Use the formula [tex]$£\left\{t^n f(t)\right\}(s)=(-1)^n \frac{d^n}{d s^n}( L \{f(s))$[/tex] to help determine the following the expressions.
(a) [tex]$L \{t \cos bt\}$[/tex]
(b) [tex]$E \left\{t^2 \cos b t\right\}$[/tex]
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(a) [tex]$L \{ t \cos bt \}( s )=$[/tex] $\square$
Answer (2)
Apply the formula L { t n f ( t )} ( s ) = ( − 1 ) n d s n d n ( L { f ( t )} ( s )) .
For L { t cos ( b t )} , calculate the first derivative of L { cos ( b t )} = s 2 + b 2 s and multiply by -1, resulting in ( s 2 + b 2 ) 2 s 2 − b 2 .
For L { t 2 cos ( b t )} , calculate the second derivative of L { cos ( b t )} = s 2 + b 2 s , resulting in ( s 2 + b 2 ) 3 2 s ( s 2 − 3 b 2 ) .
The Laplace transform of t cos ( b t ) is ( s 2 + b 2 ) 2 s 2 − b 2 and the Laplace transform of t 2 cos ( b t ) is ( s 2 + b 2 ) 3 2 s ( s 2 − 3 b 2 ) .
Explanation
Problem Setup We are given the formula L { t n f ( t )} ( s ) = ( − 1 ) n d s n d n ( L { f ( t )} ( s )) . We need to find the Laplace transform of t cos ( b t ) and t 2 cos ( b t ) . From the Laplace transform table (assumed to be known), L { cos ( b t )} = s 2 + b 2 s .
Applying the Formula for t*cos(bt) (a) For L { t cos ( b t )} , we have n = 1 and f ( t ) = cos ( b t ) . Thus, L { t cos ( b t )} = ( − 1 ) 1 d s d ( L { cos ( b t )}) . Since L { cos ( b t )} = s 2 + b 2 s , we have L { t cos ( b t )} = − d s d ( s 2 + b 2 s ) .
Calculating the Derivative Calculate the derivative: d s d ( s 2 + b 2 s ) = ( s 2 + b 2 ) 2 ( s 2 + b 2 ) ( 1 ) − s ( 2 s ) = ( s 2 + b 2 ) 2 s 2 + b 2 − 2 s 2 = ( s 2 + b 2 ) 2 b 2 − s 2 . Therefore, L { t cos ( b t )} = − ( s 2 + b 2 ) 2 b 2 − s 2 = ( s 2 + b 2 ) 2 s 2 − b 2 .
Applying the Formula for t^2*cos(bt) (b) For L { t 2 cos ( b t )} , we have n = 2 and f ( t ) = cos ( b t ) . Thus, L { t 2 cos ( b t )} = ( − 1 ) 2 d s 2 d 2 ( L { cos ( b t )}) . Since L { cos ( b t )} = s 2 + b 2 s , we have L { t 2 cos ( b t )} = d s 2 d 2 ( s 2 + b 2 s ) .
Calculating the Second Derivative We already found the first derivative: d s d ( s 2 + b 2 s ) = ( s 2 + b 2 ) 2 b 2 − s 2 . Now, calculate the second derivative: d s 2 d 2 ( s 2 + b 2 s ) = d s d ( ( s 2 + b 2 ) 2 b 2 − s 2 ) = ( s 2 + b 2 ) 4 ( − 2 s ) ( s 2 + b 2 ) 2 − ( b 2 − s 2 ) ( 2 ) ( s 2 + b 2 ) ( 2 s ) = ( s 2 + b 2 ) 3 − 2 s ( s 2 + b 2 ) − 4 s ( b 2 − s 2 ) = ( s 2 + b 2 ) 3 − 2 s 3 − 2 s b 2 − 4 s b 2 + 4 s 3 = ( s 2 + b 2 ) 3 2 s 3 − 6 s b 2 = ( s 2 + b 2 ) 3 2 s ( s 2 − 3 b 2 ) . Therefore, L { t 2 cos ( b t )} = ( s 2 + b 2 ) 3 2 s ( s 2 − 3 b 2 ) .
Final Answer (a) The Laplace transform of t cos ( b t ) is ( s 2 + b 2 ) 2 s 2 − b 2 .
(b) The Laplace transform of t 2 cos ( b t ) is ( s 2 + b 2 ) 3 2 s ( s 2 − 3 b 2 ) .
Examples
Laplace transforms are incredibly useful in electrical engineering for analyzing circuits. Imagine you have an RLC circuit (a circuit with a resistor, inductor, and capacitor). When you apply a voltage source that varies with time, like t cos ( b t ) , the Laplace transform helps you convert the differential equations that describe the circuit's behavior into algebraic equations. This makes it much easier to solve for the current and voltage in the circuit, allowing engineers to design and optimize the circuit's performance.
The Laplace transform of t cos ( b t ) is ( s 2 + b 2 ) 2 s 2 − b 2 and for t 2 cos ( b t ) it is ( s 2 + b 2 ) 3 2 s ( s 2 − 3 b 2 ) .
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