Chapter 29, Problem 55P

(a)

To determine

ToShow:

The trap filter acts to reject signals in a band of frequencies centered $\omega =\frac{1}{\sqrt{LC}}$ .

Answer to Problem 55P

The trap filter acts to reject signals in a band of frequencies centered at $\omega =\frac{1}{\sqrt{LC}}$ .

Explanation of Solution

Given:

The given diagram is shown below

  

The above circuit is a trap filter circuit. Its phasors diagram is shown below.

Calculation:

The phasor diagram for the trap filter is shown below.

  

Here ${\overrightarrow{V}}_{app}$ is the phasors for $V_{in}$ and ${\overrightarrow{V}}_L+{\overrightarrow{V}}_C$ is the phasorfor $V_{out}$ .

The projection of ${\overrightarrow{V}}_{app}$ onto the horizontal axis is $V_{app}=V_{in}$ , and the projection of ${\overrightarrow{V}}_L+{\overrightarrow{V}}_C$ onto the horizontal axis is $V_L+V_C=V_{out}$ . Assume that the impedance of the trap is zeroand the frequency at which the circuit rejects signals have to be calculated.

Define bandwidth as $\Delta \omega =\left|\omega -{\omega }_{trap}\right|$ and impedance of the trap as $\left|Z_{trap}\right|=R$ . These two will yield an expression for the bandwidth and reveal its dependence on R.

  $V_{app}$ can be represented in the sinusoidal form as

  $V_{app}=V_{apppeak}\sin \omega t$ ……(1)

Here, $V_{apppeak}=V_{peak}=I_{peak}Z$

And

  $Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}$......(2)

Output voltage $V_{out}$ can be represented in the sinusoidal form as

  $V_{out}=V_{outpeak}\sin \left(\omega t-\delta \right)$......(3)

Here, $V_{outpeak}=I_{peak}{Z}_{trap}$

And $Z_{trap}=X_L-X_C$......(4)

Now,

  $\begin{array}{c}{V}_{out}=V_{outpeak}\times \sin \left(\omega t-\delta \right)\\ V_{out}=I_{peak}\times Z_{trap}\times \sin (\omega t-\delta )\end{array}$

  $V_{out}=\frac{V_{peak}}{Z}{Z}_{trap}\sin (\omega t-\delta )$......(5)

Substitute $\sqrt{R^2+{\left(X_L-X_C\right)}^2}$ for $Z$ from equation (2) in equation (5) $V_{out}=\frac{V_{peak}}{\sqrt{R^2+{\left(X_L-X_C\right)}^2}}{Z}_{trap}\sin (\omega t-\delta )$......(6)

Again substitute $Z_{trap}$ for $X_L-X_C$ from equation (4) in equation (6)

  $V_{out}=\frac{V_{peak}}{\sqrt{R^2+Z_{trap}{}^2}}{Z}_{trap}\sin (\omega t-\delta )$.......(7)

Now,

  $V_{out}=0$ provided that $Z_{trap}=0$ .

Substitute 0 for $Z_{trap}$ in equation (4)

  $Z_{trap}=X_L-X_c=0$i.e.,

  $\begin{array}{l}\omega L-\frac{1}{\omega C}=0\\ \omega L=\frac{1}{\omega C}\\ {\omega }^2=\frac{1}{LC}\\ \boxed{\omega =\sqrt{\frac{1}{LC}}}\end{array}$

Conclusion:

Hence, it is shown that the trap filter acts to reject signals in a band of frequencies centered at $\omega =\frac{1}{\sqrt{LC}}$ .

(b)

To determine

The way, the width of the frequency band rejected depends on the resistance $R$ .

  $$

Answer to Problem 55P

The width of the frequency band rejected is directly proportional to the resistance.

Explanation of Solution

Given:

The given circuit diagram is shown below

  

The given circuit is a trap filter circuit. Its phasors diagram is shown below.

Calculation:

The required Phasor diagram is shown below

  

Here ${\overrightarrow{V}}_{app}$ is the phasors for $V_{in}$ and ${\overrightarrow{V}}_L+{\overrightarrow{V}}_C$ is the phasor for $V_{out}$ .

The projection of ${\overrightarrow{V}}_{app}$ onto the horizontal axis is $V_{app}=V_{in}$ , and the projection of ${\overrightarrow{V}}_L+{\overrightarrow{V}}_C$ onto the horizontal axis is $V_L+V_C=V_{out}$ . Assume that the impedance of the trap is zero and the frequency at which the circuit rejects signals have to be calculated.

Define bandwidth as $\Delta \omega =\left|\omega -{\omega }_{trap}\right|$......(8)

And impedance of the trap as $\left|Z_{trap}\right|=R$......(9)

These two will yield an expression for the bandwidth and reveal its dependence on R.

Substitute $X_L-X_C$ for $Z_{trap}$ in equation (9)

  $\begin{array}{l}\omega L-\frac{1}{\omega C}=R\\ {\omega }^{2}LC-1=\omega RC\end{array}$......(10)

Again,

  ${\omega }_{trap}=\frac{1}{\sqrt{LC}}$ ......(11)

Substitute ${\omega }_{trap}$ for $\frac{1}{\sqrt{LC}}$ in equation (10)

  ${\left(\frac{\omega }{{\omega }_{trap}}\right)}^2-1=\omega RC$

For ${\omega }_{trap}\approx \omega$

  $\left(\frac{{\omega }^2-{\omega }^2{}_{trap}}{{\omega }_{trap}}\right)\approx {\omega }_{trap}RC$

  $\left\{\frac{\left(\omega -{\omega }_{trap}\right)\left(\omega +{\omega }_{trap}\right)}{{\omega }_{trap}}\right\}\approx {\omega }_{trap}RC$

  $\left\{\frac{\left(\omega -{\omega }_{trap}\right)\left(\omega +{\omega }_{trap}\right)}{{\omega }_{trap}}\right\}\approx {\omega }_{trap}RC$......(12)

For the case, ${\omega }_{trap}\approx \omega$ ,

  $\left(\omega +{\omega }_{trap}\right)\approx 2{\omega }_{trap}$

Substitute $2{\omega }_{trap}$ for $\left(\omega +{\omega }_{trap}\right)$ in the equation (12)

  $2{\omega }_{trap}\left(\omega -{\omega }_{trap}\right)\approx {\omega }^2{}_{trap}RC$

  $\left(\omega -{\omega }_{trap}\right)=\frac{RC{\omega }_{trap}}{2}$

  $\Delta \omega =\left|\omega -{\omega }_{trap}\right|=\frac{R}{2\sqrt{\frac{L}{C}}}$

  $\boxed{\Delta \omega =\frac{R}{2\sqrt{\frac{L}{C}}}}$

  $\boxed{\Delta \omega \propto R}$

Conclusion:

Hence, the width of the frequency band rejected directly proportional to the resistance.