## Laplace Transform Question

### Laplace Transform Question

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

### Re: Laplace Transform Question

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

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

### Re: Laplace Transform Question

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

### Re: Laplace Transform Question

Well it's more of a substitution:<p>

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>

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

**Solving For Sin y:**<p>Rearrange EQN A, solve for cosy<p>e^jy = cosy + jsinye^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 + jsinyjsiny = 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

### Re: Laplace Transform Question

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

-Rick

-Rick

### Re: Laplace Transform Question

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

-rick

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