Solution: K m = R 1 = 10. From the original circuit, Z L1 = jωl 1 = j10 Ω. For the scaled circuit, L 1 = jk m ωl 1 = j10 10 = j100 Ω, Z L

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1 Problem 9.9 Circuit (b) in Fig. P9.9 is a scaled version of circuit (a). The scaling process may have involved magnitude or frequency scaling, or both simultaneously. If R = kω gets scaled to R = kω, supply the impedance values of the other elements in the scaled circuit. j Ω j2 Ω L C j5 Ω L 2 C 2 j5 Ω R kω R 2 2 kω (a) Original circuit L C L 2 C 2 R kω R 2 (b) Scaled circuit Figure P9.9: Circuits for Problem 9.9. Solution: From the original circuit, For the scaled circuit, K m = R R =. Z L = jωl = j Ω. Z L = jω L = jk fω Km K f L = jk m ωl = j = j Ω, Z L 2 = j2 Ω, Z C = jω C = jk mk f = j = j5 Ω, K f ωc ωc Z C 2 = j5 Ω, R 2 = K mr 2 = 2 kω = 2 kω. j Ω j2 Ω j5 Ω j5 Ω kω 2 kω Scaled circuit

2 Problem 9.2 (a) 46 db (b).4 db (c) 2 db (d) 66 db Convert the following db values to voltage ratios: Solution: (a) (46/2) = 2.3 = (b) (.4/2) =.2 =.47. (c) ( 2/2) =.6 =.25. (d) ( 66/2) = 3.3 = 5 4.

3 Problem 9.3 Generate Bode magnitude and phase plots for the following voltage transfer functions: (a) H(ω) = jω + jω.4(5+ jω)2 (b) H(ω) = ( jω) 2 (4+ j8ω) (c) H(ω) = (+ j5ω) (2+ j5ω)(2+ jω) (d) H(ω) = jω 3(+ jω) (e) H(ω) = (2+ j2ω)(+ j2ω) jω (f) H(ω) = (+ j5ω)(+ jω) 2 (2+ j2ω) (g) H(ω) = (5+ j5ω)(+ jω) Solution: (a) H(ω) = jω + jω = jω (+ jω/) = jω + jω/. Constant factor = +2 db origin Simple pole with ω c = rad/s M [db] = 2log H = 2log +2log ω 2log + jω/

4 db 4 M [db] 2 2 log ω 2 log = 2 db. 2 2 log + jω/ φ Magnitude 9 o j (b) o 9 o Phase φ(ω). H(ω) =.4(5+ jω)2 ( jω) 2 + jω/.4 25(+ jω/5)2 (+ jω/5)2 = ω 2 = ω 2. M [db] = 2log H(ω) = 2log +4log + jω/5 4log ω. Line starts at 6 db at ω = rad/s, and has slope of 4 db/decade Constant factor = 6 db origin of order 2 Simple zero with ω c = 5 rad/s, of order 2

5 6 db 2 log = 6 db 4 4 log + jω/5 M [db] db log ω 6 Magnitude φ 8 o ( + jω/5) φ(ω) 8 o 8 o Phase (c) H(ω) = 4+ j8ω 4(+ j2ω) 4(+ jω/.5) = = + j5ω (+ j5ω) (+ jω/.2). Constant factor 4 = 2 db Simple pole with ω c =.2 rad/s Simple zero with ω c =.5 rad/s M [db] = 2log H(ω) = 2log 4+2log + jω/.5 2log + jω/.2

6 db 2 2 log + jω/ db 5 M [db] log + jω/.2 Magnitude 9 o 45 o ( + jω/.5).. φ(ω) 45 o ( + jω/.2) 9 o Phase (d) H(ω) = = (2+ j5ω)(2+ jω) jω j2(+ jω/4)2(+ jω/2) ω Constant term 4 = 52 db origin = j4(+ jω/4)(+ jω/2) ω.

7 Simple zero with ω c = 4 rad/s Simple zero with ω c = 2 rad/s db M [db] 52 db 2 log + jω/4 2 2 log + jω/ log ω 6 Magnitude 9 o ( + jω/2) 45 o ( + jω/4) φ(ω). 45 o 9 o j Phase (e) H ( ω) = 3(+ jω) (2+ j2ω)(+ j2ω) = 3(+ jω/) 2 (+ jω/)(+ jω/5) =.5 3 (+ jω/) (+ jω/)(+ jω/5)

8 Constant term.5 3 = 56.5 db Simple zero with ω c = rad/s Simple pole with ω c = rad/s Simple pole with ω c = 5 rad/s db 4 2 log + jω/ log + jω/5 2 log + jω/ db Magnitude φ 9 o ( + jω/) 45 o φ(ω) 45 o 9 o ( + jω/) Phase ( + jω/5) (f) H(ω) = jω (+ j5ω)(+ jω) 2 = j 4 ω (+ jω/2)(+ jω/) 2 Constant term 4 = 8 db origin

9 Simple pole with ω c = 2 rad/s Simple pole with ω c = rad/s, of order 2 db 2 2 log ω 2 4 log + jω/ 4 2 log + jω/ M [db] 8 8 db 94 Magnitude φ 9 o j 45 o φ(ω) 45 o 9 o ( + jω/) ( + jω/2) 8 o Phase

10 (2+ j2ω) (g) H(ω) = (5+ j5ω)(+ jω) = (+ jω/) 25( + jω/)( + jω/) Constant term /25 = 48 db Simple pole with ω c = rad/s Simple zero with ω c = rad/s Simple pole with ω c = rad/s db 2 2 log + jω/, 2 log + jω/ log + jω/ M [db] 48 db φ Magnitude 9 o ( + jω/), + jω/ φ(ω) 9 o Phase + jω/

11 Problem 9.6 Determine the voltage transfer function H(ω) corresponding to the Bode magnitude plot shown in Fig. P9.6. The phase of H(ω) is 9 at ω =. M [db] 6 db 4 db 2 db Figure P9.6: Bode magnitude plot for Problem 9.6. Solution: H(ω) consists of: () A constant term K whose db value is 6 db, or K = 6/2 =. (2) A simple pole of order 3 with ω c = 5 rad/s (slope = 6 db/decade) (3) A simple zero of order 6 with ω c = 5 rad/s (slope reverses from 6 db/decade to +6 db/decade) (4) A simple pole of order 3 with ω c = 5 rad/s (slope changes to db at ω c = 5 rad/s). Hence, H(ω) = ( j)n (+ jω/5) 6 j(5+ jω)6 (+ jω/5) 3 = (+ jω/5) 3 (5+ jω) 3 (5+ jω) 3. Given that the phase of H(ω) is 9 at ω =, it follows that N =.

12 Problem 9.26 For the circuit shown in Fig. P9.26: (a) Obtain an expression for H(ω) = V o /V i in standard form. (b) Generate spectral plots for the magnitude and phase of H(ω), given that R = Ω, R 2 = 2 Ω, L = mh, and L 2 = 2 mh. (c) Determine the cutoff frequency ω c and the slope of the magnitude (in db) when ω/ω c and when ω/ω c. R V L V i L R 2 V o Figure P9.26: Circuit for Problem Solution: (a) At node V, KCL gives: Also, voltage division gives Solution is V V i + V V + =. R jωl R 2 + jωl 2 V o = R 2V R 2 + jωl 2. H(ω) = V o jωl R 2 =. V i (R + jωl )(R 2 + jωl 2 )+ jωl R To express denominator in standard form, we factor out R R 2 and expand terms: H(ω) = ( + jω L R jωl /R )( + jω L 2 R 2 ) + jω L R 2 where jωl /R = ( L + jω + L 2 + L ) ( L L 2 + jω R R 2 R 2 R R 2 jωk = + j2ξ ω/ω c +(jω/ω c ) 2, ) 2 K = L R R 2, ω c = R L L2, ξ = ω [ c L + L 2 + L ]. 2 R R 2 R 2 (b) For R = Ω, R 2 = 2 Ω, L = mh, and L 2 = 2 mh, K = 3, ω c = 3 rad/s, ξ =.25.

13 Hence, with ω c = 3 rad/s. H(ω) = j 3 ω + j2.5ω/ω c +(jω/ω c ) 2, M [db] = 2log H(ω) Spectral plots of M [db] are φ(ω) are shown in Figs. P9.26(a) and (b). 2 M (db) ω ϕ (deg) ω Figures P9.26(a) and (b) (c) Low-frequency asymptote (ω/ω c ): H(ω) j 3 ω = M [db] has slope of +2 db/decade. High-frequency asymptote (ω/ω c ): H(ω) j 3 ω2 c ω = j 3 ω = M [db] has slope of 2 db/decade.

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