I'm given a problem by the following:

Find the sinusoidal steady state response (in the time domain) of the following systems modeled by transfer function, P(s), to the input u(t). Use the Bode plot (in Matlab bode.m) of the frequency response as opposed to solving the convolution integral of the inverse Laplace transform.

$$ P(S) = 11.4/(s+1.4), u(t) = cos(5t) $$

I'm a bit confused by the question because I thought bode plot is the definition of steady state response, but it's asking me to find it in time domain. Is such thing possible? Anyways plotting this in matlab gives me the following:

$$ Y(S) = P(S)U(S) $$

where from laplace transform

$$ U(S) = s/(s^2+25) $$

Y =

           11.4 s
  s^3 + 1.4 s^2 + 25 s + 35

Continuous-time transfer function.

>> bode(Y), grid

enter image description here

This doesn't look like a typical bode plot either (Maybe because output is third order?) What can I infer from this representation of bode plot?

Edit: So this is a bode plot for just P(s)

enter image description here

At w = 5, it seems like the phase of -75 degrees and magnitude of 7db

Since magnitude in db, the final steady state response in time domain is

$$ Y_sss(t) = 2.24cos(5t - 75^{\circ}) $$

Really? This simple?


2 Answers 2


Firstly, the bode plot you are interested in is NOT the bode plot of Y, but of P. You have probably misinterpreted the problem's objective. Plot it and add resulting images to your question.

Then what you want to do is check your input signal nature. In your example, it is a pure cossine with angular frequency 5 and unitary amplitude. When looking at bode plot of P, check for the amplitude (A) and phase (theta) at such frequency. Your response Y will then be A*cos(5t + theta). [yes, easy like that]

Since you are working with a linear system, this concept can be extended to other types of input signals through proper decomposition of input by fourier series. For example, if U were a square wave, it would have all odd harmonics leading to infinity right? With the bode plot of P, I can check the gain and phase shift for each harmonic component of the resulting wave Y!

There are many others analysis to be made from bode, such as group delays and system stability. But I hope this answer leads you through the right way.

  • \$\begingroup\$ The other answerer is also correct, but failed to notice OP's misinterpretation of the problem at hand. The bode plot presented should be of P, not of Y. \$\endgroup\$ Jan 28, 2016 at 19:38
  • 1
    \$\begingroup\$ Wow really? This simple? I've made an edit in my question. Please see if my approach is correct \$\endgroup\$
    – Xiagua
    Jan 28, 2016 at 19:57
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    \$\begingroup\$ Sure. Also, you could have just calculated \$P(s)|_{s=5j}\approx 0.592-j2.114 = 2.196 \angle( -74.4 )\$ \$\endgroup\$ Jan 28, 2016 at 22:12
  • \$\begingroup\$ Yes, it is definitely correct (if you have used the cursor to check values from matlab plot). I would prefer using radian angle representation inside of sine functions, but that's nitpicking. It can be really surprising to suddenly STOP doing many calculations taught by your professors and use such a visual tool, therefore I understand your concern with the simplicity of the solution. But this is in fact the real reason for bode's existence in the first place! =) Welcome to the world of fast electrical engineering conclusions! \$\endgroup\$ Jan 29, 2016 at 7:55

It's a linear system, so a sine in is going to result in a sine out of the same frequency. The only questions are what its amplitude and phase will be.

Look at the Bode plot. The top graph gives you the gain of the system, and the second the phase shift. It's pretty much telling you what you need to know on a silver platter. With the amplitude, frequency, and phase, you should be able to plot the signal in the time domain easily.


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