[SOLVED] [Control Theory] A question about the Frequency Response method

[calenosa]

From a Control Theory book I have:

"The frequency response function describes the steady-state response of a system to sinusoidal inputs. For a linear system, a sinusoidal input of a specific frequency results in an output that is also a sinusoid with the same frequency, but with a different amplitude and phase. The frequency-response function describes the amplitude change and phase shift as a function of frequency."

The above quotation is true, however, it doesn't say anything about the dc component in the steady-state response.

When I simulated the following system in MATLAB/Simulink:

The source is a pure sine wave generator with no bias. I got the following response:


It can be seen that the response has a dc component, and I believe that the simulation result is correct. Therefore my question is can I calculate the dc component from the frequency response approach? If the answer is Yes, how can I do that?

Thank you.

 

[v_c]

Looks like the input signal (green) is \[-\sin(t)\]; therefore, the input angular frequency is \[\omega=1\text{rad/s}\].
Have you tried the Laplace transform method. The Laplace transform of the input \[-\sin(t)\] is \[-1/(s^2+1)\].
The Laplace transform of the output is the product of the input and system transforms.
\[
Y(s) = \frac{-1}{s^2+1} \, \frac{1}{s^2+s}=\frac{-1}{s(s+1)(s^2+1)}=\frac{A}{s}+\frac{B}{s+1}+\frac{Cs+D}{s^2+1}
\]
where A, B, C and D may be determined by partial fraction expansion. The time domain version of the output is of the form
\[
y(t) = A u(t) + B e^{-t} u(t)+ (\text{sinusoid \, at \, 1 rad/s}) \, u(t)
\]
so the output will definitely have a constant due to step magnitude A and and exponential decay B and a sinusoid. Does this make sense? Try for yourself and see if you can figure out the complete solution. But I agree, the solution that is simulated looks reasonable to me.

v_c

 

[calenosa]

Hi v_c,

Thank you for your answer.

Well, The simulation result makes sense to me. My problem is that I don't understand the frequency response method thoroughly. I would like to know whether the frequency response method mentions the dc component of the output signal or not? If the answer is Yes, where can I find it (in the method)?

 

[v_c]

frequency response method will only give you the sinusoidal portion of the response -- not the constant plus exponential portions

 

[ferdem]

You can put 0 for the frequency in transfer functions pertaining to frequency response. For example: 1/(1+i*w*R*C) is a transfer function for RC lowpass, if you put 0 for w, you will see its gain is 1 for DC inputs. The result you get looks strange, is that transfer function written for a real system?

 

[calenosa]

Hi v_c,

I didn't read carefully about the scope of the frequency response method. This method only deals with stable systems, which means that it cannot be applied to the simulated system.

@ferdem: No, it is just an imaginary system.

 

[albbg]

In order to calculate the DC component, you can go to the differential equation and integrate it.
y/x = 1/(S²+S) ==> y'' + y' = x where x = [M*cos(wt)+k]*1(t)
1(t) is the step function.
We need to found the homogeneous and the particular integral then sum them toghther and compensate the pulses.

homogeneous: yh'' + y' = 0 the associated equation is then λ² + λ = 0 ==> λ1=0, λ2=-1 then:

yh = A1 + A2•exp(-t)

particular integral: is in the form yp =[B1*cos(wt) + B2*sin(wt) + B3*t]*1(t)

the general solution is y = yh + yp

solving and compensating the pulses (coming from the time derivative of 1(t)) it is possible to find the constants A1, A2, B1, B2 and B3. As far as I did correct calculations I found:

B1 = -M/(w²+1)
B2 = -M/[w(w²+1)]
A1 = -B2w - k - B1
A2 = B2w + k

thus:

y = M/(w²+1)*[-2+(1+k)*exp(-t) + cos(wt) + 1/w*sin(wt)]*1(t)