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I wanted to calculate response of integrator of sinusoidal input at steady state via these two methods as mention in image but these two methods give two different answers at steady state, so where I'm going wrong? What concept I'm missing?

  • $\begingroup$ If you use two different input signals, it shouldn't be surprising that you get two different output signals. $\endgroup$
    – Matt L.
    Commented Jun 30, 2020 at 8:16
  • $\begingroup$ But I used same sinusoidal signal, one is everlasting which directly give steady state while other is same sinusoidal started at t=0, but isn't it's output should also be same as everlasting signal when t tends to infinity? $\endgroup$
    – user215805
    Commented Jun 30, 2020 at 8:21
  • 1
    $\begingroup$ The DC term in the output depends on the time the sinusoid is switched on, so in this sense you can't get a steady-state response because the DC term obviously doesn't decay. See my answer for some more detail. $\endgroup$
    – Matt L.
    Commented Jun 30, 2020 at 9:25
  • $\begingroup$ thanks for response ,isn't equation (2) seems wrong intuitively ?because if lets assume θ=0 then according to equation (2) output will be a periodic signal but its values float between [0,1/ω0 ] always and if we shift a signal at θ=π/2 then response will be −cos(ω0t)/ω0 whose values lies between[ −1/ω0 ,1/ω0] ,so output in both cases are so different that even after infinite time we can recognize the both output signals even at different times (for generality ) which seems impossible intuitively $\endgroup$
    – user215805
    Commented Jun 30, 2020 at 18:08
  • $\begingroup$ because a merely π/2 time shift cannot change the nature of output after infinite time ,so how this equation can be possible ? $\endgroup$
    – user215805
    Commented Jun 30, 2020 at 18:08

1 Answer 1


Let's take a slightly more general input signal

$$x(t)=\sin(\omega_0t)u\left(t-\frac{\theta}{\omega_0}\right),\qquad\theta\in [-\pi,\pi)\tag{1}$$

It's straightforward to show that the response of an ideal integrator to the input $(1)$ is given by


The response $y(t,\theta)$ always consists of a time-dependent component $-\cos(\omega_0t)/\omega_0$ and a DC-term $\cos(\theta)/\omega_0$, which varies between $-1/\omega_0$ and $1/\omega_0$, depending on the time when the sinusoid is switched on. Consequently, no matter how large $t$ becomes, the component that all responses to switched sinusoids have in common is always the time dependent part, but not the DC-term.

The ideal response to a sinusoid extending from $-\infty$ to $\infty$ only consists of that time-dependent component that is common to all responses to switched sinusoids. You can imagine this response as an average of $(2)$ obtained by integrating over $\theta$:


For the some discussion on the computation of the response of an ideal integrator to a sinusoidal input, take a look at this question and its answers.


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