Deleted

 

I don't think the answer is around 5 dB.

I would think Vo = (Vin * Xc) / Z

 

Hi,

Eqn for low pass RC filter

(Vout/Vin) = (1/jwc) / ((1/jwc)+R)

R = 1khz
C=3.1847 x 10^-8
f= 50khz
Vin= 2v

Find Vout express in Db

My working:

i find the magnitude of both side which gives

vout = 2/ sqrt(1+R^2 w^2 c^2)
sub in all values i get vout = 4.997 x 10^-5

This is wrong. How did you get this? Show your steps.

change to db i get : 20log(4.997x10^-5) = -86.02db which is wrong

answer shld be around 5db

This is also wrong. Where did you get it?

what when wrong?

Thks.

to admin: Sorry for the "Urgent help needed" topic......pls remove it

 

db = 20 log( vo/vi )
= 20 log( zc / (zc+r) )
= 20 log( (1/wc) / (1/wc+r) )
= 20 log( (1 / (2 pi 50k 3.18e-8) / (1 / (2 pi 50k 3.18e-8) + 1k) )
= 20 log( 100 / (100+1k) )
= -20 db

 

db = 20 log( vo/vi )
= 20 log( zc / (zc+r) )
= 20 log( (1/wc) / (1/wc+r) )
= 20 log( (1 / (2 pi 50k 3.18e-8) / (1 / (2 pi 50k 3.18e-8) + 1k) )
= 20 log( 100 / (100+1k) )
= -20 db

This calculation actually comes out to -20.83dB, which is wrong.

You forgot that the capacitor's reactance is 90 degrees out of phase with the resistance. You have to find the vector magnitude of the denominator.
Gain = 20 log( (1/wc) / sqrt((R^2) + (1/wc)^2)
Gain = -20.05dB

The OP wanted Vout when Vin=2V.
Vout=-14.03dB.

 

Ron,

I got 198.9 mV as Vo, which is approximately -20.05 dB with respect to 2 Vi.

I didn't understand your Vo being -14.03 dB

 

Ron,

I got 198.9 mV as Vo, which is approximately -20.05 dB with respect to 2 Vi.

I didn't understand your Vo being -14.03 dB

The OP wrote:

(Vout/Vin) = (1/jwc) / ((1/jwc)+R)

R = 1khz
C=3.1847 x 10^-8
f= 50khz
Vin= 2v

Find Vout express in Db

You got Vout=198.9 mV.
20 log (.1989) = -14.03dB.
You are correct in showing gain as -20.05dB, which will of course be true for any input voltage. I realize that dB is a unitless gain measurement (a ratio), but the OP said

Find Vout express in Db.

I suppose to be correct, we should say
Vout=-14.03dBV, i.e., relative to 1 volt, which identifies it as an absolute measurement rather than a ratio.

 

That is true Ron. Had you stated dBV, there wouldn't be an issue. At least you didn't use dBf as the reference ...

 

That is true Ron. Had you stated dBV, there wouldn't be an issue. At least you didn't use dBf as the reference ...

Database Format? Divorced Black Female? Decibels (frequency)? I guess I'm not up-to-date on acronyms.

 

femtowatt.

I saw it on an HP signal generator back in the early 80s. Not one of the more popular references.

 

femtowatt.

I saw it on an HP signal generator back in the early 80s. Not one of the more popular references.

That's funny! Definitely engineering humor, though. My wife doesn't think much of any of my jokes or "funny" stories. I guess I won't try that one on her.

 

If I didn't have to look it up, I would have never remembered it. It was on the old HP-8640 signal generator series.

And to think 1 uV (50 ohms) is 130 dBf. The stuff people remember is amazing sometimes.