Sinusoidal Frequency Response of a Transfer Function

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.

[Paul]

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]

@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.

Answer 1

[Star Strider]

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.

.

[Mustafa Furkan]

Is it possible for you to explain why you use the terms you use in the code? I don't fully understand, sorry.

[Star Strider]

Sure!

As I understand ‘frequency response’ (supported by a bit of online research in the last few minutes), the functions I illustrated here will plot what is generally considered to be a frequency response to your system.

In addition, I used lsim to simulate the sort of approach that you said you wanted (specifically the response to ).

My objective here is to present different approaches in order to understand the result you want, since I am not certain that I understand what that is.

[Mustafa Furkan]

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]

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.

.

Answer 2

[Paul]

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:

1. 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.
2. 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.

[Mustafa Furkan]

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]

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