# Not able to understand gain in decibels

#### aman92ullah

Joined Jan 19, 2011
12
I can get s-domain transfer functions, say for a AC coupled inverting amplifer. But I can't quite interpret it, especially when gains are in dB. I know what dB is and can interchange simple gains into dB.
But when there is a circuit involved, I get confused: how does transfer function map simple voltage gain to dB?
I was trying to solve a problem(5.25) from sedra and smith(5ed): R1 is 1k, R2 is 1M.(See attached figure)
Question is that what will be the value of coupling capacitor to ensure gain will be greater than 57dB down to 100Hz.
I am getting 10uF as answer, while the books says it's 1.6uF. I first found the transfer function, which was: Vo/Vin=(-R2/R1)*(s/(s+(1/R1C1)). I take 20 log on both sides, and solve for c1 putting s=100. Where am I wrong? Or my entire concept is botched?
Help!! Last edited:

#### t_n_k

Joined Mar 6, 2009
5,455
It appears the attachment didn't make it with your post.

#### aman92ullah

Joined Jan 19, 2011
12
I have uploaded the file now. (Sorry, first time around here. Learning)

#### t_n_k

Joined Mar 6, 2009
5,455
There's no need even to consider the 's' domain transform of the circuit.

A gain of 57dB is equivalent to a straight voltage gain of 707.95.

The amplifier gain will be given by

$$A_v=\frac{R_2}{|Z_{in}|}$$

where

$$|Z_{in}|=\sqrt{R_1^2+X_c^2}$$

If Av=707.95 and R2=1MΩ

then |Zin|=R2/Av=1MΩ/707.95=1412.5Ω

One then uses the second equation above to solve for Xc & hence C given
R1=1000Ω & |Zin|=1412.5Ω

• aman92ullah

#### aman92ullah

Joined Jan 19, 2011
12
Well, that's all right. Thanks!
But I am still stuck with the 's' domain. When we put the gain in dB, we say
Gain(dB)=20log(Vo(s)/Vin(s)) =(X)provided impedances match. My question is, what is X in the s-domain? Isn't it what I wrote in the first post?
Sorry, it might be very stupid, but I am struggling with s-domain and bode plots for some time now. Please!