# Bandpass Filter

## 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 Vout if Vin 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

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).

#### Attachments

• bandpass.jpg
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Simon Bridge
Homework Helper
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 _____."

NascentOxygen
Staff Emeritus
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.

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 _____."

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?

NascentOxygen
Staff Emeritus
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).

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:
NascentOxygen
Staff Emeritus
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 Vout to be 1∠0°, so it's the same as Vin.

Is that right?
No. Vout is dependent on frequency, it is a function of frequency.

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

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.

NascentOxygen
Staff Emeritus