In Fig. 15.7, ZS=100Ω and ZL=R+jωL, with R=10Ω and L=80μH. The available voltage is vS=VScos(ω1t), with VS=750mV and f1=20kHz. The voltage transfer function of the circuit is Hν= K//1+jf/f0, with K =100 and f0=10kHz . Use the voltage gain and phase shift to find the load voltage νL(t)

Question

In Fig. 15.7, [latex] Z_S=100 \Omega[/latex] and [latex]Z_L=R+j \omega L[/latex], with [latex]R=10 \Omega[/latex] and [latex]L=80 \mu H[/latex]. The available voltage is [latex]ν_S=V_S \cos \left(\omega_1 t ight)[/latex], with [latex]V_S=750 mV[/latex] and [latex]f_1=20 kHz[/latex]. The voltage transfer function of the circuit is

[latex] H_ u=\frac{K}{1+j f / f_0}, [/latex]

with K =100 and [latex] f_0=10 kHz [/latex] . Use the voltage gain and phase shift to find the load voltage [latex] u_L(t) [/latex]

Accepted Answer

The voltage gain and phase shift at the frequency of the source are

[latex]\begin{gathered} A_ν=\frac{K}{\sqrt{1+\left(f_1 / f_0 ight)^2}}=\frac{100}{\sqrt{5}} \cong 44.72, \\ \phi_ν=- an ^{-1}\left(\frac{f_1}{f_0} ight)=-1.107. \end{gathered} [/latex]

From (15.17), the load voltage is given by

[latex] \begin{aligned} ν_L &=A_ν V_S \cos \left[2 \pi f_1 t+\phi_ν ight] \\ &=(44.72)(750 mV ) \cos \left(2 \pi f_1 t+1.107 ight) \\ &=33.54 \cos \left(2 \pi f_1 t-1.107 ight) V , f_1=20 kHz \end{aligned} [/latex]

Question 15.4: In Fig. 15.7, ZS=100Ω and ZL=R+jωL, with R=10Ω and L=80μ...

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