9 Engineering Example 1

9.1 An RC circuit with a single frequency input

Introduction

The components in RC circuits containing resistance, inductance and capacitance can be chosen so that the circuit filters out certain frequencies from the input. A particular kind of filter circuit consists of a resistor and capacitor in series and acts as a high cut (or low pass) filter. The high cut frequency is defined to be the frequency at which the magnitude of the voltage across the capacitor (the output voltage) is $1∕\sqrt{2}$ of the magnitude of the input voltage.

Problem in words

Calculate the high cut frequency for an RC circuit is subjected to a single frequency input of angular frequency $\omega$ and magnitude $v_i$ .

1. Find the steady state solution of the equation

   $R\frac{dq}{dt}+\frac{q}{C}=v_i{e}^{j\omega t}$

   and hence find the magnitude of

   1. the voltage across the capacitor $v_c=\frac{q}{C}$
   2. the voltage across the resistor $v_R=R\frac{dq}{dt}$
2. Using the impedance method of HELM booklet  12.6 confirm your results to part (1) by calculating
   1. the voltage across the capacitor $v_c$
   2. the voltage across the resistor $v_R$ in response to a single frequency of angular frequency $\omega$ and magnitude $v_i$ .
3. For the case where $R=1k\Omega$ and $C=1\mu F,$ find the ratio $\frac{\left|v_c\right|}{\left|v_i\right|}$ and complete the table below

   $\omega$	10	$10^2$	$10^3$	$10^4$	$10^5$	$10^6$
   $\frac{\left|v_c\right|}{\left|v_i\right|}$

4. Explain why the table results show that a RC circuit acts as a high-cut filter and find the value

   of the high-cut frequency, defined as $f_{hc}={\omega }_{hc}∕2\pi ,$ such that $\frac{\left|v_c\right|}{\left|v_i\right|}=\frac{1}{\sqrt{2}}$ .

Mathematical statement of the problem

We need to find a particular solution to the differential equation $R\frac{dq}{dt}+\frac{q}{C}=v_i{e}^{j\omega t}$ .

This will give us the steady state solution for the charge $q$ . Using this we can find $v_c=\frac{q}{C}$ and $v_R=R\frac{dq}{dt}$ . These should give the same result as the values calculated by considering the impedances in the circuit. Finally we can calculate $\frac{\left|v_c\right|}{\left|v_i\right|}$ and fill in the table of values as required and find the high-cut frequency from $\frac{\left|v_c\right|}{\left|v_i\right|}=\frac{1}{\sqrt{2}}$ and $f_{hc}={\omega }_{hc}∕2\pi$ .

Mathematical solution

1. To find a particular solution, we try a function of the form $q=c_0{e}^{j\omega t}$ which means that

   $\frac{dq}{dt}=j\omega c_0{e}^{j\omega t}$ .

   Substituting into $R\frac{dq}{dt}+\frac{q}{C}=v_i{e}^{j\omega t}$ we get

   $Rj\omega c_0{e}^{j\omega t}+\frac{c_0{e}^{j\omega t}}{C}=v_i{e}^{j\omega t}\Rightarrow$ $Rj\omega c_0+\frac{c_0}{C}=v_i$

   $\Rightarrow$ $c_0=\frac{v_i}{Rj\omega +\frac{1}{C}}=\frac{C{v}_i}{RCj\omega +1}$ $\Rightarrow$ $q=\frac{C{v}_i}{RCj\omega +1}{e}^{j\omega t}$

   Thus

   1. $v_c=\frac{q}{C}=\frac{v_i}{RCj\omega +1}{e}^{j\omega t}$ and
   2. $v_R=\frac{dq}{dt}=\frac{RC{v}_{i}j\omega }{RCj\omega +1}{e}^{j\omega t}$
2. We use the impedance to determine the voltage across each of the elements. The applied voltage is a single frequency of angular frequency $\omega$ and magnitude $v_i$ such that $V=v_i{e}^{j\omega t}$ .

   For an RC circuit, the impedance of the circuit is $Z=Z_R+Z_c$ where $Z_R$ is the impedance of the resistor $R$ and $Z_c$ is the impedance of the capacitor $Z_c=-\frac{j}{\omega C}$ .

   Therefore $Z=R-\frac{j}{\omega C}$ .

   The current can be found using $v=Zi$ giving

   $v_i{e}^{j\omega t}=\left(R-\frac{j}{\omega C}\right)i\Rightarrow i=\frac{v_i{e}^{j\omega t}}{R-\frac{j}{\omega C}}$

   We can now use $v_c=z_{c}i$ and $v_R=z_{R}i$ giving

   1. $v_c=\frac{q}{C}=-\frac{j}{\omega C}\times \frac{v_i}{R-\frac{j}{\omega C}}{e}^{j\omega t}=\frac{v_i}{RCj\omega +1}{e}^{j\omega t}$
   2. $v_R=\frac{R{v}_i}{R-\frac{j}{\omega C}}{e}^{j\omega t}=\frac{RC{v}_{i}j\omega }{RCj\omega +1}{e}^{j\omega t}$

      which confirms the result in part (a) found by solving the differential equation.

3. When $R=1000\Omega$ and $C=10^{-6}F$

   $v_c=\frac{v_i}{RCj\omega +1}{e}^{j\omega t}=\frac{v_i}{10^{-3}j\omega +1}{e}^{j\omega t}$

   So $\frac{\left|v_c\right|}{\left|v_i\right|}=\left|\frac{1}{10^{-3}j\omega +1}\right|\left|e^{j\omega t}\right|=\left|\frac{1}{10^{-3}j\omega +1}\right|=\frac{1}{\sqrt{10^{-6}{\omega }^2+1}}$

   Table 1 : Values of $\left|\frac{v_c}{v_i}\right|$ for a range of values of $\omega$

   $\omega$	10	$10^2$	$10^3$	$10^4$	$10^5$	$10^6$
   $\frac{\left|v_c\right|}{\left|v_i\right|}$	0.99995	0.995	0.707	0.00995	0.0099995	0.001

4. Table 1 shows that a RC circuit can be used as a high-cut filter because for low values of $\omega$ , $\frac{\left|v_c\right|}{\left|v_i\right|}$ is approximately 1 and for high values of $\omega$ , $\frac{\left|v_c\right|}{\left|v_i\right|}$ is approximately 0. So the circuit will filter out high frequency values.

   $\frac{\left|v_c\right|}{\left|v_i\right|}=\frac{1}{\sqrt{2}}$ when $\frac{1}{\sqrt{10^{-6}{\omega }^2+1}}=\frac{1}{\sqrt{2}}\iff 10^{-6}{\omega }^2+1=2\iff 10^{-6}{\omega }^2=1\iff {\omega }^2=10^6$

   As we are considering $\omega$ to be a positive frequency, $\omega =1000.$

   So $f_{hc}=\frac{{\omega }_{hc}}{2\pi }=\frac{1000}{2\pi }\approx 159$ Hz.

Interpretation

We have shown that for an RC circuit finding the steady state solution of the differential equation with a single frequency input voltage yields the same result for $\frac{\left|v_c\right|}{\left|v_i\right|}$ and $\frac{\left|v_R\right|}{\left|v_i\right|}$ as found by working with the complex impedances for the circuit.

An RC circuit can be used as a high-cut filter and in the case where $R=1k\Omega ,C=1\mu F$ we found the high-cut frequency to be at approximately 159 Hz.

This means that the circuit will pass frequencies less than this value and remove frequencies greater than this value.

Exercises

1. Solve the equation $x^2\frac{dy}{dx}+xy=1$ .
2. Find the solution of the equation $x\frac{dy}{dx}-y=x$ subject to the condition $y\left(1\right)=2$ .
3. Find the general solution of the equation $\frac{dy}{dt}+\left(tant\right)y=cost$ .
4. Solve the equation $\frac{dy}{dt}+\left(cott\right)y=sint$ .
5. The temperature $\theta$ (measured in degrees) of a body immersed in an atmosphere of varying

    temperature is given by $\frac{d\theta }{dt}+0.1\theta =5-2.5t$ . Find the temperature at time $t$ if $\theta =60^{\∘}$ C

    when $t=0$ .

6. In an LR circuit with applied voltage $E=10\left(1-{\text{e}}^{-0.1t}\right)$ the current $i$ is given by

   $L\frac{di}{dt}+Ri=10\left(1-{\text{e}}^{-0.1t}\right).$

    If the initial current is $i_0$ find $i$ subsequently.

Answer