How to find out amplitude of harmonic after diff type of FT?
Lets find the spectrum of the signals $x_c(t) = A \cos(\Omega_0 t)$ and $x_d[n] = x_c(nT) = A\cos(\Omega T n)$

I-First consider the CTFT (continuous-time Fourier transform) $X(\Omega)$ of $x_c(t)$:
$$x_c(t) = A \cos(\Omega_0 t) = A\frac{e^{j\Omega_0 t} +e^{-j\Omega_0 t} }{2} \longleftrightarrow X_c(\Omega) = A \pi \delta(\Omega-\Omega_0) + A \pi \delta(\Omega+\Omega_0) $$ The CTFT spectrum involves two impulses at $\pm \Omega_0$. Their magnitudes are infinite whereas their weights are $A \pi$

II-Second consider the DTFT (discrete-time Fourier transform) $X_d(e^{j \omega})$ of $x_d[n] = A\cos(\Omega_0 n T)$

Approach-1: Considering directly the signal $x_d[n] = A\cos(\Omega_0 n T)$ where $\Omega_0 T = \omega_0$ and by using the DTFT properties:

$$x_d[n] = A \cos(\Omega_0 T n) = A\frac{e^{j\omega_0 t} +e^{-j\omega_0 n} }{2} \longleftrightarrow X_d(e^{j \omega}) = A \pi \delta(\omega-\omega_0) + A \pi \delta(\omega+\omega_0 ) $$ The DTFT spectrum involves two impulses (in $-\pi &lt; \omega &lt; \pi$) at $\pm \omega_0 = \Omega_0 T$. Their magnitudes are infinite and their weights are $A \pi$

Approach-2: Considering the relation between DTFT $X_c(\Omega)$ and CTFT $X_d(e^{j\omega})$ due to sampling $$\boxed{ X_d(e^{j\omega}) = \frac{1}{T} X_c(\frac{\omega}{T})}$$

Given that the CTFT $X_c(\Omega) = A \pi [\delta(\Omega-\Omega_0) + \delta(\Omega+\Omega_0)]$ we conclude on the DTFT as:
$$X_d(e^{j\omega}) =  \frac{1}{T} A \pi [\delta(\frac{\omega}{T}-\Omega_0) + \delta(\frac{\omega}{T}+\Omega_0)]$$

Recalling the property of the impulse as:
$$ \delta(\frac{\omega}{T}-\Omega_0) = \delta( \frac{\omega- \Omega_0 T }{T} ) = T \delta( \omega- \Omega_0 T )=T \delta( \omega- \omega_0 )$$ 

Yields for DTFT $X_d(e^{j\omega})$:
$$X_d(e^{j\omega}) =  \frac{1}{T} A \pi [\delta(\frac{\omega}{T}-\Omega_0) + \delta(\frac{\omega}{T}+\Omega_0)] = A \pi [\delta(\omega-\omega_0) + \delta(\omega + \omega_0)]$$ The same result of approach-1 is observed: The DTFT spectrum involves two impulses (in $-\pi &lt; \omega &lt; \pi$) at $\pm \omega_0 = \Omega_0 T$. Their magnitudes are infinite and their weights are $A \pi$

III-Third consider the $N$-point DFT of the signal $x_d[n] = A \cos(\Omega T n) = A\cos(\omega_0 n)$

Approach-1: Considering directly the $N$-point DFT sum of the discrete-time signal $x[n] = A\cos(\omega_0 n)$ for $n=0$ to $n=N-1$. 

(NOTE for the simple analysis we shall assume there's an integer $k_0$ such that $\Omega_0 T = \omega_0 = \frac{2 \pi}{N} k_0$ is satisfied; i.e., the frequency $\omega_0$ resides exactly on a DFT bin frequency for the index $k = k_0$, for $0 \leq k &lt; N$.) 

$$X[k] = \sum_{k=0}^{N-1} A \cos(\omega_0 n) e^{-j\frac{2 \pi}{N} k n } = \frac{A}{2} \sum_{k=0}^{N-1}[e^{j\omega_0 n}+e^{-j\omega_0 n}]  e^{-j\frac{2 \pi}{N} k n }  $$ Since we assume that $\omega_0 = \frac{2\pi}{N} k_0$ then we get:

$$X[k] = \frac{A}{2} \sum_{k=0}^{N-1}[e^{j\frac{2\pi}{N} k_0 n}+e^{-j\frac{2\pi}{N} k_0 n}]  e^{-j\frac{2 \pi}{N} k n }   = \frac{A}{2} \sum_{k=0}^{N-1}e^{j\frac{2\pi}{N} (k_0 -k) n} + \frac{A}{2} \sum_{k=0}^{N-1}e^{-j\frac{2\pi}{N} (k_0 +k) n}$$

We can evaluate the sums using the relation 
$$\sum_{k=0}^{N-1}e^{j\frac{2\pi}{N} (k-k_0) n} = 
\begin{cases} 
N  &amp;, \text{ for } k = k_0 \\
0 &amp;, k \neq k_0 \\
\end{cases}$$ 
Which yields for DFT $X[k]$ as:

$$X[k] = \begin{cases}
0 &amp;, \text{ for } 0 \leq k &lt; k_0 \\
\frac{A}{2} N &amp;, \text{ for } k = k_0 \\
0 &amp;, \text{ for } k_0 &lt; k &lt; N-k_0 \\
\frac{A}{2} N &amp;, \text{ for } k = N-k_0 \\
0 &amp;, \text{ for } N-k_0 &lt; k &lt; N-1 \\
\end{cases}$$

Approach-2 considering that the finite length signal $x_v[n]$ is obtained from the infinite length signal $x_d[n] = A\cos(\omega_0 n)$ by multiplying it with a window $w[n]$ of length $N$. And considering its effect on the DTFT of the windowed signal $x_v[n] = x_d[n] w[n]$ and considering that DFT $X_v[k]$ is obtained as the uniform samples of the DTFT $X_v(e^{j\omega})$ of $x_v[n]$:

Since $x_v[n] = x_d[n] w[n]$ then; 
$$X_v(e^{j\omega}) = \frac{1}{2\pi} X_d(e^{j\omega}) \star W(e^{j \omega})$$

We previously computed $X_d(e^{j \omega})$ as:
$$ X_d(e^{j \omega}) = A \pi \delta(\omega \pm \omega_0)$$

And it can be shown that the DTFT of the window is:
$$W(e^{j \omega})= e^{-j \omega ( \frac{N-1}{2} ) } \frac{ \sin(\omega N/2) }{\sin(\omega /2) }$$

And the DTFT $X_v(e^{j \omega})$ becomes:
$$ X_v(e^{j \omega}) = \frac{1}{2\pi} X_d(e^{j \omega}) \star W(e^{j \omega}) = \frac{A}{2} W(e^{j (\omega-\omega_0)}) + \frac{A}{2} W(e^{j (\omega+\omega_0)}) $$

Finally the DFT $X_v[k]$ of $x_v[n]$ will be the samples of this $V(e^{j \omega})$ at the frequencies $w_k = 2\pi k /N$. Under this special sampling we have the following: 
$$X_v[k] = X_v(e^{j \omega})|_{w_k = \frac{2 \pi}{N} k} = \frac{A}{2} W(e^{j (\frac{2 \pi}{N} k-\omega_0)}) + \frac{A}{2} W(e^{j (\frac{2 \pi}{N} k + \omega_0)}) $$

and replacing again $\omega_0$ with $\frac{2 \pi}{N} k_0$ we have:
$$X_v[k] = \frac{A}{2} W(e^{j (\frac{2 \pi}{N} (k - k_0))}) + \frac{A}{2} W(e^{j (\frac{2 \pi}{N} (k + k_0)}) $$

Similar to approach-1, and considering $W(0) = N$ we have
$$
W[k-k_0] = \begin{cases}
0 &amp;, \text{ for } 0 \leq k &lt; k_0 \\
N &amp;, \text{ for } k = k_0 \\
0 &amp;, \text{ for } k_0 &lt; k &lt; N-k_0 \\
N &amp;, \text{ for } k = N-k_0 \\
0 &amp;, \text{ for } N-k_0 &lt; k &lt; N-1 \\
\end{cases}
$$

And therefore we conclude for DFT $X_v[k]$ as
$$
X_v[k] = \begin{cases}
0 &amp;, \text{ for } 0 \leq k &lt; k_0 \\
\frac{A}{2}N &amp;, \text{ for } k = k_0 \\
0 &amp;, \text{ for } k_0 &lt; k &lt; N-k_0 \\
\frac{A}{2}N &amp;, \text{ for } k = N-k_0 \\
0 &amp;, \text{ for } N-k_0 &lt; k &lt; N-1 \\
\end{cases}
$$