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how do you even measure the frequency of a digital signal?
This may be the source of your confusion. In digital communications, data is transmitted using an analog, continuous-time waveform. This is necessary because only this kind of waveform can exist in the physical world. Like all waveforms, it has a bandwidth, and its Fourier Transform determines what ...
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You might be making this problem harder than it is by thinking in terms of frequency rather than periods. Your sinusoid is
$$x(t) = \cos\left(2\pi \frac{t}{T_p}\right)$$ and its samples are
$$x[n] = x(nT) = \cos\left(2\pi \frac{nT}{T_p}\right).$$
The samples form a discrete-time periodic sequence of period $N$ if
\begin{align}
x[n] &= x[n+N]~\forall\, ...
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You did everything correctly. I think that part of your confusion comes from the way you chose the notation. Let me use $v_k[n]$ to denote the system's response to a delayed input $x[n-k]$. $v_k[n]$ is described by the following difference equation:
$$v_k[n]=3v_k^2[n-1]-nx[n-k]+4x[n-k-1]-2x[n-k+1]\tag{1}$$
On the other hand, the delayed response $y[n-k]$ ...
1
The binomial theorem is not actually involved: the top waveform is simply pointwise raised to the 5-th power, as others have noted, and there are no missing peaks. To illustrate, consider the following crude hand-drawn figure:
I drew this in my simulation software, so it was automatically "digitized" as I drew it. Obviously, I am poor at freehand drawing ...
1
You're right, the given $x[n]$ is clearly periodic. You can show this by simply checking if
$$x[n]\stackrel{?}{=}x[n+N]\tag{1}$$
is satisfied for some positive integer $N$.
For the given $x[n]$ you get
$$\begin{align}(-1)^{n^2}&\stackrel{?}{=}(-1)^{(n+N)^2}\\&=(-1)^{n^2}(-1)^{N(N+2n)}\tag{2}\end{align}$$
From $(2)$ it follows that for $(1)$ (and ...
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The IDTFT of $X(e^{j\omega})=1$ is indeed
$$x[n]=\frac{\sin(n\pi)}{n\pi}\tag{1}$$
Now, what happens for indices $n\neq 0$? As it turns out, you can safely rewrite $(1)$ as
$$x[n]=\delta[n]\tag{2}$$
where $\delta[n]$ is the discrete-time unit impulse. (HINT: think about where the zeros of $\sin(x)$ are).
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As far as I understand, the sampling interval $T=1/F_s$ is given, and, hopefully, the period $N$ of the sampled signal. Now the only thing we can say about the period $T_p$ of the continuous-time signal is that it satisfies
$$T_p=\frac{NT}{k},\qquad k\in\mathbb{Z}^+\tag{1}$$
I.e., the period of the discrete time signal measured in units of time $(NT)$ is ...
1
You could try to Fourier transform,
$F(k) = \int_{-\infty}^{\infty} x(t) e^{-2\pi i kt} dt$
which can be expressed in terms of magnitude and phase in the following way
$F(k) = |F(k)|e^{i\Phi(k)} = a(k) + ib(k)$
with magnitude $|F(k)|$ and phase $\Phi(k) = \tan^{-1}(\frac{b}{a})$ for each frequency component $k$. Since it looks like you have a dominant $k$...
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First
(i) I know that analog signals are continuous and digital signals are
discrete. Analog signal are represented by a continuous time-varying
quantity like voltage. How are digital signals represented in actual
practice? I mean to ask how are the 0s & 1s actually
represented/transmitted over a transmission medium? Are they
represented ...
1
Let me summarize my understanding of what you're trying to do. You have a real-valued sequence $x[n]$, obtained by sampling a real-valued continuous function, and you computed its DFT $X[k]$. The sequence can be expressed in terms of its DFT coefficients:
$$x[n]=\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j2\pi nk/N},\qquad n\in[0,N-1]\tag{1}$$
where $N$ is the ...
1
We start with
$$X[k] = \sum_{n=0}^{N-1} x[n] \cdot W_{N}^{nk}$$
Let's define a new sequence of $y[r]$ of length $R=M \cdot N$, where $R, M, N \in \mathbb{Z} > 0$
$$y[r] = \begin{cases}
y[n \cdot M] = x[n] \qquad & r = n \cdot M \\
\\ 0 \qquad & \text{otherwise} \\
\end{cases}$$
Transform it
$$Y[k] = \sum_{r=0}^{R-1} y[r] \cdot W_{R}^{rk}$$
Now ...
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The images attached here has the necessary functions of opencv to solve the problem you have posed here.
The logic is that, keeping a reference line(like the red line you have drawn in the image present in the question), count how many times the intensity of the image changes either from 255 to 0 or vice versa.
Dividing this count value by 2 will yield you ...
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