3-dB Bandwidth

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anoopjose

Joined: 08 Jun 2006
Posts: 28
Helped: 1

14 Mar 2007 6:48 3-dB Bandwidth

What are the 3-dB Bandwidth of transfer functions

s/(s²+as+b) & 1/(s²+as+b) ?

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Mansour_M

Joined: 30 Jul 2006
Posts: 122
Helped: 9

14 Mar 2007 11:33 Re: 3-dB Bandwidth

Hi Dear anoopjose!

Assume d = sqrt(2). So:

***********************************************
The first one:
1/(s²+as+b) = 1/sqrt(2) = 1/d

s1 = -1/2*a+1/2*(a^2-4*b+4*c)^(1/2)
s2 = -1/2*a-1/2*(a^2-4*b+4*c)^(1/2)
***********************************************
The second one:
s/(s²+as+b) = 1/sqrt(2) = 1/d

s1 = -1/2*a+1/2*d+1/2*(a^2-2*a*d+d^2-4*b)^(1/2)
s2 = -1/2*a+1/2*d-1/2*(a^2-2*a*d+d^2-4*b)^(1/2)
***********************************************

Thanks!

demodb

Joined: 04 Oct 2005
Posts: 77
Helped: 3

14 Mar 2007 14:44 Re: 3-dB Bandwidth

think you should calculate the poles, which is done by setting the numerator equal to zero.

Kral

Joined: 28 Mar 2005
Posts: 1182
Helped: 187

14 Mar 2007 20:43 Re: 3-dB Bandwidth

anoopjose,
Make the substitution s=jw. Set the result = 1/[sqrt(2)]. In the first example, you wil get two values for w. In the second example, you will get 1 value.
Regards,
Kral