# Tag Info

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Assuming I understand the question correctly, you need to sample the string's position at least twice as fast as the maximum frequency you anticipate the string can move or the highest frequency you care about. So if the highest overtones the string produces are 16kHz, you need to measure at at least 32kHz if you want to be able to reconstruct the original ...

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Since sines and cosines can be expressed as sums of complex exponentials, you can always think of any signal as complex. For real valued signals it just works out that the imaginary parts cancel. If the sines and cosines in a real-valued signal use frequencies from 0 to B, the corresponding sum of complex exponentials will include frequencies from -B to B. ...

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Assuming your PWM above is ePWM1. 1 - Configure ePWM2 to be synchronized to ePWM1. 2 - Adjust ePWM2 period to be N times smaller. 3 - Now trigger the ADC sampling to ePWM2. Assuming you sample on the bottom and top of ePWM2 triangle, you will sample twice per ePWM2 period and 2*N times per ePWM1 period.

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For all the reasons above, an oscilloscope is inadequate to this task. Here's a couple of alternatives: Method 1: find someone with a really good spectrum analyzer, and borrow time on it. Method 2: build a really good notch filter (for a one-off you'll want to use analog components, and good ones). Characterize it (I didn't say this would be easy). Then ...

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First, how is your sine wave generated? Based on the first plot, the period seems to be about ~ 1 kHz, Second, how did you identify where the peak is? You should use a Matlab command like [value, index] = max(abs(y)) and freqFundamental = fshift(index) instead of zooming in on the plot. Third, you have a 12-bit DAC and you sample the output with an ...

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Yes and no. In principle, you can use the peak of your correlation function. However, it is not at 10000. The correlation function is symmetric around 0, so your peak is actually much closer to zero than you think. This is one of the reasons why xcorr returns two paramters, one for the lags at which the function is calculated. The correct way of plotting the ...

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I have no access to your audio files so I've downloaded: IR from here (mono/r1_omni.wav) - it's a really long one Anechoic recording from here (operatic-voice/mono/singing.wav) Resampled voice signals: Final convolved signal: As for your questions: 1. As you did the plot of IR in logarithmic scale it's clearly visible that towards its end there is ...

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Assuming that we're really talking about an analytic complex-valued signal with no negative frequency components, then a sampling frequency $\omega_s>\omega_h-\omega_l$ will guarantee that the shifted spectra don't overlap, i.e., there will be no aliasing. However, it's not guaranteed that you'll have an image of the spectrum centered at DC. This is only ...

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As was noted in the comments the convolution of $n$ rectangular functions will converge to the Gaussian impulse response. A cardinal B-spline can be defined as follows: \begin{align*} B_0(t) &= \begin{cases} 1 &\text{if } t \in [0, 1) \\0 & \text{otherwise}\end{cases}\\ B_p(t) &= (B_{p-1} * B_{0})(t) \end{align*} for $p > 1$. $B_0(t)$ is a ...

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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 ...

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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 ...

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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})&...

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