Sinusoidal Frequency Response of a Transfer Function

Question

Mustafa Furkan 2023-1-15

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编辑： Paul 2024-4-28

I have a transfer function as below. I want to plot the (FRF) response of a function like y(t)=A.sin(w.t) for this transfer function. The A and w values are constants, any value can be substituted. I think the response will be steady-state.

A = [0,1,0,0;-64,0,64,0;0,0,0,1;64,0,-128,0];
B = [0;0;0;64];
C = [1,0,0,0]; D = 0;
sys1 = ss(A,B,C,D);
sys2 = tf(sys1)
sys2 =
 
                         4096
  --------------------------------------------------
  s^4 - 1.554e-15 s^3 + 192 s^2 - 6.977e-14 s + 4096
 
Continuous-time transfer function.

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Paul 2023-1-15

编辑：Paul 2023-1-15

Hi Mustafa,

The term FRF typically refers to a property of the system, not a function. Also, by convention the input to the system is typically called u(t) and the output y(t) (that's just convention, of course you can use whatever you want). So, what I think you're asking is: "What is the output ,y(t), of the system in response to a sinusoidal input u(t) = A*sin(w*t)? I think the response will be steady-state."

However, it's not really clear what you're looking for. Do you want the output of the system in response to that input? Or do you only want the steady state output of the system in response to that input, assuming such a steady state output even exists?

Also, are you looking for a numerical solution to plot or closed-form expression?

Mustafa Furkan 2023-1-15

编辑：Mustafa Furkan 2023-1-15

@Paul I'm doing analysis on a system like in the image

Here y represents the input. I want to plot the response of the system based on this input value. As I mentioned in the question, I want to find a solution according to an equation like y(t)=A.sin(wt). Since it is coupling, steady state will be the right choice in terms of obtaining cross spectral density. That's why I specifically mentioned the steady state in the question.

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Answer 1

Star Strider 2023-1-15

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编辑：Star Strider 2023-1-15

在 MATLAB Online 中打开

If I understand your Question correctly, the bode, bodeplot, nichols, freqresp and lsim functions will do what you want, or for a specific frequency and amplitude, the lsim funciton is also an option —

A = [0,1,0,0;-64,0,64,0;0,0,0,1;64,0,-128,0];

B = [0;0;0;64];

C = [1,0,0,0]; D = 0;

sys1 = ss(A,B,C,D);

sys2 = tf(sys1)

sys2 = 4096 -------------------------------------------------- s^4 - 1.554e-15 s^3 + 192 s^2 - 6.977e-14 s + 4096 Continuous-time transfer function.

figure

stepplot(sys1)

s = stepinfo(sys1)

s = struct with fields:

RiseTime: NaN TransientTime: NaN SettlingTime: NaN SettlingMin: NaN SettlingMax: NaN Overshoot: NaN Undershoot: NaN Peak: Inf PeakTime: Inf

t = linspace(0, 100, 1001);

Ts = (t(2)-t(1))

Ts = 0.1000

Fs = 1/Ts

Fs = 10

ufcn = @(t,A,omega) A.*sin(omega.*t);

u = ufcn(t, 2.5, 150);

y = lsim(sys1, u, t);

figure

plot(t, y)

grid

xlabel('Time (s)')

ylabel('Amplitude (units)')

figure

plot(t, y)

grid

xlim([0 10])

xlabel('Time (s)')

ylabel('Amplitude (units)')

The ‘sys1’ and ‘sys2’ functions are obviously the same in this application. I just used them to illustrate the two plotting functions.

Further information is available in What is Frequency Response?

EDIT — (15 Jan 2023 at 21:46)

Eliminated everything except the lsim code, added stepplot plot and stepinfo call.

I doubt that a steady state is possible with this system.

.

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Mustafa Furkan 2023-1-15

Thank you very much for your effort @Star Strider. Your answers with Paul have yielded very close results, and there probably won't be a single answer to this question. May I ask you how can I plot the frequency response in the range of [0 500] rad/s?

Star Strider 2023-1-15

在 MATLAB Online 中打开

My pleasure!

Yes —

A = [0,1,0,0;-64,0,64,0;0,0,0,1;64,0,-128,0];

B = [0;0;0;64];

C = [1,0,0,0]; D = 0;

sys1 = ss(A,B,C,D);

sys2 = tf(sys1)

sys2 = 4096 -------------------------------------------------- s^4 - 1.554e-15 s^3 + 192 s^2 - 6.977e-14 s + 4096 Continuous-time transfer function.

omegav = linspace(0, 300, 5000);

figure

bode(sys1, omegav)

grid

[~,~,omega] = bode(sys1, omegav);

omegalimits = [min(omega) max(omega)]

omegalimits = 1×2

0 300

The ‘omegav’ vector defines the radian frequency vector that this bode call uses to calculate the frequency response. The frequency axis scale is logarithmic, however it goes from 0 to 300 rad/s, as the ‘omega’ result demonstrates.

.

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Answer 2

Paul 2023-1-15

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在 MATLAB Online 中打开

If the system were bounded-input-bounded-output (BIBO) stable, then the steady state output in response to input y(t) = A*sin(w*t) would be zss(t) = M*A*sin(wt + phi), where M and phi are determined by the magnitude and phase of the system transfer function evaluated at s = 1j*w.

However, this system has no damping and is not BIBO stable, as indicated by the eigenvalues of the A-matrix all on the imaginary axis

A = [0,1,0,0;-64,0,64,0;0,0,0,1;64,0,-128,0];
format long e
e = eig(A)
e = 
      0.000000000000000e+00 + 1.294427190999917e+01i
      0.000000000000000e+00 - 1.294427190999917e+01i
      0.000000000000000e+00 + 4.944271909999160e+00i
      0.000000000000000e+00 - 4.944271909999160e+00i

So, in this case we have to consider two cases:

w not equal to 12.94427.... and w not equal to 4.94427.... In this case, the output of the system will be the sum of zss as defined above and and two other sine waves at frequencies 12.944... an 4.94427.... The amplitude and phase of these two other sine waves would have to be determined based on A and w.
w = 12.9442.... or w = 4.94427.... In this case the output will include a term like C*t*sin(w*t), i.e., the output will grow indefinitely

Finding closed form expressions in either case is feasible, but could be painful.

In either case, you could use lsim to approximate the output via simulation; make sure the the time vector has a time separation small relative to the larger of 12.944 or w, mabye something like dt < 1/10/max(w,12.9444)

B = [0;0;0;64];

C = [1,0,0,0]; D = 0;

sys1 = ss(A,B,C,D);

% Case 1 example

w = 8; a = 1;

dt = 1/12.94/10;

t = 0:dt:10;

u = a*sin(w*t);

y = lsim(sys1,u,t);

plot(t,y)

As expected, the output is the sum of three sinusoids at the frequencies determined by w and the imaginary part of e.

Y = fft(y);

w = (0:(numel(Y)-1))/numel(Y)*2*pi/dt;

figure

plot(w,abs(Y))

xlim([0 20])

% Case 2 example

w = abs(imag(e(1))); a = 1;

dt = 1/12.94/10;

t = 0:dt:10;

u = a*sin(w*t);

y = lsim(sys1,u,t);

figure

plot(t,y),grid

This output is dominated by the growing sine wave at 12.444 rad/sec, but it does have a 4.944 rad/sec component

Y = fft(y);

w = (0:(numel(Y)-1))/numel(Y)*2*pi/dt;

figure

plot(w,abs(Y))

xlim([0 20])

This analysis would be much simpler if you added a dashpot (or damper) in parallel with two springs.

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Mustafa Furkan 2023-1-15

Thank you for your answer @Paul. I think case 1 will be the correct result for me. I want to ask one more thing, what should I do to plot the frequency response in the range of [0 500] rad/s?

Paul 2023-1-15

编辑：Paul 2023-1-15

Use bode or bodeplot to compute/plot the frequency response from the state space model.

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Answer 3

Hassan 2024-4-26

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Hi Paul,

For the case 2, what if e consist of real and imaginary part both? how to get w in that case?. I am addressing similar problem, where i found frequency that gives maximum amplitude to a system. Now i wants to plot time vs Amplitude, showing an indifinite growth, as you have shown in the example 2.

e= -683120.431219080 + 404883.368416124i; -683120.431219080 - 404883.368416124i

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Hassan 2024-4-27

编辑：Hassan 2024-4-27

I had a viscoelastic system based on voigt model, I found its Transfer function and give frequency sweep to determine the frequency corresponding to maximum amplitude. Which i believed should be the resonant frequency of the system, if it is excited by a force with that frequency, its amplitude should grow indefinitely.

Paul 2024-4-27

编辑：Paul 2024-4-28

在 MATLAB Online 中打开

I don't know what viscoelastic means nor am I familar with a voigt model.

If the the system is BIBO stable, then exciting it with a force that at the frequency of the resonant peak on the Bode plot will not caause the amplitude to grow indefinitely. Here's an example

syms s t

H(s) = 1/(s^2 + 2*0.1*1*s + 1); % lightly damped system with resonant frequency 1 rad/sec

U(s) = laplace(sin(t)*heaviside(t)); % input at 1 rad/sec

Y(s) = H(s)*U(s);

y(t) = ilaplace(Y(s),s,t)

y(t) =

fplot(y(t),[0 100])

As expected, the steady state output has frequency of 1 rad/sec and is bounded with amplitude given by

abs(H(1j*1))
ans = 5

Hard to say anything more without seeing the model of the system. Feel free to save it in a .mat file and attach it to a comment using the Paperclip icon on the Insert menu.

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