Solution+HW17

# Solution+HW17 - Problem 9.4 For the circuit shown in Fig...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Problem 9.4 For the circuit shown in Fig. P9.4, determine (a) the transfer function H = V o / V i , and (b) the frequency ϖ o at which H is purely real. C R 1 R 2 L V i + _ V o + _ Figure P9.4: Circuit for Problem 9.4. Solution: (a) KCL at mode V : V o − V i R 1 + V o parenleftbigg 1 j ϖ L + j ϖ C + 1 R 2 parenrightbigg = . Solution leads to H = V o V i = jR 2 ϖ L R 1 R 2 ( 1 − ϖ 2 LC )+ j ϖ L ( R 1 + R 2 ) . (b) Rationalizing the expression for H : H = jR 2 ϖ L R 1 R 2 ( 1 − ϖ 2 LC )+ j ϖ L ( R 1 + R 2 ) × R 1 R 2 ( 1 − ϖ 2 LC ) − j ϖ L ( R 1 + R 2 ) R 1 R 2 ( 1 − ϖ 2 LC ) − j ϖ L ( R 1 + R 2 ) = ϖ 2 R 2 L 2 ( R 1 + R 2 )+ j ϖ LR 1 R 2 2 ( 1 − ϖ 2 LC ) R 2 1 R 2 2 ( 1 − ϖ 2 LC ) 2 + ϖ 2 L 2 ( R 1 + R 2 ) 2 . The imaginary part of H is zero when ϖ = 0 (trivial resonance) or when ϖ = 1 √ LC . Problem 9.12 Convert the following dB values to voltage ratios: (a) 46 dB (b) 0.4 dB (c) − 12 dB (d) − 66 dB Solution: (a) 10 ( 46 / 20 ) = 10 2 . 3 = 199 . 5 ≃ 200. (b) 10 ( . 4 / 20 ) = 10 . 02 = 1 . 047. (c) 10 ( − 12 / 20 ) = 10 − . 6 = . 25. (d) 10 ( − 66 / 20 ) = 10 − 3 . 3 = 5 × 10 − 4 . Problem 9.13 Generate Bode magnitude and phase plots for the following voltage transfer functions: (a) H ( ϖ ) = j 100 ϖ 10 + j ϖ (b) H ( ϖ ) = . 4 ( 50 + j ϖ ) 2 ( j ϖ ) 2 (c) H ( ϖ ) = ( 40 + j 80 ϖ ) ( 10 + j 50 ϖ ) (d) H ( ϖ ) = ( 20 + j 5 ϖ )( 20 + j ϖ )...
View Full Document

{[ snackBarMessage ]}

### Page1 / 11

Solution+HW17 - Problem 9.4 For the circuit shown in Fig...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online