Find forced response of VC(t) - Can someone check my work?

[eehelp150]

Homework Statement

Find the forced response of VC(t)

Homework Equations

The Attempt at a Solution

W=1000
[tex]\frac{d^2V_C(t)}{dt^2}+\frac{R}{L}\frac{dV_C(t)}{dt}+\frac{V_C}{LC}=\frac{Vin}{LC}[/tex]
[tex]V_C(t)=Acos(1000t+\theta)[/tex]
Plug in Vc(t),R,L,C into equation 1

[tex]-10^6Acos(10^3t+\theta)-10^7Asin(10^3t+\theta)+10^7Acos(10^3t+\theta)=5*10^7cos(10^3t)[/tex]

Simplifying:
[tex](9*10^6)Acos(10^3t+\theta)-10^7Asin(10^3t+\theta)=5*10^7cos(10^3t)[/tex]

Trig ID
[tex]\sqrt{(9*10^6)^2+(10^7)^2}*Acos(10^3t+tan^-1(-\frac{9*10^6}{-10^7}))[/tex]

[tex]13.45*10^6Acos(10^3t+41.98°)[/tex]Is everything I've done so far correct? How do I find A?

 

[NascentOxygen]

Trig ID

Above this line you have an equation, below this line you should also write an equation.

Above this line your expression contains a ##\theta## term, but its simplified expression below that line has lost the ##\theta## term. Can you restore the missing ##\theta## into that simplified equivalent expression?

The amplitude of the cosinusoid on the left equates to the amplitude of that on the right, so set them equal and solve to find ##A##.

 

[eehelp150]

Above this line you have an equation, below this line you should also write an equation.

Above this line your expression contains a ##\theta## term, but its simplified expression below that line has lost the ##\theta## term. Can you restore the missing ##\theta## into that simplified equivalent expression?

The amplitude of the cosinusoid on the left equates to the amplitude of that on the right, so set them equal and solve to find ##A##.

Is this correct?
[tex]\sqrt{(9*10^6)^2+(10^7)^2}*Acos(10^3t+\theta+tan^-1(-\frac{9*10^6}{-10^7}))=5*10^7cos(10^3t)[/tex]
[tex]13.45*10^6Acos(10^3t+\theta+41.98°)=5*10^7cos(10^3t)[/tex]

 

[NascentOxygen]

I'm not able right now to verify your mathematics, but that is the form that I expect.

So equate the left side cosine magnitude to that on the right side, and solve for A.
Similarly, equate the cosine arguments, and solve for θ.​

 

[NascentOxygen]

One trap to be aware of is that tan^–1 can have multiple answers. Your calculator will give you only one, but you always need to consider whether the other/s are applicable to your particular problem.

 

[eehelp150]

I'm not able right now to verify your mathematics, but that is the form that I expect.

So equate the left side cosine magnitude to that on the right side, and solve for A.
Similarly, equate the cosine arguments, and solve for θ.​

So
13.45*10^6 A = 5*10^7
and
theta + 41.98 = 0
?

 

[eehelp150]

One trap to be aware of is that tan^–1 can have multiple answers. Your calculator will give you only one, but you always need to consider whether the other/s are applicable to your particular problem.

The formula is tan^-1(-A/C)
if A is 9 and C is -10, wouldn't it be:
tan^-1(-9/-10)
tan^-1(9/10)
=41.98

 

[NascentOxygen]

So
13.45*10^6 A = 5*10^7
and
theta + 41.98 = 0
?

That's how it's done, yes.

The formula is tan^-1(-A/C)
if A is 9 and C is -10, wouldn't it be:
tan^-1(-9/-10)
tan^-1(9/10)
=41.98

Geometrically, you need to look at all 4 quadrants when determining the answer to tan ^–1 (0.9) because there are at least 2 answers.

Mathematically, you can say there are an infinite number of solutions, but in electronic circuits, we usually consider those phase angles restricted to the range: –180°...+180°

 

[gneill]

Just a heads up, I think that somewhere along the line the 9/10 should have been 10/9. I haven't checked closely through the math to see exactly how this happened, but l suspect it would be where the "Trig ID" took place.

I say that it should have been 10/9 because solving for Vc via another method yields a phase angle of -48.01°, which corresponds to atan(-10/9).

 

[eehelp150]

Just a heads up, I think that somewhere along the line the 9/10 should have been 10/9. I haven't checked closely through the math to see exactly how this happened, but l suspect it would be where the "Trig ID" took place.

I say that it should have been 10/9 because solving for Vc via another method yields a phase angle of -48.01°, which corresponds to atan(-10/9).

Can you show the other method?
My professor wrote this trig identity on this board:
Acos(x)+Csin(x)=sqrt(A^2+C^2)*cos(x+tan^-1(-A/C))

but this website: http://myphysicslab.com/springs/single-spring/trig-identity-en.html

 

[gneill]

Can you show the other method?

I used phasors (which uses complex values for impedances) and standard circuit analysis methods. All the usual analysis methods work with phasors (Ohm's law, KVL, KCL, nodal analysis, mesh analysis, etc.), so it's just like solving a circuit with resistors except it uses complex numbers.

My professor wrote this trig identity on this board:
Acos(x)+Csin(x)=sqrt(A^2+C^2)*cos(x+tan^-1(-A/C))

but this website: http://myphysicslab.com/springs/single-spring/trig-identity-en.html
View attachment 108950

The trig identity is fine. Your professor's negative sign is merely accounting for the sign of the sine term in the problem rather than making the value of ##a## negative.

The issue will be with why you found A/C to be 9/10 rather than 10/9.

 

[eehelp150]

I used phasors (which uses complex values for impedances) and standard circuit analysis methods. All the usual analysis methods work with phasors (Ohm's law, KVL, KCL, nodal analysis, mesh analysis, etc.), so it's just like solving a circuit with resistors except it uses complex numbers.

The trig identity is fine. Your professor's negative sign is merely accounting for the sign of the sine term in the problem rather than making the value of ##a## negative.

The issue will be with why you found A/C to be 9/10 rather than 10/9.

Isn't A = 9*10^6 and C = -10^7?

 

[gneill]

I'm thinking that perhaps the issue is with applying the trig identity correctly. The underlying trig identity being invoked is:

##cos(a + b) = cos(a) cos(b) - sin(a) sin(b)##

and we want to associate cos(a) and sin(a) with the coefficients in your equation. Let's start by getting rid of the large exponents in your equation. Dividing through by ##10^7## and letting ##b = (ω t + θ)## for now yields:

##\frac{9}{10} A cos(b) - 1~A sin(b) = 5 cos(ω t)##

Then the 9/10 and 1 are associated with the cos(a) and sin(a) of the trig identity. That means that the angle a is given by:

##a = tan^{-1} \left( \frac{sin(a)}{cos(a)} \right) = tan^{-1} \left( \frac{1}{9/10} \right) = tan^{-1} \left( \frac{10}{9} \right)##

which is the 10/9 that I found by the different approach.

 

[eehelp150]

I'm thinking that perhaps the issue is with applying the trig identity correctly. The underlying trig identity being invoked is:

##cos(a + b) = cos(a) cos(b) - sin(a) sin(b)##

and we want to associate cos(a) and sin(a) with the coefficients in your equation. Let's start by getting rid of the large exponents in your equation. Dividing through by ##10^7## and letting ##b = (ω t + θ)## for now yields:

##\frac{9}{10} A cos(b) - 1~A sin(b) = 5 cos(ω t)##

Then the 9/10 and 1 are associated with the cos(a) and sin(a) of the trig identity. That means that the angle a is given by:

##a = tan^{-1} \left( \frac{sin(a)}{cos(a)} \right) = tan^{-1} \left( frac{1}{9/10} \right) = tan^{-1} \left( \frac{10}{9} \right)##

which is the 10/9 that I found by the different approach.

final answer:
Vc(t)=3.7cos(10^3t-48)