# Bandpass Filter

1. Sep 12, 2012

### hogrampage

1. The problem statement, all variables and given/known data
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 Vout if Vin has an amplitude of 1V and a phase shift of 0 degrees.

2. Relevant equations
V=RI
V=jωLI
V=$\frac{1}{jωC}$I
ω=$\frac{1}{\sqrt{LC}}$

3. The attempt at a solution
Vin = 1$e^{j(15811)}$

KVL: Vout = $e^{j(15811)}$ - jωLIm$e^{j(15811)}$ - $\frac{1}{jωC}$Im$e^{j(15811)}$

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

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2. Sep 12, 2012

### 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, "Vout has an amplitude of ______ and a phase _____."

3. Sep 12, 2012

### Staff: Mentor

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.

4. Sep 12, 2012

### hogrampage

ω=$\frac{1}{\sqrt{LC}}$

Voltage division:

Vout = $\frac{R}{R+jωL+\frac{1}{jωC}}$Vin

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

5. Sep 12, 2012

### Staff: Mentor

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).

6. Sep 12, 2012

### 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 Vout to be 1∠0°, so it's the same as Vin.

Is that right?

Last edited: Sep 12, 2012
7. Sep 12, 2012

### Staff: Mentor

It's an audio frequency, so is manageable.

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

8. Sep 12, 2012

### hogrampage

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

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

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

Since ωL - $\frac{1}{ωC}$ = 0,

Vout = $\frac{V_{in}R}{R}$

Vout = Vin

Or, would Vout be this instead?:

Vout = e$^{j(15811t)}$

Thanks

9. Sep 12, 2012

### hogrampage

Okay, so I tried this again:

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

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.

10. Sep 13, 2012

### Staff: Mentor

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.