CHAPTER 5
INTRODUCTION TO SYSTEMS
AND FREQUENCY RESPONSE FUNCTIONS
When we model the behavior of measurement systems we often break down the system into
components that are connected together in series. These components are typically transducers,
amplifiers, bridge circuits, demodulation devices, filters, analog to digital converters,
oscilloscopes and computers. Many of these components can be modeled with very simple
linear ordinary differential equations. Typically we use first and second order differential
equations to model these components. If the system is well designed, then it is very
straightforward to take models of the components and combine them to derive a model of the
whole measurement system.
In this chapter the solution of linear ordinary differential equations will be reviewed, and we
will introduce the terminology commonly used in instrumentation to describe the response of
these equations. By using the response to sinusoidal excitation, we will also describe what is
meant by the System's Frequency Response and we will show how this can be used to predict
what happens when any sinusoidal excitation is inputted to the system.
We will concentrate on two types of excitation: a step input which is what happens when we
move from one constant input to another constant input, and a sinusoidal input, which is the
simplest dynamic input to study. In Chapter 8 we will show how to take complicated signals and
break them down in a sum of sinewaves. The building blocks we use in this chapter will be used
again to predict the response of systems to these more complicated inputs.
Throughout our analysis we will be assuming that the system is linear. Having performed our
calibration (Chapter 2), we should know that we are operating in a region where the system can
be modeled as a linear system. One property of linear systems that we will use is that of
superposition. Suppose we put two separate signals, x1 and x 2 into the measurement system or
a component of the measurement system. Further suppose that the responses were y1 and y 2 ,
respectively. Now, if the input to the system is a combination of x1 (t) and x 2 (t) :
x(t)
x1 (t)
x 2 (t) ,
(1a)
y(t)
y1 (t)
y 2 (t) ,
(1b)
then the output will be:
where
and
are constants.
Another property of linear systems is observed when a sinusoidal excitation is input into the
system. The steady state response is a sinewave of the same frequency, though the amplitude and
phase may be changed by the system.
Input:
x(t) = A sin
1t
, Output:
y(t) = B sin (
1
t+
) .
5-2
Before considering the differential equation models, we will revise some complex notation
that we will use. While seeming more abstract, the complex notation approach will simplify the
mathematics considerably and help you solve the equations much faster.
Finally having considered models of components, we will show how, under certain
conditions, we can take the frequency response functions of the individual components and
multiply them together to produce an overall frequency response function for the entire system.
REVISION OF COMPLEX NOTATION
1 . A complex variable z can be written in two ways:
We will use j to denotes the
z
x
Ae j ,
jy
(2)
where x, y, A and
are all real. x is the real part of z, denoted by Real(z), and y is called the
imaginary part of x, denoted by Imag(z). A is called the magnitude, the modulus or the
amplitude, and is denoted by z .
is the phase, sometimes denoted by Arg(z). A pictorial
representation of z is shown in Figure 1.
Imaginary
x
z
A
y
Real
Figure 1: Representation of a complex variable on an Argand diagram.
From the diagram, using simple trigonometry we can see that:
A2
x2
x
A cos
y2 ,
tan
1
y
,
x
(3a and b)
,
(3c and d)
and
,
y
A sin
5-3
Sines and Cosines
Using this notation we can split up sines and cosines into exponentials. From equations (2) and
(3c and d) we can see that:
ej
t
e
j t
cos t
(4a)
j sin t ,
and similarly,
cos t
(4b)
j sin t ,
Note that we have set to equal t. This is just to be consistent with the way we will see the
sines and cosines appear later on. Adding equations (4a) and (4b) and dividing by two gives:
1 j
e
2
t
1
e
2
j t
cos t .
(5a)
Subtracting equations (4a) and (4b) and dividing by 2j gives:
1 j
e
2j
t
1
e
2j
j t
(5b)
sin t .
We will use this decomposition of sines and cosines into two complex exponentials later on in
this chapter.
Differentiating and Integrating
In the simple use that we will make of complex variables, we can usually just treat j as a
constant. So if we differentiate exp( j t ) with respect to t, we will obtain:
d j
e
dt
t
j ej
t
,
(6a)
Similarly, if we integrate we will get:
t
j t
e dt
0
ej
j
t t
0
ej
j
t
1
.
j
(6b)
Notice, as always, the introduction of the constant (1/ j , in this case) after substituting in the
initial condition at t = 0.
5-4
Functions of Complex Variables
Consider the complex function, F(z). Using z = x + jy, we can express F(z) in its real and
imaginary form thus:
F(z)
where u, v, A and
u(x, y)
A(x, y) e j
j v(x, y)
(x,y)
,
(7)
are real functions of x and y.
Often in this course, the complex function we are considering will be something we call a
frequency response function and is typically a function of frequency, . If you wish, you could
think of this as having come from a function of a complex variable, x + jy, where x = 0 and y =
. We most often look at this function in its modulus and phase form, and typically, the form of
the complex function is a polynomial in j divided by a polynomial in j . For example:
bo b1( j )
1 a1( j ) a 2 ( j )2
bo
(1 a 2
jb1
) ja1
(8)
.
2
When calculating the magnitude and phase of this function, it is easier to put the numerator and
denominator into magnitude and phase form and use the relationship:
B ej b
A ej a
B j(
e
A
a)
b
.
(9)
Following from this, the example complex function's magnitude and phase are:
Magnitude
b2o (b1 )2
2 2
(1 a 2
)
(a1 ) 2
Phase
tan
b1
bo
1
tan
1
a1
1 a2
2
.
(10)
Complex Conjugates
The complex conjugate of a complex quantity is found by replacing j by -j in the quantity. So if
we have:
f
u
jv
Ae j
, f*
u
jv
Ae
j
.
(11)
Notice that we use a * to denote the complex conjugate. Also, taking the conjugate has no effect
on the magnitude (A) and changes the sign of the phase. Multiplying a complex quantity by its
complex conjugate gives:
f f*
u2
v2
A2 .
5-5
Use the expressions in (11) to prove this result.
Example
If T = (2 + 4j)/(1 - 3j), calculate the magnitude, the phase and the complex conjugate of T.
Solution
Magnitude = Magnitude of top / Magnitude of bottom.
Magnitude2
(22 42 ) /(12 ( 3)2 )
20 /10 2
Magnitude = 2 1.414
Phase = Phase of the top - Phase of the bottom.
Phase = tan 1(4 / 2) tan 1( 3/1)
tan 1 2 tan 1 3 2.356 rads 135
Complex conjugate: replace j by -j.
(2-4j)/(1+3j) = 1.414 e j2.356 .
Using MATLAB to Evaluate Complex Functions
When you are evaluating complex functions, MATLAB understands j or i to mean 1 , unless
you use those letters to denote another variable. It will also calculate the magnitude (abs) and
phase (phase). Below is an m-file example of how to evaluate the function above.
%ch4ex1.m program to evaluate a given complex function
%
%Specify coefficients
b0=1;b1=0.9; a1=0.25;a2=0.01;
omega=input('Input frequency at which you wish to evaluate function ')
F=(b0+j*omega*b1)/(1-omega*omega*a2+j*omega*a1);
[abs(F) phase(F)]
We can also plot this as a function of for a set of values of . Suppose we wanted to plot the
function for
0 to 100 in steps of 0.5 rad/s. The m-file for this would be:
%ch4ex2.m program to evaluate a given complex function and plot.
%
%Specify coefficients
b0=1;b1=0.9;a1=0.25;a2=0.01;
%
%Specify omega
omega=(0:0.5:100);
%
%Evaluate the function, notice the use of the "." to specify
%element by element calculations.
F=(b0+j*omega*b1)./(1-omega.*omega*a2+j*omega*a1);
%
%Now plot the magnitude in dB and the phase in degrees.
subplot(211);
plot(omega,20*log10(abs(F)));
5-6
xlabel('Frequency rad/s')
ylabel('Magnitude in dB')
title('Example Frequency Response')
subplot(212)
plot(omega,phase(F)*180/pi)
ylabel('Phase in degrees')
xlabel('Frequency rad/s')
The result of running this m-file is shown in Figure 2.
Figure 2: Plot of a complex function, magnitude and phase
5-7
SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS
The modeling of the components in the measurement system will usually lead to linear, ordinary
differential equations. The most general form of such an equation is:
an
dny
dt n
an
d n 1y
1
dt n 1
a1
dy
dt
y
bm
d mx
dt m
bm
d m 1x
1
dt m 1
b1
dx
dt
box(t)
(13)
y is the response of the measurement system component,
x is the input to the measurement system component,
n is the order of the system.
Many measurement system components can be modeled by either first or second order
differential equations. For example:
dy(t)
y(t) K x(t)
(14)
dt
could be an equation that models the dynamic behavior of a thermocouple.
is called the time
constant and is related to how quickly the system responds, and K is the static sensitivity, which
is an indication of the gain of the system. In this example its units would be V/ C. It is the same
as the sensitivity calculated from the static calibration.
1 d2 y(t)
2
dt 2
n
2 dy(t)
dt
n
y(t)
K x(t)
(15)
could be the differential equation of a strain-gage accelerometer.
n is called the natural
frequency, is called the damping ratio, and K is again the static sensitivity. n and affect
how quickly the system responds and how quickly it settles down after a perturbation.
How do we find y(t) if we know x(t)?
There are two parts to the solution, which come from the superposition principle. The input, x(t),
can be thought of as x(t) + 0.0, and so we need to find a solution with the right hand side as
given, and add to it the solution with the terms involving x(t) set to 0. Since by convention we
put all the terms in x(t) on the right hand side (RHS) of the equation, then this second solution is
formed by setting the RHS to zero.
The first solution is called the particular solution, and will have a form very similar to x(t).
The second solution is called the homogeneous solution. The homogeneous solution will
involve some unknown constants (number equal to the order of the system), which are found by
substituting known initial conditions into the final solution. So the solution steps are outlined
below. Also given is the measurement systems terminology for the particular and homogeneous
solutions.
5-8
Stage 1:
Particular Solution (Steady State Solution)
Pick a solution that looks like the function x(t) but is in its most general form. For
example:
If x(t) = A cos 5 t , then choose yp (t)
B cos (5 t
).
So here the solution has the same frequency of the input but has undergone an
amplitude and a phase change. B and are found by substituting the expressions for
x(t) and y(t) into the differential equation and comparing coefficients.
Another example:
If x(t) = t 2 then choose y p (t)
ao
a1t a 2 t 2 .
Again substitute into the differential equation and compare coefficients.
Stage 2:
Homogeneous Solution (Transient Solution)
Set the RHS of the differential equation to zero. Choose a solution of the form:
yh (t) e
t
(16)
Substitute into the differential equation (RHS = 0), and divide by e t . This will yield
a polynomial in . Consider equation (13) as an example. The resulting polynomial
will be:
an
n
an
n 1
a1
1
(17)
1 0
Use a root solver on the computer to find the roots. Let's denote these by:
1,
2,
n.
There are actually n possible homogeneous solutions and we combine them to form
the full homogeneous part of the solution. Since these are all solutions to an
excitation of 0, we can multiply them by constants, which we will need to determine.
So,
y h (t)
Stage 3:
A1 e
1t
A2 e
2t
A3 e
3t
An e
nt
Total Solution
To form the total solution, add the particular and the homogeneous solution. In
measurement language, add the steady state and the transient solution.
(18)
5-9
y(t)
ysteady state
Stage 4:
(19)
yp (t) yh (t)
ytransient
Evaluating the constants from initial conditions
To find the total solution you need to evaluate the constants that arose in the
homogeneous solution. Initial conditions are conditions given at t = 0. You need n
initial conditions, where n is the order of the system. We assume that we know x(t),
the input, for all time, t.
1st Order system:
2nd Order system:
3rd Order system:
etc.
y(0) should be known.
y(0) and y(0) should be known.
y(0), y(0) and y(0) should be known.
So take your total solution and differentiate it (n-1) times, substitute in t = 0 and
equate to known initial conditions. You should now have n equations and n
unknowns. Solve for A1, A 2 , A n .
Example
Consider a thermocouple with a time constant of .1 seconds and a static sensitivity of 3 mV/ C.
Initially the output is 0.01 Volt. The temperature is suddenly changed to 100 C? What would the
response be?
Solution
The differential equation is:
dy
dt
y
Kx
where
t = 0s.
0.1 and K
Stage 1:
The input is a constant and so the steady state solution (particular solution) will also
be a constant. Set y = C and substitute into the above differential equation. When we
differentiate a constant the result will be zero. The input is 100 after t = 0s. Hence,
C
Stage 2:
0.003V/ C . Let's assume that the temperature change occurred at time:
K 100
0.3Volts.
Hence, ysteady state =0.3Volts.
Set the RHS of the differential equation to 0. Assume a solution of the form e t and
substitute into the homogeneous form of the differential equation. Divide through by
e t . This results in
1 0. Hence,
1
1
0.1
10.
5-10
The transient solution (homogeneous solution) is therefore:
t
y transient
Stage 3:
Ae
10t
The total solution is the sum of the transient and steady state solutions:
y(t)
Stage 4:
Ae
0.3 Ae
10t
The initial condition (only one needed because its a 1st order system) is y(0) = 0.01
Volts. From the expression for the total solution with t = 0 substituted in, we see that,
0.01 = 0.3 + A, implying that A = - 0.29.
The total solution is therefore:
y(t)
0.3 0.29e
10t
Volts, for t
0,
and
y(t) = 0.01 Volts for t < 0.
The response of this system is shown in Figure 3.
Figure 3: The response of a first order system to be a step change in constant input
5-11
Example: Including bias effects
Suppose that a thermocouple similar to the one in the previous example is being used. The time
constant of .1 seconds and a static sensitivity of 3 mV/ C. Initially the output was 0.01 Volt
when the temperature was 18 C. The temperature is suddenly changed to 100 C? What would
the response be?
Solution
Note that K.18 = 0.054 does not equal 0.01 Volts, i.e., there is a bias in the measurement system.
The differential equation describing this system is:
dy
y b Kx
dt
where
0.1 and K = 0.003V/ C. b is the bias, which we can determine from knowing that the
output was a constant 0.01 Volts when the temperature was 18 C, prior to the step input. From
the differential equation, this gives:
0.01 b 0.003 18 Volts
Which yields b = 0.044.
We can follow the same solution strategy as before if we do a substitution of variables: z = y – b.
Because b is a constant, z y . The differential equation in z is therefore:
dz
dt
Stage 1:
z
K x.
The input is a constant and so the steady state solution (particular solution) will also
be a constant. Set z = C and substitute into the above differential equation. When we
differentiate a constant the result will be zero. The input is 100 C after t = 0s. Hence,
C = K 100 = 0.3 Volts. Hence, zsteady state
Stage 2:
0.3 Volts .
Set the RHS of the differential equation to 0. Assume a solution of the form e t and
substitute into the homogeneous form of the differential equation. Divide through by
e t . This results in
The transient solution (homogeneous solution) is therefore:
z transient
Stage 3:
Ae
10t
The total solution is the sum of the transient and steady state solutions:
z(t) 0.3 A e
10t
.
5-12
Stage 4:
The initial condition (only one needed because it’s a 1st order system) is y(0) = 0.01
Volts and hence z(0) = 0.054. From the expression for the total solution with t=0
substituted in, we see that,
0.054 = 0.3 + A, implying that A = - 0.246 .
The total solution is therefore:
y(t)
b z(t)
0.044 0.3 0.246 e
10t
Volts,for t
0.
and
y(t) = 0.01 Volts for t < 0.
Example
An LRC circuit can be modeled as a 2nd order differential equation with a natural frequency of
1000 rad/s and a damping ratio of 0.01. In a calibration test the sensitivity was found to be 2
Volts/Volt. Before t = 0s the input is a constant voltage of 0 Volts. At t = 0 the input is changed
to a sinusoidal voltage of amplitude 6 Volts peak to peak and of frequency 10 Hz: 3 sin(100 t).
Calculate the response.
Solution
The differential equation describing this system is equation (15):
1 d 2 y(t)
2
dt 2
n
2 dy(t)
n dt
y(t)
K x(t)
which becomes, after substituting in the values for the natural frequency, damping ratio and
sensitivity,
d 2 y(t)
dy(t)
2 10 5
y(t) 2 x(t)
2
dt
dt
We can approach forming the steady state solution in two ways:
(1) Assume a solution of the form: B sin (100 t ) and substitute into the
differential equation. Compare the coefficients of sin(100 t ) and of cos(100 t )
to find B and .
10
Stage 1
6
(2) Recognize that we can split the sine into two exponentials:
x(t)
3 j100
e
2j
t
3
e
2j
j100 t
5-13
and using the principle of superposition, we can calculate the response to each
exponential separately. We can then combine them to form ysteady state (t).
We will adopt the second approach here, because we use a similar approach when
determining system frequency response functions. For ease of notation, let 100
x(t) e j
t
produces y(t) T1 e j t ,T1 may be complex.
3 j
e
2j
Hence: x(t)
x(t) e
j t
t
j t
produces y(t) T2 e
3
e
2j
hence: x(t)
3
T1 e j
2j
produces y(t)
j t
t
.
,
3
T2 e
2j
produces y(t)
j t
.
Therefore, x(t) = 3sin( t) produces
ysteady state (t)
3
T1 e j
2j
3
T2 e
2j
t
j t
,
We need to find T1 and T2 . Substituting
x(t)
ej
t
T1 e j
, y(t)
t
into the differential equation gives:
T1 10 6 ( j )2
2 10
5
(j )
1 ej
t
2 ej
t
,
leading to:
T1
2
10
Recall that
T1
6
(j )
2
2 10
5
.
(j ) 1
is 100 . Substitution into the above equation yields:
2
1 0.0987
j 0.0063
.
The response to the first exponential is:
.
5-14
3
T1 e j100
2j
ysteady state1
t
Following exactly the same procedure for x(t) e
T2
2
6
10
( j )
2
2 10
5
j t
and y(t) T2 e
j t
we get:
T1*.
( j ) 1
That is, T2 is the complex conjugate of T1 .
3 *
T1 e
2j
ysteady state2
j100 t
.
Combining the two solutions gives:
ysteady state
3 T1 sin(100 t phase(T1 ) ),
where:
T1
6.66
phase(T1 )
Stage 2:
tan
1
0.0063
0.9013
0.007 radians.
Now we form the homogeneous or transient solution. Set x terms to 0, assume a
solution of the form e t , substitute into the equation, and divide through by e t . This
yields:
10
6 2
2 10
5
1 0.
The roots of this are:
1,2
10
100 1,000,000
10
j999.95
Therefore the transient (homogeneous) solution is:
y transient
Stage 3:
A1e(
10 j999.95) t
A 2e (
10 j999.95) t
The total solution is the sum of the steady state and transient solutions.
y(t) 6.66sin(100 t 0.007) A1e(
10 j999.95) t
A2e(
10 j999.95) t
5-15
Stage 4:
Now use the initial conditions. The system was at rest prior to the application of the
sinusoidal input and so y(0) and y(0) both equal 0. This leads to the two equations:
0
0
6.66 sin( 0.007) A1 A 2
666 cos( 0.007) ( 10 j999.95) A1 ( 10
A1* and that:
Solving these equations, we find that A2
A1 1.046 e j1.549 , A2 1.046e
j1.549
j999.95) A 2 .
.
To do this calculation you could use the following MATLAB m-file.
ch4ex3.m Program to solve two simultaneous equations.
%
y = [ 6.66*sin(0.007) , -666*pi*cos(0.007) ]';
X = [ 1 1
-10+j*999.95 , -10-j*999.95 ];
a=X\y
Combining these exponential terms we obtain a damped cosine wave.
y transient
2.092e
10t
2.092 e
10t
1 j(999.95 t
e
2
1
e
2
1.549)
j(999.95 t 1.549)
,
and hence
y transient
cos(999.95t 1.549).
The total solution is therefore:
y(t) 6.66sin(100 t 0.007) 2.092 e
10t
cos(999.95t 1.549).
The response of this system is shown in Figure 4.
5-16
Figure 4: The response of a second order system to a sinusoidal excitation, turned on at t = 0.
(top) Total solution, (middle) transient solution, (bottom) steady state solution.
5-17
FREQUENCY RESPONSE FUNCTIONS
When characterizing the behavior of measurement systems, we are interested in whether they can
respond fast enough to track the signals we wish to measure. The faster the fluctuations in a
signal, the higher the frequencies are in the signals. This is illustrated in Figure 5.
Figure 5: (a) Slower fluctuations, e.g. a 5 Hz sine wave,
(b) faster fluctuations, e.g., a 100 Hz sine wave
To characterize this, we examine how the steady state response of the system varies with the
frequency of a sine wave input. The system will affect the amplitude and the phase of the
sinewave and hence if the input is
x(t)
Asin( t
),
(20)
the output will be
y(t)
A T sin( t
phase(T)).
(21)
T e j phase(T) is complex and varies with . We often write it as T( j ) to emphasize this
frequency dependence. | T | is the magnitude of what is called the frequency response function
and phase (T) is the phase of the frequency response function.
T
It will become clear that knowledge of T( j ) will give us complete knowledge of the system.
That is, if we know T( j ) , we can also write down the differential equation model of the
measurement system. For the types of measurement system components we are interested in
T( j ) is often in the form:
5-18
T( j )
bo b1( j ) b2 ( j )2
1 a1( j ) a 2 ( j )2
b NB ( j ) NB
a NB ( j ) NA
(22)
Furthermore, NB = 0 or 1 and NA = 1, 2 or 3, are typically values for the types of measurement
component systems that we study.
The Ideal Frequency Response Function
Before going on to calculate frequency response functions, let us discuss what its ideal properties
would be. Ideally we would like the shape of the input signal, x(t), to be preserved as it passes
through the measurement system. This means that we do not mind if x(t) is scaled or time
shifted, but all frequency components present in the signal must be affected in the same way, i.e.,
ideally:
y(t)
K x(t t o ) ,
where t o is the delay and K is a constant. If we plotted the ideal frequency response gain versus
frequency, it would be a constant. How does the time delay affect the phase of each frequency
component? If,
x(T) A sin( t) , then y(t)
K Asin( t
to)
(23)
The time delay introduces a phase that is a linear function of frequency:
an ideal frequency response function is either 0 (t o
t o . So the phase of
0) or a linear function of frequency.
It is unlikely that a system would exhibit these ideal characteristics for all frequencies. When
we purchase or develop our own measurement system we look for the region of frequencies
where these ideal characteristics are almost satisfied. We define that region as the operating
frequency range of the measurement system. That is, all frequency components in the signals we
wish to measure should lie in that region.
Calculating the Frequency Response Function from the Differential Equation
We will derive this frequency response function for the general system model (without bias) as
given in equation (13). We adopt an approach seen in a previous example where we split a
sinewave excitation into two complex exponential terms.
Consider only one of the exponentials:
y(t)
x(t)
e j t , and denote the steady state response by
T e j t . Alternatively, we could consider the steady state response of the system to an
excitation x(t) X e j t to be y(t) Y e j t . Because this is a linear system and the excitation is
a single frequency, we know that Y = X.T, and hence we can find T, the frequency response
5-19
function of the system, from Y/X. Substitute these expressions for x(t) and y(t) into equation
(13).
a n ( j )n Y e j
t
a n 1( j ) n 1 Y e j
t
Yej
t
b oX e j
b1( j )Xe j
t
bm ( j )m Xe j
t
t
Comparing coefficients of e j t and collecting terms in Y and X we obtain
a n ( j )n
a n 1( j ) n
1
a n 2 ( j )n n
2
a1( j ) 1 Y
bo b1( j )
bm ( j ) m X
Dividing by X times the polynomial in the brackets in front of Y, we can express T, the system
frequency response function, as:
T( j )
Y
X
bo b1( j ) b2 ( j )2
1 a1( j ) a 2 ( j )2
bm ( j ) m
a n ( j )n
(24)
Derivatives in the Differential Equation and (jω) Terms in the Frequency Response Function
Each time we took a derivative of e j t we increased the power of the (jω) in the term.
dmy
dt
( j )m Y e j t
So by seeing a term 6( j )5 , which is equal to 6j
5
etc.
(25)
, in the denominator of the frequency
5
d y
term in the differential equation. Similarly, if
dt 5
we see a term 2.3 2 , which is equal to –2.3 ( j )2 , in the numerator of the frequency response
response function we know that there was a 6
function we know that there was a 2.3
d2x
term on the RHS of the differential equation.
dt 2
Example
What is the differential equation model relating the excitation, f(t), to the response, z(t), of a
linear system, if the frequency response function, H( j ) , is given by:
4j
1
2
3
0.02 j
.
5-20
Solution
Expressing f(t) and z(t) as Fe j t and Ze j t , respectively, then H( j )
2
1
Z ej
t
Z ej
0.02 j
( j )2 Z e j
t
t
0.02( j ) Z e j
Z / F. Therefore:
4j
3 F ej
t
.
t
4( j ) F e j
t
3 F ej
t
.
Therefore the differential equation is:
z z 0.02z
3f .
4f
Response to sine waves
If we have an input e j
e
j 1t
1t
, the response is T( j
1) e
j 1t
.
By similar argument, if we have an input
, the response is
T( j
1)
e
j 1t
T( j
*
1)
e
j 1t
T( j
1)
e
j phase(T( j 1))
e
j 1t
,
where * denotes complex conjugate. So the response to
x(t)
B sin( 1t
(26)
)
is the sum of the responses to
B j j
e e
2j
1t
and
B
e
2j
j
e
j 1t
which is
y(t)
T( j 1)
B j j phase(T( j 1)) j 1t
e e
e
2j
T( j 1)
B j
e
e j phase(T( j 1))e j 1t
2j
Combining these complex exponentials into a sinewave we have
y(t)
T( j 1) Bsin
1t
phase(T( j 1)) .
(27)
The frequency response function tells us how the amplitude and phase of a sinewave change as
the sinewave passes through a system. So once we have the frequency response function, the
steady state response of the system, to any sinusoidal excitation, can be calculated.
Example
5-21
Calculate the steady state response to x(t) = 3 sin 100t of the system whose behavior is
described by:
5y y
40x
Solution
Step 1:
Calculate the frequency response function, G( j ).
Let x(t)
X e j t and y(t)
5( j )Y e j t
Y
X
G( j )
Step 2:
100 rad/s.
40
1 j 500
Calculate the Magnitude and Phase of G( j100).
40
G( j100)
2
1
phase (G( j100))
Step 4:
40 X e j t
40
5j 1
Evaluate G( j ) at
G( j100)
Step 3:
Yej t
Y e j t . Substitute into the differential equation.
2
500
1
0.0800
2
500
0 tan 1
1
1.5688 rads.
Write down the steady state response to x(t) = 3 sin 100t.
y(t)
3 0.0800sin(100t 1.5688)
0.2400sin(100t 1.5688).
Example
The frequency response of a thermocouple plus an AC amplifier measurement system is
T( j )
(1
K 3j
1 j ) (1
2 j ) (1 j 3 )
(28)
(a) What was the differential equation model of the system?
(b) What is the steady state response of this system to a sinusoidal fluctuation about a mean
temperature,
x(t) 20 5sin 30t C
5-22
Solution
(a) Expand factors in the frequency response function and multiply both sides of (28) by the
denominator of the frequency response function.
1 ( 1
2
3) j
( 1 2
2 3
1 3 )( j
)2
( 1 2 3 )( j )3
Y
X
K 3( j )
Multiply both sides by X e j t and use the results shown in equation (22) to give the differential
equation:
y ( 1
dy
( 1 2
3)
dt
2
2 3
1 3)
d2y
dt
2
d3y
1 2 3
dt
K 3
2
(b) Split this input up into the steady state response to the constant: x1(t)
dx
dt
20 C and the steady
state response to the sinewave: x 2 (t) 5sin 30t C, giving y1(t) and y2 (t) , respectively.
The total steady state response is then
y1(t) y2 (t)
Steady state response to x1(t)
Either, consider x1(t) to be 20 cos t where
y1(t)
20 T( j0) cos(0t phase (T(0))
0, hence the steady state response is
dx
0 and that the steady state
0. Or, notice that
dt
response to a constant is a constant, and therefore
dmy
0 for all m 1. So the L.H.S. of the
dt m
differential equation becomes y and the right hand side is zero, giving a solution y1 0.
Steady state response to x 2 (t)
Here the frequency is 30 rad/s so the steady state response is
y2 (t) 5 T( j30) sin(30t phase(T( j30)))
Evaluate the T( j ) at
30 rad/s, the frequency of the excitation.
T( j30)
T( j30)
(1
K 3 j30
1 j30) (1
2 j30) (1 j 3 30)
30 K 3
1
1
1
1 (30 1)2 2 1 (30 2 )2 2 1 (30 3 )2 2
phase(T(j30)) = phase (top) – phase (bottom)
5-23
30 3
tan 1
0
2
(tan 1 30 1 tan 1 30 2
(tan 1 30 1 tan 1 30 2
tan 1 30 3 )
tan 1 30 3 )
Note that we calculated the phase of each of the factors in the denominator:
phase{1 j
tan 1
i}
i
1
and summed them up. This is faster than expanding out the denominator. We are using the
property of products of complex variables. If z i
z1z 2z 3
A1e j
1
A 2e j
2
A 3e j
Aie j i then
(A1A 2A 3 )e j(
3
so z1z 2z3 A1A2A3 i.e., z1 z 2 z3 and phase(z1z 2z 3 )
phase(z1) phase(z2 ) phase(z3).
1
1
2
3)
2
3
i.e.,
So the total response is
y(t)
y1(t) y2 (t) 0 5 T( j30) sin 30t phase(t( j30))
where T( j30) and phase(T( j30)) are given above.
Example
An accelerometer can be modeled as a second order system:
m
d2v
dt
2
c
dv
dt
kv
Ba
(29)
where v is the output voltage and a is the acceleration. m, c and k are the physical properties of
the accelerometer: mass, damping coefficient and stiffness respectively.
(a) What is the natural frequency, damping ratio and sensitivity of the accelerometer?
(b) Calculate the frequency response function.
(c) Plot the frequency response function. Plot the magnitude in dB, the phase in degrees and
plot both against a log frequency axis. This allows us to better see trends and makes it
easier to combine plots of different frequency response functions. Decibels (dB) are defined
as 20 log10 (magnitude).
5-24
Plot over a range that highlights all the features of the frequency response function. Choose
.7, .1, and .001 and K 10mV/(m/s2 ).
n 10,000 rad/s,
Solution
(a) The standard form for a second order differential equation is
1 d 2 v(t)
2
n
dt
2 dv(t)
n dt
2
v(t)
(30)
K a(t)
where n is the undamped natural frequency, is the damping ratio and K is the
sensitivity. To put (28) in this form divide through by k. Equating coefficients we get:
m
k
Therefore
n
k
m
1
2
n
c n
2k
c
k
,
c
2 km
2
B
k
and
n
K
B
.
k
and K
(b) To calculate the frequency response function assume that a(t) A e j t and v(t) V e j t .
The frequency response function is then given by T = V/A. Substitute into (30) to get
1
2
n
( j )2
2
(j ) 1 V ej t
K Ae j t
n
Therefore,
T
V
A
K
(31)
2
1
2
n
j
2
n
(c) Since the natural frequency is 10,000 rad/s, let's plot from
will notice that when
n
102 to
106 rad/s. You
104 rad/s the real part of the denominator is zero and hence
the phase of T is 90 . Also T is close to a maximum here. As
gets much larger than
n the terms in the denominator become large and T becomes very small. Hence, by
plotting from 2 orders of magnitude below n to 2 orders of magnitude above
should see all the interesting features of the frequency response diagram.
Let's plot 5000 points of the frequency response function. To make the
spaced on a log scale, we choose the i so that
n , we
values equally
5-25
log10
2
i 10
i
i log10 (106 102 )
, i
4999
102
0, 1,
4999
(106 102 )(i / 4999)
(32)
In general,
i
min
( max
min )
i /(N 1)
In MATLAB a function logspace creates, the
i
0, 1,
N 1
(33)
i for you. To create a vector, om, of 5000
2
6
frequency values, equally spaced on a logarithmic scale from 10 to 10 type:
>>om=logspace(2,6,5000);
Since we will be calculating T for different values of , and also because we will be using
this form of the frequency response function many times, it would be a good idea to create
a MATLAB function to calculate the frequency response given K, n , and the frequency
vector om. We can then call this function from other m-files. To do this create an m-file of
the same name as the function as shown below.
%sndord.m function to calculate the value of a second order
%
system at values in the array om.
%
function [T] = sndord(K,omn,zeta,om);
T=K./((1-(om/omn).^2)+j*(2*zeta*om/omn));
The function (m-file) has to be in a directory MATLAB can find, e.g. the directory you are
working in. The MATLAB m-file to do the plotting will now look something like that
shown below.
% ch5fig6.m program to calculate a second order function
%
for different values of zeta.
%
Function: sndord.m is used
%
omn=10000; K=0.01;
zeta=[ 0.7 0.1 0.001 ];
om=logspace(2,6,5000);
t1=sndord(K,omn,zeta(1),om);
t2=sndord(K,omn,zeta(2),om);
t3=sndord(K,omn,zeta(3),om);
subplot(211)
semilogx(om,20*log10(abs(t1)),om,20*log10(abs(t2)),om,20*log1
0(abs(t3)));
axis(([100 1000000 -120 20])
set(gca,'YTick',-120:20:20)
grid
5-26
xlabel('Frequency - rad/s')
ylabel('Mag (dB - V/(m/s^2))')
subplot(212)
semilogx(om,angle(t1)*180/pi,om,angle(t2)*180/pi,om,angle(t3)
*180/pi);
axis([100 1000000 -180 0])
set(gca,'YTick',-180:30:0)
grid
xlabel('Frequency - rad/s')
ylabel('Phase - degrees')
The result of running this m-file is shown in Figure 6. Notice that as
decreases the peak
in the magnitude plot becomes more pronounced and the majority of the 90 phase shift
0.7 the peak in the frequency response
occurs over a shorter region of frequencies. At
magnitude has nearly disappeared.
Figure 6: Second order system frequency response for different
values of damping ratio. (a) magnitude ref 1mV/ (m/s2 ) , (b) phase.
Characteristics of a second order frequency response diagram
5-27
(a) Peak in the magnitude
A peak will occur when the magnitude of the frequency response denominator in (31) is at a
minimum. At this value of , the square of this quantity will also have a minimum. To
find this minimum differentiate with respect to
and set the result to zero. That is,
differentiate
2 2
1
2
2
2
n
n
to get
2
2 1
2
n
2
n
Setting this equal to zero and solving for
2
8 2
2
2
n
gives
(1 2 2 ) 2n
This gives the resonant natural frequency
1 2 2
res
n
When 2 1/2 (
.7071) this becomes imaginary, and a peak no longer appears. If
very small res approaches n .
is
(b) Phase change
2
tan 1 2 n 2 , if K 0 . When
n the phase change is 90 or
( n
)
/2 rads. This is often used as a method of finding the natural frequency of a second order
system.
The phase is
As
the phase tends to 180 or
radians.
(c) Roll-off in magnitude at high frequencies
A decade is an increase in frequency by a factor of 10. When
the transfer function is dominated by the 2 / 2n term, i.e.
T( j )
and
K2 4n
2
n
4
is large the magnitude of
5-28
10log10 T( j )2
A decade above this (
n
20log10 K 2n 40log10
n
20log10 K 2n
40log10 10
20log10 K 2n
40log10
40log10 10
20log10 K 2n
40log10
40
10 ) ,
10log10 T( j10 ) 2
So at high frequencies the magnitude drops by 40 dB every time we increase the frequency
by a factor of 10. We say that the roll-off is 40 dB/decade.
Example for the Reader
A first order system (RTD + bridge circuit) may be modeled by the differential equation
shown below,
dy
dt
y
x(t)
(a) What are the time constant ( ) and static sensitivity (K)?
(b) Derive the frequency response function in terms of
and K.
(c) If the cut-off frequency ( c ) is defined as the point where the frequency response function
magnitude drops 3 dB from its value at
0 , show that c 1/ . Prove that the phase of
the frequency response function is 45 at the cut-off frequency.
(d) Plot the frequency response function over a region that illustrates all its main features.
.05 seconds and K 35mV/ C).
(Choose
(e) At high frequencies, prove that the roll-off is 20 dB/decade. Check on your plot that this is
so.
(f) What is the steady state response of this system to
x(t) 5sin 2t 20sin 300t C?
Calculating a useful operating frequency range
We have noted earlier that to determine which signals we can measure accurately, we must
determine the operating frequency range of a system. That is, where the frequency response
function gain is constant and the phase is zero or a linear function of frequency. To simplify
5-29
matters, consider operating frequency regions where the phase is close to zero, i.e., systems
without time delays. Signals we wish to measure must only have frequency content in the
operating frequency range, if we wish to be able to measure the signals without distortion.
Usually a system will not exhibit this desirable behavior exactly in any frequency region, and
we must define an acceptable error in gain and phase that we are willing to accept. We then find
at what frequencies this measurement system lies within these error bounds; this defines the
operating frequency range. This is illustrated in Figure 7, where the measurement system is a
second order system and in this example we are setting the acceptable error to be that the gain is
flat within 1.0%. Note: though not shown here, there is a significant phase error in the region
defined by this error bound on the gain.
Figure 7: Acceptable error bounds on the frequency response of a
second order system, used to define an operating frequency range
0.1, (b)
0.9 and (c)
0.66
(a)
The "flat" part of the frequency response function of a second order system is approximately
equal to K, the static sensitivity and the value of the frequency response function gain at zero
5-30
rad/s. We wish to find at which frequency the magnitude becomes 0.99 K or 1.01 K, whichever
is the lower frequency. If the damping ratio is less than 0.7071 then the frequency response gain
will peak and thus it is likely that by solving T( j ) 1.01 K we will find the upper limit of the
operating frequency region. If the peak value of the frequency response function is less than
1.01 K or the damping ratio is larger than 0.7071, we will find the upper frequency limit by
solving T( j ) 0.99 K .
What is an acceptable error depends on the application and the cost of measurement errors.
While 1% error in the gain was used here, it is by no means a standard definition of an
acceptable error.
Example
A measurement system behaves as a first order system with a frequency response function:
20
1 j0.01
G( j )
The operating frequency range is defined as the frequency region where the gain error is less
than 0.5% and the phase error is less than 1 . What is the operating frequency range?
Solution
A first order system frequency response gain drops from its value at low frequencies. To find the
upper frequency limit as defined by the gain error, we solve:
202
1 (0.01 )
2
(0.995 20)2
Note that we are using the square of the frequency response gain. Inverting both sides of this
equation and multiplying by 400 gives:
1 0.0001 2
0.995 2 ,
giving a solution of:
100 0.995 2 1 10.04 rad/s.
Now consider a phase error of 1 0.0175 radians. The phase of a first order system drops from
0 at DC (0 rad/s) to 90 at high frequencies. Hence, to find the upper frequency limit defined
by this phase error, we solve:
0.01
0.0175 0 tan 1
1
This results in:
100 tan(0.0175) 1.75 rad/s
5-31
So the phase error limit restricts the operating frequency range much more than the gain error
limit. The operating frequency range for this first order system, as defined by the error criteria
specified, is 0 to 1.75 rad/s.
PUTTING SYSTEMS TOGETHER
T(t)
G1 (jω)
Thermocouple
y1(t)
G2 (jω)
Amplifier
y2(t)
G3 (jω)
Filter
v(t)
Figure 8: Block diagram of the measurement system
Suppose we know the frequency response of each of the components: G1,G2 and G3. Can we
say what the frequency response function of the entire system is?
When we connect the components together the voltage we measure at the output of each system
will drop. So,
>
y1
disconnected
from amp
disconnected
from filter
and
connected
to amp
>
y2
,
y1
,
y2
connected
to filter
This is caused by current flowing from the circuitry in one system to the next system. This is
discussed more fully in Chapter 7. If the subsystems are designed properly we can reduce the
voltage change so that the disconnected and connected voltages are approximately equal. If this
is the case, then the output of the subsystem component, when disconnected, becomes the input
to the next subsystem component, when it is connected.
Consider the steady state response of the measurement system to e j t . Passing this input
through the first component (thermocouple), the steady state response will be y1(t) G1 e j t .
We will assume connection effects are negligible and this becomes the input to the second
component. e j t input into the second component produces a steady state output G 2 e j t .
From linear system theory, G1 e j t input into the second stage will produce a steady state output
G1 times that produced by e j t , and so,
y2 (t)
G1 G2 e j t
5-32
Again we will assume that connection effects are negligible and this becomes the input to the
third component (filter). e j t input to the filter produces a steady state output G 3 e j t .
Therefore, G1 G2 e j t will produce a steady state output G1 G2 times this, i.e.,
v(t)
G1G 2 G 3 e j t
So we input e j t into the thermocouple, and G1 G2 G3 e j t is the steady state response of the
measurement system, if we can ignore connection effects. The frequency response function of
the entire system is therefore
G( j ) G1( j ) G2 ( j ) G3( j ) ,
i.e. the product of the individual frequency response functions. Note also, therefore,
G( j )
G1( j ) G 2 ( j ) G3( j )
and
phase (G( j )) phase(G1( j )) phase(G2 ( j ))
The steady state response of the entire system to T(t)
v(t)
AG
A G1( j o ) G2 ( j o ) G3( j o ) sin
o
ot
phase(G3( j ))
Asin o is
sin( ot phase(G)
o
)
phase(G1( j o )) phase(G 2 ( j o )) phase(G3 ( j o ))
This approach to putting systems together can be applied to many problems, not only
measurement system design. It is attractive to look at simpler subcomponents and then connect
them together using this block diagram approach. However, caution must be exercised when
doing this. The fundamental assumption, that there is no interaction between the systems when
connected, may not be true. Below is an example of using this block diagram approach to an
office noise problem.
Example
A printer in an office is making a noise that can be modeled by the sum of two sinewaves.
p(t) 0.25 sin 1t 0.125 sin( 2t
) Pascal
720 rad/s and 2 1080 rad/s. This is measured with a microphone that has a constant
frequency response function with a sensitivity of 50 mV/Pascal attached to an amplifier with a
constant frequency response function = 100 V/V. (This is not true at very high frequencies, but
is a sufficient model for this problem.)
1
John and Felicity share an office that is separated from the printer by a partition. The
transmission path from the printer to John and Felicity's desks is approximately modeled by the
frequency response function.
5-33
1
T( j )
1 j / 800
Pascal/Pascal
In addition, Felicity is wearing headphones with a frequency response function,
H( j )
0.9
(1 j / 600
Pascal/Pascal
Calculate the steady state sound pressure that Felicity will sense from the printer. If you
measured this with a microphone placed inside the headphones, what output voltage would you
measure. Assume as you used the same microphone to make this measurement, as you used to
measure the printer noise.
Solution
There are two systems here; (1) the sound pressure measurement and (2) the transmission path
through the partition and the headphones.
(1)
p(t)
(2)
mic
K=50mv/pascal
v(t)
Amp
K=100V/V
p1(t)
v1(t)
p(t) Transmission
through
partition
T(jω)
Headphone
H(jω)
p1(t)
Figure 9: Block diagrams of two systems. (1) MIC plus amp,
(2) transmission path plus headphones
The first system is attached to the second to make the measurement of the sound at Felicity's ear.
Assuming that no interaction effects occur between the systems when connected, the frequency
response function relating p1(t) to p(t) is
G1( j ) T( j )H( j )
1
.9
1 j / 800 1 j / 600
Pascal/Pascal
and the frequency response function relating p1(t) to v1(t) is
G2 ( j )
T( j ) H( j ) 0.05 100
4.5
V/Pascal
(1 j / 800) (1 j / 600)
and
G2 ( j ) 5 G1( j )
Since we are interested in the steady state response to sinewaves at 720 rad/s and 1080 rad/s, let's
evaluate the frequency response function at these values.
0.9
G1( j720)
,
(1 j720 / 800) (1 j720 / 600)
5-34
0.9
G1( j720)
(1
phase(G1( j720))
1
0.9 ) 2 (1 1.22 ) 2
2
1
0.4283 ,
tan 1 0.9 tan 1 1.2
1.6089 rads
Similarly,
G1( j1080)
G1( j1080)
phase(G1( j1080))
0.9
,
(1 j1080 / 800) (1 j1080 / 600)
.9
1
(1 1.35 ) 2 (1 1.82 ) 2
2
1
tan 1 1.35 tan 1 1.8
0.2602 ,
1.9969 rads
The steady state response of G1 to the input is therefore
p1(t) 0.25 0.4283 sin (720t 1.6089) 0.125 0.2602 sin (1080t
9.9969)
Pascal
The measured signal will be 5p1(t) since the microphone plus amplifier's response is considered
frequency independent in this example. So the measured voltage is
v1(t) 0.5353 sin(720t 1.6089) 0.1626 sin(1080t
1.9969) Volts
Note how the partition and headphones had a more significant effect on the 1080 rad/s
component than they did on the 720 rad/s component.
When solving these "wordy" problems, sketching out a block diagram of the system components
and their connections will help you to sort out what is actually happening. Associated with each
of the blocks should be enough information to write down or derive a frequency response
function. Once you have connected the system and derived the overall frequency response
function you know everything about the system. From this you can write down the differential
equation of the system, or you can quickly calculate the steady state response to a sinusoidal
excitation.
5-35
SUMMARY
In this chapter we have reviewed basic complex variable relationships and the solution to linear
ordinary differential equations. In particular, we have focused on steady state response
(particular solution) to sinusoidal and step inputs, though we have derived transient
(homogeneous) solutions as well.
Building on the response to sinusoidal excitation we derived the frequency response function
associated with a system by solving for the steady state response to e j t . This function is
directly related to the differential equation model of the system and its magnitude and phase tell
us how a particular frequency component is affected by the system, i.e.
x(t)
Asin ot, y(t)
A T( j o ) sin
ot
phase(T( j o ))
where x(t) is the input, y(t) the response and T( j ) the frequency response function.
We also noted that, for measurement systems, the useful frequency operating range was the
range where the gain of the frequency response function was flat and the phase was either zero or
a linear function of frequency, the latter corresponding to the system introducing a time shift into
the measurement. It is therefore very important to measure the frequency response function of
the measurement system to determine the useful frequency range.
At the end of the chapter we briefly looked at combining frequency response functions of
components of measurement systems to generate an overall system frequency response function.
If the components are well designed and voltages between subcomponents in the measurement
system do not drop when connections are made, then the frequency response function of a
measurement system, consisting of subsystems connected in series with frequency response
functions: G1( j ),G2 ( j ),G3( j ),and G4 ( j ) , will be
G( j ) G1( j ) G2 ( j ) G3( j ) G4 ( j )
This allows us to look at the simpler system components in isolation and model them separately.
Then, by putting them together we can derive the more complicated frequency response function
(model) of the whole system.