5

For continuous-time systems, a pole at location $s_0=\sigma_0+j\omega_0$ will create a time-domain contribution of the form $$e^{s_0t}=e^{\sigma_0t}e^{j\omega_0t}\tag{1}$$ which is a damped oscillation if the pole is in the left half-plane (i.e., $\sigma_0<0$), and if the pole is not on the real axis (i.e., $\omega_0\neq 0)$. For $\omega_0=0$ there is ...


3

Since $y[n-n_0] = y[n] \ast \delta [n - n_0]$ and convolution is both associative and distributive, is it not the case that $$ y[n-n_0] = \left(x[n] \ast r[n] \right) \ast \delta[n-n_0] = x[n] \ast r[n-n_0] = x[n-n_0] \ast r[n]? $$


2

Note that the frequencies of the FFT grid are given by $$f_k=\frac{kf_s}{N},\qquad k=0,1,.\ldots,N-1\tag{1}$$ where $N$ is the FFT length. Now observe that the two frequencies of the given signal fall exactly on the grid. So you only have to determine these two frequencies, and then use $(1)$ to figure out the corresponding indices. EDIT: It's always ...


2

Why not taking the definition of convolution and see what happens? $$y[n-n_0] = \sum_k x[k]r[n-n_0-k]$$ which gives $$y[n-n_0] = x[n]*r[n-n_0]$$ Also, since $x[n]*r[n] = r[n]*x[n]$, $$y[n-n_0] = \sum_k r[k]x[n-n_0-k]$$ which gives $$y[n-n_0] = r[n] * x[n-n_0]$$ So both (3) and (4) are correct.


2

So called spectral leakage always occurs when transforming any finite length observation. That's because it's an artifact of windowing, and any non-infinite length observation (-t/2 ... +t/2) requires some sort of window, inherent rectangular or otherwise. A window function allow you to choose your artifact: from window "distortion" to Sinc effects, and/or ...


2

I would recommendate the lib NWaves, not so mature, but enough for most of cases in life.


1

Here a few hints to get you started: Define a sequence $x[k]=\langle g_n, g_{n-k} \rangle$ and write it as a convolution. From that it should be easy to find its DTFT $X(e^{j\omega})$ in terms of $G(e^{j\omega})$. Next note that the given sequence is just a downsampled version of $x[k]$, so its spectrum is an aliased version of $X(e^{j\omega})$. That's how ...


1

Please see this post that details most of the noise considerations for an ADC What are advantages of having higher sampling rate of a signal? In particular the relationship for the noise floor given a full scale sine wave: $$SNR = 6.02 \text{ dB/bit} + 1.76 \text{ dB}$$ Using your numbers: 1.5Vpp sinewave. If we assume this sinewave is full scale (any ...


1

In the first case you have a pole at the location (-0.7). In the second case, your pole is at 0.7. Having a pole at -0.7 means that the natural frequency of your system is fs/2, that's why you have an oscillation at fs/2. Since the pole is stable, i.e. inside the unit circle, the oscillation eventually dies down. Edit : You happen to have a pole at fs/2, ...


1

The problem with representing a discrete time just as a list like: $$x[n]=[6, 4, 0, 2]$$ is that you don't have a time reference. The first index $n=0$ or the sample in $\textbf{bold}$ or $\underline{\mathrm{underline}}$ is usually taken as the start of the sequence. For example, $$x[n]=[\mathbf{6}, 4, 0, 2]$$ or $$x[n]=[\underline{6}, 4, 0, 2]$$ So $x[...


1

Time align the quantized signal with your input signal and subtract to create an error signal (match the amplitudes such as to minimize the error signal). The SNR can then be determined as the ratio of the standard deviation of the input signal (after amplitude matching) to the standard deviation of the error signal (20Log10 of this ratio to be in dB). The ...


1

Discrete time sequence have frequency $$-\dfrac{1}{2}\leq f\leq \dfrac{1}{2}\tag{1}$$ Continuous time signal $x(t)=A\cdot\sin(2\pi Ft)\tag{2}$ The discrete sequence can be obtained from CT signal through periodic sampling. $$t=n\cdot T= \dfrac{n}{F_s}\tag{3}$$ Plugging the above relation in $(2)$ we get $$f=F/F_s\tag{4}$$ Again plugging the $f$ in $(1)...


1

The maximum frequency of a discrete-time signal is half the sampling frequency. A signal with maximum frequency in discrete time is an alternating sequence, and since its period is two sample intervals, its frequency is $f_s/2$. All frequencies in the discrete domain are normalized by that maximum frequency. This implies that a discrete-time system's ...


1

The reason one uses a complex signal at baseband is so that one can distinguish between positive and negative frequency...otherwise, when the signal is mixed to a higher frequency, it will have two identical sidebands, taking up twice the spectrum. Without knowing what you're modulating, I can only offer general advice...take your real baseband signal ...


1

When $n=0$ you may need to know $y[-1]$. The current output $y(n)$ depends on the current input $x(n)$ and previous output $y(n-1)$ scaled.


1

There are a few things going on here: Does this thought process look correct? Strictly, No. The main reason for this is that you are not performing a resampling step that is the differentiating point between the Discrete Fourier Transform (DFT) and order analysis. Is it correct to say that the final fft plot is in the frequency domain? From this link,...


1

TL;DR Your statements are inaccurate. The concept of frequency is $\frac{1}{T}$ where $T$ is the period time. Be careful with the continuous-time and discrete-time definition for a perioud. discrete time sinusoids have frequency −π<ω<π or $−\frac{1}{2}<f<\frac{1}{2}$ So, this statement is not accurate. A frequency is defined as the number ...


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