Download 27.02.2012 ECE 232 Lab2 Second Order Circuits Preliminary Work

27.02.2012
ECE 232 Lab2
Second Order Circuits
Preliminary Work:
1) Consider the series RLC circuit of Fig.1.
R = 10kΩ pot.
C = 4.7 nF
L = 0.1 H
Vin(t) =Vpeak. u(t) volts
Fig. 1
a. Show that the differential equation for Vc(t) is given as,
d2
R d
1
1
V (t ) 
VC (t ) 
VC (t ) 
Vin (t )
2 C
L dt
LC
LC
dt
The characteristic equation for this differential equation is,
The roots of the characteristic equation are denoted by s1 and s2 when they are distinct and by s0 when the
roots are repeated.
b. When the roots if the characteristic equation are repeated the response is called critically damped. Determine
the value of the resistor R=R0 for the critically damped response.
c. Determine the values of the currents and voltages at t  0  , t  0  and t   for all circuit elements.
d. For R>R0 (overdamped response):
i. Solve the differential equation and show that the variables Vc(t) and iL(t) are given below.
VC (t )  V peak 
i L (t ) 
V peak
s1  s 2
( s 2 e s1t  s1e s2t ) ,
1 V peak
(e s1t  e s2t )
L s1  s 2
where s1 and s2 are the real distinct roots of the characteristic equation.
ii. Determine VL(t), VR(t), ic(t), iR(t) using the above results.
iii. Determine and sketch VL(t), Vc(t), VR(t), ic(t), iL(t) and iR(t) for R=3R0.
e. For R=R0 (critically damped response):
i. Show that the variables Vc(t) and iL(t) are given as,
VC (t ) V peak V peak (s0 t 1) e s0t ,
i L (t ) 
V peak
L
t e s0 t
where s0 is the only (repeated) root of the characteristic equation.
ii. Repeat Part 1.d.ii for R=R0.
iii. Repeat Part 1.d.iii for R=R0.
f.
For R<R0 (underdamped response):
i. Show that the variables Vc(t) and iL(t) are given as,
VC (t )  V peak  V peak
i L (t ) 
V peak
dL
 0 t
d
e sin ( d t  arctan
),
d

e t sin ( d t )
where s1     j d and s 2     j d are the complex conjugate roots of the characteristic equation
and  0 
1
.
LC
ii. Repeat Part 1.d.ii.
iii. Repeat Part 1.d.iii for R=R0/4.
2) Consider the parallel RLC circuit of Fig. 2.
R1 = 10 KΩ
C = 0.1 μF
L = 0.1 H
Fig. 2
a. Obtain the differential equation for Vc(t).
b. Determine the value of R=R0 for the critically damped response.
c. Determine the value of Vc(t) at t  0  , t  0  and t   .
EXPERIMENTAL WORK
1) Set up the circuit of Fig. 1. Adjust the square wave output of the signal generator so that Vin(t) is a 2 Vp-p
square wave with 150Hz frequency.
a. Determine experimentally R=R0 for the critically damped response.
b. Observe and plot (Ac components only) Vin(t), VL(t), VR(t) and VC(t), also note the DC levels for the following
cases:
i. R=3R0 ,
ii. R=R0 ,
iii. R=R0/4
2) Set up the circuit of Fig.2. and repeat Part 1.
3) Set up the following circuit for f = 10kHz. Observe and plot V0(t) for the following cases:
a. S1 and S2 open. b. S1 closed and S2 open. c. S1 and S2 closed.
R1 = 10 KΩ
R2 = 10 KΩ
C = 0.1 μF
L = 0.1 H
1a.
C = 4.7 nF
L = 0.1 H
Vin(t) =Vpeak. u(t) volts
,
KVL around the loop gives;
,
,



or we can write this equation in the operator format

Call
which has the characteristic equation
,
,
,

1b. If the roots are repeated;

For L=0.1H, C=4.7nF 
1c. For
there is no initial condition stated here for any circuit element, therefore
, also since step fuction u(t) is used in this
experiment,it is zero until t reaches zero. Furthermore since u(t) is a bounded function, there is no
change in the initial current of the inductor and initial voltage of the capacitor. Both initial current of the
inductor and initial voltage of the capacitor. Both must be continuous.
1d. I.
(overdamped response)
The characteristics equation of the homogenous eqn. is
,
,

so we have 2 real roots.
,

soln. of the homogenous equation
a particular soln.
put into eqn., to get;
,



&

,
II. Determine
using the above results.


III. Determine and sketch
,
and
, R=3Ro=27875.9Ω, C = 4.7 nF, L = 0.1 H
The plots are as in experimental work part.
1e. I.
for R=3Ro.
(critically damped response)

,
,
put into eqn., to get;


;
II.
III.
, R=Ro=9225.3Ω, C = 4.7 nF, L = 0.1 H
1f. I.
(underdamped response)

,
,

Complex roots:
 if we put in diff eqn.,

=0
0
II.
III.
, R=Ro/4=2306.3Ω,

2a.
R1 = 10 KΩ
C = 0.1 μF
L = 0.1 H
Vin(t) = Vpeak. u(t) volts
Firstly,

b.Characteristic eqn.
,
For critically damped;
c. For
For
,
,
For
, so at steady-state circuit diagram will behave as
Nothing was said about the initial condition. Since there is no impulsive source on the circuit, the
capacitor voltage would be continuous.