# -3dB Point Calculation

Discussion in 'Homework Help' started by simply_me, Feb 26, 2012.

1. ### simply_me Thread Starter Member

May 6, 2010
50
0
Hello,

In the question I suppose to find the -3dB point of the amplifier stages (current). I found an expression for the transfer function, and now I need to find the -3dB point. My transfer function expression is 1/(constant+Cs/gm). My idea to find the -3dB point is:
1. Assuming a low pass filter, I'll find the value of the transfer function when s=0, convert to dB, and subtract 3dB from it. Then convert back to decimal.
2. Then equate the obtained value to the general transfer function to find the value for s. But I'm not sure if I should square both sides (for magnitude) or not.
I'm not sure if this is the correct path, any thoughts?

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
You appear to have a first order transfer function of the form

G(s)=K/(s+a)

which has a -3dB point at ω=a [radians/sec]

So you need to re-cast your transfer into the form shown above and the rest is relatively simple.

3. ### simply_me Thread Starter Member

May 6, 2010
50
0
Hi t_n_k,

Sure I can change the form, but how do I show that the -3dB point is at omega=a (instead of stating it)?

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
OK - your method will work.

Perhaps this can be done somewhat more efficiently as follows:

If

$G(s)=\frac{K}{s+a}$

Then

$|G(0)|=\frac{K}{a}$

$|G(j\omega)|=\frac{K}{\sqrt{\omega^2 +a^2}}$

and the ratio R is given

$R=\frac{|G(j\omega)|}{|G(0)|}$

or

$R=\frac{a}{ \sqrt{ \omega^2 + a^2}}$

One then solves

$20log(R)=-3 \ $dB$$

to find the ω value corresponding to the -3dB point.

The latter relationship (strictly speaking) reduces to

$R=\frac{1}{\sqrt{2}}$

Last edited: Feb 26, 2012
simply_me likes this.
5. ### simply_me Thread Starter Member

May 6, 2010
50
0
Thank you very much for the clarification.