Homework Assignment No. 3 - Solutions

Transcription

1 ECE 6440 Summer 2003 Page 1 Homework Aignment o. 3 Problem 1 (10 point) Aume an LPLL ha F() 1 and the PLL parameter are 0.8V/radian, K o 100 MHz/V, and the ocillation frequency, f oc 500MHz. Sketch the control voltage at the output of the phae detector if the input frequency jump from 500MHz to 650MHz. Find the tranfer function from the input frequency, f in, to the output of the phae detector, v d. ω 1 v d V d (θ 1 θ 2 ) θ 1 K o V d ω 2 K o V d 1 K o θ 1 K ω 1 d SU03H03S1A V d ω 1 K o V d () K o ω 1 () K o ω 1 k 1 k 2 K o By partial fraction expanion we can how that k 1 k 2 ω 1 K o ω 1 K v 1.5V ote the unit of ω 1 K are (V/rad)(rad/ec) v 1/ec V and K v (2π 100MHz/V)(0.8V/rad.) x10 6 (1/ec.) v d (t) ω 1 K v (1e K vt ) 1.5(1e x106t ) A plot of v d (t) i hown below. 1.5V v d (t) 1.0V 0.5V 0V Time (n) Fig. SU03H03S1B

2 ECE 6440 Summer 2003 Page 2 Problem 2 (10 point) A Type I PLL incorporate a VCO with K o 100MHz/V, a phae detector with 1V/rad, and a firtorder, lowpa filter with ω LPF 2π x10 6 radian/ hown below. A divider of 100 ha been placed in the feedback path to implement a frequency yntheizer. (a.) Find the value of the natural damping frequency, ω n, and the damping factor, ζ, for the tranfer function φ out ()/φ in (), for thi PLL. (b.) If a tep input of φ in i applied at t 0, what i the teadytate phae error at the output of the phae detector, φ e? The teadytate error i evaluated by multiplying the deired phae by and letting 0. (a.) φ out K o Phae Detector Filter VCO φ in φ e 1 V c K o φ out ω LPF 1 Divider 1/100 F02E2P3 1 K d φ in φ out φ out 1 K o 1 ω LPF 1 ω LPF φ out () K o φ in () 1 ω LPF K o K o ω LPF 2 ω LPF K o ω LPF Thu, ω n 2 K o ω LPF 2πx106 2πx π 2 x10 12 ω n 2πx10 6 ζ ω LPF 2ω n 2 ω LPF K o ω LPF ω n 2πx10 6 and ζ ω LPF K o 1 2 (b.) Firt we mut olve for φ e () which i found a 1 ω 1 LPF ω LPF φ e () K φ o out () K o 100 2πx πx K o ω LPF 2 ω LPF K o ω LPF If φ in () φ in, then we can write φ e () ( 2 ω LPF ) φ in 2 ω LPF K o ω LPF Therefore, we ee that the teadytate error i φ(t ) 0. K o φ in 1 ω LPF ω 2 n 2 2ζω n ω 2 n φ in ()

3 ECE 6440 Summer 2003 Page 3 Problem 3 (10 point) Modify the active filter hown of Problem 4 of Homework 2 to deign the laglead loop filter hown below. The capacitor can be no larger than 10pF. Give the value of 1, 2, C 1 and C 2. 0dB F(jω) db 10K 100K ω(rad/ec.) V in 10kΩ 1 C 2 2 _ V out SU03H03P3A 20dB S03H03P3 The tranfer function correponding to the above Bode plot i, F() C 1 C 2 The modification of the filter i v d hown where from Prob. 4 of Homework 2, T1 Ti 2 i 2 i The tranfer function of thi filter i found a, F() V c () V d () T2 T1 C 1 1 T1 T2 C 2 1 T2 T1 T, T C and T C We ee if T2 T1, then C 2 10C 1. Chooing C 2 10pF give C 1 1pF. Thi give T 104 C T 2 i 2 i x x x x Ω Therefore, Ω, C 1 1pF and C 2 10pF The realization i completed by replacing each of the T reitor with the following equivalent: T1 10kΩ 10kΩ T2 Loop Filter SU03H03S2A vc SU03H03S3B 10.02Ω

4 ECE 6440 Summer 2003 Page 4 Problem 4 (10 point ) Uing the filter of Problem 3, find the value of ω n and ζ of the PLL if 1V/radian, K o 2Mradian/V ec. What i the teady tate phae error in degree if a frequency ramp of 10 9 radian/ec. 2 i applied to the PLL? Uing the definition give in the note for the time contant of the paive laglead filter we get, F() τ 2 1 (τ 1 τ 2 ) 1 τ ec. and τ 1 9x10 5 ec. ω n K o τ 1 τ 2 2x x x10 3 radian/ec. ζ ω n 2 τ 2 1 K o K d 2x x Auming the PLL ha a high loop gain, then the teadytate phae error can be found a θ e ( ) ω ω n x radian 2.86

5 ECE 6440 Summer 2003 Page 5 Problem 5 (10 point) Solve for the croover frequency of the PLL of Problem 3 and 4 and find the phae margin. Ue SPICE to find the openloop frequency repone of the PLL and from your plot determine the croover frequency and phae margin and compare with your calculated value. The croover frequency can be found a, ω c ω n 2ζ 2 4ζ x x10 5 (1.6089) 2.275x10 5 radian/ec kHz The open loop tranfer function i given a LG() K v 1τ 1 1τ 2x The phae margin can be written a, PM tan 1 ω c 10 5 tan1 ω c SPICE eult: Problem H3P5Open Loop epone of an LPLL with LeadLag Filter VS 1 0 AC K * Loop bandwidth Kv 2xE6 Tau11E4 Tau21E5 ELPLL 2 0 LAPLACE {V(1)} {(2E6/(S0.001))*((11E5*S)/(11E4*S))} * ote: The added to S in the denominator i to prevent * blowup of the problem at low frequencie K *Steady tate AC analyi.ac DEC K.PIT AC VDB(2) VP(2).POBE.ED db or Degree F(jω) Phae Margin 69 Phae SU03H03S5 Frequency (Hz) ωc 36kHz The imulation reult agree well with the calculated reult.