Everywhere in theory (books , online) there is this statement "For sinusoidal inputs, any LTI has a sinusoidal output with the gain of $|H(s)|$, the same frequency, and a phase shift equal to $\angle {H(s)}$". So they have something like this: $$x(t) = A\cos(\omega t + \phi), |H(s)|=B , \angle{H(s)} = \phi' \\\implies y(t)=(AB)\cos(\omega t+ \phi + \phi')$$

Now, my question is why nobody talks about $sin$ as input. Let's say we have $$x(t) = A\sin(\omega t + \phi) = A\cos(\omega t + \phi -\pi/2) \\\implies y(t) = AB\cos(\omega t + \phi + \phi' -\pi/2)= AB\sin(\omega t + \phi + \phi')$$

So exactly the same holds for sin right?

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    $\begingroup$ Correct, it holds for any phase in $\cos(\omega t + \varphi)$ which includes the sine wave. In fact it holds for all complex sinusoidals as well $e^{j\omega + \varphi}$. $\endgroup$
    – Hilmar
    Sep 19, 2023 at 12:00
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    $\begingroup$ It's just a 90° difference in phase. Otherwise $\cos(\cdot)$ and $\sin(\cdot)$ are identical in every respect. The reason $\cos(\cdot)$ is used most often is that the Fourier Transform is entirely real and while for $\sin(\cdot)$, the Fourier Transform is entirely imaginary. It's just a convention that sometimes allows us to forget about an imaginary unit $j$ floating around somewhere. $\endgroup$ Sep 19, 2023 at 19:30

3 Answers 3


It's important to understand the response of an LTI system to a complex exponential

$$x(t)=e^{j(\omega_0 t+\phi)}\tag{1}$$

(as pointed out in Dan's answer).

The response to the input signal $(1)$ is given by the convolution integral

$$\begin{align}y(t) & =\int_{-\infty}^{\infty}x(t-\tau)h(\tau)d\tau\\&=\int_{-\infty}^{\infty}e^{j\omega_0(t-\tau)}e^{j\phi}h(\tau)d\tau\\&=e^{j(\omega_0 t+\phi)}\int_{-\infty}^{\infty}h(\tau)e^{-j\omega_0\tau}d\tau\\&=e^{j(\omega_0 t+\phi)}H(\omega_0)\\&=|H(\omega_0)|e^{j(\omega_0 t+\phi+\arg\{H(\omega_0)\})}\tag{2}\end{align}$$

where $h(t)$ is the system's impulse response, and $H(\omega)$ is the corresponding frequency response.

From $(2)$ we see that the amplitude of a complex exponential at the input is multiplied by the magnitude of $H(\omega)$ at the given frequency, and the phase of $H(\omega)$ is added to the phase of the input signal.

If you know and understand $(2)$, then by taking the real part or the imaginary part you can see what happens to a cosine or a sine at the input. Unsurprisingly, it's the same thing: the magnitude is multiplied by $|H(\omega_0)|$, and $\arg\{H(\omega_0)\}$ is added to the phase.


Either sin, cos, or exp can be used and they are all related by Euler's formula as:

$$e^{j\omega t} = \cos(\omega t) + j \sin(\omega t)$$

My guess as to why educators may favor using cosine is that it has a simpler result in the frequency domain, such that subsequent details can be described without yet introducing complex numbers (personally, I teach everything with $e^{j\omega t}$ from the start) : given the cosine is a even function in the time domain, it is a real function in the frequency domain as two impulses each with magnitude half. Any real even function in time will be real and even in frequency. Any real odd function in time will be odd and imaginary in frequency.

Fourier Transform of Sinuosoids

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    $\begingroup$ I always like your figures, they make everything very clear. I just wanted to point out a minor detail: if you use $\omega$ as the frequency variable, the constants associated with the Dirac impulses in the frequency domain get multiplied by $2\pi$. The constants you used are correct if you use $f$ as the frequency variable. But I don't know if it's worth changing the figures ... $\endgroup$
    – Matt L.
    Sep 19, 2023 at 13:54
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    $\begingroup$ @MattL. thank you and excellent point. Yes I think it is worth updating; I will do that! $\endgroup$ Sep 19, 2023 at 16:47
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    $\begingroup$ Dan, there's a reason why the "ordinary frequency" definition (or convention) of the continuous Fourier Transform is preferred. I almost never use the angular frequency convention: $$ X(j\omega) \triangleq \mathscr{F} \Big\{ x(t) \Big\}$$ unless my intention is to go back and forth between that and the Laplace Transform. And then you will always see a "$j$" attached to the "$\omega$" in the argument of $X(\cdot)$. You will never (or very rarely) see an "$X(\omega)$" (without the "$j$") coming from me unless I am quoting someone else and running with that. $\endgroup$ Sep 19, 2023 at 19:37
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    $\begingroup$ Yes thank you @robertbristow-johnson ! I agree with you in the convenience and simplicity of going back and forth without that 2π showing up. Also to your point the clarity with using X(ejω) for discrete time transforms. Plot updated. $\endgroup$ Sep 19, 2023 at 23:20

There is a trigonometric identity

To this problem it brings much clarity

To shift from sine to cosine

add degrees 10 times nine

Then your problem answers itself, uh, ity.


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