Monday, May 30, 2016

Day 21:: Sinusoidal Analysis/ Phasor Lab

Today, we discuss how to apply the circuit analysis technique we have studied before and the phasor techniques to find the steady-state response of circuits with sinusoidal sources.
Part 1: 
Nodal Analysis
Nodal analysis can apply to the sinusoidal sources circuit the same way we did before. Only difference is we may calculate with complex number. This 
write down nodal equations at two essential nodes. 

do the algebra work to find the voltage at two essential nodes. 

Mesh Analysis


write down KVL for each loop. I3 is given, then solve the system to find each current. 


Part 2: Phasor Lab 

The cut of frequency wc=R/L, 10*wc , (1/10)*wc. 
In RL circuit, VL leads I, f increases, then the phase shift increases. 

The cut off frequency, phase shift between Vl and I is about 45, and the Vl leads the current. 

The results from experiments.

wc (w-cut-off)

phase shift (i-vl) = -45

10*wc

phase shift = -84.28

wc/10

phase shift= -5.7


Superposition 

Since ac circuits are linear, the superposition theorem applies to ac circuits the same way it applies to dc circuits.  The theorem becomes important if the circuit has sources operating at different frequencies. In this case, since the impedances depend on frequency, we must have a different frequency-domain circuit for each frequency.  The total response must be obtained by adding the individual responses in the time domain. It is incorrect to try to add the responses in the phasor or frequency domain.

Source transformation 
the review of source transformation
Thevernin and Norton Equivalent circuits

We transform the circuit to the impedance form. 
To find V Th , we apply KCL at node 1 to find I0.  Then apply KVL to the right hand loop. Vth = 55, -90. 
To find Zth, remove the independent source and connect an arbitrary fixed current source (In this case 3A since it makes the math easy) to terminals a and b, 


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