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The CTFT of an analog signal is a representation of that analog signal in terms of the frequency parameter of sinusoidal (cosine specifically) functions whose weighted sum make up that signal. The cosines are continuous functions of time and the amplitude of a particular cosine at a given frequency is given by the amplitude of the spectrum at that frequency.

Is the interpretation for the DTFT of a sampled signal analogous where the spectrum indicates the frequency of (discrete-time/sampled) cosines that sum to the sampled signal, the spectrum amplitude indicates the magnitude of that cosine, etc.?

For instance, if $F=100\textrm{ Hz}$ and $|X(F=100\textrm{ Hz})| = 50$, then we know we have the component cosine signal: $50\cdot\cos(100n)$. Assuming $X(F)$ is the DTFT of $x(n)$.

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  • $\begingroup$ I think you forgot the $2\pi$ in the $\cos$ argument in the question. $\endgroup$ – Gilles Mar 31 '16 at 9:00
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Let's first consider the continuous-time case. From the formula for the inverse Fourier transform

$$x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\Omega)e^{j\Omega t}d\Omega\tag{1}$$

you can see that at a given frequency $\Omega$ (in rad/s), the value of the spectrum $X(\Omega)$ can be interpreted as the "strength" with which the complex exponential $e^{j\Omega t}$ contributes to the signal $x(t)$. You cannot conclude (as you seem to suggest) that the signal contains a sinusoid with amplitude $|X(\Omega)|$. This would only be the case if the signal $x(t)$ were periodic and, consequently, the integral in $(1)$ could be written as a sum, namely the Fourier series. Also note that even in that case the sinusoid is not just a cosine with zero phase, because the Fourier coefficients are generally complex and, hence, determine the amplitude as well as the phase of the respective sinusoid.

With this in mind, we can interpret the formula for the inverse discrete-time Fourier transform (IDTFT):

$$x[n]=\frac{1}{2\pi}\int_{-\pi}^{\pi}X(\omega)e^{jn\omega}d\omega\tag{2}$$

Note that I use a different variable here because $\omega$ in $(2)$ is not the same as $\Omega$ in $(1)$ (see below). Also note that the limits of the integral in $(2)$ are $\pm \pi$ instead of $\pm \infty$. The reason for the latter is that the sampling frequency sets an upper limit for the frequencies that can be uniquely represented by the DTFT (check out Aliasing and the Sampling Theorem if this is new to you). Apart from that, the structures of $(1)$ and $(2)$ are the same. So you can interpret the DTFT of $x[n]$ at a certain "frequency" $\omega$ as the "strength" with which the complex exponential $e^{jn\omega}$ is represented in the sequence $x[n]$. Again, it's generally not the amplitude of some sinusoid contained in $x[n]$, for the same reasons as outlined above.

The last thing that remains is the explanation of the difference between the frequency variables $\Omega$ and $\omega$. $\Omega$ is simply the frequency in radians per second, so we have $\Omega=2\pi f$, where $f$ is the frequency in Hertz. Since in discrete-time all frequencies must be seen as relative to the sampling frequency $f_s$, the variable $\omega$ is a normalized frequency in radians: $\omega=2\pi f/f_s$. So the integration limits in $(2)$ correspond to $f=\pm f_s/2$. The frequency $f_s/2$ is called the Nyquist frequency, and frequencies above the Nyquist frequency cannot be represented uniquely for a given sampling frequency.

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