Table 5.1: Assigned ValuesObjective:The objective of this laboratory exercise was to analyze the response of a system to a sinusoidal voltage source by using sinusoidal steady state phasor analysis techniques.Phasors:The concept of phasors comes from the analysis of sinusoidal currents and voltages. These sinusoidal currents are called “alternating current” or AC for short. When a current or voltage can take the form:)cos()( tAti[A] (5.1)or)cos()( tAtv[V] (5.2)then corresponding phasors I and V are written as:AI  [A] (5.3)andAV  [V] (5.4)Since phasors are really complex numbers, I and V can also be written as:)sin()cos(jAAI  [A] (5.5)and)sin()cos(jAAV  [V] (5.6)where j2 = -1.Resistors, Capacitors, and Inductors in Phasor Domain:In phasor domain, a resistor with a sinusoidal current going across it will exhibit asinusoidal voltage. Since ohm’s law states that v=iR, for our sinusoidal current i(t) = Acos(t+), it follows that v(t)=RAcos(t+). The phasor form of this voltage is:1RIjRAARRAV  )sin()cos((5.7)A capacitor in time domain can be analyzed using i(t)=Cdv/dt. It can be shown that the derivative of a phasor is that phasor multiplied by j. Therefore, taking the voltage to be sinusoidal gives i(t)=C*jAcos(t+). The phasor form of this is:CVjI(5.8)or,ICjV1(5.8b)An inductor in time domain can be analyzed using v(t)=Ldi/dt. Assuming a sinusoidal current gives v(t)=L*jAcos(t+). The phasor form of this is:CIjV(5.9)In phasor domain, the property of impedance is much like resistance in time domain. It can be added when in series, and added like resistors in parallel. The value ofimpedance for a circuit component is V/I, which is R for resistors, jL for inductors, and –j/(C)When analyzing circuits in phasor domain, the same rules apply as in time domain. Kirchhoff’s laws are satisfied. Voltage and current divider work the same way. Mesh and node-voltage analysis also works.Analysis:The circuit analyzed in lab is shown in figure 5.1.KCL at the negative input of the first op-amp gives:201)2sin((RflVdtftAdC(5.10)This can be simplified to:)2cos(21 ftfjRflCAV(5.11)Also, by converting the cosine to sine:)2cos(21 ftfRflCAV(5.12)Now both the input signal and the are in terms of the same sine function. Taking this function as the reference phasor allows the elimination of the time dependance.KCL at the negative input of the second op-amp gives:2112)2sin(RfVoRVRftA(5.13)Simplifying for Vo:1212)2sin(2RRfVRftARfVo (5.13b)Substituting eq. 5.12 into 5.13b gives:12)2cos(22)2sin(2RRfftfRflCARftARfVo(5.14)This equation can be cleaned up by setting:22RRfAR (5.15)and3122RRffRflCAI(5.16))(tan180122RIIRjIRVo(5.17)In lab, all values were preset. These values are found in table 5.1.Table 5.1: Assigned ValuesVariable ValueR2 10kRf2 10kRf1 1kR1 7.5kC .12uFA 10Vf 1kHzThese values lead to a value of R = 10V and I = 10.053V. Therefore, |Vo|=14.18V and =134.85.Build and