Practice Problems (KEY)

Practice Problems (KEY)#

Problem 1.#

Calculate the cutoff frequency for a circuit with:

a. \(R = 1\ \text{k}\Omega\) and \(C = 3\ \text{nF}\)

Start with:

\[ f_{cutoff}=\frac{1}{2\pi RC} \]

Substitute values:

\[ f_{cutoff}=\frac{1}{2\pi(1000)(3\times10^{-9})} \]

Compute:

  • \(RC = (1000)(3\times10^{-9}) = 3\times10^{-6}\)

  • \(2\pi RC = 2\pi(3\times10^{-6}) \approx 1.884\times10^{-5}\)

So:

\[ f_{cutoff}\approx \frac{1}{1.884\times10^{-5}} \approx 5.31\times10^{4}\,\text{Hz}=53\,\text{kHz} \]

b. \(R = 5\ \text{k}\Omega\) and \(C = 1.5\ \mu\text{F}\)

Start with:

\[ f_{cutoff}=\frac{1}{2\pi RC} \]

Substitute values:

\[ f_{cutoff}=\frac{1}{2\pi(5000)(1.5\times10^{-6})} \]

Compute:

  • \(RC = (5000)(1.5\times10^{-6}) = 7.5\times10^{-3}\)

  • \(2\pi RC = 2\pi(7.5\times10^{-3}) \approx 4.712\times10^{-2}\)

So:

\[ f_{cutoff}\approx \frac{1}{4.712\times10^{-2}} \approx 21.22\,\text{Hz} \]

Problem 2.#

What should the resistor value be for the following filters:

a. \(f_{cutoff} = 1.5\ \text{kHz}\) and \(C = 500\ \text{nF}\)

\[ R=\frac{1}{2\pi(1500)(500\times10^{-9})} \]

Compute:

  • \(fC = (1500)(500\times10^{-9}) = 7.5\times10^{-4}\)

  • \(2\pi fC \approx 2\pi(7.5\times10^{-4}) \approx 4.712\times10^{-3}\)

So:

\[ R\approx \frac{1}{4.712\times10^{-3}} \approx 212\,\Omega \]

b. \(f_{cutoff} = 417\ \text{Hz}\) and \(C = 56\ \text{nF}\)

\[ R=\frac{1}{2\pi(417)(56\times10^{-9})} \]

Compute:

  • \(fC = (417)(56\times10^{-9}) = 2.3352\times10^{-5}\)

  • \(2\pi fC \approx 2\pi(2.3352\times10^{-5}) \approx 1.467\times10^{-4}\)

So:

\[ R\approx \frac{1}{1.467\times10^{-4}} \approx 6.8\times10^{3}\,\Omega=6.8\,\text{k}\Omega \]

c. \(f_{cutoff} = 2\ \text{kHz}\) and \(C = 500\ \mu\text{F}\)

\[ R=\frac{1}{2\pi(2000)(500\times10^{-6})} \]

Compute:

  • \(fC = (2000)(500\times10^{-6}) = 1000\times10^{-3} = 1\)

  • \(2\pi fC = 2\pi(1)=2\pi\)

So:

\[ R=\frac{1}{2\pi}\approx 0.159\,\Omega=159\,\text{m}\Omega \]

Problem 3.#

Are the circuits below high or low pass filters? How do you know?

Key idea#

Evaluate the transfer function at:

  • \(f=0\) (DC / very low frequency)

  • \(f\rightarrow\infty\) (very high frequency)

If gain is near 1 at low frequency and near 0 at high frequency → LPF.
If gain is near 0 at low frequency and near 1 at high frequency → HPF.

3(a)#

Given:

\[ \frac{v_0(t)}{v_s(t)}=\frac{1}{1+\frac{R}{j2\pi f L}} \]

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At \(f=0\): the term \(\frac{R}{j2\pi f L}\rightarrow\infty\), so denominator \(\rightarrow\infty\):

\[ \frac{v_0}{v_s}\rightarrow 0 \]

At \(f\rightarrow\infty\): the term \(\frac{R}{j2\pi f L}\rightarrow 0\), so denominator \(\rightarrow 1\):

\[ \frac{v_0}{v_s}\rightarrow 1 \]

Conclusion: High Pass Filter (HPF).

3(b)#

Given:

\[ \frac{v_0(t)}{v_s(t)}=\frac{R}{j2\pi f L + R} \]

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At \(f=0\):

\[ \frac{v_0}{v_s}=\frac{R}{0+R}=1 \]

At \(f\rightarrow\infty\): \(j2\pi f L\) dominates, so magnitude grows large and:

\[ \frac{v_0}{v_s}\rightarrow 0 \]

Conclusion: Low Pass Filter (LPF).

3(c)#

Given:

\[ \frac{v_0(t)}{v_s(t)}=\frac{1}{j2\pi f C R + 1} \]

At \(f=0\):

\[ \frac{v_0}{v_s}=\frac{1}{0+1}=1 \]

At \(f\rightarrow\infty\): \(j2\pi fCR\) dominates, so:

\[ \frac{v_0}{v_s}\rightarrow 0 \]

Conclusion: Low Pass Filter (LPF).

3(d)#

Given:

\[ \frac{v_0(t)}{v_s(t)}=\frac{1}{1+\frac{1}{j2\pi fRC}} \]

At \(f=0\): the term \(\frac{1}{j2\pi fRC}\rightarrow\infty\), so denominator \(\rightarrow\infty\):

\[ \frac{v_0}{v_s}\rightarrow 0 \]

At \(f\rightarrow\infty\): the term \(\frac{1}{j2\pi fRC}\rightarrow 0\), so denominator \(\rightarrow 1\):

\[ \frac{v_0}{v_s}\rightarrow 1 \]

Conclusion: High Pass Filter (HPF).

Problem 4.#

Calculate the cutoff frequency of the following systems.

a. A transmission line modeled as an R-L circuit with \(R = 4\ \Omega\) and \(L = 5\ \mu\text{H}\).

Use:

\[ f_{cutoff}=\frac{R}{2\pi L} \]

Substitute:

\[ f_{cutoff}=\frac{4}{2\pi(5\times10^{-6})} \]

Compute:

  • \(2\pi L = 2\pi(5\times10^{-6}) \approx 3.1416\times10^{-5}\)

So:

\[ f_{cutoff}\approx \frac{4}{3.1416\times10^{-5}}\approx 1.27\times10^{5}\,\text{Hz}=127\,\text{kHz} \]

b. An R-C low pass filter with \(R = 60\ \Omega\) and \(C = 5\ \text{nF}\).

\[ f_{cutoff}=\frac{1}{2\pi RC}=\frac{1}{2\pi(60)(5\times10^{-9})} \]

Compute:

  • \(RC=(60)(5\times10^{-9})=3\times10^{-7}\)

  • \(2\pi RC \approx 2\pi(3\times10^{-7}) \approx 1.885\times10^{-6}\)

So:

\[ f_{cutoff}\approx \frac{1}{1.885\times10^{-6}}\approx 5.31\times10^{5}\,\text{Hz}=531\,\text{kHz} \]

c. A C-R high pass filter with \(R = 100\ \Omega\) and \(C = 8\ \mu\text{F}\).

\[ f_{cutoff}=\frac{1}{2\pi RC}=\frac{1}{2\pi(100)(8\times10^{-6})} \]

Compute:

  • \(RC=(100)(8\times10^{-6})=8\times10^{-4}\)

  • \(2\pi RC \approx 2\pi(8\times10^{-4}) \approx 5.027\times10^{-3}\)

So:

\[ f_{cutoff}\approx \frac{1}{5.027\times10^{-3}}\approx 199\,\text{Hz} \]

Problem 5.#

Your communications radio has a lower frequency bound of \(800\ \text{kHz}\). You know it has a capacitor value of \(100\ \text{nF}\), but what is the resistor value?

Rearrange:

\[ R=\frac{1}{2\pi f_{cutoff}C} \]

Substitute:

\[ R=\frac{1}{2\pi(800000)(100\times10^{-9})} \]

Compute:

  • \(fC=(800000)(100\times10^{-9})=0.08\)

  • \(2\pi fC = 2\pi(0.08)\approx 0.50265\)

So:

\[ R\approx \frac{1}{0.50265}\approx 1.99\,\Omega \]

Problem 6.#

Design a high pass filter to get rid of a DC bias (\(0\ \text{Hz}\)) using a \(100\ \Omega\) resistor you have available.

DC bias corresponds to \(f=0\) Hz, so we choose a practical cutoff (given): \(f_{cutoff}=10\) Hz.

Use:

\[ f_{cutoff}=\frac{1}{2\pi RC} \]

Solve for \(C\):

\[ C=\frac{1}{2\pi f_{cutoff}R} \]

Substitute:

\[ C=\frac{1}{2\pi(10)(100)} \]

Compute:

  • \(2\pi(10)(100)=2000\pi\approx 6283.19\)

So:

\[ C\approx \frac{1}{6283.19}\approx 1.59\times10^{-4}\,\text{F}=159\,\mu\text{F} \]

Problem 7.#

For the circuit below, what is the magnitude of the gain, \(\left|\frac{v_{o}}{v_{in}}\right|\), at \(60\ \text{Hz}\)?

Given transfer function:

\[ \frac{v_0(t)}{v_s(t)}=\frac{1}{1+\frac{1}{j2\pi fRC}} \]

Substitute the provided values (as shown in the key):

\[ \frac{v_0(t)}{v_s(t)}= \frac{1}{1+\frac{1}{j2\pi(60)(5)(0.0015)}} \]

The evaluated complex result is:

\[ 0.889 + j0.314 = 0.943\angle 19.5^\circ \]

Magnitude is:

\[ \left|\frac{v_0}{v_{in}}\right| = 0.942 \]

Interpretation:

  • The filter passes \(94.2\%\) of the input at 60 Hz, so 60 Hz is in the passband.

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