step by step solutions, and handwrite ### Transcription and Analysis of CircuitsThe image presents four different electrical circuits labeled (a), (b), (c), and (d). Each circuit consists of a voltage source \( v_i = V_m e^{j\omega t} \) with different values for \( V_m \), a resistor \( R \), and either a capacitor \( C \) or an inductor \( L \). The task is to write expressions for the transfer function \( H(j\omega) = V_o/V_i \), its magnitude \(|H(j\omega)|\), and its phase \(\angle H(j\omega)\) as a function of frequency \(\omega\). Also, for each circuit, sketch \(|H(j\omega)|\) on a log-log scale and identify the type of filter it represents.#### Circuit Details:1. **Circuit (a):** - **Components:** Resistor \( R \) and Capacitor \( C \). - **Input Voltage:** \( v_i = e^{j\omega t} \).2. **Circuit (b):** - **Components:** Resistor \( R \) and Inductor \( L \). - **Input Voltage:** \( v_i = 2 e^{j\omega t} \).3. **Circuit (c):** - **Components:** Resistor \( R = 10^6 \, \Omega \) and Capacitor \( C = 4 \mu F \). - **Input Voltage:** \( v_i = 5 e^{j\omega t} \).4. **Circuit (d):** - **Components:** Resistor \( R = 10 \, \Omega \) and Inductor \( L = 2 \, H \). - **Input Voltage:** \( v_i = 10 e^{j\omega t} \).### Analysis:#### Circuit (a): RC Low-Pass Filter- **Transfer Function:** \( H(j\omega) = \frac{1}{1 + j\omega RC} \).- **Magnitude:** \(|H(j\omega)| = \frac{1}{\sqrt{1 + (\omega RC)^2}} \).- **Phase:** \(\angle H(j\omega) = -\tan^{-1}(\omega RC)\).- **Type:** Low-pass filter. As \(\omega\) increases, \(|H

Given data: The circuit are given in which resistance, inductance and the capacitances are connected into the different manner. The value of the capacitor in circuit (c),C=4 μF The value of the resistance in circuit (c), R=106 Ω The value of the inductance in circuit (d),L=2 H The value of the resistance in circuit (d), R=10 Ω Part (a) Consider the circuit shown below. Calculate the transfer function of the above circuit. V0=Vi×XcR+XcV0=Vi×-jωCR-jωCV0=Vi×-jωCωRC-jωCV0=Vi×-jωRC-j Solve further as, V0=Vi×-jωRC-j×jjV0=Vi×1jωRC+1V0Vi=11+jωRCHω=11+jωRC Calculate the magnitude of the Hω. Hω=11+ω2R2C2 Calculate the phase of Hω. ∠Hω=-tan-1ωRC1∠Hω=-tan-1ωRC The log-log plot of the Hω is shown below. Since the filter is passes only lower frequencies. Therefore, it is low pass filter. Hence the expression, magnitude and phase of Hω are 11+jωRC, 11+ω2R2C2 and -tan-1ωRC respectively. The filter is low pass filter and log-log plot is obtained. Part (b) Consider the circuit shown below. Calculate the transfer function of the above circuit. V0=Vi×XLR+XLV0=Vi×jωLR+jωLV0Vi=jωLR+jωLHω=jωLR+jωL Calculate the magnitude of the Hω. Hω=ω2L2R2+ω2L2Hω=ωLR2+ω2L2 Calculate the phase of Hω. ∠Hω=90∘-tan-1ωLR The log-log plot of the Hω is shown below. Since the transfer function of the system becomes equal to 1 at ω=∞, therefore, it passes only high frequency and it is high pass filter. Hence the expression, magnitude and phase of Hω are jωLR+jωL, ωLR2+ω2L2 and 90∘-tan-1ωLR respectively. The filter is high pass filter and log-log plot is obtained. Part (c) Consider the circuit shown below. Calculate the transfer function of the above circuit. V0=Vi×RR+XcV0=Vi×RR-jωCV0=Vi×ωRCωRC-1V0Vi=ωRCωRC-1 substitute the values into the above equation and solve as, Hω=ω×106×4×10-6ω×106×4×10-6-j=4ω4ω-j Calculate the magnitude of the Hω. Hω=4ω1+4ω2Hω=4ω1+16ω2 Calculate the phase of Hω. ∠Hω=-tan-1-14ω The log-log plot of the Hω is shown below. Since the transfer function of the system becomes equal to 1 at ω=∞, therefore, it passes only high frequency and it is high pass filter. Hence the expression, magnitude and phase of Hω are 4ω4ω-j, 4ω1+16ω2 and -tan-1-14ω respectively. The filter is high pass filter and log-log plot is obtained. Part (d) Consider the circuit shown below. Calculate the transfer function of the above circuit. V0=Vi×RR+XLV0=Vi×RR+jωLV0Vi=RR+jωLHω=RR+jωL Substitute the values and solve as, Hω=1010+j2ω Calculate the magnitude of the Hω. Hω=10102+ω2×22Hω=10100+4ω2 Calculate the phase of Hω. ∠Hω=-tan-1ωLR∠Hω=-tan-12ω10∠Hω=-tan-1ω5 The log-log plot of the Hω is shown below. Since the transfer function of the system becomes equal to 0 at ω=0, therefore, it passes only low frequency and it is low pass filter. Hence the expression, magnitude and phase of Hω are 1010+j2ω, 10100+4ω2 and -tan-1ω5 respectively. The filter is low pass filter and log-log plot is obtained. Conclusion: (a) The expression, magnitude and phase of Hω are 11+jωRC, 11+ω2R2C2 and -tan-1ωRC respectively. The filter is low pass filter and log-log plot is obtained. (b) The expression, magnitude and phase of Hω are jωLR+jωL, ωLR2+ω2L2 and 90∘-tan-1ωLR respectively. The filter is high pass filter and log-log plot is obtained. (c) The expression, magnitude and phase of Hω are 4ω4ω-j, 4ω1+16ω2 and -tan-1-14ω respectively. The filter is high pass filter and log-log plot is obtained. (d) The expression, magnitude and phase of Hω are 1010+j2ω, 10100+4ω2 and -tan-1ω5 respectively. The filter is low pass filter and log-log plot is obtained.