# How to verify the voltage and frequency response in a simulated equalizer?

I am working on a simple equalizer, intended to filter bass, midband and treble frequencies.

I'm wiring the bode plotter of Multisim in a wrong way,perhaps.

The circuit is:

The idea is to measure the input (this is done with XSC2) the output of each band (XSC3) the final output (XSC1) and the frecuency response of the filters (bass (U2A), treble (U1A) and midband (U3A).

The input is simulated with senoidal signals of 0.5Vpp and 60,400,2500 and 10000 Hertz.

The Bode plot is giving this response:

For the low (bass) filter the design values are: $$\R_{Ab}=R_{Bb}=R_{2b}=1\,k\Omega\$$

$$\f_{C}=\frac{1}{2\pi RC}=500\,Hz\$$

$$\C_{Ab}=C_{Bb}=C=0.318\,\mu F\$$

For the high (treble) filter the values calculated are: $$\R_{At}=R_{Bt}=R_{2t}=1\,k\Omega\$$

$$\f_{C}=\frac{1}{2\pi RC}=5\,kHz\$$

$$\C_{At}=C_{Bt}=C=0.0318\,\mu F\$$

The bandpass (midband) filter is calculated as: $$\f_{c}=\frac{1}{2\pi C}\sqrt{\frac{R_{1m}+R_{3m}}{R_{1m}R_{2m}R_{3m}}}\$$

$$\Q=\pi f_{C}CR_{2m}=1.57\$$

$$\A_{0}=\frac{R_{2m}}{2R_{1m}}=1\$$

$$\R_{3m}=\frac{Q}{2\pi f_{c}C(2Q^{2}-A_{0})}=\frac{1.57}{2\pi(2.5\times10^{3})(0.01\mu)(1.57^{2}-1)}=6.8\,k\Omega\$$

$$\BW=\frac{f_{0}}{Q}=1.6\,kHz\$$

The capacitor values in the model are used as calculated to verify the operation, later when implementing the board I will use values of $$\0.33\$$ and $$\0.033\,\mu F\$$.
Then how the bode tools must be wired to correcty check the filters response?

UPDATE
Following the information of audioguru, I have fixed to a higher value the resistors (10k) and updated the unbiased + input of the U1A
but I can't tell how this is working (I mean interpret, guessing this is a good result and is filtering the low frequencies), like in the next figure, the turquoise signal is the input with all the components and the red signal is the output (filtered?) of the bass stage.