English: $Insertion\ loss\ (\alpha)\ dB = 20\left| \frac{1}{S_{21}} \right|\ = 20\frac{A + \frac{B}{Z_{0}} + CZ_{0} + D}{2}\ = \ $ $20\left| \left( \frac{1 + \frac{R + jX}{Z_{0}} + 0 + 1}{2} \right) \right|\ = 20\left| \left( \frac{2 + \frac{R + jX}{Z_{0}}}{2} \right) \right|\ = 20\left| 1 + \frac{R}{2Z_{0}} + j\frac{X}{2Z_{0}} \right|\ $ $\alpha\ (dB) = 20\left\lbrack \left( \left( 1 + \frac{R}{2Z_{0}} \right)^{2} + \left( \frac{X}{2Z_{0}} \right)^{2} \right) \right\rbrack^{\frac{1}{2}}\ $ $\alpha\ (dB) = 10\left\lbrack \left( \left( 1 + \frac{R}{2Z_{0}} \right)^{2} + \left( \frac{X}{2Z_{0}} \right)^{2} \right) \right\rbrack\ $ $\ (\alpha)\ dB\ = \ 20\left| \frac{1}{S_{21}} \right|\ 20\left| \frac{1 + \frac{R\ + \ jX}{Z_{0}} + \ 0\ + \ 1}{2} \right|$ $20\left| \left( \frac{1 + \frac{R\ + \ jX}{Z_{0}} + \ 0\ + \ 1}{2} \right) \right|\ = \ 20\left| \left( \frac{2 + \frac{R\ + \ jX}{Z_{0}}}{2} \right) \right|\ = \ 20\left| 1 + \frac{R}{2Z_{0}} + \ j\frac{X}{2Z_{0}} \right|\ $ $\alpha\ (dB) = \ 20\left\lbrack \left( \left( 1 + \frac{R}{2Z_{0}} \right)^{2} + \left( \frac{X}{2Z_{0}} \right)^{2} \right) \right\rbrack^{\frac{1}{2}}\ $ $\alpha\ (dB) = \ 10\left\lbrack \left( \left( 1 + \frac{R}{2Z_{0}} \right)^{2} + \left( \frac{X}{2Z_{0}} \right)^{2} \right) \right\rbrack\ $ So, for given values of R = R_(f) , X = ωL, you can find out what is the insertion loss of this particular switch.
Telugu: $Insertion\ నష్టం\ (\alpha)\ dB = 20\ఎడమ| \frac{1}{S_{21}} \కుడి|\ = 20\frac{A + \frac{B}{Z_{0}} + CZ_{0} + D}{2}\ = \ $ $20\ఎడమ| \లెఫ్ట్( \frac{1 + \frac{R + jX}{Z_{0}} + 0 + 1}{2} \కుడి) \కుడి|\ = 20\ఎడమ| \లెఫ్ట్( \frac{2 + \frac{R + jX}{Z_{0}}}{2} \కుడి) \కుడి|\ = 20\ఎడమ| 1 + \frac{R}{2Z_{0}} + j\frac{X}{2Z_{0}} \కుడి|\ $ $\alpha\ (dB) = 20\ఎడమ\lbrack \ఎడమ( \ఎడమ( 1 + \frac{R}{2Z_{0}} \కుడి)^{2} + \ఎడమ( \frac{X}{2Z_{0}} \కుడి)^{2} \కుడి) \కుడి\rbrack^{\frac{1}{2}}\ $ $\alpha\ (dB) = 10\లెఫ్ట్\lbrack \ఎడమ( \ఎడమ( \frac{R}{2Z_{0}} \కుడి)^{2} + \ఎడమ( \frac{X}{2Z_{0}} \కుడి)^{2} \కుడి) \కుడి\rbrack\ $\ (\alpha)\ dB\ = \ 20\lbrack \ఎడమ( \ఎడమ( 1 + \frac{R}{2Z_{0}} \కుడి)^{2} +\ఎడమ( \frac{X}{2Z_{0}} \కుడి)^{2} \కుడి) \కుడి\ rbrack\ \frac{1}{S_{21}} \కుడి|\ 20\ఎడమ| \frac{1 + \frac{R\ + \ jX}{Z_{0}} + \ 0\ + \ 1}{2} \కుడి|$ $20\ఎడమ| \లెఫ్ట్( \frac{1 + \frac{R\ + \ jX}{Z_{0}} + \ 0\ + \ 1}{2} \కుడి) \కుడి|\ = \ 20\ఎడమ| \ఎడమ( \frac{2 + \frac{R\ + \ jX}{Z_{0}}}{2} \కుడి) \కుడి|\ = \ 20\ఎడమ| 1 + \frac{R}{2Z_{0}} + \ j\frac{X}{2Z_{0}} \కుడి|\ $ $\alpha\ (dB) = \ 20\ఎడమ\lbrack \ఎడమ( \ఎడమ( 1 + \frac{R}{2Z_{0}} \కుడి)^{2} + \లెఫ్ట్( \frac{X}{2Z_{0}} \కుడి)^{2} \కుడి)\కుడి) \కుడి\rbrack^{frac{1}{2}}\ $ $\alpha\ (dB) = \10\లెఫ్ట్\lbrack \ఎడమ( \ఎడమ( 1 + \frac{R}{2Z_{0}} \కుడి)^{2} +\ఎడమ( \frac{x}{2Z_{0}} \కుడి){2}^కుడి\rbrack\\  R = R_(f) , X = ωL యొక్క ఇవ్వబడ్డ విలువల కొరకు, ఈ నిర్ధిష్ట స్విచ్ యొక్క చొప్పించే నష్టం ఏమిటో మీరు కనుగొనవచ్చు.