Lab Experiment 7
Characterization a 2nd Order System
by Frequency Response
Prelab Experiment 7
In the last lab we used the step response of the system to find its transfer function. In this experiment we
obtain the frequency response of the same system to again find its transfer function.
The transfer function for the DC servomotor that we used in the last lab experiment is:
G( s) 
K2
s( s  1)
(7.1)
where:
K2

= proportionality constant
= time constant
The frequency response of a system is the steady-state response of the system to sinusoidal inputs. By
replacing s with j  the transfer function becomes a frequency response function. The frequency response
function is complex so there is a resulting phase shift between the input and output sinusoid as well as a
change in the amplitude. The frequency response for transfer function in equation (7.1) is:
K2
G ( j ) 
j ( j

 1)
1
(7.2)
where,  = 1/1. A Bode diagram is a plot of the magnitude and phase of the transfer function. The
function “bode” in MATLAB  creates a Bode plot given the transfer function. For example, if K2 is = 48
and  is ¼ (i.e., 1 = 4), then the Bode plot for G(j) is shown in Figure 7.1. The straight-line
approximation to the slope of the magnitude plot for frequencies less than 4 rad/sec is –20 dB/decade. For
Figure 7.1 Bode Plot of Equation 7.2
1
frequencies greater than 4 rad/sec, the straight-line approximation is –40 dB/decade. Given the Bode
magnitude plot, the time constant can be found by finding frequency at which the straight-line
approximations to where the two slopes intersect. The time constant, , is equal to 1/(the angular frequency
of the intersection = 1/1).
Consider applying a sinusoidal input to a system and observing the output shown below.
1.5
inupt
output
1
0.5
t
Tp
0
-0.5
-1
-1.5
0
2
4
6
8
10
12
14
By inputting sinusoids of various frequencies and measuring the ratio of the output amplitude to the input
and converting to dB using:
 Output Magnitude 
Magnitude( dB)  20 log10 

 Input Magnitude 
(7.3)
the magnitude curve can be created. The phase curve can be measured by determining the phase difference
between input and output sinusoids. The phase difference is:
where
Phase Lag 
t
Tp
t 360
Tp
= time difference between the same two
points on each sinusoid for example the
zero crossing point
= period of sinusoid = 1/f where f is
frequency in Hz
2
(7.4)
Prelab Assignment
1. Using K1 = 3.3 and  = 0.25 sec, have MATLAB create a Bode plot for:
G( s) 
K1
s  1
Draw straight-line approximations to the magnitude plot to show that the lines intersect at 1/ .
Calculate  to show that it agrees with the value used to generate the Bode plot. Notice how the low
frequency asymptote intersects the vertical axis at 20 Log10(K1). Continue the line through 1/ onto the
phase plot. What phase could also be used to determine  ?
2. Calculate the closed-loop transfer function, T(s), if the plant given in Eq. (7.1), is placed in a negative
unity-feedback position control system.
3. Using MATLAB, calculate the Bode plot for the T(s) found in (2.) with K = 48 and  = 0.25 sec.
4. The value of K2 can be obtained from the Bode plot of T(s) by finding M peak , the maximum
magnitude of the frequency response curve, using equation (7.5) to find , and then finding K2 from
equation (7.6) (see last Lab Eqn. 1.11). Use the Bode magnitude plot from (3.) to find M peak and then
work backwards to find K2 thus verifying your understanding of the procedure. Compare the calculated
K2 to the value used in (3) to generate the Bode plot.
M peak 
 
1
2 1   2
1 1
2 K 2
3
(7.5)
(7.6)