phasors
10. PHASOR ANALYSIS
10.1 INTRODUCTION
When a system is stimulated by an input it will respond. Initially there is a substantial transient response, that is eventually replaced by a steady state response. Techniques for finding the combined steady state and transient responses were covered in earlier chapters. These include the integration of differential equations, and numerical solutions. Phasor analysis can be used to find the steady state response only. These techniques involve using the phasor transform on the system transfer function, input and output.
10.2 PHASORS FOR STEADY-STATE ANALYSIS
When considering the differential operator we can think of it as a complex number, as in Figure 10.1 Transient and steady-state parts of the differential operator. The real component of the number corresponds to the natural decay (e-to-the-t) of the system. But, the complex part corresponds to the oscillations of the system. In other words the real part of the number will represent the transient effects of the system, while the complex part will represent the sinusoidal steady-state. Therefore to do a steady-state sinusoidal analysis we can replace the 'D' operator with
, this is the phasor transform.
Figure 10.1 Transient and steady-state parts of the differential operator
An example of the phasor transform is given in Figure 10.2 A phasor transform example. We start with a transfer function for a mass-spring-damper system. In this example numerical values are assumed to put the equation in a numerical form. The differential operator is replaced with jω, and the equation is simplified to a complex number in the denominator. This equation then described the overall response of the system to an input based upon the frequency of the input. A generic form of sinusoidal input for the system is defined, and also converted to phasor (complex) form. (Note: the frequency of the input does not show up in the complex form of the input, but it will be used later.) The steady state response of the system is then obtained by multiplying the transfer function by the input, to obtain the output.
Figure 10.2 A phasor transform example
To continue the example in Figure 10.2 A phasor transform example values for the sinusoidal input force are assumed. After this the method only requires the simplification of the complex expression. In particular having a complex denominator makes analysis difficult and is undesirable. To simplify this expression it is multiplied by the complex conjugate. After this, the expression is quickly reduced to a simple complex number. The complex number is then converted to polar form, and then finally back into a function of time.
Figure 10.3 A phasor transform example (cont'd)
Figure 10.4 Calculations in polar notation
The cartesian form of complex numbers seen in the last section are well suited to operations where complex numbers are added and subtracted. But, when complex numbers are to be multiplied and divided these become tedious and bulky. The polar form for complex numbers simplifies many calculations. The previous example started in Figure 10.2 A phasor transform example is redone using polar notation in Figure 10.5 Correcting quadrants for calculated angles. In this example the input is directly converted to polar form, without the need for calculation. The input frequency is substituted into the transfer function and it is then converted to polar form. After this the output is found by multiplying the transfer function by the input. The calculations for magnitudes involve simple multiplications. The angles are simply added. After this the polar form of the result is converted directly back to a function of time.
Figure 10.5 Correcting quadrants for calculated angles
Consider the circuit analysis example in Figure 10.6 Phasor analysis of a circuit. In this example the component values are converted to their impedances, and the input voltage is converted to phasor form. (Note: this is a useful point to convert all magnitudes to powers of 10.) After this the three output impedances are combined to a single impedance. In this case the calculations were simpler in the cartesian form.
Figure 10.6 Phasor analysis of a circuit
The analysis continues in Figure 10.7 Phasor analysis of a circuit (cont'd) as the output is found using a voltage divider. In this case a combination of cartesian and polar forms are used to simplify the calculations. The final result is then converted back from phasor form to a function of time.
Figure 10.7 Phasor analysis of a circuit (cont'd)
Phasor analysis is applicable to systems that are linear. This means that the principle of superposition applies. Therefore, if an input signal has more than one frequency component then the system can be analyzed for each component, and then the results simply added. The example considered in Figure 10.2 A phasor transform example is extended in Figure 10.8 A example for a signal with multiple frequency components (based on the example in Figure 10.2). In this example the input has a static component, as well as frequencies at 0.5 and 20 rad/s. The transfer function is analyzed for each of these frequencies components. The output components are found by multiplying the inputs by the response at the corresponding frequency. The results are then converted back to functions of time, and added together.
Figure 10.8 A example for a signal with multiple frequency components (based on the example in Figure 10.2 A phasor transform example)
10.3 VIBRATIONS
Oscillating displacements and forces in mechanical systems will cause vibrations. In some cases these become a nuisance, or possibly lead to premature wear and failure in mechanisms. A common approach to dealing with these problems is to design vibration isolators. The equations for transmissibility and isolation is shown in Figure 10.9 Transmissibility. These equations can compare the ratio of forces or displacements through an isolator. The calculation is easy to perform with a transfer function or Bode plot.
Figure 10.9 Transmissibility
Figure 10.10 Drill problem: Select a K value
10.4 PROGRAMS
- solving a problem using Scilab
Figure 10.11 Scilab Analysis for Phasors
- solving a problem using C
Figure 10.12 C Program for Analysis for Phasors
- solving a problem using C
Figure 10.13 C Program for Analysis for Phasors
10.5 SUMMARY
· Phasor transforms and phasor representations can be used to find the steady state response of a system to a given input.
· Vibration analysis determines frequency components in mechanical systems.
10.6 PRACTICE PROBLEMS
1. Develop a transfer function for the system pictured below and then find the response to an input voltage of Vi = 10sin(1,000,000 t) using phasor transforms.
2. A single d.o.f. model with a weight of 1.2 kN and a stiffness of 340 N/m has a steady-state harmonic excitation force applied at 95 rpm (revolutions per minute). What damper value will give a vibration isolation of 92%? Use phasors to do the calculations.
3. Four helical compression springs are used, one at each corner of a piece of equipment. The spring rate is 240 N/m for each spring and the vertical static deflection of the equipment is 10mm. Calculate the mass of the equipment and determine the amount of isolation the springs would afford if the equipment operating frequency is twice the natural frequency of the system.
4. a) For the circuit below find the transfer function and the steady state response for an input of Vi=5sin(1000t)V.
b) Verify the results in part a) by explicitly solving the differential equation.
10.8 ASSIGNMENT PROBLEMS
1. Given an input of F=5sin(62.82t), find the output, x, using phasors for the following transfer functions.