Handwrite and step by step solutions A cross-over circuit takes a single input signal and splits it into two or more output signals at different frequencies or frequency ranges. One example is shown in the circuit below, which represents a cross-over circuit for a sound system, where \( v_1(t) \) and \( v_2(t) \) represent speakers such as a tweeter or a woofer each with an impedance of \( Z = 6\Omega \).1. Find the transfer function of each output \( v_1(t) \) and \( v_2(t) \), and determine which one acts as the tweeter (high frequencies) and woofer (low frequencies).2. Identify the values of \( L \) and \( C \) such that the cross-over frequency (frequency at which the two transfer functions cross) is equal to 3 kHz. Plot the system gain for each output using the values you identified for \( L \) and \( C \).**Circuit Explanation:**The circuit schematic consists of:- An input voltage source \( v_i(t) \).- A series inductor \( L \) connected between the input and the first output \( v_1(t) \).- A parallel capacitor \( C \) connected to the second output \( v_2(t) \).- Two load impedances, both \( 6\Omega \), connected across \( v_1(t) \) and \( v_2(t) \).This configuration is a simple LC filter network designed to separate audio frequencies for a tweeter and woofer in a sound system.

Given data: Input supply voltage, vit The voltage across the resistor which is in series with the inductor is vo1t. The voltage across the resistor which is in series with a capacitor is vo2t. The resistance value is, R=6 Ω. The frequency at which both the transfer functions are equal, f=3 kHz. We have to find the transfer of each output and to determine which one acts as a tweeter and woofer. We have to also find out the values of L and C so that the crossover frequency is equal to 3 kHz. 1. Calculation of the impedance Consider the circuit drawn below; Let us assume that the branch-1 impedance is Z1 and branch-two impedance is Z2. Z1=jωL+6 Here ω is the angular frequency and L is the inductance. Z2=jωC+6=6-jωCCalculation of the current through each branch The current through the branch-1 is, I1=VinZ1 Substitute the above-calculated values I1=Vin6+jωL The current through the branch-2 is, I2=VinZ2 Substitute the above-calculated values I2=Vin6-jωCDetermination of the expression of the transfer function for the branch-1 The expression for the voltage across the resistor of branch-1 v01t=I1×6 Substitute the above-calculated values of I1 v01t=Vin6+jωL×6v01tVin=66+jωL When ω→0, v01tVin=1 When ω→∞, v01tVin=0 As above transfer allows only low values of ω. So this transfer function is LPF (Low pass filter, woofer.) Determination of the expression of the transfer function for the branch-2 The expression for the voltage across the resistor of branch-1 v02t=I2×6 Substitute the above-calculated values of I1 v02t=Vin6-jωC×6v02tVin=66-jωC When ω→0, v02tVin=2 When ω→∞, v02tVin=1 As above transfer allows only high values of ω. So this transfer function is HPF (High pass filter, Tweeter.) 2.Determination of the value of L and C The point at which transfer functions are equal (magnitudes are equal), v01tVin=v02tVin Substitute the above-calculated values of transfer function, 66+jωL=66-jωC 62+ωL2=62+1ωC2 On solving the above equation, ωL=1ωCω=LC2πf=LC squaring both sides, LC=2πf2 substitute f=3 kHz LC=2π×3×1032=3.5×108 Form the above ratio, C=10 pF and L=3.5 mH Therefore the value of capacitance and inductance are, C=10 pF and L=3.5 mHDetermination of cross-over frequency for both the transfer function The time constant for the first transfer function is, τ1=LR1 So, the expression for the frequency of first transfer function is, f1=1τ1=RL Substitute L=3.5 mH and R=6 Ω f1=63.5×10-3=1714.28 Hz Therefore the frequency of the first transfer function is, f1=1714.28 Hz. The time constant for the second transfer function is, τ2=RC So, the expression for the frequency of first transfer function is, f1=1τ1=1RC Substitute C=10 pF and R=6 Ω f2=16×10×10-12=1.66×1010 Hz Therefore the frequency of the second transfer function is, f2=1.66×1010 Hz.The plot of system gain for each transfer function is shown below, The plot of system gain for the first transfer function is shown below, The plot of system gain for the second transfer function is shown below,