𝑋𝑋 ( 𝑠𝑠 ) = ( 𝑠𝑠−𝑏𝑏 1 ) ⋅⋅⋅ ( 𝑠𝑠−𝑏𝑏 𝑚𝑚 ) ( 𝑠𝑠−𝑎𝑎 1 ) 𝑘𝑘 ( 𝑠𝑠−𝑎𝑎 𝑘𝑘+1 ) ⋅⋅⋅ ( 𝑠𝑠−𝑎𝑎 𝑛𝑛 ) , o Then 𝑋𝑋 ( 𝑠𝑠 ) = 𝐴𝐴 11 𝑠𝑠−𝑎𝑎 1 + + 𝐴𝐴 1𝑘𝑘 ( 𝑠𝑠−𝑎𝑎 1 ) 𝑘𝑘 + 𝐴𝐴 𝑘𝑘+1 𝑠𝑠−𝑎𝑎 𝑘𝑘+1 + + 𝐴𝐴 𝑛𝑛 𝑠𝑠−𝑎𝑎 𝑛𝑛 o Here, 𝐴𝐴 1ℓ = lim 𝑠𝑠→𝑎𝑎 1 𝑑𝑑 ( 𝑘𝑘−ℓ ) ( 𝑑𝑑𝑑𝑑 ) ( 𝑘𝑘−ℓ ) ( 𝑠𝑠−𝑎𝑎 1 ) 𝑘𝑘 𝑋𝑋 ( 𝑠𝑠 ) ( 𝑘𝑘−ℓ )! for = 1, … , 𝑘𝑘 and 𝐴𝐴 𝑖𝑖 = lim 𝑠𝑠→𝑎𝑎 𝑖𝑖 ( 𝑠𝑠 − 𝑎𝑎 𝑖𝑖 ) 𝑋𝑋 ( 𝑠𝑠 ) for 𝑖𝑖 = 𝑘𝑘 + 1, … , 𝑛𝑛 If 𝑚𝑚 ≥ 𝑛𝑛
EECS 216 – Discussion #12 (2020-04-10) – [email protected] First perform a polynomial division ) ( ) ( s N s D in which case you get a quotient 𝑄𝑄 ( 𝑠𝑠 ) and a remainder 𝑅𝑅 ( 𝑠𝑠 ) 𝑁𝑁 ( 𝑠𝑠 ) 𝐷𝐷 ( 𝑠𝑠 ) = 𝑄𝑄 ( 𝑠𝑠 ) + 𝑅𝑅 ( 𝑠𝑠 ) 𝐷𝐷 ( 𝑠𝑠 ) , where the degree of 𝑅𝑅 ( 𝑠𝑠 ) will be less than that of 𝐷𝐷 ( 𝑠𝑠 ) Then you can use partial fractions for 𝑅𝑅 ( 𝑠𝑠 ) 𝐷𝐷 ( 𝑠𝑠 ) In this course, we will probably only encounter at most the case of 𝑚𝑚 = 𝑛𝑛 , in which case quotient 𝑄𝑄 ( 𝑠𝑠 ) would simply be a constant o This technique, and why it works, will be illustrated through the following examples o Example #4 a) Find 𝑥𝑥 ( 𝑡𝑡 ) when 𝑋𝑋 ( 𝑠𝑠 ) = 1 𝑠𝑠 2 +3𝑠𝑠+2 𝑋𝑋 ( 𝑠𝑠 ) = 1 ( 𝑠𝑠+1 )( 𝑠𝑠+2 ) = 𝐴𝐴 𝑠𝑠+1 +