elc 106.2 lab 4 - complete response of second order circuits

8/12/2019 ELC 106.2 Lab 4 - Complete Response of Second Order Circuits

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III. DISCUSSION A ND R ESULTS

Solving the circuit in Fig. 1,

Fig. 1 Given Circuit

Since, there is no initial stored energy, the initial

conditions for the circuit is is:

(8)

(9)

Using nodal analysis at node v1 and v-, we can solve for

the differential equation for vout.

(10)

Solving for the natural response, the equation in the left

side will equate to zero. Then, vout will transform to

characteristic equation. After finding the roots, the natural

response equation will be:

[ ] (11)

In solving the forced response, it is easier to solve when

using the phasor equivalent circuit as shown in Fig. 2.

Fig. 2 Phasor equivalent circuit

Using nodal analysis,KCL @ node v1

(

) ( )

(12)

(13)

KCL @ node v-

( ) (

)

(14)

Substituting (14) in (13)

( )

(15)

Solving for vout,f (t),

√ (16)

Total response:

[ ] √

(17)

Using initial conditions, since vout =v2 ,

vout ( 0 )=v2(0)=0= A+ √

A= -√

A= -2

[ ] [ ]

√ (18)

Using equation (14), since vout = v2

( )

(19)

Substituting (19) to (18),

√ B = 6

Substituting the values of A and B to (17),

[ ] √

Fig. 3 Vin and Vout at sine wave with 318 Hz



Fig. 4 Vin and Vout at sine wave with 2 kHz

Fig. 5 Vin and Vout at square wave with 2 kHz

Fig. 6 Vin and Vout at square wave with 100 kHz (revised)

In fig. 3, it is the graph of both input voltage:

5cos(2000t), and the sinusoidal output voltage; however, the

output was shifted upward; thus, it is harder to distinguish the behavior of the output voltage from the input voltage. Channel

1’s volts/div value is 5, while 2mV for the channel 2. These

settings are further set to fig. 4, and fig. 5. Fig. 6 is the

revision of the square wave, now set to AC coupled, and has

0.1 ms/div time division and 50 mV/div. Fig. 7 possessed 100

kHz frequency in the signals, and doubled for the fig. 8. From

these figures, it can be seen that the output signals becomes

smaller as the input signal increases. As the frequency

increases, more noise can be observed in the graphs of both

input and output signals.



IV.

CONCLUSION

From the constructed circuit, consisting of two capacitors

and an op-amp, the variation of input signal was determined as

well as a change from sine wave to square wave and varying

frequencies.

It is easier to use the phasor equivalent circuit to find the

forced response of the circuit when the input signal is

sinusoidal.

It can be concluded, from the experiment, that as the

frequency increases, the output signal becomes smaller and

noisier.

R EFERENCES

[1] "Second Order Circuits." Eastern Mediterranean University Open

Courseware. N.p.. Web. 30 Jun 2013.

<http://opencourses.emu.edu.tr/pluginfile.php/602/mod_resource/conten

t/0/Lecture_Notes/second_order_circuits.pdf>.

http://opencourses.emu.edu.tr/pluginfile.php/602/mod_resource/content/0/Lecture_Notes/second_order_circuits.pdf




elc 106.2 lab 4 - complete response of second order circuits

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