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I am studying about Fourier series from book"Signals and Systems Laboratory with MATLAB"

I came across topic "Orthogonality of Complex Exponential Signals"

I am confused in case when m=k, will the two signals still orthogonal in this case ?(highlighted yellow)

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    $\begingroup$ When m=k, the two signals are the same signal. They cannot be orthogonal to themselves. $\endgroup$ – Juancho May 7 at 18:16
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Well, when $m=k$ the integral is: $$ \int_0^T e^{j(m-k)\Omega_0t} dt = \int_0^T e^{j \cdot 0 \cdot\Omega_0t} dt = \int_0^T dt = T $$

So as Juancho says in the comments, it's the same signal and so can't be orthogonal to itself.

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The text is a bit cumbersome, in that it states "things" before defining them. And the zero-signal (or vector) is considered orthogonal to every other vector. This is why the sentence "If $I=0$ for..." seems unnecessarily complicated to me. And the $m$ and $k$ indices introduced at the beginning come out of nowhere until we understand they are integer complex exponential indices. Moreover, introducing $x$ and $y$ at the very end is quite inconsistent with the previous $x_m$ and $x_k$ notations.

In a mood for conciseness and precision, let me rewrite the text (with missing hypotheses, like $\Omega_{0}\neq 0 $), hoping it will be more direct.

Let us consider in the following complex-valued continuous-time periodic signals with period $T>0$. The inner product of $x_1$ and $x_2$ is defined as:

$$I(x_1,x_2)=\int_{t_{0}}^{t_{0}+T} x_1(t) x_2^{*}(t) d t$$

where $x^{*}$ denotes the complex conjugate of $x$. Two signals are orthogonal if their inner product is zero. Suppose now that $(m,k)\in \mathbb{Z}^2$, $\Omega_{0}\neq 0 $ and define $x_{1}(t)=e^{j m \Omega_{0} t}$ and $x_{2}(t)=e^{j k \Omega_{0} t}$. Since: $$I(x_1,x_2)=\int_{0}^{T} e^{j m \Omega_{0} t_{e}-j k \Omega_{0} t} d t=\int_{0}^{T} e^{j(m-k) \Omega_{0} t} d t=\left\{\begin{array}{ll}{T,} & {k=m} \\ {0,} & {k \neq m}\end{array}\right.\,,$$ then the complex exponentials $e^{j m \Omega_{0} t}$ and $e^{j k \Omega_{0} t}$ are orthogonal if $m\neq k$.

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