Figure 141$10^{4} \mathrm{rad} / \mathrm{s}$ , and a Maximum Deviation in the Passband Transmission Of Low-Pass

Question

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The Answer is (a)
$(a) Figure 14.1.2 Figure 14.1.2 shows the s -plane locations of the poles. Here, \begin{aligned} \omega_{p} & \equiv ext { frequency of the passband edge }=10^{4} \mathrm{rad} / \mathrm{s} \ \epsilon & \equiv ext { passband ripple factor }=1 \end{aligned} The pair of complex-conjugate poles p_{1} and p_{1}^{*} have \omega_{01}=\omega_{p}=10^{4} \mathrm{rad} / \mathrm{s} and \begin{aligned} \frac{\omega_{01}}{2 Q_{1}} & =\omega_{01} \cos 72^{\circ} \ \Rightarrow Q_{1} & =\frac{1}{2 \cos 72^{\circ}}=1.618 \end{aligned} The pair of complex-conjugate poles p_{2} and p_{2}^{*} have \omega_{02}=\omega_{p}=10^{4} \mathrm{rad} / \mathrm{s} and Q_{2}=\frac{1}{2 \cos 36^{\circ}}=0.618 The real pole p_{3} has a frequency \omega_{03}=\omega_{p}=10^{4} \mathrm{rad} / \mathrm{s} Thus, the transfer function of the fifth-order lowpass filter is T(s)=\frac{10^{20}}{\left(s+10^{4} ight)\left(s^{2}+s \frac{10^{4}}{1.618}+10^{8} ight)\left(s^{2}+s \frac{10^{4}}{0.618}+10^{8} ight)} where the numerator was found by enforcing the condition T(0)=1 (b) Refer to Figure 14.1.3 Figure 14.1.3 Each of the pair of complex-conjugate poles will be realized utilizing the low-pass circuit shown in Fig. 14.1.3. Here, \begin{aligned} C & =10 \mathrm{nF} \ \omega_{0} & =10^{4}=\frac{1}{C R} \ R & =\frac{1}{\omega_{0} C}=\frac{1}{10^{4} imes 10 imes 10^{-9}} \ & =10 \mathrm{k} \Omega \end{aligned} For the circuit realizing \left(p_{1}, p_{1}^{*} ight) , Q R=1.618 imes 10=16.18 \mathrm{k} \Omega For the circuit realizing \left(p_{2}, p_{2}^{*} ight) , Q R=0.618 imes 10=6.18 \mathrm{k} \Omega Each circuit has a unity de gain. Figure 14.1.4 Figure 14.1.4 shows the circuit for realizing the real-axis pole, p_{3} . Here C=10 \mathrm{nF} \quad ext { and } \quad R=\frac{1}{\omega_{0} C}=10 \mathrm{k} \Omega The feedback resistance is selected equal to R so as to obtain a unity de gain. Placing the three sections in cascade provides the realization of the fifth-order Butterworth lowpass filter. The resulting circuit is shown in Fig. 14.1.5 (c) \begin{aligned} |T| & =\frac{1}{\sqrt{1+\left(\omega / \omega_{p} ight)^{2 N}}} \ A & =10 \log \left[1+\left(\omega / \omega_{p} ight)^{2 N} ight], \mathrm{dB} \end{aligned} where \omega_{p}=10^{4} \mathrm{rad} / \mathrm{s} \quad ext { and } \quad N=5 At \omega=\omega_{S}=2 imes 10^{4} \mathrm{rad} / \mathrm{s} , \begin{aligned} A & =10 \log \left(1+2^{10} ight) \ & =30.1 \mathrm{~dB} \end{aligned} Figure 14.1.6$

Figure 14.1.2
Figure 14.1.2 shows the

s

-plane locations of the poles. Here,

\begin{aligned}\omega_{p} & \equiv ext { frequency of the passband edge }=10^{4} \mathrm{rad} / \mathrm{s} \\epsilon & \equiv ext { passband ripple factor }=1\end{aligned}

The pair of complex-conjugate poles

p_{1}

and

p_{1}^{*}

have

\omega_{01}=\omega_{p}=10^{4} \mathrm{rad} / \mathrm{s}

and

\begin{aligned}\frac{\omega_{01}}{2 Q_{1}} & =\omega_{01} \cos 72^{\circ} \\Rightarrow Q_{1} & =\frac{1}{2 \cos 72^{\circ}}=1.618\end{aligned}

The pair of complex-conjugate poles

p_{2}

and

p_{2}^{*}

have

\omega_{02}=\omega_{p}=10^{4} \mathrm{rad} / \mathrm{s}

and

Q_{2}=\frac{1}{2 \cos 36^{\circ}}=0.618

The real pole

p_{3}

has a frequency

\omega_{03}=\omega_{p}=10^{4} \mathrm{rad} / \mathrm{s}

Thus, the transfer function of the fifth-order lowpass filter is

T(s)=\frac{10^{20}}{\left(s+10^{4} ight)\left(s^{2}+s \frac{10^{4}}{1.618}+10^{8} ight)\left(s^{2}+s \frac{10^{4}}{0.618}+10^{8} ight)}

where the numerator was found by enforcing the condition

T(0)=1

(b) Refer to Figure 14.1.3
$(a) Figure 14.1.2 Figure 14.1.2 shows the s -plane locations of the poles. Here, \begin{aligned} \omega_{p} & \equiv ext { frequency of the passband edge }=10^{4} \mathrm{rad} / \mathrm{s} \ \epsilon & \equiv ext { passband ripple factor }=1 \end{aligned} The pair of complex-conjugate poles p_{1} and p_{1}^{*} have \omega_{01}=\omega_{p}=10^{4} \mathrm{rad} / \mathrm{s} and \begin{aligned} \frac{\omega_{01}}{2 Q_{1}} & =\omega_{01} \cos 72^{\circ} \ \Rightarrow Q_{1} & =\frac{1}{2 \cos 72^{\circ}}=1.618 \end{aligned} The pair of complex-conjugate poles p_{2} and p_{2}^{*} have \omega_{02}=\omega_{p}=10^{4} \mathrm{rad} / \mathrm{s} and Q_{2}=\frac{1}{2 \cos 36^{\circ}}=0.618 The real pole p_{3} has a frequency \omega_{03}=\omega_{p}=10^{4} \mathrm{rad} / \mathrm{s} Thus, the transfer function of the fifth-order lowpass filter is T(s)=\frac{10^{20}}{\left(s+10^{4} ight)\left(s^{2}+s \frac{10^{4}}{1.618}+10^{8} ight)\left(s^{2}+s \frac{10^{4}}{0.618}+10^{8} ight)} where the numerator was found by enforcing the condition T(0)=1 (b) Refer to Figure 14.1.3 Figure 14.1.3 Each of the pair of complex-conjugate poles will be realized utilizing the low-pass circuit shown in Fig. 14.1.3. Here, \begin{aligned} C & =10 \mathrm{nF} \ \omega_{0} & =10^{4}=\frac{1}{C R} \ R & =\frac{1}{\omega_{0} C}=\frac{1}{10^{4} imes 10 imes 10^{-9}} \ & =10 \mathrm{k} \Omega \end{aligned} For the circuit realizing \left(p_{1}, p_{1}^{*} ight) , Q R=1.618 imes 10=16.18 \mathrm{k} \Omega For the circuit realizing \left(p_{2}, p_{2}^{*} ight) , Q R=0.618 imes 10=6.18 \mathrm{k} \Omega Each circuit has a unity de gain. Figure 14.1.4 Figure 14.1.4 shows the circuit for realizing the real-axis pole, p_{3} . Here C=10 \mathrm{nF} \quad ext { and } \quad R=\frac{1}{\omega_{0} C}=10 \mathrm{k} \Omega The feedback resistance is selected equal to R so as to obtain a unity de gain. Placing the three sections in cascade provides the realization of the fifth-order Butterworth lowpass filter. The resulting circuit is shown in Fig. 14.1.5 (c) \begin{aligned} |T| & =\frac{1}{\sqrt{1+\left(\omega / \omega_{p} ight)^{2 N}}} \ A & =10 \log \left[1+\left(\omega / \omega_{p} ight)^{2 N} ight], \mathrm{dB} \end{aligned} where \omega_{p}=10^{4} \mathrm{rad} / \mathrm{s} \quad ext { and } \quad N=5 At \omega=\omega_{S}=2 imes 10^{4} \mathrm{rad} / \mathrm{s} , \begin{aligned} A & =10 \log \left(1+2^{10} ight) \ & =30.1 \mathrm{~dB} \end{aligned} Figure 14.1.6$

Figure 14.1.3
Each of the pair of complex-conjugate poles will be realized utilizing the low-pass circuit shown in Fig. 14.1.3. Here,

\begin{aligned}C & =10 \mathrm{nF} \\omega_{0} & =10^{4}=\frac{1}{C R} \R & =\frac{1}{\omega_{0} C}=\frac{1}{10^{4} imes 10 imes 10^{-9}} \& =10 \mathrm{k} \Omega\end{aligned}

For the circuit realizing

\left(p_{1}, p_{1}^{*} ight)

,

Q R=1.618 imes 10=16.18 \mathrm{k} \Omega

For the circuit realizing

\left(p_{2}, p_{2}^{*} ight)

,

Q R=0.618 imes 10=6.18 \mathrm{k} \Omega

Each circuit has a unity de gain.
$(a) Figure 14.1.2 Figure 14.1.2 shows the s -plane locations of the poles. Here, \begin{aligned} \omega_{p} & \equiv ext { frequency of the passband edge }=10^{4} \mathrm{rad} / \mathrm{s} \ \epsilon & \equiv ext { passband ripple factor }=1 \end{aligned} The pair of complex-conjugate poles p_{1} and p_{1}^{*} have \omega_{01}=\omega_{p}=10^{4} \mathrm{rad} / \mathrm{s} and \begin{aligned} \frac{\omega_{01}}{2 Q_{1}} & =\omega_{01} \cos 72^{\circ} \ \Rightarrow Q_{1} & =\frac{1}{2 \cos 72^{\circ}}=1.618 \end{aligned} The pair of complex-conjugate poles p_{2} and p_{2}^{*} have \omega_{02}=\omega_{p}=10^{4} \mathrm{rad} / \mathrm{s} and Q_{2}=\frac{1}{2 \cos 36^{\circ}}=0.618 The real pole p_{3} has a frequency \omega_{03}=\omega_{p}=10^{4} \mathrm{rad} / \mathrm{s} Thus, the transfer function of the fifth-order lowpass filter is T(s)=\frac{10^{20}}{\left(s+10^{4} ight)\left(s^{2}+s \frac{10^{4}}{1.618}+10^{8} ight)\left(s^{2}+s \frac{10^{4}}{0.618}+10^{8} ight)} where the numerator was found by enforcing the condition T(0)=1 (b) Refer to Figure 14.1.3 Figure 14.1.3 Each of the pair of complex-conjugate poles will be realized utilizing the low-pass circuit shown in Fig. 14.1.3. Here, \begin{aligned} C & =10 \mathrm{nF} \ \omega_{0} & =10^{4}=\frac{1}{C R} \ R & =\frac{1}{\omega_{0} C}=\frac{1}{10^{4} imes 10 imes 10^{-9}} \ & =10 \mathrm{k} \Omega \end{aligned} For the circuit realizing \left(p_{1}, p_{1}^{*} ight) , Q R=1.618 imes 10=16.18 \mathrm{k} \Omega For the circuit realizing \left(p_{2}, p_{2}^{*} ight) , Q R=0.618 imes 10=6.18 \mathrm{k} \Omega Each circuit has a unity de gain. Figure 14.1.4 Figure 14.1.4 shows the circuit for realizing the real-axis pole, p_{3} . Here C=10 \mathrm{nF} \quad ext { and } \quad R=\frac{1}{\omega_{0} C}=10 \mathrm{k} \Omega The feedback resistance is selected equal to R so as to obtain a unity de gain. Placing the three sections in cascade provides the realization of the fifth-order Butterworth lowpass filter. The resulting circuit is shown in Fig. 14.1.5 (c) \begin{aligned} |T| & =\frac{1}{\sqrt{1+\left(\omega / \omega_{p} ight)^{2 N}}} \ A & =10 \log \left[1+\left(\omega / \omega_{p} ight)^{2 N} ight], \mathrm{dB} \end{aligned} where \omega_{p}=10^{4} \mathrm{rad} / \mathrm{s} \quad ext { and } \quad N=5 At \omega=\omega_{S}=2 imes 10^{4} \mathrm{rad} / \mathrm{s} , \begin{aligned} A & =10 \log \left(1+2^{10} ight) \ & =30.1 \mathrm{~dB} \end{aligned} Figure 14.1.6$

Figure 14.1.4
Figure 14.1.4 shows the circuit for realizing the real-axis pole,

p_{3}

. Here

C=10 \mathrm{nF} \quad ext { and } \quad R=\frac{1}{\omega_{0} C}=10 \mathrm{k} \Omega

The feedback resistance is selected equal to

R

so as to obtain a unity de gain.
Placing the three sections in cascade provides the realization of the fifth-order Butterworth lowpass filter. The resulting circuit is shown in Fig. 14.1.5
$(a) Figure 14.1.2 Figure 14.1.2 shows the s -plane locations of the poles. Here, \begin{aligned} \omega_{p} & \equiv ext { frequency of the passband edge }=10^{4} \mathrm{rad} / \mathrm{s} \ \epsilon & \equiv ext { passband ripple factor }=1 \end{aligned} The pair of complex-conjugate poles p_{1} and p_{1}^{*} have \omega_{01}=\omega_{p}=10^{4} \mathrm{rad} / \mathrm{s} and \begin{aligned} \frac{\omega_{01}}{2 Q_{1}} & =\omega_{01} \cos 72^{\circ} \ \Rightarrow Q_{1} & =\frac{1}{2 \cos 72^{\circ}}=1.618 \end{aligned} The pair of complex-conjugate poles p_{2} and p_{2}^{*} have \omega_{02}=\omega_{p}=10^{4} \mathrm{rad} / \mathrm{s} and Q_{2}=\frac{1}{2 \cos 36^{\circ}}=0.618 The real pole p_{3} has a frequency \omega_{03}=\omega_{p}=10^{4} \mathrm{rad} / \mathrm{s} Thus, the transfer function of the fifth-order lowpass filter is T(s)=\frac{10^{20}}{\left(s+10^{4} ight)\left(s^{2}+s \frac{10^{4}}{1.618}+10^{8} ight)\left(s^{2}+s \frac{10^{4}}{0.618}+10^{8} ight)} where the numerator was found by enforcing the condition T(0)=1 (b) Refer to Figure 14.1.3 Figure 14.1.3 Each of the pair of complex-conjugate poles will be realized utilizing the low-pass circuit shown in Fig. 14.1.3. Here, \begin{aligned} C & =10 \mathrm{nF} \ \omega_{0} & =10^{4}=\frac{1}{C R} \ R & =\frac{1}{\omega_{0} C}=\frac{1}{10^{4} imes 10 imes 10^{-9}} \ & =10 \mathrm{k} \Omega \end{aligned} For the circuit realizing \left(p_{1}, p_{1}^{*} ight) , Q R=1.618 imes 10=16.18 \mathrm{k} \Omega For the circuit realizing \left(p_{2}, p_{2}^{*} ight) , Q R=0.618 imes 10=6.18 \mathrm{k} \Omega Each circuit has a unity de gain. Figure 14.1.4 Figure 14.1.4 shows the circuit for realizing the real-axis pole, p_{3} . Here C=10 \mathrm{nF} \quad ext { and } \quad R=\frac{1}{\omega_{0} C}=10 \mathrm{k} \Omega The feedback resistance is selected equal to R so as to obtain a unity de gain. Placing the three sections in cascade provides the realization of the fifth-order Butterworth lowpass filter. The resulting circuit is shown in Fig. 14.1.5 (c) \begin{aligned} |T| & =\frac{1}{\sqrt{1+\left(\omega / \omega_{p} ight)^{2 N}}} \ A & =10 \log \left[1+\left(\omega / \omega_{p} ight)^{2 N} ight], \mathrm{dB} \end{aligned} where \omega_{p}=10^{4} \mathrm{rad} / \mathrm{s} \quad ext { and } \quad N=5 At \omega=\omega_{S}=2 imes 10^{4} \mathrm{rad} / \mathrm{s} , \begin{aligned} A & =10 \log \left(1+2^{10} ight) \ & =30.1 \mathrm{~dB} \end{aligned} Figure 14.1.6$

(c)

\begin{aligned}|T| & =\frac{1}{\sqrt{1+\left(\omega / \omega_{p} ight)^{2 N}}} \A & =10 \log \left[1+\left(\omega / \omega_{p} ight)^{2 N} ight], \mathrm{dB}\end{aligned}

where

\omega_{p}=10^{4} \mathrm{rad} / \mathrm{s} \quad ext { and } \quad N=5

At

\omega=\omega_{S}=2 imes 10^{4} \mathrm{rad} / \mathrm{s}

,

\begin{aligned}A & =10 \log \left(1+2^{10} ight) \& =30.1 \mathrm{~dB}\end{aligned}

$(a) Figure 14.1.2 Figure 14.1.2 shows the s -plane locations of the poles. Here, \begin{aligned} \omega_{p} & \equiv ext { frequency of the passband edge }=10^{4} \mathrm{rad} / \mathrm{s} \ \epsilon & \equiv ext { passband ripple factor }=1 \end{aligned} The pair of complex-conjugate poles p_{1} and p_{1}^{*} have \omega_{01}=\omega_{p}=10^{4} \mathrm{rad} / \mathrm{s} and \begin{aligned} \frac{\omega_{01}}{2 Q_{1}} & =\omega_{01} \cos 72^{\circ} \ \Rightarrow Q_{1} & =\frac{1}{2 \cos 72^{\circ}}=1.618 \end{aligned} The pair of complex-conjugate poles p_{2} and p_{2}^{*} have \omega_{02}=\omega_{p}=10^{4} \mathrm{rad} / \mathrm{s} and Q_{2}=\frac{1}{2 \cos 36^{\circ}}=0.618 The real pole p_{3} has a frequency \omega_{03}=\omega_{p}=10^{4} \mathrm{rad} / \mathrm{s} Thus, the transfer function of the fifth-order lowpass filter is T(s)=\frac{10^{20}}{\left(s+10^{4} ight)\left(s^{2}+s \frac{10^{4}}{1.618}+10^{8} ight)\left(s^{2}+s \frac{10^{4}}{0.618}+10^{8} ight)} where the numerator was found by enforcing the condition T(0)=1 (b) Refer to Figure 14.1.3 Figure 14.1.3 Each of the pair of complex-conjugate poles will be realized utilizing the low-pass circuit shown in Fig. 14.1.3. Here, \begin{aligned} C & =10 \mathrm{nF} \ \omega_{0} & =10^{4}=\frac{1}{C R} \ R & =\frac{1}{\omega_{0} C}=\frac{1}{10^{4} imes 10 imes 10^{-9}} \ & =10 \mathrm{k} \Omega \end{aligned} For the circuit realizing \left(p_{1}, p_{1}^{*} ight) , Q R=1.618 imes 10=16.18 \mathrm{k} \Omega For the circuit realizing \left(p_{2}, p_{2}^{*} ight) , Q R=0.618 imes 10=6.18 \mathrm{k} \Omega Each circuit has a unity de gain. Figure 14.1.4 Figure 14.1.4 shows the circuit for realizing the real-axis pole, p_{3} . Here C=10 \mathrm{nF} \quad ext { and } \quad R=\frac{1}{\omega_{0} C}=10 \mathrm{k} \Omega The feedback resistance is selected equal to R so as to obtain a unity de gain. Placing the three sections in cascade provides the realization of the fifth-order Butterworth lowpass filter. The resulting circuit is shown in Fig. 14.1.5 (c) \begin{aligned} |T| & =\frac{1}{\sqrt{1+\left(\omega / \omega_{p} ight)^{2 N}}} \ A & =10 \log \left[1+\left(\omega / \omega_{p} ight)^{2 N} ight], \mathrm{dB} \end{aligned} where \omega_{p}=10^{4} \mathrm{rad} / \mathrm{s} \quad ext { and } \quad N=5 At \omega=\omega_{S}=2 imes 10^{4} \mathrm{rad} / \mathrm{s} , \begin{aligned} A & =10 \log \left(1+2^{10} ight) \ & =30.1 \mathrm{~dB} \end{aligned} Figure 14.1.6$

Figure 14.1.6

Figure 141 $10^{4} \mathrm{rad} / \mathrm{s}$ , and a Maximum Deviation in the Passband Transmission Of Low-Pass

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Figure 141 104rad/s10^{4} \mathrm{rad} / \mathrm{s}104rad/s , and a Maximum Deviation in the Passband Transmission Of Low-Pass

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Figure 141 $10^{4} \mathrm{rad} / \mathrm{s}$ , and a Maximum Deviation in the Passband Transmission Of Low-Pass

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