Circuit Theory/Lossy Tuned Circuits/Band-stop filter

Example 1
Problem: Compute the magnitude and phase of the frequency response of the circuit below, where R=1&Omega;, L=0.025H and C=0.4F. Find both of the frequency boundaries of the stop-band.

We use our method of considering the circuit as a voltage divider to find the frequency response:




 * $$H(j \omega)$$||$$= \frac{V_o}{V_i}$$
 * ||$$ = \frac{\frac{1}{LC}-\omega^2}{\frac{1}{LC}-\omega^2+j \omega \frac{R}{L}}$$
 * ||$$ = \frac{100-\omega^2}{100-\omega^2+40j\omega}$$
 * }
 * ||$$ = \frac{100-\omega^2}{100-\omega^2+40j\omega}$$
 * }

Let us first consider the general shape of the graph before we plot it.


 * At &omega;=0, the imaginary part of the denominator disappears and the response is unity (1∠0°). At &omega;=∞, the quadratic dominates and the numerator and denominator cancel and the response is unity (1∠0°). Therefore, the magnitude of the response tends to 1 at &omega;=0 and &omega;=∞, and the phase tends to 0.
 * At &omega;=10, the magnitude falls to zero. However, if we approach this frequency from below, we find that at &omega;=10–, the response amplitude is 0∠–90°, but from above, at &omega;=10+, the response amplitude is 0∠90°, indicating a very rapid phase flip.

We see that the graph shows exactly this: