Let's solve the problem step by step. --- ## Given: The filter's transfer function is: \[ H(j\omega) = \frac{17j\omega}{15 + 8j\omega} \] ### (a) Is this filter Low Pass or High Pass? #### **Step 1: Analyze the Transfer Function** - For low frequencies (\(\omega \to \)): \[ H(j\omega) \approx \frac{17j \cdot }{15 + } = \] (Output is close to zero at low frequencies.) - For high frequencies (\(\omega \to \infty\)): \[ H(j\omega) \approx \frac{17j\omega}{8j\omega} = \frac{17}{8} \] (Output approaches a constant at high frequencies.) #### **Conclusion:** - **This filter is a High Pass filter**, as it attenuates low frequencies and passes high frequencies. --- ### (b) If an input \(u(t) = 2\sin(3t) - 3\sin(14t)\), what is the output? #### **Step 1: Express the input as a sum of sinusoids** \[ u(t) = 2\sin(3t) - 3\sin(14t) \] #### **Step 2: Find the frequency response for each frequency** - For \(\sin(3t)\), \(\omega_1 = 3\) - For \(\sin(14t)\), \(\omega_2 = 14\) #### **Step 3: Find output for each frequency using \(H(j\omega)\)** For a sinusoidal input \(A\sin(\omega t)\), the output is \(|H(j\omega)|A \sin(\omega t + \arg H(j\omega))\). --- #### **For \(\omega_1 = 3\):** \[ H(j3) = \frac{17j \cdot 3}{15 + 8j \cdot 3} = \frac{51j}{15 + 24j} \] - Convert denominator to polar: \(15 + 24j\) - Magnitude: \(\sqrt{15^2 + 24^2} = \sqrt{225 + 576} = \sqrt{801}\) - Phase: \(\tan^{-1}(24/15)\) So, - Magnitude: \(|H(j3)| = \frac{51}{\sqrt{801}} \approx 1.804\) - Phase: - Numerator: \(+90^\circ\) - Denominator: \(\tan^{-1}(24/15) \approx 58^\circ\) - Total phase: \(90^\circ - 58^\circ = 32^\circ\) So, \[ \text{Output for } 2\sin(3t):\ 2 \times 1.804 \sin(3t + 32^\circ) \approx 3.608\sin(3t + 32^\circ) \] --- #### **For \(\omega_2 = 14\):** \[ H(j14) = \frac{17j \cdot 14}{15 + 8j \cdot 14} = \frac{238j}{15 + 112j} \] - Magnitude: \(\sqrt{15^2 + 112^2} = \sqrt{225 + 12544} = \sqrt{12769} = 113\) - \(|H(j14)| = \frac{238}{113} \approx 2.106\) - Phase: Numerator \(+90^\circ\), denominator \(\tan^{-1}(112/15) \approx 82.4^\circ\), so phase difference \(90^\circ - 82.4^\circ = 7.6^\circ\) So, \[ \text{Output for } -3\sin(14t):\ -3 \times 2.106 \sin(14t + 7.6^\circ) \approx -6.318\sin(14t + 7.6^\circ) \] --- #### **Total Output:** \[ y(t) = 3.61\sin(3t + 32^\circ) - 6.32\sin(14t + 7.6^\circ) \] --- ### (c) What is the cut-off frequency? The cutoff frequency (\(\omega_c\)) is where the magnitude drops to \(1/\sqrt{2}\) (about .707) of its maximum. \[ |H(j\omega)| = \frac{17\omega}{\sqrt{15^2 + (8\omega)^2}} \] Set this equal to the maximum value (\(\frac{17}{8}\)) times \(\frac{1}{\sqrt{2}}\): \[ \frac{17\omega_c}{\sqrt{225 + 64\omega_c^2}} = \frac{17}{8} \cdot \frac{1}{\sqrt{2}} = \frac{17}{8\sqrt{2}} \] Solving for \(\omega_c\): \[ \frac{\omega_c}{\sqrt{225 + 64\omega_c^2}} = \frac{1}{8\sqrt{2}} \] \[ \omega_c = \frac{\sqrt{225 + 64\omega_c^2}}{8\sqrt{2}} \] \[ 8\sqrt{2} \omega_c = \sqrt{225 + 64\omega_c^2} \] \[ (8\sqrt{2} \omega_c)^2 = 225 + 64\omega_c^2 \] \[ 128 \omega_c^2 = 225 + 64\omega_c^2 \] \[ 128 \omega_c^2 - 64\omega_c^2 = 225 \] \[ 64\omega_c^2 = 225 \] \[ \omega_c^2 = \frac{225}{64} \] \[ \omega_c = \frac{15}{8} \] --- ## **Summary of Answers** **a.** High Pass Filter **b.** Output: \[ y(t) \approx 3.61\sin(3t + 32^\circ) - 6.32\sin(14t + 7.6^\circ) \] **c.** Cutoff frequency: \[ \omega_c = \frac{15}{8} \approx 1.875\ \text{rad/s} \]

# Step-by-Step Solution ## Given: The filter's transfer function is: \[ H(j\omega) = \frac{17j\omega}{15 + 8j\omega} \] ## (a) Is this filter Low Pass or High Pass? ### Step 1: Analyze the Transfer Function - **Low Frequencies (\(\omega \to 0\))**: \[ H(j\omega) \approx \frac{17j \cdot 0}{15 + 8j \cdot 0} = 0 \] (Output is close to zero at low frequencies.) - **High Frequencies (\(\omega \to \infty\))**: \[ H(j\omega) \approx \frac{17j\omega}{8j\omega} = \frac{17}{8} \] (Output approaches a constant at high frequencies.) ### Conclusion: - **This filter is a High Pass filter**, as it attenuates low frequencies and passes high frequencies. --- ## (b) If an input \(u(t) = 2\sin(3t) - 3\sin(14t)\), what is the output? ### Step 1: Express the Input \[ u(t) = 2\sin(3t) - 3\sin(14t) \] ### Step 2: Find Frequency Response for Each Frequency - For \(\sin(3t)\), \(\omega_1 = 3\) - For \(\sin(14t)\), \(\omega_2 = 14\) ### Step 3: Find Output for Each Frequency Using \(H(j\omega)\) For a sinusoidal input \(A\sin(\omega t)\), the output is: \[ y(t) = |H(j\omega)| A \sin(\omega t + \arg H(j\omega)) \] --- ### For \(\omega_1 = 3\): \[ H(j3) = \frac{17j \cdot 3}{15 + 8j \cdot 3} = \frac{51j}{15 + 24j} \] - **Magnitude**: \[ |H(j3)| = \frac{51}{\sqrt{15^2 + 24^2}} = \frac{51}{\sqrt{801}} \approx 1.804 \] - **Phase**: - Numerator: \(+90^\circ\) - Denominator: \(\tan^{-1}(24/15) \approx 58^\circ\) - Total phase: \(90^\circ - 58^\circ = 32^\circ\) So, the output for \(2\sin(3t)\): \[ y_1(t) \approx 2 \times 1.804 \sin(3t + 32^\circ) \approx 3.608\sin(3t + 32^\circ) \] --- ### For \(\omega_2 = 14\): \[ H(j14) = \frac{17j \cdot 14}{15 + 8j \cdot 14} = \frac{238j}{15 + 112j} \] - **Magnitude**: \[ |H(j14)| = \frac{238}{\sqrt{15^2 + 112^2}} = \frac{238}{113} \approx 2.106 \] - **Phase**: - Numerator: \(+90^\circ\) - Denominator: \(\tan^{-1}(112/15) \approx 82.4^\circ\) - Total phase: \(90^\circ - 82.4^\circ = 7.6^\circ\) So, the output for \(-3\sin(14t)\): \[ y_2(t) \approx -3 \times 2.106 \sin(14t + 7.6^\circ) \approx -6.318\sin(14t + 7.6^\circ) \] --- ### Total Output: Combining both outputs: \[ y(t) \approx 3.608\sin(3t + 32^\circ) - 6.318\sin(14t + 7.6^\circ) \] --- ## (c) What is the Cut-off Frequency? The cutoff frequency (\(\omega_c\)) is where the magnitude drops to \(1/\sqrt{2}\) (about 0.707) of its maximum value. ### Step 1: Set Up the Equation The magnitude of the transfer function is: \[ |H(j\omega)| = \frac{17\omega}{\sqrt{15^2 + (8\omega)^2}} \] Setting this equal to the cutoff condition: \[ \frac{17\omega_c}{\sqrt{225 + 64\omega_c^2}} = \frac{17}{8\sqrt{2}} \] ### Step 2: Solve for \(\omega_c\) Cross-multiplying and squaring both sides leads to: \[ (8\sqrt{2} \omega_c)^2 = 225 + 64\omega_c^2 \] \[ 128 \omega_c^2 = 225 + 64\omega_c^2 \] \[ 64\omega_c^2 = 225 \] \[ \omega_c^2 = \frac{225}{64} \] \[ \omega_c = \frac{15}{8} \] --- ## Summary of Answers **a.** High Pass Filter **b.** Output: \[ y(t) \approx 3.608\sin(3t + 32^\circ) - 6.318\sin(14t + 7.6^\circ) \] **c.** Cutoff frequency: \[ \omega_c = \frac{15}{8} \approx 1.875\ \text{rad/s} \]