Circuit analysis contained in a lab report indicates that the network function is H(\omega)=\frac{1+j \frac{\omega}{630}}{10(1+j \frac{\omega}{6300}}

opatovaL

opatovaL

Answered question

2021-09-28

Circuit analysis contained in a lab report indicates that the network function of a circuit is H(ω)=1+jω63010(1+jω6300}. This lab report contains the following frequency response data from measurements made on the circuit. Do these data seem reasonable?
ω,rad/.s.200.400.795.1585.3162|H(ω)|.0.105.0.12.0.16.0.26.0.460ω,rad/.s.6310.12,600.25,100.50,000.100,000|H(ω)|.0.71.1.0.1.0.1.0.1.0

Answer & Explanation

Clara Reese

Clara Reese

Skilled2021-09-29Added 120 answers

Step 1.
|H(ω)|=1+j63063010(1+j6306300}=110(12+1212+0.12=0.14
Comparing the calculated value with the values from the given table and it seems reasonable.
Step 2.
Calculating |H(ω)|=|1+j630063010(1+j63006300|=110(12+10212+12)=0.707
Comparing the calculated value with the values from the given table and it seems reasonable.

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