# Tag Info

3

Even though I'm not much of a wavelet expert, I can answer with a definitive YES!! Now, do you have any messy details to complain about, like the size of the box, the power that it consumes, or the number of frames of delay between the taking of the image and its display? Because that's your real problem. Take the particular wavelet transform you want to ...

2

I'd try a very "tinkery" approach here: Erode the image, so that the black area is shrunk by a fixed radius of pixels from its border (say, 5px). Dilate the resulting image by the same amount measure the amount of difference between original and processed image. The idea is that something that is a locally convex border doesn't suffer through erosion (it's ...

2

50 KBaud means the symbol rate is 50 KHz (and in the OP's case the bit rate with 1 bit per symbol). The units of Hz is 1/sec. So the duration of each bit (the bit period) is simply $$t_p = \frac{1}{50,000} = 20 \mu\text{s}$$

2

Based in my comment, here are quite accurate coefficients for A-Weighting filter to be used with 33kHz sampling only: b = [0.759941332414235 -1.931718891800229 0.5144488202077094 2.040523943539552 -1.283344336756084 -0.5073918947145036 0.3880093863750892 0.03272320276137633 -0.01319156202714533] a = [1 -2.840785425332851 1.688214038430988 1.889562806030876 -...

1

Strictly: No such $B$ exists. You could simply have a sufficient streak of "bad luck" and draw positive $\eta>\epsilon>0$ continuously, for example. Obviously, $B<\lim_{N\to\infty}\sum_{n=-N}^{N-1}\epsilon_n<\lim_{N\to\infty}\left\lvert\sum_{n=-N}^{N-1}\eta_n\right\rvert\,\forall B\in \mathbb R$. Granted, the event that every $\eta_n>... 1 One of the radio stations around here used to have the slogan "It may not be your favorite song, but it has a lot of the same notes." A DFT analysis can tell you what notes are present, and at what intensity. With calibration, you can even measure the real world level. This doesn't get anywhere near psycho-acoustics though. What you could ... 1 Is what I am doing correct? Yes If not, how would I reproduce the SINAD function? By replicating the exact same algorithm that the Matlab function uses. Your answer is more correct, but you also "cheated". You have access to the original sine and noise signals. sinad() ONLY has access to the sum of both. Hence it needs to do an estimation using ... 1 Strict real-time can be complicated, but it definitely is doable with a controllable delay. Some call that practically real-time or almost real-time. You will easily find wavelet FPGA implementations for video compression or processing for instance. I would like to underline that overcomplete or redundant wavelets can be a bit more complicated to implement,... 1 If you sample a finite-power continuous time WSS random process$x(t)$, the auto-correlation of the sampled process$y[k]=x(kT)$equals the sampled auto-correlation of the continuous-time process: $$y[k]=x(kT)\;\Longleftrightarrow \;R_y[k]=R_x(kT)\tag{1}$$ Since the power spectrum is the Fourier transform of the auto-correlation, the power spectrum of the ... 1 Hmm. For standard deviation, I see https://dsp.stackexchange.com/a/8632/829 which states for uniform power spectral density$N_0/2$, the standard deviation is $$\sigma^2 = \int_{-\infty}^\infty \frac{N_0}{2}|H(f)|^2\,\mathrm df$$ which in my case would yield$\$\begin{align} \sigma^2 &= \int_{-\infty}^\infty \frac{N_0}{2}|H(f)|^2\,df \\ &...

Only top voted, non community-wiki answers of a minimum length are eligible