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4

It's your observation interval which creates the main problem. Your reasoning based on the Nyquist sampling theorem is ok; of course with a pure sine wave at the exact Nyquist frequency you will have troubles and therefore it's wise to relax the sampling frequency (slightly) above that of Nyquist rate, such 2.2 Hz instead of a strict 2 Hz... So this is one ...

3

As others said in the comments, this looks like numerical error. 3rd-order filters are not typically prone to this, but the higher your sampling frequency, the closer the poles move to +1: You might benefit from splitting this into second-order sections. The easiest way is to use output='sos' and sosfreqz() and sosfilt(), which handle the splitting ...

3

To rephrase your question: Is there a process that halves the number of samples of a full-bandwidth signal without significantly affecting the appearance of the spectrogram, except for scaling of time or frequency axes? Resampling does not do that. The only such process I know of is time stretching / pitch scaling followed by decimation. Actually, one ...

3

Since you stated you are still confused, the following graphics may help further understand the other two good answers: First be sure that you understand what the sampling process (which is multiplying an analog signal by a stream of impulses in the time domain) looks like in both time and frequency; a stream of impulses in the time domain is a stream of ...

3

Assume the original sampling rate is $F_s$ and that the filters are perfect brickwall Halfband filters. After the low pass filtering the frequency content is from $0 \rightarrow \frac{F_s}{4}$. Because of the new bandwidth, the sampling rate can be reduced to $F_s/2$. For the high-pass filter, the frequency content lies in the range $\frac{F_s}{4} \... 3 The sampling theorem requires a perfectly bandlimited signal, bandlimited to below twice the sampling frequency. The problem with this is that only an infinite length signal (e.g. exists before the big bang) can be perfectly bandlimited. This is from the Fourier theorem regarding any domain with finite support. Thus all real-world signals are ... 3 Is the rate of 2B exclusive? Yes. The sampling theorem states that the signal must be band limited to half the sample rate. That implies that the energy at the Nyquist frequency must be zero. In practice you need a healthy margin between the highest usable frequency and the Nyquist frequency. There is always some "transition band" that you need to get the ... 3 I think you're confusing two different (but related) terms. Nyquist says that in a channel of bandwidth$B$you can transmit up to$2B$orthogonal pulses per second. So,$R_p \leq 2B$, where$R_p$is the pulse rate. To achieve$R_p = 2B$, the pulses need to be sinc-shaped. Other, more practical pulses achieve slightly less than that. For example, raised ... 3 It really boils down to aliasing. In continuous-time, if you have any two signals$x_1(t) = \sin(2 \pi F_1 t)$and$x_2(t) = \sin(2 \pi F_2 t)$, then as long as$F_1$and$F_2$are distinct, the signals are, too. But consider sampling at some time interval$T_s$, so that the sampled signals are$x_1(k) = \sin(2 \pi F_1 T_s k)$and$x_2(k) = \sin(2 \pi F_2 ...

2

The Fourier transform of any strictly real time domain signal will be conjugate symmetric (in the frequency domain). That means that a strictly real pure sinusoid with a frequency of just below Fs/2 will also have a conjugate symmetric component just above -Fs/2 in the frequency domain, thus spanning nearly the full range Fs. If you try to FT a real signal ...

2

I don't think the exam question is stated properly, unless there is additional context that the student is supposed to know. First, the bandwidth depends on the encoding. The question seems to assume binary encoding, but PCM does not specify this. Second, the bandwidth depends on the pulse shape. Usually one assumes rectangular unless otherwise specified, ...

2

The sampling is indeed analogous to mixing as to my understanding. In the sampling process, we multiply the time domain signal with an impulse train - the impulses in time are represented as impulses in frequency at integer multiples of the sampling rate. So instead of one or two (for a real sine wave) impulses in frequency, we have an infinite number but ...

2

With an IIR filter, the higher the ratio is of sampling frequency to bandpass filter frequency, the closer the poles are to the complex unit circle in the Z-transform. Poles close to the unit circle can produce numerical instabilities (ill conditioned computations) when computing the filters response. This is because any numerical quantization or "rounding ...

2

The "alias" frequencies you see appearing above the Nyquist limit $f_s/2$ do not exist in a proper interpretation of the digitised signal; they are appearing due to incorrect interpolation of the samples during reconstruction. If you correctly interpolate the signal from the samples, those frequences will not be there. Remember, a key condition of the ...

2

Let's take specific numbers instead of symbols. Suppose that the signal $x(t)$ being sampled is a pure sinusoid (extending from $-\infty$ to $\infty$) at $1200$ Hz, say $\cos(2\pi\cdot 1200 t)$, $-\infty < t < \infty$. (For the record, the Fourier transform of this signal is a pair of impulses at $\pm 1200$ Hz but we won't be needing to think about ...

2

If it is a real signal, as you are doing, then the negative spectrum will be the complex conjugate of the positive spectrum. Since there are N samples in frequency given N samples in time, and the frequency axis extends from 0 to N-1 where 0 represents DC and N represents the sampling rate (with the upper half of this from N/2 to N-1 equally representing the ...

2

the lomb scargle in matlab, (plomb) returns a frequency vector as the second output. i would be surprised if the routine you are using doesn’t do the same. there is a common tendency for python signal processing libraries to be functionally equivalent to matlab conventions. your plots show symmetry around a “center” frequency which is the same as what a ...

2

I'm top-editing this since it answers the question directly. The sinc series is fundamentally a $C/x$, so you can extract as many absolutely convergent series out of it as you want, but what is left over is still only conditionally convergent. Also, you can rescale $x$ and it is still a $C/x$ series. Saying you have a summation to or from infinity is an ...

2

A quick answer, but as human hearing does not go past 20kHz bandwidth, 44.1 kHz is enough for storing and transmitting audio. The problem is that the analog antialiasing filter before ADC must be extremely sharp to pass 20 kHz enough and block 22.05 kHz enough and this is just needs many components with good performance and tolerance. When sampling at higher ...

2

Considering basic audio applications, the digital to analog conversion reconstruction filter (aka interpolation filter) is a low pass analog filter that removes all the image spectrum at the output before it goes to loudspeakers and retains only the baseband spectrum that resides in the filter's passband: inside its cutoff frequency of the lowpass filter. ...

1

Sampling in the time domain is equivalent to periodic repetition in the frequency domain with $1/T_s$ as the period. So formally you'd have something like $$R_{sampled}(f) = \sum_{n=-\infty}^{\infty}R_z(f-2\cdot n)$$ If you evaluate this, you actually get $R(f) = 2$. It's easier to see if you draw it: take the initial spectrum, shift it by -6,-4,-2,0,2,4,6.....

1

How comfortable are you with the concept of negative frequencies? Your 5 kHz cosine consists of two lines in the spectrum, one at +5kHz one at -5kHz. If you made your first diagram such that it goes from $-fs/2$ to $+fs/2$, you'd see exactly these two. I find this way of displaying the spectrum much more intuitive. Outside this band, you will see periodic ...

1

If your 1 second clip of sound is repeated continuously, then you have a periodic waveform having 1 Hz as it's fundamental frequency. But to capture the sound as we would hear it, it must be sampled a rate that exceeds twice of the highest frequency we can expect to hear. This is why you see 44100 Hz and 48000 Hz for music and maybe 8000 Hz for just voice. ...

1

It might be better to think of Nyquist sampling as shooting fish in a bucket. The bucket limits what is in the bucket by being only so large. This is like a band limit. We can only have so much variation. Two samples is the very least number of samples we need to specify that 1 sec of 1 Hz sine wave, although intuition tells us we are on the edge. Three ...

1

We don't ignore poles on the contour. As mentioned in a comment, poles are avoided by modifying the contour as shown in the figure below, where a contour appropriate for a pole at $s=0$ is shown. Fig. 1: Nyquist contour for a pole at $s=0$ (from "Modern Control Engineering" by K. Ogata). The contour moves around the pole along a semi-circle centered at the ...

1

This is a phenomenon called aliasing https://en.wikipedia.org/wiki/Aliasing Usually, there are anti-aliasing filters that will filter out signals with frequencies higher than fs/2. However, in your case I think there isno anti-aliasing filter. Hence your 14 kHz-signal aliases to 2 kHz. Since f > fs/2, then apparent frequency = 16 kHz - 14 kHz = 2 kHz

1

Since nobody answered my I will answer using what I read on the internet, "Nyquist frequency" is the max frequency we can get using given sampling rate. the $f_s$ of convolution between 2 function is indeed the min of their max frequency, but I'm not sure yet about the part which stated that this is an upper bound because frequency may cancel each other. ...

1

The equation in your question is correct. If we assume that $x(t)$ is an ideally band-limited signal with bandwidth $B$, then, according to the sampling theorem, we have $$x(t)=\sum_{n=-\infty}^{\infty}x(nT)\frac{\sin\left[\pi (t-nT)/T\right]}{\pi (t-nT)/T}\tag{1}$$ where the sampling rate $1/T$ satisfies $1/T>2B$. If we assume that the integral over $... 1 By the Convolution Theorem multiplication in Time / Spatial domain is equivalent of Convolution in the Frequency Domain. The sampling rate (In its classic interpretation) is proportional to the support of a function in the frequency domain. So if a function has a certain support in frequency, what would be its support after convolution with itself? Indeed ... 1 So apply the following principals to get to a solution: Shifts in the time domain don't affect the bandwidth, so the$-4$and the$+3$can be ignored. Reversal of the time domain axis doesn't affect the bandwidth, so you can ignore the negative sign on the$-2t$. The$\mathrm{sinc}()\$ function has a Fourier transform that is a rectangle function of a ...

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