In Fig. 16.11, the op amp is ideal, C=220pF,R2=100kΩ, and νS=A[cos(2πf0t)+cos(10πf0t)+cos(100πf0t)] with A=10mV and f0=50kHz. (a) Express the excitation as an exponential Fourier series and obtain the Fourier coefficients for the load voltage νL. (b) Express the load voltage in amplitude-phase

Question

In Fig. 16.11, the op amp is ideal, [latex]C=220 \mathrm{pF}, R_2=100 \mathrm{k} \Omega[/latex], and

[latex] \begin{aligned} u_S=& A\left[\cos \left(2 \pi f_0 t ight)+\cos \left(10 \pi f_0 t ight) ight.\\ &\left.+\cos \left(100 \pi f_0 t ight) ight] \end{aligned} [/latex],

with [latex]A=10 \mathrm{mV}[/latex] and [latex]f_0=50 \mathrm{kHz}[/latex]. (a) Express the excitation as an exponential Fourier series and obtain the Fourier coefficients for the load voltage [latex] u_L [/latex]. (b) Express the load voltage in amplitude-phase form.

Accepted Answer

(a) The fundamental frequency is [latex]f_0[/latex] and the harmonics present in the excitation have frequencies [latex]f_0, 5 f_0[/latex], and [latex]50 f_0[/latex]. Euler's identity (or Table 16.1) gives

Table 16.1 Relations among coefficients of the three forms of a Fourier series
from / To	Exponential	Amplitude-Phase	Quadrature
Exponential		[latex] \begin{aligned} &A_k=2\left\|X_k ight\|, k=1,2, \cdots \ &A_0=\left\|X_0 ight\| \ & heta_k=∡ X_k \end{aligned} [/latex]	[latex] \begin{aligned} &a_k=2 \operatorname{Re}\left(X_k ight), k=1,2, \cdots \ &b_k=-2 \operatorname{Im}\left(X_k ight), k=1,2, \cdots \ &a_0=X_0, b_0=0 \end{aligned} [/latex]
Amplitude-Phase	[latex] \begin{aligned} &X_{\pm k}=\frac{A_k}{2} \angle \pm heta_k \ &k=1,2, \cdots \ &X_0=A_0 \cos \left( heta_0 ight)=x_{d c} \end{aligned}[/latex]		[latex] \begin{aligned} &a_k=A_k \cos \left( heta_k ight), \ &b_k=-A_k \sin \left( heta_k ight), \ &k=1,2, \cdots \end{aligned} [/latex]
Quadrature	[latex] \begin{aligned} &X_{\pm k}=\frac{1}{2}\left(a_k \mp j b_k ight), \&k=1,2, \cdots \ &X_0=a_0 \end{aligned} [/latex]	[latex] \begin{aligned} &A_k=\sqrt{a_k^2+b_k^2} , k=1,2, \cdots \ &A_0=\left\|a_0 ight\| \ & heta_k=∡\left(a_k-j b_k ight) \end{aligned} [/latex]

[latex] \begin{aligned} u_S=& \frac{A}{2}\left[e^{j 2 \pi f_0 t}+e^{-j 2 \pi f_0 t}+e^{j 10 \pi f_0 t}+e^{-j 10 \pi f_0 t} ight.\\ &\left.+e^{j 100 \pi f_0 t}+e^{-j 100 \pi f_0 t} ight] \\ =& \sum_{k=-\infty}^{\infty} X_k e^{j 2 \pi k f_0 t}, \end{aligned} [/latex]

with

[latex] X_k= \begin{cases}\frac{A}{2}, & k=\pm 1, \pm 5, \pm 50, \\ 0, & ext { otherwise. }\end{cases} [/latex]

We need the voltage transfer function for the circuit. Kirchhoff's current law gives

[latex] \begin{aligned} &-j 2 \pi f C ilde{V}_s=\frac{ ilde{V}_L}{R_2} \\&\Rightarrow H_ν=\frac{ ilde{V}_L}{ ilde{V}_S}=-j 2 \pi f C R_2=-j 2 \pi f au \end{aligned} [/latex]

with [latex] au=R_2 C=22 \mu s [/latex] . The Fourier coefficients for the load voltage are

[latex] \begin{aligned} Y_k &=\left\{\begin{array}{ll} H_ u\left(k f_0 ight) \frac{A}{2}, & k=\pm 1, \pm 5, \pm 50 \\ 0, & ext { all other } k \end{array} ight\}\\ &= \begin{cases}-j \pi k f_0 au A, & k=\pm 1, \pm 5, \pm 50,\\ 0, & ext { all other } k,\end{cases} \end{aligned} [/latex]

where [latex]f_0=50 \mathrm{kHz}, au=22 \mu s[/latex], and [latex]A=10 \mathrm{mV}[/latex]. Thus,

[latex] Y_k=\left\{\begin{array}{cl} -j 35 \mathrm{mV}, & k=1 \\ -j 173 \mathrm{mV}, & k=5 \\ -j 1.728 \mathrm{~V}, & k=50 \\ Y_k^*, & k=-1,-5,-50 \\ 0, & ext { all other } k \end{array} ight. [/latex]

(b) From Table 16.1 , the amplitude-phase parameters for the load voltage are

[latex] \begin{aligned} &B_k=2\left|Y_k ight|=\left\{\begin{array}{cc} 69.1 \mathrm{mV}, & k=1 \\ 345 \mathrm{mV}, & k=5 \\ 3.456 \mathrm{~V}, & k=50 \\ 0, & ext { all other } k \end{array} ight\}, \\ & heta_k=∡ Y_k=\left\{\begin{array}{cc} -\frac{\pi}{2}, & k=1, 5, 50 \\ 0, & ext { all other } k \end{array} ight\} \end{aligned} [/latex].

and the amplitude-phase series for the load voltage is

[latex] \begin{aligned} u_L=& 0.069 \cos \left(2 \pi f_0 t-\frac{\pi}{2} ight) \\ &+0.345 \cos \left(2 \pi 5 f_0 t-\frac{\pi}{2} ight) \ &+3.456 \cos \left(2 \pi 50 f_0 t-\frac{\pi}{2} ight) \mathrm{V} \end{aligned}[/latex].

The components of the source voltage have equal amplitudes. The highest-frequency component of the load voltage is 25 times larger than the lowest-frequency component, illustrating that differentiation enhances the highfrequency components of a signal.

Question 16.12: In Fig. 16.11, the op amp is ideal, C=220pF,R2=100kΩ, and ν...

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