# Estimate the Bandwidth Help!

1. Mar 23, 2010

### mmmboh

Hi, this is a problem I am trying to do:

So I found that at 3dB, V0=0.7079Vi
and I wasn't sure if I should use real numbers or complex numbers, I did both, I'll show the real one:

Vo/Vi=0.7079=R/(R+1/wc)=Rwc/(Rwc+1)...I can solve for w, but this will only give one answer, when I use complex numbers I get 2 answers, but I'm pretty sure it's wrong...

Edit: I think I am suppose to find the total impedance by squaring each term in the denominator of R/(R+1/wc) and then square rooting it...I can then solve for f and I get +/- 2.08x106Hz, then the bandwidth would just be the positive value times 2....but I think this may still be wrong...

Am I doing this completely wrong?

Thanks!

Last edited: Mar 23, 2010
2. Mar 23, 2010

### mmmboh

anyone have any hints?

3. Mar 23, 2010

### satchmo05

From what I remember, the impedance of a resonant network is purely Z(jw) = R; therefore, I do not understand why you are implementing all of these equations.

4. Mar 27, 2010

### mmmboh

Then I guess I'm not sure what to do...

5. Mar 27, 2010

### rl.bhat

Here the step voltage V is a dc voltage. When you apply this step voltage to the RC circuit, Vo will increase until it reaches the applied voltage V. So in the problem where is frequency?
If the applied square wave is a continuous wave from, you have to give the period of this wave form. So that you will be knowing the duration of charging and the discharging through the capacitor.

6. Mar 27, 2010

### mmmboh

The frequency is 1/RC isn't it?

7. Mar 27, 2010

### elect_eng

You are basically on the right path here. Basically, you want the magnitude of the complex frequency response Vo/Vi to be 0.707. This occurs at a frequency f=1/(2piRC).

In general, you replace capacitors with impedance -j/(2pi f C), where j is the square root of negative one. Then you can use the voltage divider equation to get the frequency response, which is a complex function of frequency. You take the magnitude of this complex function and set it equal to 0.707 and solve for f.

8. Mar 27, 2010

### elect_eng

Actually, I should retract some of my previous post. This comment is good in that it points out that the question is a little confusing. I believe that you need to find the bandwidth of the output waveform. In the frequency domain, the input signal is a 1/s, or 1/jw spectrum. Note that w is angular frequency w=2*pi*f, and j is the square root of negative one. This input function is filtered by a high pass filter (RC type). The net effect is a frequency response function with a peak response and a 3 dB bandwidth that can be solved for. My previous post gives the hints for doing this calculation. The key is to take the magnitude of the complex response function and then solve for the 3 dB points.

9. Mar 27, 2010

### mmmboh

Hm, won't V0 at first be equal to V because the resistor comes after the capacitor and all the voltage will be across it, but eventually all the voltage will be across the capacitor so V0 will be zero? So isn't V0 decreasing not increasing?

This is directed towards Rl.Bhat's comment.

10. Mar 27, 2010

### mmmboh

Ok what I did was find the Vo/Vi which = jwRC/(1+jwRC) and I set that equal to 0.7079.
Then I multiplied the numerator and denominator by the conjugate of the denominator so I could find the magnitude, so $$V_0/V_i=(jwRC+w^2R^2C^2)/(1+w^2R^2C^2)$$...I then separated the numerator in to 2 parts, squared each part, added them together, square rooted that, and set that equal to 0.7079, I then squared both sides so I could solve for w...I ended up getting a crazy equation so I used wolfram alpha...one of the solutions for f is of course $$1/(2\pi RC)$$, the other is a complex number....I don't think this is what you meant for me to do

I guess I am finding the wrong frequency? Am I finding the frequency of the input waveform and not the output waveform? I'm not really sure what I am talking about...first time I've done a question like this.

11. Mar 28, 2010

### elect_eng

Well, I think you are doing the first thing I told you to do. Unfortunately, as I mentioned in the next post, I think this is wrong. I believe that the question is poorly worded, but I seems to be asking you to find the bandwidth of the output signal, which is the net effect of the input signal passing through a high pass filter.

Note that the input spectrum is 1/(jw) which decreases with frequency. The filter effect you already calculated, and it is a high pass filter which has increasing response with frequency. The combination of the input signal with the filter creates an output spectrum with a peak in the middle frequency band. You need to solve for the high and low frequency at the 3 dB point. The difference between these frequencies is the bandwidth.

12. Mar 28, 2010

### mmmboh

Hm maybe it's because it's really late over here, but I am not sure how I am suppose to set up the equations to solve for the 3 dB points for the output wave :S

13. Mar 28, 2010

### rl.bhat

During the step voltage capacitor charges and discharges.
You can find the time taken to reach Vo = Vi*0.7079 during charging and discharging.
During charging Vo = Vi*e^-t/RC where Vo is the voltage across R.
e^t1/RC = Vi/Vo = 1.414. RC is given. Find t1. Then 1/t1.
During discharging
Vo =Vi(1 - e^-t/RC)
e^-t2/RC = 1 - Vo/Vi . Find t2 and 1/t2.
The band width may be the difference between 1/t1 and 1/t2.
From the given data only this much can be done.

14. Mar 28, 2010

### elect_eng

I agree with rl.bhat that this question is a little confusing. I can't say his approach is wrong, but I think it is possible to work this problem in the frequency domain.

The output waveform will have a spectrum given by

Vo/Vi = (jwRC/(1+jwRC))/(jw)

This is the combination of the input spectrum with the filter. Note that the spectrum of a step function is 1/(jw), and the high pass filter response you already specified before.

If you plot the magnitude response of this, I believe you will see a the response and can establish a 3 dB point. What I see is the net effect of a low pass filter response.

Vo/Vi = (RC)/(1+jwRC)

So, I think you get the same answer as before f = 1 / (2 pi R C)

Last edited: Mar 28, 2010
15. Mar 28, 2010

### mmmboh

Thanks guys! One thing, if I do it elec_eng way....don't I only have one 3 db frequency...so I can't have a bandwidth in that case? and what would the frequency spectrum look like then?

16. Mar 28, 2010

### mathman44

I'm stuck here as well.

17. Mar 28, 2010

### elect_eng

For a low-pass filter spectrum (which seems to be the case here), the low frequency limit is assumed to be 0 Hz. Hence, the bandwidth is equal to the upper 3 dB frequency.

I want to be clear that I'm not positive my recommended approach is the correct answer, but this would be the way I'd answer the question if it were put to me.