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The equality $$\frac{1}{T}\sum_{k = -\infty}^{\infty}X(j(\omega - k\frac{2\pi}{T})) = \sum_{k = -\infty}^{\infty}x(kT)e^{-j\omega kT}\tag{1}$$ is an instance of Poisson's sum formula. The term on the right-hand side of $(1)$ is just the Fourier series representation of the periodic function on the left-hand side of $(1)$. The samples $x(kT)$ of the time ...

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The condition $$x(nT)=\delta[n]\tag{1}$$ is called the Nyquist criterion for zero intersymbol interference (ISI). It is important for the design of transmit pulses in digital communication systems. Condition $(1)$ can be expressed in the frequency domain as $$\frac{1}{T}\sum_{k=-\infty}^{\infty}X\left(\omega-\frac{2\pi k}{T}\right)=1\tag{2}$$ where $X(\omega)... 2 Yes absolutely there is also the equivalent of aperture jitter results in a DAC. This is due to the effects of jitter in the sampling clock itself, and variation in the electronics of the time duration for the translation from the different digital levels to analog levels in the output. In either case the jitter is not necessarily just due to a S/H circuit ... 1 You're right that a triangular function is not band-limited and that it can't be sampled without introducing aliasing. However, this doesn't mean that linear interpolation can't be useful. The usefulness of linear interpolation depends on the data to be interpolated and on the desired interpolation factor. If the data have a low pass character, i.e., if they ... 1 Are these Tx chain and Rx chains the correct way to implement this transmission? No. When you want to transmit a real cosine in passband, then you need a single complex tone in baseband. Not a real one. Otherwise, you'll have two real tones on the air, one at the mixing frequency + cosine frequency, and one at the mixing frequency - cosine frequency. But ... 1 This is just as it turns out when you do the math. The discrete-time Fourier transform (DTFT) of the sampled continuous-time impulse response$h(t)is $$H_d(e^{j\omega T})=\sum_nh(nT)e^{-jn\omega T}\tag{1}$$ With $$h(nT)e^{-jn\omega T}=\int_{-\infty}^{\infty}h(t)e^{-j\omega t}\delta(t-nT)dt\tag{2}$$ this can be written as \begin{align}H_d(e^{j\omega T})&... 1 That \frac{1}{T_s} factor in your sampled signal expression is \frac{20}{\pi}.\omega_s = 40 \ rad/sec2\pi f_s = 40f_s = \frac{40}{2\pi} = \frac{20}{\pi}$$The sampled signal is basically given by :$$X_{sampled}(f) = f_s \sum^{\infty}_{k=-\infty} X(f - kf_s)$Meaning : Sampling a signal with impulse train at sampling frequency$f_s$gives ... 1 From my limited experience in signal processing, I know that some signals cannot be represented in a digital machine if they are in a band-limited form. The signal takes on a sequence of discrete sample values so that it takes the replicated form. So the question is how is that possible? It boils down to a simple yet very important equation :$X(n)=sin(...

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If you have a pure tone signal that has a frequency higher than the Nyquist frequency, it doesn't "disappear", it looks like it has a different frequency (mirror image around the Nyquist). This is called aliasing. If it happens to be exactly at Nyquist, it could disappear if your sampling happens to hit the zero crossings. The same is true for a ...

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Nyquist didn't say that. The Nyquist-Shannon sampling theorem says that the sampling rate asymptotically approaches twice the bandwidth of the signal, more or less no matter where the signal's center frequency is. So you care about the width of the signal, not its maximum frequency. But where you got the expression wrong in one direction pertaining to ...

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Probably no, but... maybe. The classical sampling theorem is often misstated: if a continuous signal is band-limited, it can be recovered exactly (in theory) if the sampling frequency is twice its maximum frequency. However, there are cases where signals can be recovered above the Nyquist limit, provided that we have more information, because ambiguities may ...

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You won't get the discrete $y_d[m]$ by re-arranging the equations, but by sampling the $y_c(t)$ at integral multiples of $T$. When you sample the RHS in equation1 exactly at $t=mT$ instances, you get $y_d[m]$. You need to remember that $sinc(m-n) = 1$ for $m=n$ and $0$ otherwise, and (BTW in equation 1, in $sinc$ argument $\pi$ won't be there). Anyway, take ...

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