# Why cosine is not an eigen signal?

According to this website:

If the output of a system has the same type as its input signal, then the input signal is referred to as the eigen function of the system.

but in this question it is saying only option $(c)$ is eigen function. My question is why can't it be option $(b)$ and $(d)$ as it will also obey the defintion of eigen function as given in the website?

PS: Please let me if know my question is not clear i will edit it but please don't downvote.

• If the question was "which function is always an eigenfunction of an LTI system", then (c) would be the only correct answer. With the current formulation "which of the following can be the eigen signal of an LTI system", all answers are correct because each is an eigenfunction of the trivial identity system for example. – Jazzmaniac Aug 9 '17 at 17:05
• @Jazzmaniac I remember we had a discussion of this before :-) Form the question itself, as you said, yes all of them can be an eigenfunction of an (some) LTI system. But I belive they wanted to say which of them can be eigenvalues of arbitrary (therefore of every) LTI systems in general... – Fat32 Aug 9 '17 at 17:37
• @Fat32, if the wanted to say that then they should have said that. ;-) – Jazzmaniac Aug 9 '17 at 18:12

That definition is rubbish.

An eigenfunction of a system is a function that passes that system, only scaled by a scalar value.

Now, suppose your system is a simple complex band pass filter, which has a pass band at $f_0$, but stop bands everywhere else, especially at $-f_0$.

Now your cosine $\cos(2\pi f_0 t)$ has spectral components at both $f_0$ and $-f_0$. The output of that filter contains only a multiple if the positive part. Hence, the output is not a scalar multiple of the input cosine. Done.

• Muller if you are saying that definition is not correct then show me referece please what is the correct defintion because even in books its given like that – Rohit Aug 7 '17 at 7:32
• also from your logic it will be true for option c also because if the option c is at 2f0 then output will be zero then how it will satify please help – Rohit Aug 7 '17 at 7:38
• Rohit I really don't need to reference anything when I say that eigenfunctions pass a system simply scaled. That is what mathematicians mean when they use the word "Eigen". No, by my logic option c would not be true, because 0 is never an eigenvalue. – Marcus Müller Aug 7 '17 at 9:10
• Muller Sir what i am trying to say is that when you take option c in the frequency of $2f_0$ and the filter range is upto only$f_0$ then output will be zero then i will not get scaled input.plz help me whats wrong i am thinking? – Rohit Aug 7 '17 at 9:40
• @MarcusMüller, I'm afraid you got that mixed up. Zero is never an eigenvector, but it can be an eigenvalue. In fact the eigensubspace with associated eigenvalue 0 is so important that it even goes under a special name. It's called the nullspace of an endomorphism. – Jazzmaniac Aug 9 '17 at 18:25

Given a continuous time LTI system with impulse response $h(t)$ and determined with the transform $\mathcal{T}\{\cdot\}$, we define an input/output relationship as follows:

$$y(t) = \mathcal{T}\{ x(t) \}$$ which can be evaluated based on the convolution integral as: $$y(t) = \mathcal{T}\{ x(t) \} = x(t) \star h(t) = \int_{-\infty}^{\infty} h(\tau) x(t-\tau) d\tau$$

Now we call a signal (function) $x(t)$ as an eigenfunction of such a system if $$y(t) = \mathcal{T}\{ x(t) \} = K_x x(t)$$

Where $K_x$ is a complex constant (the eigenvalue corresponding to the eigenfunction) dependent on the system and input parameters. Note that the output $y(t)$ must be equal to the input waveform, scaled by K, for all $t$.

From this definition point of view, lets consider whether $x(t) = e^{j\omega_0 t}$ is an eigenfunction of LTI systems in general or not?

$$y(t) = \mathcal{T}\{ e^{j\omega_0 t} \} = \int_{-\infty}^{\infty} h(\tau) e^{j \omega_0 (t-\tau)} d\tau = e^{j \omega_0 t} \left( \int_{-\infty}^{\infty} h(\tau) e^{-j \omega_0 \tau} d\tau \right) = K \cdot e^{j \omega_0 t}$$

Where the complex constant $K$ (the eigenvalue) is recognized as the continuous-time Fourier transform, $H(\omega)$ evaluated at the frequency $\omega_0$, of the impulse response $h(t)$ of the system, which is also called as the Frequency response. Expressing $K$ as $K=H(\omega_0) = |H(\omega_0)| e^{j \phi_0}$ we can rewrite the output as $y(t) = K e^{j\omega_0 t} = |H(\omega_0)| e^{j \phi_0} e^{j\omega_0 t} = |H(\omega_0)| e^{j\omega_0 (t + \phi / \omega_0)} = |H(\omega_0)| e^{j\omega_0 (t - t_0)}$ where $t_0 = - \phi/\omega_0$.

We can show the relation as $$x(t) \longleftrightarrow |H(\omega_0)| e^{j\phi_0} x(t)$$ Therefore we conclude that $x(t)=e^{j\omega t}$ in general is an eigenfunction of arbitrary LTI systems.

What about the function $x(t)=e^{j \omega_1 t} + e^{j \omega_2 t}$ ? Using linearity property of LTI systems we can show that the respective outputs for each added term will be $y_1(t) = |H(w_1)| e^{j\phi_1} e^{j\omega_1 t}$ and $y_2(t) = |H(w_2)| e^{j\phi_2} e^{j\omega_2 t}$ therefore we have $$y(t) = y_1(t) + y_2(t) = |H(w_1)| e^{j\phi_1} e^{j\omega_1 t} + |H(w_2)| e^{j\phi_2} e^{j\omega_2 t} = K_1 e^{j \omega_1 t} + K_2 e^{j \omega_2 t} \neq K \left( e^{j \omega_1 t} + e^{j \omega_2 t} \right)$$

Hence unless we have $K = K_1 = K_2$, the signal $x(t)=e^{j \omega_1 t} + e^{j \omega_2 t}$ is not an eigenfunction of LTI systems in general.

Finally note that for the signal $x(t)=\cos(w_0 t)$ there exist LTI systems having the property that their impulse responses, $h(t)$, are real and even will accept them as eigenfunctions, but not every LTI in general will accept $\cos(w_0 t)$ as an eigenfunction, hence $\cos(w_0 t)$ is not an eigenfunction of LTI systems in general. On the other hand $e^{j \omega_0 t}$ or $e^{s t}$ with complex $s$ in general are eigenfunctions of every LTI system.

• Thanks sir... Please answer this question too no one answered yet dsp.stackexchange.com/q/42940/29743 – Rohit Aug 7 '17 at 14:50
• ok let me look at it. – Fat32 Aug 7 '17 at 15:09
• @Rohit, while reading the answer I recognised that I forgot to mention about $\cos(2t)$ , and while adding that into the answer, I have also changed the body which can be important for your understanding. Please read it again to see for yourself. – Fat32 Aug 9 '17 at 16:56
• sir can eigen function be non periodic ? as in every book its saying that complex exponential can be eigen function and we know complex exponential are periodic that why i am asking.like e^-2t can be eigen function ? – Rohit Aug 24 '17 at 13:05
• @Rohit yes it can. You may ask a new question with the title "can eigen functions of LTI be non-periodic ?" to see why. – Fat32 Aug 24 '17 at 13:34

[EDIT: a quite similar question, Are complex exponentials the only eigenfunctions of LTI systems, was already answered to by @Jazzmaniac]

Generic LTI systems can indeed be characterized by their eignefunctions and eigenvalues. You ought to check the clear overview in Eigenfunctions of LTI Systems.

Complex exponentials are always eigenfunctions of LTI systems, but they are not necessarily the only eigenfunctions. However, having the sum of eigenfunctions still be an eigenfunction requires conditions, like they have the same eigenvalues.

Hence, the question here seems ill-posed to me.

Which of the following can be the eigen signal of an LTI system?

Well, a, b, c, and d: any of them can be the eigen signal of a well-chosen LTI system. For instance, a $0$-output system accepts any non-zero input as an eigenfunction. We can find less trivial examples.

For d), @Fat32 suggested that there exist LTI systems admitting them as eigenfunctions. An instance can be obtained remembering that the second derivative of a cosine is a cosine:

$$(\cos 2t)^{''} = -4 \cos 2t$$

Hence, the second derivative operator being linear and time-invariant, $t\to\cos 2t$ is an eigen signal with eigenvalue $-4$. Even-order derivatives would work as well. Similarly, you could find under which conditions an LTI is such that it admits other eigenfunctions, and build your own examples. For a), any lazy system that multiplies the input by a constant does the job.

Allowing only answser c) suggest that the eigen signals are thought for "generic" LTI systems. The question could be:

Which of the following can be the eigen signal of generic LTI systems?

• sir i never thought about option A.can you please enlighten me ? – Rohit Aug 10 '17 at 12:29
• @Rohit Some details have been added – Laurent Duval Aug 11 '17 at 13:18
• @LaurentDuval can eigen function be non periodic ? as in every book its saying that complex exponential can be eigen function and we know complex exponential are periodic that why i am asking. – Rohit Aug 24 '17 at 13:01
• An eigenfunction could be non periodic. A pure amplifier admits every vector as eigen, thus possibly a non periodic vector. However, you can more properties with structured matrices. For instance, persymmetric matrices admit either symmetric or antisymmetric eigenvectors, in a number that can be predicted. See for instance Properties of the Eigenvectors of Persymmetric Matrices with Applications to Communication Theory, which is used for instance in the derivation of Lapped Orthogonal transforms – Laurent Duval Aug 24 '17 at 15:47
• with distinct eignevalues... – Laurent Duval Aug 24 '17 at 15:57