### 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

$\phantom{\rule{2em}{0ex}}R\frac{dq}{dt}+\frac{q}{C}={v}_{i}{e}^{j\omega t}$

and hence find the magnitude of

1. the voltage across the capacitor ${v}_{c}=\frac{q}{C}$
2. the voltage across the resistor ${v}_{R}=R\phantom{\rule{1em}{0ex}}\frac{dq}{dt}$
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 $R=1\phantom{\rule{1em}{0ex}}k\Omega$ and $C=1\phantom{\rule{1em}{0ex}}\mu F,$ find the ratio $\frac{\left|{v}_{c}\right|}{\left|{v}_{i}\right|}$ and complete the table below

 $\omega$ 10 $1{0}^{2}$ $1{0}^{3}$ $1{0}^{4}$ $1{0}^{5}$ $1{0}^{6}$ $\frac{\left|{v}_{c}\right|}{\left|{v}_{i}\right|}$
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 $R\phantom{\rule{1em}{0ex}}\frac{dq}{dt}+\frac{q}{C}={v}_{i}{e}^{j\omega t}$ .

This will give us the steady state solution for the charge $q$ . Using this we can find ${v}_{c}=\frac{q}{C}$ and ${v}_{R}=R\phantom{\rule{1em}{0ex}}\frac{dq}{dt}$ . 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

$\phantom{\rule{2em}{0ex}}\frac{dq}{dt}=j\omega {c}_{0}{e}^{j\omega t}$ .

Substituting into $R\phantom{\rule{1em}{0ex}}\frac{dq}{dt}+\frac{q}{C}={v}_{i}{e}^{j\omega t}$ we get

$Rj\omega {c}_{0}{e}^{j\omega t}+\frac{{c}_{0}{e}^{j\omega t}}{C}={v}_{i}{e}^{j\omega t}\phantom{\rule{1em}{0ex}}⇒$ $Rj\omega {c}_{0}+\frac{{c}_{0}}{C}={v}_{i}$

$⇒$ ${c}_{0}=\frac{{v}_{i}}{Rj\omega +\frac{1}{C}}=\frac{C{v}_{i}}{RCj\omega +1}$ $⇒$ $q=\frac{C{v}_{i}}{RCj\omega +1}\phantom{\rule{1em}{0ex}}{e}^{j\omega t}$

Thus

1. ${v}_{c}=\frac{q}{C}=\frac{{v}_{i}}{RCj\omega +1}\phantom{\rule{1em}{0ex}}{e}^{j\omega t}$ and
2. ${v}_{R}=\frac{dq}{dt}=\frac{RC{v}_{i}j\omega }{RCj\omega +1}\phantom{\rule{1em}{0ex}}{e}^{j\omega t}$
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

${v}_{i}{e}^{j\omega t}=\left(R-\frac{j}{\omega C}\right)\phantom{\rule{1em}{0ex}}i\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}i=\frac{{v}_{i}{e}^{j\omega t}}{R-\frac{j}{\omega C}}$

We can now use ${v}_{c}={z}_{c}i$ and ${v}_{R}={z}_{R}i$ giving

1. ${v}_{c}=\frac{q}{C}=-\frac{j}{\omega C}×\frac{{v}_{i}}{R-\frac{j}{\omega C}}\phantom{\rule{1em}{0ex}}{e}^{j\omega t}=\frac{{v}_{i}}{RCj\omega +1}\phantom{\rule{1em}{0ex}}{e}^{j\omega t}$
2. ${v}_{R}=\frac{R{v}_{i}}{R-\frac{j}{\omega C}}\phantom{\rule{1em}{0ex}}{e}^{j\omega t}=\frac{RC{v}_{i}j\omega }{RCj\omega +1}\phantom{\rule{1em}{0ex}}{e}^{j\omega t}$

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

3. When $R=1000\phantom{\rule{1em}{0ex}}\Omega$ and $C=1{0}^{-6}F$

${v}_{c}=\frac{{v}_{i}}{RCj\omega +1}\phantom{\rule{1em}{0ex}}{e}^{j\omega t}=\frac{{v}_{i}}{1{0}^{-3}j\omega +1}\phantom{\rule{1em}{0ex}}{e}^{j\omega t}$

So $\frac{\left|{v}_{c}\right|}{\left|{v}_{i}\right|}=\left|\frac{1}{1{0}^{-3}j\omega +1}\right|\phantom{\rule{1em}{0ex}}\left|{e}^{j\omega t}\right|=\left|\frac{1}{1{0}^{-3}j\omega +1}\right|=\frac{1}{\sqrt{1{0}^{-6}{\omega }^{2}+1}}$

Table 1 : Values of $\left|\frac{{v}_{c}}{{v}_{i}}\right|$ for a range of values of $\omega$

 $\omega$ 10 $1{0}^{2}$ $1{0}^{3}$ $1{0}^{4}$ $1{0}^{5}$ $1{0}^{6}$ $\frac{\left|{v}_{c}\right|}{\left|{v}_{i}\right|}$ 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.

$\phantom{\rule{2em}{0ex}}\frac{\left|{v}_{c}\right|}{\left|{v}_{i}\right|}=\frac{1}{\sqrt{2}}$ when $\frac{1}{\sqrt{1{0}^{-6}{\omega }^{2}+1}}=\frac{1}{\sqrt{2}}\phantom{\rule{1em}{0ex}}⇔\phantom{\rule{1em}{0ex}}1{0}^{-6}{\omega }^{2}+1=2⇔1{0}^{-6}{\omega }^{2}=1⇔{\omega }^{2}=1{0}^{6}$

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 $R=1\phantom{\rule{1em}{0ex}}k\Omega ,C=1\phantom{\rule{1em}{0ex}}\mu F$ 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 ${x}^{2}\frac{dy}{dx}+x\phantom{\rule{1em}{0ex}}y=1$ .
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 $\frac{dy}{dt}+\left(tant\right)\phantom{\rule{1em}{0ex}}y=cost$ .
4. Solve the equation $\frac{dy}{dt}+\left(cott\right)\phantom{\rule{1em}{0ex}}y=sint$ .
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 $\theta =6{0}^{∘}$ 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

$\phantom{\rule{2em}{0ex}}L\frac{di}{dt}+Ri=10\left(1-{\text{e}}^{-0.1t}\right).$

If the initial current is ${i}_{0}$ find $i$ subsequently.

1. $y=\frac{1}{x}lnx+\frac{C}{x}$
2. $y=xlnx+2x$
3. $y=\left(t+C\right)cost$
4. $y=\left(\frac{1}{2}\phantom{\rule{1em}{0ex}}t-\frac{1}{4}sin2t+C\right)\phantom{\rule{0.3em}{0ex}}$ cosec $\phantom{\rule{0.3em}{0ex}}t$
5. $\theta =300-25t-240{\text{e}}^{-0.1t}$
6. $i=\frac{10}{R}-\left(\frac{100}{10R-L}\right){\text{e}}^{-0.1t}+\left[{i}_{0}+\frac{10L}{R\left(10R-L\right)}\phantom{\rule{1em}{0ex}}\right]{\text{e}}^{-Rt∕L}$