π ω π ω π

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HW#3
ME 106/EE106 Mechatronics
Dr. Du
1. Given a RLC circuit with values show on the right,
+
Vin
L
R
R = 1 k
L = 0.1 H
C = 0.1 F
Vout
C
(1) Derive the transfer function of the RLC circuit.
(2) Plot the Bode diagram using MATLAB.
(3) Find the cutoff (corner) frequency.
(4) Is it a high-pass or a low-pass filter?
Problem 2 (This problem has been partially solved, please finish the solution by filling in the blanks) The
signal from a sensor looks like the figure below.
(1) Design an RC filter that removes (i.e., attenuates by 90%) the high frequency component (noise) from
the signal. Assume that your filter will drive a load of 20 k.
(2) What is the corner frequency of the filter you designed?
Solution:
The given signal is the superposition of a main signal (the lower frequency one) and a noise signal (the higher
frequency one) – a typical output signal from a sensor. A low-pass filter can be designed to remove (actually
attenuated) the noise signal.
The frequency of each signal can be found as follows (attention: the unit of the given period is 1 ms = 0.001 s in
the graph):
 main  2 f main 
2

Pmain
 noise  2 f noise 
Dr. Winncy Du, E310F, MAE Dept., SJSU, Tel.:408-924-3866,E-mail: [email protected]
Course Web: www.engr.sjsu.edu/wdu/ME106Spring2010
To design a RC filter is to choose a proper R and C values so that the useful signal (the lower frequency signal
in this case) can pass through, while the noise signal (with the higher frequency) will be attenuated. Assume that
we want the noise signal to be attenuates by 90%, that means
Vout , noise
Vin , noise
1

1 
2
noise
R 2C 2
 0.1
RC 
then,
How do we decide what value of R or C should be? One consideration is that we try to keep
Vout , main
Vin , main
close to
1 as possible as we can. That means the main (useful) signal is kept to its original value as possible. In the real
practice, we often send the filter output to a device (called “load” here). Assume a load of 20 k is added to the
circuit (see the graph below). To a low frequency input, the capacitor is near to be open (recall the voltage and
current relationship of a capacitor: V (t ) 
dV (t )
1
dV (t ) 1
is very small
I (t )dt or
 I (t ) . If the change

dt
C
dt
C
(slow), the current I(t) will be very small, near to zero).
Thus, there is a voltage division will occur between R
and RL
i.e.,
R
+
-
Vin
C
RL
Vout
Vout 
RL
Vin
R  RL
We don’t want R to get too much voltage because there
will be nothing left for the load Vout .
That is, we want Vout / Vin is close to 1:
Vout 
RL
Vin
R  RL

Vout
RL
1


Vin R  RL R / RL  1
That means the ratio R / RL must be as small as possible. Usually R / RL < 10% is good enough. Let us take
R / RL = 5%, thus (assume a load of 20 k),
R
Then the capacitor can be determined as
C
Please continue to finish this problem by drawing the Bode diagram, and then find its corner frequency.
Dr. Winncy Du, E310F, MAE Dept., SJSU, Tel.:408-924-3866,E-mail: [email protected]
Course Web: www.engr.sjsu.edu/wdu/ME106Spring2010
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