Module-7: Frequency Domain Analysis of Linear Systems
Akshya Swain
Department of Electrical, Computer & Software Engineering, The University of Auckland,
Auckland, New Zealand.
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 1 / 30
Learning Outcome of This Module
I After completion of this module, the student must be able to do the following:
1. Analytically compute the frequency response of a second order system.
2. Determine various frequency domain specifications such as bandwidth, resonant
frequency etc
3. Could correlate the frequency domain specifications with time domain response.
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 2 / 30
What is Frequency Response of a System
I It is the steady-state response of a system to a sinusoidal input.
1. In frequency-response methods, we vary the frequency of the input signal over a
certain range and study the resulting response.
I For linear systems, the frequency of input and output signal remains the same i.e. in linear systems, (energy transfer from input to output occurs at same frequency, while the ratio of magnitude of output signal to the input signal and phase between two signals may change.
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 3 / 30
Root Locus Method vs Frequency Response Method
Advantages of Root locus Method:
i. Good indicator of transient response;
ii. Explicitly shows location of all closed-loop poles;
iii. Trade-offs in the design are fairly clear. Disadvantages of Root Locus Method
i. Requires a transfer function model (poles and zeros); ii. Difficult to infer all performance metrics;
iii. Hard to determine response to steady-state (sinusoids) iv. Hard to infer stability margins
I Advantages of Frequency Response Methods
i. Frequency response methods are a good complement to the root locus techniques: ii. Can infer performance and stability from the same plot
iii. Can use measured data rather than a transfer function model iv. Design process can be independent of the system order
v. Time delays are handled correctly
vi. Graphical techniques (analysis and synthesis) are quite simple.
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 4 / 30
Computing Steady State Response of Linear System to Sinusoidal Inputs-1
Proof:
I Consider the stable,linear, time-invariant system with transfer function G(s) shown in Figure.
I The input and output of the system are denoted by x(t) and y(t), respectively.
I If the input x(t) is a sinusoidal signal, the steady-state output will also be a sinusoidal signal of the same frequency, but with possibly different magnitude and phase angle.
The steady-state output of a transfer function system can be obtained directly from the sinusoidal transfer function, that is, the transfer function in which s is replaced by jω, where ω is frequency.
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 5 / 30
Computing Steady State Response of Linear System to Sinusoidal Inputs-2
Let us assume that the input signal is given by
x(t) = Xsinωt
I Suppose that the transfer function G(s) can be written as a ratio of two polynomials in s as:
G(s) = p(s) = p(s)
q(s) (s+p1)(s+p2)…(s+pn)
I Without loss of generality, it is assumed that the transfer function has only distinct poles located at −p1, −p2, . . . , −pn.
I The Laplace transform of the output Y (s) is expressed as:
Y (s) = G(s)X(s) = p(s)X(s). q(s)
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 6 / 30
Computing Steady State Response of Linear System to Sinusoidal Inputs-3
I It is worth noting that the steady-state response of a stable, linear, time-invariant system to a sinusoidal input does not depend on the initial conditions. (Thus, we can assume the zero initial condition.)
I The Laplace transform of the input signal x(t) = Xsinωt is given by X(s)= ωX
I Thus
s2 +ω2
Y (s) = G(s)X(s) = G(s) ωX s2 +ω2
= a + a + b1 + b2 s+jω s−jω (s+p1) (s+p2)
+…+ bn (s+pn)
where where a and the bi, (where i = 1, 2, ., n) are constants and a is the complex conjugate of a.
I Taking the inverse Laplace transform gives
y(t) = ae−jωt + aejωt + b1e−p1t + b2e−p2t + . . . + bne−pnt
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 7 / 30
Computing Steady State Response of Linear System to Sinusoidal Inputs-4
I The output response to the sinusoidal input is expressed as
y(t) = ae−jωt + aejωt + b1e−p1t + b2e−p2t + . . . + bne−pnt
I For a stable system, the poles lie in the left half of the s-plane.
I Therefore, as time approaches infinity, the terms e−p1 t ,e−p2 t ,. . .,e−pn t approach
zero.
I Thus, all the terms on the right hand side of above equation, except the first
two, drop out at steady state.
I The steady state response of the system is given by
yss(t) = ae−jωt + aejωt
where the constants a and a can be evaluated as:
a =G(s) ωX (s+jω)
a =G(s) ωX (s+jω) s2+ω2
=− =−
X G ( − j ω ) 2j
X G ( − j ω ) 2j
s2+ω2
s=−j ω
s=−j ω
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 8 / 30
Computing Steady State Response of Linear System to Sinusoidal Inputs-5
I Since G(jω) is a complex quantity, it can be written in the following form G(jω) = |G(jω)|ejφ
where |G(jω)| )l represents the magnitude and φ represents the angle of G(jω); that is,
φ = \G(jω) = tan−1 Imaginary part of G(jω) Real part of G(jω)
I The expression for G(−jω) is written as:
G(−jω) = |G(−jω)|e−jφ = |G(jω)e−jφ
I But as derived before
a = G(s)s2+ω2 (s + jω)
ω X a = G(s)s2+ω2 (s + jω)
X G ( − j ω ) 2j
X G ( − j ω ) 2j
ω X
= =
s=−j ω
s=−j ω
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 9 / 30
Computing Steady State Response of Linear System to Sinusoidal Inputs-6
I The expression for steady state output can now be simplified as: yss(t) = ae−jωt + ae−jωt
= X|G(jω)|ej(ωt+φ) − e−j(ωt+φ)
2j = X|G(jω)|sin(ωt + φ)
= Y sin(ωt + φ)
where Y = X|G(jω)|.
I Thus, it is concluded that
a. If a stable, linear, time-invariant system is excited by a sinusoidal input will, then at steady state, the output will be sinusoidal.
b. The frequency of the output will be equal to that of the input.
c. But the amplitude and phase of the output will, in general, be different from
those of the input.
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 10 / 30
Computing Steady State Response of Linear System to Sinusoidal Inputs-6
I The amplitude of the output is given by the product of that of the input and |G(jω)|, while the phase angle differs from that of the input by the amount φ = \G(jω).
I An example of input and output sinusoidal signals is shown in Figure
I : For sinusoidal inputs
Y (jω) Y (jω) Y (jω)
|G(jω)| = |X(jω)|, \G(jω) = \X(jω), G(jω) = X(jω) where G(jω) is called the sinusoidal transfer function.
I The sinusoidal transfer function of any linear system is obtained by substituting jω for s in the transfer function of the system.
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 11 / 30
Example:Computing Steady State Response of Linear System to Sinusoidal Inputs
I Consider the system shown in Figure.
I The transfer function G(s) is
G(s) = K
1+Ts
I For sinusoidal input x(t) = Xsinωt, , the steady-state output yss(t) can be
found as follows:
I Substituting jω for s in G(s) yields
G(jω) = K 1+jωT
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 12 / 30
Example:Computing Steady State Response of Linear System to Sinusoidal Inputs
I The amplitude ratio of the output to the input |G(jω)| and the phase angle φ is given by
|G(jω)| = √ K , φ = \G(jω) = −tan−1ωT 1+ω2T2
I Thus, for the input x(t) = Xsinωt, the steady-state output yss(t) is given as: yss(t)= √ XK sinωt−tan−1ωT
1+ω2T2
I Note that for small values of ω, the amplitude of the steady-state output yss(t) is almost equal to K times the amplitude of the input.The phase shift of the output is small for small values of ω.
I For large ω, the amplitude of the output is small and almost inversely proportional to ω.The phase shift approaches −900 as ω approaches infinity.
I This is a phase-lag network.
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 13 / 30
Frequency Response of Closed Loop Systems-1
I The closed loop transfer function of a single loop feedback control system is expressed as:
M(s) = C(s) = G(s) R(s) 1 + G(s)H(s)
I Under the sinusoidal steady state, s = jω and the above equation becomes M(jω) = C(jω) = G(jω)
R(jω) 1 + G(jω)H(jω)
I The sinusoidal steady-state transfer function M(jω) may be expressed in terms
of its magnitude and phase as:
where
M(jω) = |M(j(ω)|\M(jω) = M(ω)\φm(ω)
G ( j ω ) | G ( j ω ) | M(ω) = 1 + G(jω)H(jω) = |1 + G(jω)H(jω)|
φm(jω) = \G(jω) − \1 + G(jω)H(jω)
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 14 / 30
Frequency Response of Closed Loop Systems-2
Significance and Desired Characteristics of M(ω)
I M(ω) can be regarded as magnification of the feedback control system.
I It is similar to the gain or amplification of an electronic amplifier.
I Ideally, an amplifier must have a flat gain for all frequencies;realistically,it should have a flat gain in the audio frequency range.
I In control systems the ideal design criterion is similar.
I However, if it is desirable to keep the output C(jω) identical to the input R(jω)
at all frequencies, then M(jω) must be unity for all frequencies.
I However, M(jω) can be unity only when G(jω) is infinite, while H(jω) is finite
and nonzero.
I An infinite magnitude for G(jω) is impossible to achieve in practice, nor would it be desirable because
a. Most control systems become unstable when its loop gain becomes very high. b. Furthermore, all control systems are subjected to noise. Thus, in addition to
responding to the input signal, the system should be able to reject and suppress noise and unwanted signals.
I This means that the frequency response of a control system should have a cutoff characteristic in general, and sometimes even a band-pass characteristic.
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 15 / 30
Frequency Response of Closed Loop Systems-3
I The figure below shows the gain and phase characteristics of an ideal low-pass filter which has a sharp cut-off frequency at ωc. It is impossible to realize it physically.
I Typical gain and phase characteristics of a feedback control system are shown in below figure.
I Note that,the great majority of control systems have the characteristics of a low-pass filter, so the gain decreases as the frequency increases.
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 16 / 30
Frequency Domain Characteristics & Specifications
I In the design of linear control systems using the frequency-domain methods, it is necessary to define a set of specifications so that the performance of the system can be identified.
I Specifications such as the maximum overshoot, damping ratio, and the like used in the time domain can no longer be used directly in the frequency domain.
I The following frequency domain specifications are often used in practice. Resonant Peak Mr
a. The resonant peak Mr is the maximum value of |M(jω)|.
b. The magnitude of Mr gives indication on the relative stability of a stable closed-loop system.
c. Normally, a large Mr corresponds to a large maximum overshoot of the step response.
d. In practice the desirable value of Mr should be between 1.1 and 1.5.
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 17 / 30
Frequency Domain Characteristics & Specifications
Resonant Frequency ωr
– The resonant frequency ωr, is the frequency at
which the peak resonance Mr occurs.
Bandwidth BW
– The bandwidth BW is the frequency at which |M(jω)| drops to 70.7% of, or 3dB down from, its zero-frequency value.
a. The bandwidth of a control system gives indication on the transient response properties in the time domain.
i. A large bandwidth corresponds to a faster rise time, since higher-frequency signals are more easily passed through the system.
ii. Conversely, if the bandwidth is small, only signals of relatively low frequencies are passed, and the time response will be slow and sluggish.
iii. Thus the bandwidth and rise time are inversely proportional to each other.
b. Bandwidth also indicates the noise-filtering characteristics and the robustness of the system.
i. The robustness represents a measure of the sensitivity of a system to parameter variations.
ii. A robust system is one that is insensitive to parameter variations.
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 18 / 30
Frequency Domain Characteristics and Specifications
Cutoff Rate
a. Often, bandwidth alone is inadequate to indicate the ability of a system in distinguishing signals from noise.
b. Sometimes it may be necessary to look at the slope of |M(jω)|, which is called the cutoff rate of the frequency response, at high frequencies.
c. Apparently, two systems can have the same bandwidth, but the cutoff rates may be different.
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 19 / 30
Resonant Peak (Mr),Resonant Frequency (ωr) and Bandwidth (BW) of a Second Order System
I Consider the closed-loop transfer function of a second-order system C ( s ) ω n2
M(s) = R(s) = s2 +2ζωns+ωn2 (1) I At sinusoidal steady state, s = jω, Eq. (1) becomes
C ( j ω ) ω n2 1
M(jω) = R(jω) = (jω)2 + 2ζωn(jω) + ωn2 = 1 + j2(ω/ωn)ζ − (ω/ωn)2 (2)
I Letting u = ω/ωn, Eqn(2) can be expressed as:
M(ju) = 1 (3)
1 + j2uζ − u2
I The magnitude and phase of M(ju) are respectively given by
|M(ju)| = 1 (4) 1
\M(ju) = φm(ju) = −tan−1 2ζu (5) 1−u2
[(1 − u2)2 + (2ζu)2]2
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 20 / 30
Resonant Peak (Mr),Resonant Frequency (ωr) and Bandwidth (BW) of
a Second Order System
I The resonant frequency is determined by setting the derivative of |M(ju)| with respect to u to zero. Thus,
=0 (6)
(7)
(8)
2 d|M(ju)|=−1(1−u2)2+(2ζu)2−3 4u3−4u+8uζ2
du 2 This gives
4u3 −4u+8uζ2 = 4u(u2 −1+2ζ2) = 0 I TherootsofEq. (7)areur =0and
ur =1−2ζ2
I The solution of ur = 0 merely indicates that the slope of the |M(ju)|-versus-ω curveiszeroatω=0;itisnotatruemaximumifζ islessthan0.707. Eq. (8) gives the resonant frequency
ωr = ωn1 − 2ζ2 (9)
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 21 / 30
Resonant Peak (Mr),Resonant Frequency (ωr) and Bandwidth (BW) of a Second Order System
I Note that the frequency is a real quantity.
I Hence Eq. (9) is meaningful only for 2ζ2 < 1 or ζ < 0.707.
I This implies that, for all values of ζ greater than 0.707, the resonant frequency is ωr = 0 and Mr = 1.
I Substituting Eq. (8) into Eq. (4) for u and simplifying, we get 1
Mr = , ζ ≤ 0.707 (10) 2ζ 1−ζ2
a. It is important to note that, for the prototype second-order system, Mr is a function of the damping ratio ζ only, and ωr is a function of both ζ and ωn.
b. Furthermore, although taking the derivative of |M(ju)| with respect to u is a valid method of determining Mr and ωr, for higher-order systems, this analytical method is quite tedious and is not recommended.
c. ForsecondordersystemMr =1andωr =0whenζ≥0.707.
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 22 / 30
Resonant Peak (Mr),Resonant Frequency (ωr) and Bandwidth (BW) of a Second Order System
I√ From the definition of bandwidth, set the value of |M(ju)| to 1
≈ 0.707. This (11)
(12)
(13)
gives
Then
which gives
11
|M(ju)|=
1 =√ [(1 − u2)2 + (2ζu)2]2
2
2
22 21√ (1−u ) +(2ζu) 2 = 2
u2 =1−2ζ2±4ζ4 −4ζ2 +2
I The plus sign should be chosen in equation (13), since u must be a positive real quantity for any ζ. Therefore, the bandwidth of the prototype second-order system is determined from Eq. (13) as:
1
BW =ωn 1−2ζ2+4ζ4 −4ζ2 +2 2 (14)
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 23 / 30
Relationships Between the Time-domain Response and the Frequencydomain Response.
1. The maximum overshoot of the unit step response in the time domain depends upon ζ only. Eq (15)
2. The resonance peak of the closed-loop frequency response Mr depends upon ζ only Eq(16).
3. The rise time increases with ζ, and the bandwidth decreases with the increase of ζ, for a fixed ωn , Eq.(18), Eq(17).
- Therefore, bandwidth and rise time are inversely proportional to each other.
4. Bandwidth is directly proportional to ωn.
5. Higher bandwidth corresponds to larger Mr .
− √πζ
Mp = e 1−ζ2 (15)
1−2ζ2+4ζ4 −4ζ2 +2 2 (17)
1 + 1.1ζ + 1.4ζ2
tr ≈ ω (18)
n
1
2ζ 1−ζ2
Mr=
1
(16)
BW =ωn
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 24 / 30
Example-1:Computation of Frequency Domain Specifications
Consider a unity feedback system with forward-path transfer function
G(s) = K
s(s + 6.54)
Analytically find the resonance peak Mr,resonant frequency ωn, and bandwidth BW of the closed loop system for the following values of K:
a. K = 5
b. K = 21.39
c. K=100 Solution:
I We know the standard form of G(s) is G ( s ) =
ω n2
s(s + 2ζωn)
I Comparing with the given G(s), we find ωn2 = K. This implies ωn = √K. I Further
6.54 6.54 2ζωn=6.54, =⇒ζ=2ω=√
n2K
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 25 / 30
Example-1:Computation of Frequency Domain Specifications(contd)
I We know that
1
Mr= 2, ωr=ωn 1−2ζ2,tr≈
2ζ 1−ζ
1
BW =ωn 1−2ζ2+4ζ4 −4ζ2 +2 2
1 + 1.1ζ + 1.4ζ2
ω
n
a. For K = 5
ωn =√5=2.24rad/sec, ζ= 6.54 =1.46, Mr =1.0,ωr =0rad/sec
4.48 BW = 0.863 rad, tr = 2.5 sec
b. For K = 21.39
ωn = √21.39 = 2.24 rad/sec, ζ = 6.54 = 0.707, Mr = 1.0, ωr = 0.0648 rad/sec
9.24
BW = 4.62 rad, tr = 0.5357 sec c. For K = 100
ωn = √100 = 10 rad/sec, ζ = 6.54 = 0.327, Mr = 1.618, ωr = 0.0648 rad/sec 20
Akshya Swain r Module-7: Frequency Domain Analysis of Linear Systems 26 / 30 BW=14.34rad,t =0.15sec
Example-2:Computation of Frequency Domain Specifications
For a second order unity-feedback control system with the close loop transfer function
G(s) = C(s) = ωn2
R(s) s2 + 2ζωns + ωn2
it is desired that the maximum overshoot must not exceed 10% and the rise time must be less than 0.1sec. Analytically, find the corresponding limiting values of Mr andBW.
Solution:
I We will use the following formula which gives the relationship between the percentage overshoot Mp and damping ζ.
−√πζ
1−ζ2
1
−ln(%M /100) p
Mp(%) = e
Mr= 2, ωr=ωn 1−2ζ2,tr≈
, ζ = π2 + ln2(%Mp/100)
1 + 1.1ζ + 1.4ζ2
ω
2ζ 1−ζ
1
BW =ωn 1−2ζ2+4ζ4 −4ζ2 +2 2
n
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 27 / 30
Example-2:Computation of Frequency Domain Specifications (contd)
Given Mp = 0.1. This gives ζ = 0.59 I Using these relations we get
1
Mr = = 1.05
2ζ 1−ζ2
Given tr = 0.1sec, we determine the value of ωn from the relation
1 + 1.1ζ + 1.4ζ2
tr ≈ ω = 0.1sec
n
I Thus the minimum ωn = 17.77rad/sec
1
I The minimum BW = ωn 1−2ζ2+4ζ4 −4ζ2 +2 2 = 20.56rad/sec
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 28 / 30
Example-3:Computation of Frequency Domain Specifications
The forward path transfer function of a system with integral controller H(s)=K is
s
G(s) = 1 10s+1
a. Find K when the closed loop resonant peak Mr = 1.4.
b. Determine the frequency at resonance , overshoot for step input.
Solution:
The closed loop transfer function is
Given
C(s)=GH= K = K
R(s) 1+GH
10s2 +s+K s2 +0.1s+0.1K
=1.4 =⇒ ζ=0.387
Mr=
1
2ζ 1−ζ2
I Comparing the closed loop transfer function with the standard one, we get ζωn = 0.1. This gives ωn = 0.129 rad/sec.
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 29 / 30
Example-3:Computation of Frequency Domain Specifications (contd)
Further, since
ωn2 = K, =⇒ K = 10ωn2 = 0.16669. The resonant frequency
I The peak overshoot
ωr = ωn1 − 2ζ2 = 0.1 rad/sec
πζ
Mp=exp − =0.268
1−ζ2
Akshya Swain
Module-7: Frequency Domain Analysis of Linear Systems 30 / 30