Laplace Transform Question

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ele1200
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Laplace Transform Question

Post by ele1200 » Sun Jul 13, 2003 12:36 pm

Can someone show me the derivation for the laplace transform of cos wt?<p>Thanks

greg123
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Re: Laplace Transform Question

Post by greg123 » Sun Jul 13, 2003 1:05 pm

See if this makes sense....<p>
Remember the relationship sin wt = ½j (e^jwt - e^-jwt) EQN C<p>***see below for this derivation*****<p>
Also note that the laplace transform for e^jwt is 1/(s-jw) and the laplace transform for e^-jwt is 1/(s+jw). We will now substitute this into the above relationship. <p>sin wt = ½j (e^jwt - e^-jwt)
L sin wt = ½j ( 1/(s-jw) - 1/(s+jw) ) *now multiply through
L sin wt = ½j ( (s+jw)-(s-jw) / (s+jw)(s-jw) )
L sin wt = ½j ( 2jw /( s² + w² )
L sin wt = w /( s² + w² )<p>We can derive cos wt using the first derivative of sin wt.<p>L cos wt = L [ 1/w ( d/dt sin wt uT )]
L cos wt = 1/w [ s [ w /( s² + w² )]]
L cos wt = s /( s² + w² )<p>Note from above that d/dt is equal to s.<p>Now we can also use EQN D and substitute the transforms:<p>cos wt = ½ ( e^jy + e^-jy )
L cos wt = ½ ( 1/(s-jw) + 1/(s+jw) )
L cos wt = s /( s² + w² )<p>
Thats it!<p>Greg<p>
**********Derivation************<p>For the following Ø indicates phi or the combination of theta and -90º.<p>If we connect a sinusoidal source of Vmcos (wt + theta) to a network we have a response of Imcos(wt + theta). Since the applied source is a combination of real and imaginary, we will construct the imaginary forcing function by multipling by the j operator.<p>jVmsin(wt + Ø) and the response would be jImsin(wt + Ø)<p>By using the superposition theorum, our real and imaginary function would give an input of real + imaginary OR<p>Vmcos(wt + Ø) + jVmsin(wt + Ø) <p>would yield the response of <p>Imcos(wt + Ø) + jImcos(wt + Ø)<p>As you can see these expressions can be rather troublesome. We wil use Eulers identities to simplify the relationship.<p>Euler states that <p>e^jy = cosy + jsiny EQN A
e^-jy = cosy - jsiny EQN B<p>If we subtract A from B, solving for sin, we have<p>siny = ½j ( e^jy - e^-jy) Which is EQN C<p>This is the identity we used first.<p>If we subtract A from B, solving for cos, we have<p>cosy = ½ ( e^jy + e^-jy ) EQN D

Will
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Re: Laplace Transform Question

Post by Will » Mon Jul 14, 2003 12:39 pm

Great stuff ! greg - Hope I'm not putting my foot in it here (How else can one learn things ?) - but, in the second last line do you not mean " If we add A & B ?
BB

greg123
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Re: Laplace Transform Question

Post by greg123 » Mon Jul 14, 2003 2:19 pm

Well it's more of a substitution:<p>Solving For Sin y:<p>Rearrange EQN A, solve for cosy<p>e^jy = cosy + jsiny
e^jy - jsiny = cosy<p>Substitute into EQN B<p>e^-jy = cosy - jsiny
e^-jy = (e^jy - jsiny) - jsiny
e^-jy - e^jy = -2jsiny * -1
(e^jy - e^-jy = 2jsiny )/2j
siny = ½j(e^jy - e^-jy)<p>Solving For Cosy:<p>Solve EQN A for siny:<p>e^jy = cosy + jsiny
jsiny = e^jy - cosy<p>Sub into EQN B:<p>e^-jy = cosy - jsiny
e^-jy = cosy - (e^jy - cosy)
e^-jy + e^jy = 2cosy
cosy = ½(e^-jy + e^jy)<p>Thats it, substitution.<p>greg

rosborne
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Re: Laplace Transform Question

Post by rosborne » Thu Jul 17, 2003 6:41 pm

Another method using the definition of the unilateral Laplace Transform isImage I've been trying to figure out how to post this so, sorry if it is no longer topical.
-Rick

rosborne
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Re: Laplace Transform Question

Post by rosborne » Thu Jul 17, 2003 6:44 pm

Wow, that didn't come out the way I expected, sorry about that. Jeez, that's big, but at least it's somewhat readable.
-rick

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