How do I see the gain from this Bode plot? (simple)

[jean28]

This is the Bode plot of a Broadband Bandreject Op Amp amplifier. The exercise says this:

From the Bode magnitude plot in Fig. 15.14, we see that the bandreject filter has cutoff frequencies of 100 rads/s and 2000 rad/s and a gain of 3 in the passbands.

I know that in order to get both cutoff frequencies I need to subtract 3 dB from the plot, but how do I get the gain? I would think that the gain is 3 dB because it is also the distance between the top of the graph and the cutoff frequencies, but wouldn't that imply that the gain would ALWAYS be 3?

What am i doing wrong here? Thanks in advance.

EDIT

Whoops, forgot the graph. Here it is:

http://i1226.photobucket.com/albums/ee410/jean28x/e0660bbbc8da983dbd0530bbf6c195ce.jpg

 

[0xDEADBEEF]

It's a gain of 3 not a gain of 3 dB! They just say that you have about 9.5dB on the left and the right so the signal is multiplied approximately by a factor of 3.

 

[yungman]

This is the Bode plot of a Broadband Bandreject Op Amp amplifier. The exercise says this:
I know that in order to get both cutoff frequencies I need to subtract 3 dB from the plot, but how do I get the gain? I would think that the gain is 3 dB because it is also the distance between the top of the graph and the cutoff frequencies, but wouldn't that imply that the gain would ALWAYS be 3?

What am i doing wrong here? Thanks in advance.

EDIT

Whoops, forgot the graph. Here it is:

http://i1226.photobucket.com/albums/ee410/jean28x/e0660bbbc8da983dbd0530bbf6c195ce.jpg

You want to look at the pass band gain, just read off the graph. Look at the graph at the low frequency and high frequency where the plot level off at about +9dB. Gain in dB is 20 log G, where G is linear gain. So +9dB give G≈2.81. But I just roughly read +9dB, so it is likely gain of 3, I like to stay with simple number of +9dB rather +9.54dB or something.

The notch filter is the -3dB on each side, so if you consider pass band gain is +9dB, then look at the two points of the graph crossing +9dB-3dB=+6dB.

 

[jean28]

You want to look at the pass band gain, just read off the graph. Look at the graph at the low frequency and high frequency where the plot level off at about +9dB. Gain in dB is 20 log G, where G is linear gain. So +9dB give G≈2.81. But I just roughly read +9dB, so it is likely gain of 3, I like to stay with simple number of +9dB rather +9.54dB or something.

The notch filter is the -3dB on each side, so if you consider pass band gain is +9dB, then look at the two points of the graph crossing +9dB-3dB=+6dB.

Ok I see it now.

9.54 = 20 log₁₀ (G)
0.477 = log₂₀ (G)
10 ^0.477 = G
2.99 = G

Thank you very much.

 

[alphysicist]

The gain is represented by the y-axis on the Bode plot. In this case, the gain is shown in decibels (dB) and can be read directly from the graph. In the passbands, the gain is approximately 3 dB, which means there is a 3 times increase in voltage at those frequencies.

To get the exact value of the gain, you can use the formula 20log(A), where A is the voltage gain. In this case, A = 3, so the gain is 20log(3) = 9.54 dB.

You are correct that the distance between the top of the graph and the cutoff frequencies is 3 dB, but this does not necessarily mean that the gain will always be 3 dB. The gain can vary depending on the design of the amplifier and the frequencies being passed through it. In this case, the gain is specifically designed to be 3 in the passbands.

To summarize, the gain can be read directly from the y-axis on the Bode plot and is shown in decibels. In this case, the gain is approximately 3 dB in the passbands and can be calculated using the formula 20log(A), where A is the voltage gain.