opatovaL

2021-09-28

Circuit analysis contained in a lab report indicates that the network function of a circuit is $H\left(\omega \right)=\frac{1+j\frac{\omega }{630}}{10\left(1+j\frac{\omega }{6300}\right\}}$. This lab report contains the following frequency response data from measurements made on the circuit. Do these data seem reasonable?
$\begin{array}{llllll}\omega ,\mathrm{rad}/.\mathrm{s}& .200& .400& .795& .1585& .3162\\ |\mathbf{H}\left(\omega \right)|& .0.105& .0.12& .0.16& .0.26& .0.460\\ \omega ,\mathrm{rad}/.\mathrm{s}& .6310& .12,600& .25,100& .50,000& .100,000\\ |\mathbf{H}\left(\omega \right)|& .0.71& .1.0& .1.0& .1.0& .1.0\end{array}$

Clara Reese

Step 1.
$|H\left(\omega \right)|=\mid \frac{1+j\frac{630}{630}}{10\left(1+j\frac{630}{6300}\right\}\mid =\frac{1}{10}\left(\sqrt{\frac{{1}^{2}+{1}^{2}}{{1}^{2}+{0.1}^{2}}}=0.14}$
Comparing the calculated value with the values from the given table and it seems reasonable.
Step 2.
Calculating $|H\left(\omega \right)|=|\frac{1+j\frac{6300}{630}}{10\left(1+j\frac{6300}{6300}}|=\frac{1}{10}\left(\sqrt{\frac{{1}^{2}+{10}^{2}}{{1}^{2}+{1}^{2}}}\right)=0.707$
Comparing the calculated value with the values from the given table and it seems reasonable.

Do you have a similar question?