Sinusoidal Steady State

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(Adapted by notes from T.F. Weiss)

Introduction and Outline

Previously we showed that for systems described sets of linear differential equations, we can easily derive the system's transfer function, H(s), in the form of a rational function in s (a ratio of polynomials). That is,

H(s) = {\sum_{\tilde{l}=0}^{\tilde{L}} b_{\tilde{l}} s^{\tilde{l}} \over \sum_{l=0}^L a_l s^l} = \frac{n_h(s)}{d_h(s)},
where L is the order of the system (assuming a_L \ne 0), the L values of s for which \sum_{l=0}^L a_l s^l = d_h(s) =0 are its poles or natural frequencies, and the \tilde{L} values of s for which \sum_{\tilde{l}=0}^{\tilde{L}} b_{\tilde{l}} s^{\tilde{l}} = n_h(s) = 0 are its zeros.

We also showed that given an input u(t) = U e^{s_0t} and a set of initial conditions, the output, \mathbf{y}, is given by

y(t) = \underbrace{\sum_{l=1}^L A_l e^{\lambda_l t}}_{homogeneous} + \underbrace{H(s_0)Ue^{s_0t}\rule[-1em]{0em}{2em}}_{particular} \;\;for\;\; t>0,

where the L coefficients \{A_1,A_2,\ldots,A_l\} are determined from L initial conditions, and the \lambda_l's are the roots of the denominator polynomial, or the natural frequencies, or the poles, of H(s).

Our focus so far has been on the the roots of the denominator polynomial, the poles or natural frequencies, \lambda_1,...\lambda_L. They are important because they determine a system's stability. That is, if

\Re(\lambda_i) < 0\;\;\;\; i \in \{1,...,L\}

then given any bounded input of the form u(t) = U e^{s_0 t}, \Re(s_0) < 0, the output y(t) remains bounded. But what happens when H(s) is used in a control system?

Suppose H(s) = \frac{n_h(s)}{d_h{s}} is used in a feedback system with a gain, K_0. Then from Black's formula, the closed-loop system function, G(s), would be

G(s) = \frac{K_0 H(s)}{1 + K_0 H(s)} = \frac{K_0 n_h(s)}{d_h(s) + K_0 n_h(s)} = \frac{n_g(s)}{d_g{s}}.

The stability of G(s) depends on its natural frequencies, which are the roots of d_g(s), given by

d_g(s) = d_h(s) + K_0 n_h(s).

So, the stability of H(s) depends only on its denominator polynomial, d_h(s), but the stability of a feedback system that includes H(s) depends on BOTH H(s)'s denominator AND numerator polynomial. How much of each depends on the feedback gain, K_0.

If we design a more complicated controller,

K_0 K(s) = K_0 \frac{n_k(s)}{d_k(s)},
and then use K_0K(s)H(s) in a feedback system, the natural frequencies, and hence the stability, of that feedback system will depend on the roots of
d_k(s)d_h(s) + K_0 n_k(s)n_h(s).
If we try to design K(s) so that all the natural frequencies of the associated closed-loop transfer function are dominated by very negative real parts, we are designing by "placing the poles" of the closed-loop transfer function. This is exactly the strategy we used in discrete time, with differences. In CT, we pick \lambda_i so that e^{\lambda_i t} goes to zero as fast as possible, but in the DT, we pick \lambda_i so that \lambda_i^n goes to zero as quickly as possible. In CT, \lambda_i's dominated by very negative real parts correspond to fast decay, and in DT, \lambda_i's with magnitudes close to zero correspond to fast decay.

Rather than placing poles, a process that often defies intuition, we will look at a different approach based on using an alternative representation of our system functions. Thanks to a result in complex analysis (whose formal proof goes beyond the scope of this class), we can characterize H(s), a complex function of complex variable, by examining H for purely imaginary values of s. That is,

H(s)|_{s=j \omega} = H(j \omega) \;\;\; \omega \in [-\infty,\infty],
completely characterizes H(s).

Since s is a complex number, its real and imaginary parts form a two-dimensional plane over which one evaluates H. But we are now saying you can learn everything you need to know about H by sampling s on a SINGLE LINE in the two-dimensional plane. Kind of amazing, until you think about the form of H(s) as a rational function with a finite set of real coefficients.

As we will see below, H(j \omega) is the "frequency response" of our system, meaning it tells us about the response to input sinusoids as a function of sinusoid frequency \omega = 2 \pi f, where f is frequency in Hertz. We can also characterize a feedback system's frequency response the same way,

G(j \omega ) = \frac{K_0 K(j \omega) H(j \omega)}{1 + K_0 K(j \omega) H(j \omega)}.
If G(j \omega) tells us something about the response of our feedback system to input sinusoids, we might be particularly concerned if there is a frequency \omega such that
K_0 K(j \omega) H(j \omega) = -1,
because then G(j \omega) = \infty. Perhaps we should design K(s) to avoid this circumstance.

While it is simple to substitute j \omega for s in formulas for H(s), the meaning of that substitution bears more study. In what follows, we try to do just that.

Examples

When we relate Y = H(s) U

Examples:

I(s) = Y(s) V(s) = \frac{1}{Ls +\frac{1}{Cs} + R} or V(s) = Z(s) I(s) = (Ls +\frac{1}{Cs} + R) I(s)

Y(s) = H(s) U(s) = \frac{1}{s^2 + A} U(s)

Composing Systems

W(s) = H_1(s) U(s) \;\;\;\; Y(s) = H_2(s) W(s) \;\;\;\; Y(s) = H_2(s) H_1(s) U(s)

Careful when modeling systems with bidirections coupling (like electrical circuits). A single block in our block diagrams is uni-directional (though if feedback is used, the entire diagram might not be uni-directional). That a block is unidirection means that the inputs to a block diagram are not directly effected by the block's ouput. So for example,

is NOT NOT NOT described by

V_2(s) = \left(\frac{1}{R_1 C_1 s + 1}\right) \left( \frac{1}{R_2 C_2 s + 1} \right) V(s)

because the left R-C circuit "loads" the right RC circuit.

The system function H(s)

Real and complex poles

$H(s)$ is a complex function of a complex variable $s$. The plots show $|H(s)|$.

Effect of a zero

Interpretation of $H(s)$ by pole and zero vectors

In general, an $H(s)$ that is a rational function has the form $$ H(s) = K {(s - z_1)(s-z_2)\cdots(s-z_M) \over (s - p_1)(s-p_2)\cdots(s-p_N)} . $$ Thus, $H(s)$ consists of products and quotients of the form $$ (s- s_k). $$ Each of these terms are vectors in the complex $s$ plane.
The vector (s - s_k) points from s_k to s . It can be expressed in polar form as (s - s_k) = |s- s_k|e^{j \angle (s - s_k)} .

Therefore,

H(s) = K {(s - z_1)(s-z_2)\cdots(s-z_M) \over (s - p_1)(s-p_2)\cdots(s-p_N)} ,

consists of products and quotients of vectors, i.e.,

H(s) = K {|s - z_1|e^{j\angle (s - z_1)}|s-z_2|e^{j\angle (s-z_2)}\cdots|s-z_M|e^{j\angle (s-z_M)} \over |s - p_1|e^{j\angle (s - p_1)}|s-p_2|e^{j\angle (s-p_2)}\cdots|s-p_N|e^{j\angle (s-p_N)}} ,

so that

|H(s)| = |K| {|s - z_1||s-z_2|\cdots|s-z_M| \over |s - p_1||s-p_2|\cdots|s-p_N|} ,

and

\angle H(s) = \angle K + \angle (s - z_1) + \angle (s-z_2) \cdots + \angle (s-z_M) - \angle (s - p_1) - \angle (s-p_2) \cdots - \angle (s-p_N) ,

and

H(s) = |H(s)|e^{j\angle H(s)} .

Frequency response

Relation to system function

Note that for

u(t) = {Ue^{j\omega t} + Ue^{-j\omega t} \over 2} = U\cos(\omega t) ,
where in this case, u(t) is REAL and the output in generates must be real also.
y(t) = {UH(j\omega)e^{j\omega t} + UH(-j\omega)e^{-j\omega t} \over 2} ,
y(t) = {UH(j\omega)e^{j\omega t} \over 2} + \left({UH(j\omega)e^{j\omega t} \over 2} \right)^* ,
y(t) = 2 \Re \left\{ {UH(j\omega)e^{j\omega t} \over 2} \right\} ,
y(t) = \Re \{ U|H(j\omega)|e^{j \angle H(j\omega)} e^{j\omega t} \} ,
y(t) = U|H(j\omega)| \cos(\omega t + \angle H(j\omega)) .

H(s) along the imaginary axis

We shall examine H(s) for s = j\omega or H(j\omega) where \omega is the radian frequency.

H(s) evaluated along the imaginary axis is called the frequency response.

Radian frequency and frequency

  • \omega is the radian frequency in units of radians/second.
  • \omega can be expressed as \omega = 2\pi f.
  • f is the frequency in units of cycles per second which is called Hertz (Hz).
  • The frequency response H(j2\pi f) is often plotted versus f.

Measurement of H(j\omega)

To measure H(j\omega) of a test system we can use the system shown below.

However, when the sinusoid is turned on the response contains both a transient and a steady-state component. For example, suppose

H(s) = {5s \over (s+1)^2 + 10^2} .
The response of this system, y(t), to the input u(t) = \cos(2 \pi t)\,u(t) is shown below.

Note the transient at the onset which damps out after a few cycles of the sinusoid so that the response approaches the particular solution, i.e., this is the steady-state response.

Therefore, to measure H(j\omega) we turn on the oscillator and wait till steady state is established and measure the input and output sinusoid. At each frequency, the ratio of the magnitude of the output to the magnitude of the input sinusoid defines the magnitude of the frequency response. The angle of the output minus that of the input defines the angle of the frequency response.

More elaborate systems are available for measuring the frequency response rapidly and automatically.

Relation of time waveforms, vector diagrams, and frequency response

Consider and LTI system with system function

H(s) = {1 \over s + 1},
which has the frequency response

H(j\omega) = {1 \over j\omega + 1} .

The input is

u(t) = \cos(\omega t) ,

and the response is

y(t) = |H(j\omega)|\cos(\omega t +\angle H(j\omega)) ,

y(t) = {1 \over \sqrt{\omega^2 + 1}}\cos(\omega t - \tan^{-1}\omega).

Use MATLAB software (sysfun.m) to relate time waveforms, vector diagrams, and frequency response.

Different ways of plotting the frequency response

The frequency response $H(j\omega) = 1/(j\omega + 1)$, is plotted in linear coordinates on the left and in mostly log coordinates (the Bode diagram) on the right

Bode diagrams

Definition and rationale

Frequency responses are commonly plotted as Bode diagrams which are plots of

20 \log_{10} |H(j\omega)| is plotted versus \log_{10} \omega (log-log).

\angle H(j\omega) is plotted versus \log_{10} \omega (log x - linear y).

The reasons are:

  • Logarithmic coordinates are useful when the range of |H(j\omega)| and/or \omega is large.
  • Asymptotes to the frequency response are easily plotted.
  • When the poles and zeros are on the real axis, the asymptotes are excellent approximations to the frequency response.

Pole-zero and time-constant form

Consider an LTI system whose system function has poles and zeros on the negative, real axis. It can be displayed in pole-zero form as follows

H(s) = C {(s + z_1)(s+z_2)\cdots(s+z_M) \over (s + p_1)(s+p_2)\cdots(s+p_N)} .
This system function can be evaluated along the j\omega axis to yield
H(j\omega) = C {(j\omega + z_1)(j\omega+z_2)\cdots(j\omega+z_M) \over (j\omega + p_1)(j\omega+p_2)\cdots(j\omega+p_N)} .
By dividing the numerator by each zero and the denominator by each pole, the frequency response can be put in time-constant form as follows
H(j\omega) = K {(j\omega\tau_{z1} + 1)(j\omega\tau_{z2} + 1)\cdots(j\omega\tau_{zM} + 1) \over (j\omega\tau_{p1} + 1)(j\omega\tau_{p2} + 1)\cdots(j\omega\tau_{pN} + 1)} .

Magnitude and angle

The magnitude as a function of frequency $\omega$ can be expressed as a ratio of products,

|H(j\omega)| = |K| {|j\omega\tau_{z1} + 1||j\omega\tau_{z2} + 1| \cdots |j\omega\tau_{zM} + 1| \over |j\omega\tau_{p1} + 1||j\omega\tau_{p2} + 1|\cdots|j\omega\tau_{pN} + 1|},

and the angle can be represented as the difference of sums

\angle H(j\omega) = \angle K + \angle (j\omega\tau_{z1} + 1) + \angle (j\omega\tau_{z2} + 1) \cdots +\angle (j\omega\tau_{zM} + 1) - \left(\;\; \angle (j\omega\tau_{p1} + 1) + \angle (j\omega\tau_{p2} + 1) \cdots + \angle (j\omega\tau_{pN} + 1) \;\;\right)

Logarithmic magnitude

Taking twenty times the logarithm of the magnitude yields

20\log_{10}|H(j\omega)| = 20\log_{10}|K| \;\;\;+
20\log_{10}|j\omega\tau_{z1} + 1| + 20\log_{10}|j\omega\tau_{z2} + 1| \cdots +20\log_{10}|j\omega\tau_{zM} + 1| - 20\log_{10}|j\omega\tau_{p1} + 1| - 20\log_{10}|j\omega\tau_{p2} + 1| \cdots -20\log_{10}|j\omega\tau_{pN} + 1|

Note than both the logarithmic magnitude and the angle are expressed as sums of terms of the form

\pm 20\log_{10}|j\omega\tau + 1| \;\; and \;\; \pm \angle (j\omega\tau + 1) .
Therefore, to plot the frequency response we need to add terms of the above form.

Decibels

It is common to plot frequency responses as Bode diagrams whose magnitude is expressed in decibels. The decibel, denoted by dB, is defined as $20 \log_{10} |H| $. Decibel equivalents for a few quantities.
  • |H| \rightarrow 20\log_{10}|H| dB
  • 10^n \rightarrow 20n dB
  • 2 \rightarrow \approx 6.02 \approx 6 dB
  • 5 \rightarrow \approx (20 - 6) =14 dB
  • \sqrt{2} \rightarrow \approx 6/2 = 3 dB
  • 1.12 \rightarrow \approx 1 dB

How many decibels correspond to |H| = 50? Express |H| = 100/2. Then 20\log_{10}(100/2) = 20\log_{10}100 - 20\log_{10}2 \approx 40 -6 = 34 dB.

Corner frequency

At $\omega = \omega_c = 1/\tau$, called the corner or cut-off frequency,
  • the low- and high-frequency asymptotes intersect,
  • the magnitude is
    \pm 20\log_{10}|1 + j\omega \tau| = \pm 20\log_{10}|1 + j1| = \pm 20\log_{10} \sqrt{2} \approx \pm 3 \; dB,
  • the angle is
    \pm \angle (1 + j\omega \tau) = \pm \angle (1 + j) = \pm \pi/4 .

Example --- first-order lowpass system

First-order lowpass systems arise in a large variety of physical contexts. For example,

For the parameters M=B=R=C=1, the frequency responses for the two systems are

H(j\omega) = {V(j\omega) \over F(j\omega)} = {1 \over j\omega + 1} \;\; and \;\; H(j\omega) = {V_o(j\omega) \over V_i(j\omega)} = {1 \over j\omega + 1}
so that both frequency responses are first-order lowpass filters with a corner frequency of 1 rad/sec.

Magnitude

For $$ H(j\omega) = {1 \over j\omega +1} , $$ the low-frequency asymptote has a slope of 0 and an intercept of 0 dB and the high-frequency asymptote has a slope of -20 dB/decade and an intercept of 0 dB at the corner frequency.
The corner frequency is 1 rad/sec and the bandwidth is 1 rad/sec. The two asymptotes intersect at \omega = 1 where 20 \log_{10}|H(j\omega)| = -3 dB.}

Angle

For $$ H(j\omega) = {1 \over j\omega +1} , $$ the low- and high-frequency asymptotes of the angle of the frequency response are 0 and $-90^{\circ}$ ($-\pi/2$ radians), respectively. The angle is $-45^{\circ}$ at the corner frequency (1 rad/sec).
A line drawn from the low frequency asymptote a decade below the corner frequency to the high frequency asymptote a decade above the corner frequency approximates the angle of the frequency response.

Physical interpretations

With M=B=1, the frequency response is

H(j\omega) = \frac{V(j\omega)}{F(j\omega)} = \frac{1}{j\omega + 1}.
At low frequencies, |H(j\omega)| \rightarrow 1 and \angle H(j\omega) \rightarrow 0. The inertia of the mass is negligible, and the damping force dominates so that the external force is proportional to velocity.

At high frequencies, |H(j\omega)| \rightarrow \frac{1}{\omega} and \angle H(j\omega) \rightarrow -90^{\circ}. The inertia of the mass dominates so that the acceleration is proportional to external force and the velocity decreases as frequency increases.

With R=C=1, the frequency response is

H(j\omega) = {V_o(j\omega) \over V_i(j\omega)} = {1 \over j\omega + 1}.

At low frequencies, |H(j\omega)| \rightarrow 1 and \angle H(j\omega) \rightarrow 0. The impedance of the capacitance is large so that all the input voltage appears at the output.

At high frequencies, |H(j\omega)| \rightarrow 1/\omega and \angle H(j\omega) \rightarrow -90^{\circ}. The impedance of the capacitance is small so that the current is determined by the resistance and the output voltage is determined by the impedance of the capacitance which decreases as frequency increases.

Miniproblem

Given

H(s) = {s \over (s+1)(s+10)} .
Determine the magnitude of the frequency response in the form of a Bode diagram.

H(j\omega) = {1 \over 10}\left({j\omega \over (j\omega+1)(j\omega/10 +1)}\right) .

Summary

The frequency response

  • is the system function evaluated along the j\omega-axis,
  • has a simple geometric interpretation in the s-plane,
  • describes the sinusoidal steady state response of an linear time-invariant system to an input sinusoid,
  • has asymptotes that are easily sketched in a Bode diagram when the poles and zeros lie on the negative real axis.

Summary Questions

A linear, time-invariant, continuous-time system is described by its system function:

H(s)={s+1\over s^2+5s+6}

Is this system stable?

Does this system exhibit resonance?

Is the step response of this system a monotonic function of time?

Enter the final value of the step response: