Solving for Vout: Challenging Lab Problems

[hogrampage]

Homework Statement

I am not sure why our professor is making us do labs that are 6 chapters above where we are, but here it is:

Compute V_out if V_in has an amplitude of 1V and a phase shift of 0 degrees.

Homework Equations

V=RI
V=jωLI
V=\frac{1}{jωC}I
ω=\frac{1}{\sqrt{LC}}

The Attempt at a Solution

ω = 15811 rad/s
V_in = 1e^{j(15811)}

KVL: V_out = e^{j(15811)} - jωLI_me^{j(15811)} - \frac{1}{jωC}I_me^{j(15811)}

First of all, is that right? Second, I have no idea where to go from that (if it's right).

 

[Simon Bridge]

Where did you get that frequency from?
Your final answer should be in the same terms as the problem was described. In this case, "V_out has an amplitude of ______ and a phase _____."

 

[NascentOxygen]

V=RI
V=jωLI
V=\frac{1}{jωC}I
ω=\frac{1}{\sqrt{LC}}

It is a series circuit, so the current is common to all elements. Determine the voltage across R, and divide it by the voltage across all 3 elements, to get Vo/Vin.

 

[hogrampage]

Where did you get that frequency from?
Your final answer should be in the same terms as the problem was described. In this case, "V_out has an amplitude of ______ and a phase _____."

ω=\frac{1}{\sqrt{LC}}

Voltage division:

V_out = \frac{R}{R+jωL+\frac{1}{jωC}}V_in

Is that correct? Also, my value for ω seems awfully high. Am I using the wrong formula for it?

 

[NascentOxygen]

The imaginary terms in the denominator can be grouped, and the denominator expressed as:

R + j(...)

Then convert from this complex number to polar form (as magnitude and angle).

 

[hogrampage]

What about my value for ω? It just doesn't seem right (it's really high). I mean, maybe it is correct, I've just never seen a frequency that high in previous labs.

I found V_out to be 1∠0°, so it's the same as V_in.

Is that right?

 

[NascentOxygen]

What about my value for ω? It just doesn't seem right (it's really high). I mean, maybe it is correct, I've just never seen a frequency that high in previous labs.

It's an audio frequency, so is manageable.

I found V_out to be 1∠0°, so it's the same as V_in.

Is that right?

No. V_out is dependent on frequency, it is a function of frequency.

 

[hogrampage]

I don't understand. Most of the stuff cancels out, so I get V_in:

V_out = \frac{R*V_{in}}{R+jωL-\frac{j}{ωC}}\frac{R-j(ωL-\frac{1}{ωC})}{R-j(ωL-\frac{1}{ωC})}

V_out = \frac{V_{in}R(R-j(ωL+\frac{1}{ωC})}{R(R+ωL-\frac{1}{ωC})}

Since ωL - \frac{1}{ωC} = 0,

V_out = \frac{V_{in}R}{R}

V_out = V_in

Or, would V_out be this instead?:

V_out = e^{j(15811t)}

Thanks

 

[hogrampage]

Okay, so I tried this again:

V_out = \frac{R∠0°}{\sqrt{R^{2}+(ωL-\frac{1}{ωC})^{2}}∠tan^{-1}(\frac{ωL-\frac{1}{ωC}}{R})}

Is that right? lol

Obviously, I would plug in the values for R, L, and C. So, if I want the voltage at, say, 1000Hz, I just plug that into ω.

 

[NascentOxygen]

That's close. You have an angle in the numerator, and an angle in the denominator—these can be combined into one angle in the numerator.

You have determined that at the resonant frequency, this bandpass filter has a gain of unity. (At other frequencies, you would find it has a gain < 1, otherwise it's not doing the job of a bandpass filter. )

Probably advisable to also express ω₀ in Hertz.

If this is preparation for lab work, consider plotting v₀/vᵢ for a range of frequencies below and above ω₀, and also plot ɸ vs. f so that you know what to be looking for in your lab measurements.