(a) Obtain expressions for the voltage transfer function, the current transfer function, the transimpedance, and the transadmittance of the circuit (amplifier) in Fig. 15.3. (b) Let ZS=100Ω,Zi=10kΩ,Zo=10Ω, ZL=5kΩ, and g= g0/1+jf/f0 ;g0=0.2 S,f0=2kHz Find the load voltage for

Question

(a) Obtain expressions for the voltage transfer function, the current transfer function, the transimpedance, and the transadmittance of the circuit (amplifier) in Fig. 15.3.

(b) Let [latex] Z_S=100 \Omega, Z_i=10 \mathrm{k} \Omega, Z_o=10 \Omega , Z_L=5 \mathrm{k} \Omega[/latex], and

[latex] g=\frac{g_0}{1+j f / f_0} ; g_0=0.2 \mathrm{~S}, f_0=2 \mathrm{kHz} [/latex]

Find the load voltage for

[latex] \begin{aligned} u_S=& V_0+V_1 \cos \left(2 \pi f_1 t ight) \\ &+V_2 \cos \left(2 \pi f_2 t+0.7 ight) \end{aligned}, [/latex]

where [latex]V_0=100 \mathrm{mV}, V_1=800 \mathrm{mV}, V_2=400 \mathrm{mV}, f_1=1 \mathrm{kHz}, f_2=5 \mathrm{kHz}[/latex]

Accepted Answer

(a) The load and the output subcircuit comprise a current divider, for which

[latex] \begin{aligned} ilde{I}_L &=\frac{Z_o g ilde{V}_{i n}}{Z_o+Z_L} \Rightarrow ilde{V}_L= ilde{I}_L Z_L \\ &=\frac{Z_L Z_o g ilde{V}_{i n}}{Z_o+Z_L} . (15.7) \end{aligned} [/latex]

The source and the input subcircuit comprise a voltage divider, for which

[latex] ilde{V}_{i n}=\frac{Z_i ilde{V}_S}{Z_i+Z_S} . [/latex] (15.8)

Using the right side of (15.8) to replace [latex] ilde{V}_{i n} [/latex] in the right side of (15.7) gives

[latex] ilde{V}_L=\frac{Z_i Z_o Z_L g ilde{V}_S}{\left(Z_o+Z_L ight)\left(Z_i+Z_S ight)} ext {. }[/latex]

From (15.3), the voltage transfer function is given by

[latex] H_ u=\frac{ ilde{V}_L}{ ilde{V}_S} [/latex] (15.3)

[latex] H_ u=\frac{Z_i Z_o Z_L g}{\left(Z_o+Z_L ight)\left(Z_i+Z_S ight)} ext {. }[/latex] (15.9)

From (15.4-15.6), we obtain the current transfer function

[latex] H_i=\frac{ ilde{I}_L}{ ilde{I}_S}=\frac{ ilde{V}_L / Z_L}{ ilde{V}_S / Z_S}=\frac{Z_S}{Z_L} H_ u, [/latex] (15.4)

[latex] H_y=\frac{ ilde{I}_L}{ ilde{V}_S}=\frac{ ilde{V}_L / Z_L}{ ilde{V}_S}=\frac{1}{Z_L} H_ u [/latex] (15.6)

[latex] H_i=\frac{Z_S}{Z_L} H_ u=\frac{Z_i Z_o Z_S g}{\left(Z_o+Z_L ight)\left(Z_i+Z_S ight)}, [/latex] (15.10)

the transimpedance

[latex] H_z=Z_S H_ u=\frac{Z_S Z_i Z_o Z_L g}{\left(Z_o+Z_L ight)\left(Z_i+Z_S ight)}, [/latex]

and the transadmittance

[latex] H_y=\frac{1}{Z_L} H_ u=\frac{Z_i Z_o g}{\left(Z_o+Z_L ight)\left(Z_i+Z_S ight)} . [/latex]

(b) We use superposition. From part (a), the voltage transfer function is

[latex]H_ u=\frac{ ilde{V}_L}{ ilde{V}_S}=\frac{Z_i Z_o Z_L g}{\left(Z_o+Z_L ight)\left(Z_i+Z_S ight)}, [/latex]

where (in this example) the impedances are all resistive. Only the intrinsic transconductance g depends upon the frequency of the excitation.

For the parameter values given in the problem statement,

[latex] \begin{aligned} H_ u &=\frac{Z_i Z_o Z_L g}{\left(Z_o+Z_L ight)\left(Z_i+Z_S ight)} \\ &=\frac{\left(10^4 \Omega ight)(10 \Omega)(5,000 \Omega)}{(5,010 \Omega)(10,100 \Omega)}\ \ & imes \frac{0.2 \mathrm{~S}}{1+j f / f_0}=\frac{1.976}{1+j f / f_0} ; f_0=2 \mathrm{kHz} . \end{aligned} [/latex]

For each sinusoidal component of the excitation, the associated phasor load voltage is given by

[latex] ilde{V}_L=H_ u ilde{V}_S [/latex], (15.11)

where [latex] ilde{V}_S[/latex] is the phasor for the component having frequency f . We need only (1) calculate the value of the voltage transfer function [latex]H_ u[/latex] for each frequency represented in the excitation, (2) use (15.11) to find the corresponding component of the load voltage, and (3) add the results (superposition). Table 15.1 summarizes the required calculations. The load voltage is given by

[latex] \begin{aligned} u_L(t)=& 0.198+1.414 \cos \left(2 \pi f_1 t-0.464 ight) \\ &+0.294 \cos \left(2 \pi f_2 t-0.490 ight) \mathrm{V} \end{aligned} [/latex]

Table 15.1 See Example 15.2
f (Hz)	[latex] f / f_0 [/latex]	[latex] H_ν=1.976 /\left(1+j f / f_0 ight) ; f_0=2 \mathrm{kHz} [/latex]	[latex] ilde{V}_S(\mathrm{~V}) [/latex]	[latex] ilde{V}_L(\mathrm{~V})=H_ν ilde{V}_S [/latex]
0	0	1.976	0.1	0.198
1000	0.5	[latex] 1.768 \angle-0.464 [/latex]	[latex] 0.8 \angle0 [/latex]	[latex] 1.414 \angle -0.464 [/latex]
500	2.5	[latex] 0.734 \angle-1.190 [/latex]	[latex] 0.4 \angle 0.7 [/latex]	[latex] 0.294 \angle -0.490 [/latex]

with [latex]f_1=1 \mathrm{kHz}, f_2=5 \mathrm{kHz}[/latex]

Question 15.2: (a) Obtain expressions for the voltage transfer function, th...

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