# Tag Info

## Hot answers tagged bandpass

9

"Zero-Mean" is the word that's commonly used to describe signals and signals with a zero average. "This is a zero-mean filter." If you really mean a filter that is specifically meant to cancel the DC component, a "DC blocker" is a name for that.

4

You're right that the multiplication of a low pass and a high pass filter results in a band pass filter, as long as the cut-off frequency of the low pass is higher than the cut-off frequency of the high pass. The problem with that approach is that low pass and high pass filters with magnitude responses that are optimal according to some chosen criterion (...

3

In complement to Marcus, I have read the term "zero-sum": "zero-sum window", "zero-sum filter", "zero-sum kernel", the latter being more frequent. It is similar to "unit-sum windows", ie windows whose amplitudes sum to one. "Zero-average" can be found in image processing: Further note that applying ...

3

The order is classically the maximum delay (in samples) that is used to produce each output (Filter order by JOS). For FIR filters, the length is "one plus order": $$y[n] = x[n]-x[n-1]+x[n-2]$$ would be 3-tap and order 2. If the filter is IIR, it would have "more coefficients" than "one plus order". I am not sure one can say "taps" for IIR filter. This ...

3

Your basic problem is that filtfilt (and most other linear filtering routines) take filters that are designed for infinitely long time expanses, and apply them to a chunk of data as if the data were extended infinitely in both directions with zeros. So you have a legitimate bandpass filter, and it's "seeing" a legitimate jump in the signal at the ...

2

Will we be able to recover the complex modulated symbol based on $s_2(t)$ too? Yes, no information is lost between the two. How can we prove that mathematically? One way to think about it is to write out each individually and see what you have. Let $a(t)=a_I(t)+ja_Q(t)$, expanding $s_1(t)$ and $s_2(t)$ we have: \begin{align} s_1(t) &= a_I(t)\text{cos}(...

2

To add, the reason it is called the order specifically is because it is the order of the polynomial as given by the transfer function of the filter- which for an FIR filter is just the coefficients with increasing powers of $z^{-1}$: $$H(z) = \sum_{n=0}^{N-1}h_n z^{-n}$$ For example a three "tap" FIR filter with coefficients [0.5 1.5 0.5] is 2nd order with ...

2

Note that for the complex noise envelope $z(t)=x(t)+jy(t)$, the autocorrelation $R_z(\tau)$ is defined by (cf. Eq. $(4.1.47)$ in Proakis) $$R_z(\tau)=\frac12E\big\{z^*(t)z(t+\tau))\big\}=\frac12\big[R_x(\tau)+R_y(\tau)\big]+j\frac12\big[R_{xy}(\tau)-R_{yx}(\tau)\big]\tag{1}$$ As shown in the chapter you refer to, for the real-valued bandpass noise $n(t)$ to ...

2

The Japanese link actually implies how to derive Eq.5, and ironically they also refer to an existing dsp.se answer at the bottom. Derivation of Eq.5 is as follows: Consider a moving average filter of length $N$, with the impulse response $h[n]$: $$h[n] = \begin{cases} ~~~1/N~~~,~~~n=0,1,...,N-1 \\ ~~~0~~~,~~~ \text{otherwise} \end{cases} \tag{1}$$ Magnitude ...

2

The answer to this question could be tricky. For a simple answer you may take the bandwidth of a bandpass signal as $\omega_2 - \omega_1$ (highest minus the lowest frequencies), and don't care about what happens in between the two. Hence your bandwidth allocation is not fully efficient, and a sampling based on this bandwidth will include redundant ...

2

The filter order equals the number of poles, that's a fact. How people call and use parameters in software is a totally different question. Many routines that design standard IIR filters use frequency transformations to design bandpass or bandstop filters from a prototype lowpass filter. Often, the specified filter order is the order of the prototype lowpass ...

1

The formulas for cutoff frequencies of a bandpass filter are different for different filter designs and can be calculated from the transfer function $H(ω)$. For a "basic" series RLC filter $$H(ω) = {\frac {ωRC} {\sqrt{(1-ω^2LC)^2 + (ωRC)^2}} }$$ and this filter can be made narrowband if implemented with a high quality factor (see later in this ...

1

In audio applications, this would be a low-cut filter. The term is often used synonymous with high pass, though that would not accurately describe general zero-mean filters. A unit-mean filter meanwhile is indeed a lowpass filter.

1

From one of your comments it appears that you've actually already answered the question. Here are a few remarks and questions that should help you gain some more understanding: If you have a transfer function $H(z)$ (and if you assume that the corresponding system is stable), then the frequency response is obtained by choosing $z=e^{j\omega}$. For $H(z)=z^{-... 1 Multiplication in the frequency domain is the equivalent of convolution (digital filtering) in the time domain - one of the fundamental properties of the Fourier transform. Using an FFT in practice to filter requires stitching together overlapping chunks and a large delay if in real-time. However in the frequency domain you can immplement brick-wall filters ... 1 The reason the bandpass filter eliminated the problem is because it removed the DC component. This can also be done by simply subtracting the average before taking the DFT. However there is much more to understand in this question that should be of interest to the OP, and after reading and understanding the detail given below, this first statement should ... 1 Clearly it doesn't make a difference if you use the real part or the imaginary part (as long as you know what you're doing), because $$\textrm{Im}\left\{a(t)e^{j2\pi f_ct}\right\}=\textrm{Re}\left\{b(t)e^{j2\pi f_ct}\right\}\tag{1}$$ with $$b(t)=-ja(t)=a_Q(t)-ja_I(t)\tag{2}$$ So you basically just exchange the in-phase and quadrature components (apart from a ... 1 okay, the magnitude frequency response for a general$N$th-order Butterworth lowpass filter is: $$\Big|H_\mathrm{LP}(f)\Big|^2 = \frac{1}{1 + \left(\frac{f^2}{W^2} \right)^N}$$ for bandpass filtering, centered at frequency$f_0$, this substitution is made to the LPF: $$f \leftarrow \frac{f}{f_0} - \frac{f_0}{f}$$ so then it comes out as:$\$ \Big|H_\...

1

sox is doing exactly what you tell it to do – it plots a spectrogram of the bandwidth you can represent with your sampling rate. The fact that most of your picture shows no energy just shows your filter is working, and that's exactly what one would want to see! However, as you can see, your signal is solidly oversampled: You can downsample it to a much ...

1

Typically if you're designing to an amplitude response, you set the phase at zero. In processing time-domain signals, this is to keep the group delay of the filter constant. In processing images, this will keep the filter effects centered, rather than displacing different spectral components of the image with respect to one another. At least in time-domain ...

1

I recently developed a function to easily perform octave and fractional octave filtering, it is available on github: PyOctaveBand It uses the SOS coefficients and performs downsampling to filter correctly at low frequencies.

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