# Ee 31113 Tutorial 1

 EE 31113 Tutorial 1 1 The block diagram of a control system is shown in Fig. 3P-7. Draw an equivalent SFG for the system. Find the following transfer functions by applying the gain formula of the SFG directly to the block diagram. Compare the answers by applying the gain formula to the equivalent SFG.     2 Phase – locked loops are control systems used for precision motor-speed control. The basic elements of a phase-locked-loop system incorporating a dc motor are shown in Fig. 4P-20(a). An input pulse train represents the reference frequency or desired output speed. The digital encoder produces digital pulses that represent motor speed. The phase detector compares the motor speed and the reference frequency and sends an error voltage to the filter (controller) that governs the dynamic response of the system. Phase detector gain = , encoder gain = , counter gain = 1/N, and dc-motor torque constant = . Assume zero inductance and zero friction for the motor. Derive the transfer function of the filter shown in Fig. 4P-20(b). Assume that the filter sees infinite impedance at the output and zero impedance at the input. Draw a functional block diagram of the system with gains or transfer functions in the blocks. Derive the forward – path transfer function when the feedback path is open. Find the closed-loop transfer function . Repeat parts (a), (c), and (d) for the filter shown in Fig. 4P-20(c). The digital encoder has an output of 36 pulses per revolution. The reference frequency is fixed at 120 pulses/sec. Find in pulses/rad. The idea of using the counter N is that with fixed, various desired output speeds can be attained by changing the value of N. Find N if the desired output speed is 200rpm. Find N if the desired output speed is 1800 rpm. EE 31113 Tutorial 1(Solution) 1 Solution       2 Solution Transfer function: Block Diagram: Forward – path transfer function: Closed – loop transfer function: Forward – path transfer function: Closed – loop transfer function:  =36 pulses / rev = 36 /2 pulses /rad = 5.73 pulses /rad. (f)   pulses/sec Thus, N=1. For =1800RPM, 120=N(36/2 )1800(2 /60)=1080N. Thus, N=9.Download 13.31 Kb.Share with your friends: