The Continuous-Time Fourier Series (CTFS), often called just Fourier Series (FS), tells us that we can write a signal as a linear combination of orthogonal functions, or bases. In general, we can use any bases of orthogonal functions, but we will mainly use complex exponentials. 

\begin{align*} 
x(t) &= \sum \limits_{k=-\infty}^{\infty} X_k e^{ik\omega_0 t} \\
x(t) &= \sum \limits_{k=-\infty}^{\infty} X_k \Psi_k(t) \\
\end{align*}

This brings us to our synthesis equation:
$$
x = \sum \limits_{k=-\infty}^{\infty}X_k\Psi_k
$$
We use the inner product to find the spectral coefficient $X_k$, which brings us to our analysis equation:

$$
X_k = \frac{\langle x,\Psi_k\rangle}{\langle \Psi_k,\Psi_k\rangle}
$$

But how do we define our inner product? In discrete-time, we used

\begin{nicebox}
$$
\langle f,g\rangle = \sum \limits_{n=\langle p\rangle} f(n)g^*(n) \quad \mbox{(Discrete-time inner product)}
$$
\end{nicebox}

We can simplify this using linear algebraic language. 

\begin{align*}
f &= \left[ \begin{array}{c} f(0) \\ f(1) \\ \vdots \\ f(p-1) \\ \end{array} \right] \\
g &= \left[ \begin{array}{c} g(0) \\ g(1) \\ \vdots \\ g(p-1) \\ \end{array} \right] 
\end{align*}

Therefore, we can write $\langle f,g \rangle$ as multiplication of a tranposed vector with another vector:

$$ \langle f,g\rangle = f^Tg^* $$

But now that we are in continuous-time, we need more. We have a continuum of values, not a discrete, countable set. Hence, we define the inner product as an integral:

\begin{nicebox}
$$
\langle f,g\rangle = \int_{\langle p\rangle} f(t)g^*(t) dt \quad \mbox{(Continuous-time inner product)}
$$
\end{nicebox}

Now lets look at $\langle \Psi_k, \Psi_\ell\rangle$:
\begin{align*}
\langle\Psi_k, \Psi_\ell\rangle &= \int_{\langle p\rangle} \Psi_k(t)\Psi_\ell^*(t) dt \\
\langle\Psi_k, \Psi_\ell\rangle &= \int_{\langle p\rangle} e^{ik\omega_0 t}e^{-i\ell \omega_0 t} dt \\
\langle\Psi_k, \Psi_\ell\rangle &= \int_{\langle p\rangle} e^{i(k-\ell)\omega_0 t} dt \\
\end{align*}

We have two cases, where the exponent is zero when $k=\ell$, and then when it is non-zero and $k\neq \ell$.


I) $k=\ell$


This one is more simple. We are integrating 1 over a contiguous interval of length $p$, hence the value is $p$:

\begin{align*}
\langle \Psi_\ell, \Psi_\ell\rangle &= \int_{\langle p\rangle} dt = p
\end{align*}


II) $k\neq\ell$


This case requires a bit more work. Lets integrate.

\begin{align*}
\langle\Psi_k, \Psi_\ell\rangle &= \int_{\langle p\rangle} e^{i(k-\ell)\omega_0 t} dt \\
\langle\Psi_k, \Psi_\ell\rangle &= \left.\frac{e^{i(k-\ell)\omega_0 t}}{i(k-\ell)\omega_0} \right|_{\langle p\rangle} \\
\langle\Psi_k, \Psi_\ell\rangle &= \frac{e^{i(k-\ell)\omega_0 p}- e^{0}}{i(k-\ell)\omega_0} \\
\langle\Psi_k, \Psi_\ell\rangle &= \frac{e^{i(k-\ell)2\pi}- e^{0}}{i(k-\ell)\omega_0} \\
\langle\Psi_k, \Psi_\ell\rangle &= \frac{0}{i(k-\ell)\omega_0} = 0\\
\end{align*}

Lets look at this in another perspective and change to sines and cosines instead of integrating:
\begin{align*}
\langle\Psi_k, \Psi_\ell\rangle &= \int_{\langle p\rangle} e^{i(k-\ell)\omega_0 t} dt \\
\langle\Psi_k, \Psi_\ell\rangle &= \int_{\langle p\rangle} \cos((k-\ell)\omega_0 t) + i\sin((k-\ell)\omega_0 t) dt \\
\langle\Psi_k, \Psi_\ell\rangle &= \int_{\langle p\rangle} \cos((k-\ell)\omega_0 t) dt + \int_{\langle p\rangle}i\sin((k-\ell)\omega_0 t) dt \\
\end{align*}
Consider the case where $k-\ell=1$. We are summing over a full period of the cosine function:

% \img{images/trigonometric/cos1/cos.ps}

Notice that there is an equal amount of positive and negative area, hence the value is zero. If $k-\ell=2$, then we are integrating $\cos(2\omega_0 t)$, which has a period $p/2$, hence we would integrate over two full periods of the cosine function giving us zero:

% \img{images/trigonometric/cos2/cos.ps}

If you were to let $k-\ell=N$, where $N\in \Z$, then you would be integrating over $N$ periods since the function would have a period of $p/N$. This same phenomenon occurs with the sine function.

% \img{images/trigonometric/sin1/sin.ps}

Therefore, any time we have $k-\ell \neq 0$, we are integrating over integer multiples of the period, hence

$$\langle \Psi_k, \Psi_\ell\rangle = p\delta(k-\ell)$$

Looking back at the analysis equation, we have

$$ X_k = \frac{\langle x,\Psi_k\rangle}{\langle\Psi_k, \Psi_k\rangle} = \frac{1}{p} \int_{\langle p\rangle}x(t)e^{-ik\omega_0 t} dt $$

Note that since $\omega_0$ and $k$ are both required, the equation relies on them, and sometimes you may see $X(k\omega_0)$.

To sum everything up, we have two important equations regarding the continuous-time fourier series (FS):

\begin{nicebox}
\begin{align*}
x(t) &= \sum \limits_{k=-\infty}^{\infty}X_ke^{ik\omega_0t} \quad &\mbox{(synthesis equation)} \\
X_k &= \frac{1}{p} \int_{\langle p\rangle}x(t)e^{-ik\omega_0t}dt \quad &\mbox{(analysis equation)}
\end{align*}
\end{nicebox}

Because we can write $x(t)$ as an infinite number of $X_k$s, $X_k$ is not necessarily periodic, but $x(t)$ is $p$-periodic in continuous-time. Note that we can write $x(t)$ using any orthogonal set of basis vectors:

\begin{align*}
x(t) &= \sum \limits_{k=-\infty}^{\infty}X_k\Psi_k(t) \\
x(t) &= \sum \limits_{k=0}^{\infty}A_k \cos(k\omega_0 t) + \sum \limits_{k=1}^{\infty}B_k \sin(k\omega_0 t)
\end{align*}

Consider the following bases:
\begin{align*}
\Psi_k(t) &= \cos(\omega_0 t) \quad k\in\Z_{\oplus} \\
\Phi_\ell(t) &= \sin(\omega_0 t) \quad \ell \in \Z_+ 
\end{align*}
In order to use these as bases, we need to show the following:

\begin{align*}
\langle \Psi_k, \Psi_m\rangle &= 0 \quad \mbox{if } k\neq m \\
\langle \Psi_k, \Phi_\ell\rangle &= 0 \quad \forall\mbox{ } k,\ell \\
\langle \Phi_\ell, \Phi_r\rangle &= 0 \quad \mbox{if } \ell\neq r \\
\end{align*}

Then to determine the coefficients, we can use the inner product.

Continuous-time Fourier Series

The Continuous Fourier Series is defined as

\begin{nicebox}
\begin{align*}
x(t) &= \sum \limits_{k=-\infty}^{\infty}X_ke^{ik\omega_0t} \quad &\mbox{(synthesis)} \\
X_k &= \frac{1}{p} \int_{\langle p\rangle }x(t)e^{-ik\omega_0t}dt \quad &\mbox{(analysis)}
\end{align*}
\end{nicebox}

\begin{example}
1) Given an inpulse train $x(t)$ of Dirac deltas every $p$ samples, find the spectral coefficients and corresponding frequencies within the signal.


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\begin{align*}
X_k &= \frac{1}{p} \int_{\langle p\rangle }x(t)e^{-ik\omega_0t}dt \\
X_k &= \frac{1}{p} \int_{-p/2}^{p/2}x(t)e^{-ik\omega_0t}dt \\
X_k &= \frac{1}{p} \int_{-p/2}^{p/2}\delta(t)e^{-ik\omega_0t}dt \\
X_k &= \frac{1}{p} \cdot e^{0} \\
X_k &= \frac{1}{p} \\
\end{align*}

Therefore, we can determine that $\forall k$, $X_k = \frac{1}{p}$, and the signal is made up of every harmonic of $\omega_0$, and so we may write the signal in its continuous-time fourier series expansion as

$$ x(t) = \frac{1}{p} \sum \limits_{k=-\infty}^{\infty}e^{ik\omega_0 t} $$

Note that there are infinitely many harmonics of the fundamental frequency, not $p$ harmonics. This is a unique feature of the impulse train in continuous time. How do we know that $e^{ik\frac{\omega_0}{2}t}$ doesn't contibute to the signal? For $k=1$, $e^{i\frac{\omega_0}{2}t}$ has a period $2p$ which contradicts the periodicity of the signal $x$.

\end{example}


\begin{example}
2) Consider a periodic signal in continuous defined by $x(t) = \begin{cases} 1 & -T/2 \leq t \leq T/2  \\ 0 & \mbox{otherwise} \end{cases}$, where $T \in \R$, and $T/2 \langle  p/2$. 


%\includegraphics{images/idealpass/periodicpass.ps}


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\begin{align*}
X_k &= \frac{1}{p} \int_{\langle p\rangle }x(t)e^{-ik\omega_0 t}dt \\
X_k &= \frac{1}{p} \int_{-T/2}^{T/2}e^{-ik\omega_0 t}dt \\
X_k &= \frac{1}{p} \lr{ \left. \frac{e^{-ik\omega_0 t}}{-ik\omega_0} \right|_{-T/2}^{T/2}} \\
X_k &= \frac{1}{p} \frac{e^{-ik\omega_0 T/2} - e^{ik\omega_0 T/2}}{-ik\omega_0}  \\
X_k &= \frac{1}{p} \frac{e^{ik\omega_0 T/2} - e^{-ik\omega_0 T/2}}{ik\omega_0}  \\
X_k &= \frac{1}{p} \frac{2i\sin(k\omega_0 T/2)}{ik\omega_0}  \\
X_k &= \frac{1}{k\pi} \sin(k\omega_0 T/2)  \\
\end{align*}

Seems as though we have found a link between the analysis equation and frequency response of the system. 

$$ X_k = \left. X(\omega) \right|_{w=k\omega_0} \Rightarrow X(\omega) = \frac{2}{p}\frac{\sin(\omega T/2)}{\omega} $$

Since $X_k$ has a dependence on $k$, not all frequencies contribute equally. As $k\to \infty$, $\abs{X_k}\to 0$.

\end{example}

FS Examples

Given a periodic signal $x$ in CT, we can estimate the signal using the formula: 

$$ \hat{x}(t) = \sum \limits_{k=-N}^{N}\alpha_k e^{ik\omega_0 t}   $$

How can we determine the coefficients that make the estimates closest in terms of error energy? Let $W\in\R^{2N+1}$ be a subspace spanned by the set of orthogonal basis vectors $\Psi_0, \Psi_1, \Psi_{-1} \dots, \Psi_{N}, \Psi_{-N}$. If $x$ is a vector that is not in the column space of $W$, then we can project $x$ onto the column space of $W$, producing an approximation vector $\hat{x}$, which has a distance of $\abs{\left| \mathcal{E}_N\right|}$ from the vector $x$. The vectors $\mathcal{E}_N$ and $\hat{x}$ are orthogonal.

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We want to minimize $\abs{\left| \mathcal{E}_N\right|}$ to make the best approximation. $\mathcal{E}_N = x - \hat{x}_N$, hence $\abs{\left| \mathcal{E}_N\right|} = \abs{\left| x-\hat{x}_N\right|}$. Since $\mathcal{E} \perp W$, then we can use the inner product and the properties of orthogonality to solve for the coefficients. Since $W \in \R^{2N+1}$, we have $2N+1$ equations and $2N+1$ coefficients.

\begin{align*}
\langle \hat{x}_N, \Psi_\ell\rangle  &= \langle \sum \limits_{k=-N}^{N}\alpha_k e^{ik\omega_0 t} , \Psi_\ell\rangle  \\
\langle \hat{x}_N, \Psi_\ell\rangle  &= \sum \limits_{k=-N}^{N}\alpha_k \langle e^{ik\omega_0 t} , \Psi_\ell\rangle  \\
\langle \hat{x}_N, \Psi_\ell\rangle  &= \alpha_\ell \langle e^{i\ell\omega_0 t} , \Psi_\ell\rangle  \\
\langle \hat{x}_N, \Psi_\ell\rangle  &= \alpha_\ell \langle \Psi_\ell , \Psi_\ell\rangle  \\
\alpha_\ell &= \frac{\langle \hat{x}_N, \Psi_\ell\rangle }{\langle \Psi_\ell , \Psi_\ell\rangle } \\
\end{align*}

if $\hat{x}(t) = \sum \limits_{k=0}^{M-1}\alpha_k \Psi_k$, where $x$ is a $p$-periodic signal, which $\Psi_k$'s would you choose? Since we are taking a subset, the larger exponentials are better since they dominate. Otherwise, using $\{0,1,\dots,M\}\in\Z$ is not using very much. Pick the largest in magnitude FS coefficient, and then go down from there, but it doesn't have to be continguous. 


Approximations

Recall that we can write the frequency response of a system as

$$ H(\omega) = \sum \limits_{n\in \Z} h(n) e^{-i\omega n}  $$

Also, recall the $2\pi$-periodicity of the frequency response:

$$ H(\omega + 2\pi) = H(\omega) \forall \mbox{ } \omega  $$

The fourier series was written as $ x(t) = \sum \limits_{k=-\infty}^{\infty}X_ke^{ik\omega_0t} $, and $x(t+p) = x(t)$. We can combine these principles to come up with the DTFT. $x$ was $p$-periodic, $H$ is $2\pi$-periodic. 

\begin{align*}
x &\leftrightarrow e^{ik\omega t} \leftrightarrow X_n \\
H &\leftrightarrow e^{-i\Omega_0\omega n} \leftrightarrow h(n) \quad \mbox{where } \Omega_0 = 1 \\
\end{align*}

$X_n$ and $h(n)$ are the coefficients. $x$ is a periodic signal and so is $H$. $x(t)$ and $H(\omega)$ are functions over continuous variables.

\begin{align*}
x(t) & = \sum \limits_{n=-\infty}^{\infty}X_ne^{in\omega_0t} = \sum \limits_{n=-\infty}^{\infty} X_n\Psi_n\\
H(\omega) &=  \sum \limits_{n=-\infty}^{\infty}h(n)e^{-in\Omega_0\omega} = \sum \limits_{n=-\infty}^{\infty} H_n\Phi_n 
\end{align*}

Assume for now that $\langle \Phi_n, \Phi_m\rangle  = 0 $ when $n\neq m$. We want to find $h(n)$ in terms of $H(\omega)$. So we can project $H$ onto $\Phi_m$ to determine $h(m)$, or $H_m$:

\begin{align*}
\langle H, \Phi_m\rangle  &= \langle \sum_n\in\Z h(n)\Phi_n, \Phi_m\rangle  \\
\langle H, \Phi_m\rangle  &= \sum_n\in\Z h(n)\langle \Phi_n, \Phi_m\rangle  \\
\langle H, \Phi_m\rangle  &= h(m)\langle \Phi_m, \Phi_m\rangle  \\
h(m) &= \frac{\langle H, \Phi_m\rangle }{\langle \Phi_m, \Phi_m\rangle } \\
\end{align*}

We can always rely on the ratio of two inner products. $H(\omega)$ is a $2\pi$-periodic function over a continuous variable $\omega$, and $\Phi_m(\omega)$ is also a $2\pi$-periodic function over a continuous variable $\omega$.

\begin{align*}
\langle H, \Phi_m\rangle  &= \int_{\langle 2\pi\rangle }H(\omega)\Phi_m^*(\omega)d\omega \\
\langle H, \Phi_m\rangle  &= \int_{\langle 2\pi\rangle }H(\omega)e^{i\omega m}d\omega \\
\end{align*}

We still need to prove orthogonality and determine the value of $\langle \Phi_m, \Phi_m\rangle $:

\begin{align*}
\langle \Phi_n, \Phi_m\rangle  &= \int_{\langle 2\pi\rangle }\Phi_n(\omega)\Phi_m^*(\omega)d\omega \\
\langle \Phi_n, \Phi_m\rangle  &= \int_{\langle 2\pi\rangle }e^{-i\omega n}e^{i\omega m}d\omega \\
\langle \Phi_n, \Phi_m\rangle  &= \int_{\langle 2\pi\rangle }e^{i\omega(m-n)}d\omega \\
\end{align*}


We have two cases, where the exponent is zero when $m=n$, and then when it is non-zero and $m\neq n$. The first is more simple. We are integrating 1 over a contiguous interval of length $2\pi$, hence the value is $2\pi$:
\begin{align*}
\langle \Phi_m, \Phi_m\rangle  &= \int_{\langle 2\pi\rangle } dt = 2\pi
\end{align*}

This case requires a bit more work. Lets integrate.

\begin{align*}
\langle \Phi_n, \Phi_m\rangle  &= \int_{\langle 2\pi\rangle } e^{i(m-n)\omega} d\omega \\
\langle \Phi_n, \Phi_m\rangle  &= \left.\frac{e^{i(m-n)\omega}}{i(m-n)} \right|_{\langle 2\pi\rangle } \\
\langle \Phi_n, \Phi_m\rangle  &= \frac{e^{i(m-n)2\pi}- e^{0}}{i(m-n)} \\
\langle \Phi_n, \Phi_m\rangle  &= \frac{0}{i(m-n)} = 0\\
\end{align*}

Now we can come to a conclusion regarding the DTFT. 


\begin{nicebox}
\begin{align*}
h(n) &= \frac{1}{2\pi} \int_{\langle 2\pi\rangle } H(\omega)e^{i\omega n}d\omega \quad &\mbox{(synthesis equation)} \\
H(\omega) &= \sum \limits_{n=-\infty}^{\infty}h(n)e^{-i\omega n} \quad &\mbox{(analysis equation)}
\end{align*}
\end{nicebox}

$H(\omega)$ is $2\pi$-periodic, but $h(n)$ is not necessarily periodic. We can make a meaningful interpretation of the synthesis equation if we write the synthesis equation as a linear combination of complex exponentials:

$$ h(n) = \int_{\langle 2\pi\rangle }\lr{\frac{H(\omega)d\omega}{2\pi}}e^{i\omega m}  $$

A discrete-time function can be broken down into a linear combination of complex exponentials. How much of each complex exponential is present in the function is determined by the coefficient. What distinguished this coefficient is $H(\omega)$. So the fourier transform motivates the term \emph{filter} since it can break signals up into their constinuent frequencies and filter them in or out of the output signal.


DTFT

\begin{example}

1) Find the DTFT of $x(n) = \alpha^n u(n)$ where $\abs{\alpha} \lt 1$. 

\begin{align*}
X(\omega) &= \sum \limits_{n=-\infty}^{\infty}\alpha^n u(n)e^{-i\omega n} \\
X(\omega) &= \sum \limits_{n=0}^{\infty}\alpha^n e^{-i\omega n} \\
X(\omega) &= \frac{1}{1-\alpha e^{-i\omega}} \\
\end{align*}

If $\alpha \gt 0$, this is a low-pass filter:

If $\alpha \lt 0$, this is a high-pass filter:

What about if $\alpha \in \C$? Lets consider an example where $\abs{\alpha} \lt 1$, and $\alpha$ is in the $\pi/4$ direction. First, analyze the vectors on the unit circle:

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Notice that the denominator, $e^{i\omega} - \alpha$ is at a maximum at $-3\pi/4$, giving the magnitude response a minimum. The same vector has its minimum at $\pi/4$, giving the magnitude response a maximum at $\pi/4$.

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\rput(3.9269908125, -0.3){$\frac{5\pi}{4}$}
% end x-axis labels:

% y-axis
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% begin x-axis labels:
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\psline(-0.1, 3)(0.1, 3)
% end y-axis labels:

\end{pspicture}
\end{center}

Remember that we only need to graph over on period of $2\pi$, but it demonstrates the periodicity of the magnitude response.

\end{example}

\begin{example}

2) Consider an ideal low-pass filter that kills all frequencies except frequencies in $[-A,A]$. What is the impulse response $h(n)$?


% \includegraphics{images/idealpass/idealpass.ps}

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\begin{pspicture}(-1,-1)(4,1.5)

%axis
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% 0 
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Use the discrete-time fourier transform.
\begin{align*}
h(n) &= \frac{1}{2\pi} \int_{\langle 2\pi\rangle}H(\omega)e^{i\omega n} d\omega \\
h(n) &= \frac{1}{2\pi} \int_{-A}^{A}e^{i\omega n} d\omega \\
h(n) &= \frac{1}{2\pi} \lr{\left. \frac{e^{i\omega n}}{in}\right|_{-A}^{A}} \\
h(n) &= \frac{1}{2\pi} \lr{\frac{e^{iAn}-e^{-iAn}}{in}} \\
h(n) &= \frac{1}{2\pi} \frac{2i\sin (An)}{in} \\
h(n) &= \frac{1}{\pi} \frac{\sin (An)}{n} \\
\end{align*}

Note that we could have also written this as
\begin{align*}
h(n) &= \frac{1}{2\pi}\left[ \int_{-A}^{A}\cos(\omega n) d\omega + i  \int_{-A}^{A}\sin(\omega n) d\omega\right] \\
h(n) &= \frac{1}{2\pi}\int_{-A}^{A}\cos(\omega n) d\omega  \\
h(n) &= \left. \frac{1}{2\pi} \sin(\omega n) \right|_{-A}^{A} \\
h(n) &= \frac{1}{\pi} \frac{\sin (An)}{n} \\
\end{align*}

Since sine is an odd function, then integrating from $-A$ to $A$ gives us zero, so we are left with integrating the cosine only. What happens to the impulse response $h(n)$ as $A$ approaches $\pi$?

$$ \lim_{A\to\pi} \frac{\sin (An)}{\pi n} =  \frac{\sin (\pi n)}{\pi n}  $$

Now for all $n \neq 0$, $h(n) = 0$, because we are feeding a sine function multiples of $\pi$, and $h(0)=1$ by l'Hopital's rule. This means that as $A$ approaches $\pi$, the impulse response becomes the identity element of convolution: 

$$  \lim_{A\to\pi} \frac{\sin (An)}{\pi n} = \delta(n) $$

Another important thing to note is that this signal cannot exist in reality. This is because signals that are absolutely summable have DTFTs that are continuous functions of $\omega$. That means

$$ \sum \limits_{n=-\infty}^{\infty}\abs{x(n)} \lt \infty $$

Since $\frac{1}{\pi} \frac{\sin (An)}{n}$ decays on the order of $\frac{1}{n}$, it is not absolutley summable. But, the absolute sum of its energy is finite. 

$$ \sum \limits_{n=-\infty}^{\infty}\abs{x(n)}^2 \lt \infty $$

This is because $\sum\lr{\frac{1}{n}}^2$ converges. Functions that are not absolutely summable but are square summable (have finite energy) have DTFT's that are discontinuous. You cannot use the analysis equation, but you can use the synthesis equation.

\end{example}

\begin{example}

3) Find the DTFT of $x(n)=e^{i\omega_0 n}$. 

\begin{align*}
\sum \limits_{n=-\infty}^{\infty}\abs{x(n)} \\ 
\sum \limits_{n=-\infty}^{\infty}\abs{e^{i\omega_0 n}} \\ 
\sum \limits_{n=-\infty}^{\infty}1 = \infty \\
\end{align*}

This is not absolutely summable or square summable. The discontinuity gets worse and $X(\omega)$ has deltas!

We know that this represents a dirac delta in the frequency domain centered at $\omega_0$. This gives us a frequency response of $A\delta(\omega -\omega_0)$ for some constant $A$.


% \includegraphics{images/pulsetrains/freqpulse.ps}

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\begin{center}
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\end{pspicture}
\end{center}

So we can use the synthesis equation over an interval of $2\pi$:

\begin{align*}
x(n) &= \frac{1}{2\pi} \int_{\langle 2\pi\rangle} X(\omega) e^{i\omega n}d\omega \\
e^{i\omega_0 n} &= \frac{1}{2\pi} \int_{\langle 2\pi\rangle} A \delta(\omega-\omega_0) e^{i\omega n}d\omega \\
e^{i\omega_0 n} &= \frac{A}{2\pi} \int_{-\pi}^{\pi} \delta(\omega-\omega_0) e^{i\omega n}d\omega \\
e^{i\omega_0 n} &= \frac{A}{2\pi} e^{i\omega_0 n} \\
A &= 2\pi \\
\end{align*}

This tells us that $X(\omega) = 2\pi\delta(\omega - \omega_0)$, but we know that $X(\omega + 2\pi k) = X(\omega) $ $ \forall k \in \Z$, so we must write the frequency response in the form:

$$ X(\omega) = \sum \limits_{k=-\infty}^{\infty} 2\pi \delta(\omega-\omega_0 +2\pi k) $$

If we restricted the domain, for example, $-\pi \leq \omega \lt \pi$, then $X(\omega) = 2\pi\delta(\omega-\omega_0)$. 

\end{example}

4 Week 4