# Tag Info

11

Your professor is right, and you're almost right too. The filter is clearly an FIR filter, but because its frequency response can be expressed as a geometric series, a recursive implementation is possible. If you write the transfer function as a rational function you get $$H(z)=2\frac{1-z^{-12}}{1+z^{-2}}\tag{1}$$ which is almost the same as you got, apart ...

5

You can equalize magnitude and phase simultaneously by defining a desired complex frequency response $$D(\omega)=M(\omega)e^{j\phi(\omega)}\tag{1}$$ with magnitude $M(\omega)$ and phase $\phi(\omega)$ chosen such that they compensate for the given magnitude and phase distortions. An FIR filter approximating $(1)$ can be designed by using the following error ...

4

This is unfortunately a difficult problem. A microphone doesn't have a single frequency response. It has a different one for each direction of incidence. Especially for second order microphones (dipoles, cardioids, etc.), the frequency response also depends heavily on the distance and radiation impedance of the source. So first you need to decide: what ...

4

Or another way, visually, with the plot of the magnitude response of your filter shown below you see that it goes to zero at $\pi/2$. So what you could do is use the relationship between the angular frequency $\omega$ in [radians/sample] and the relative frequency $f$ in [cycles/sample] $$\omega = 2\pi f = 2\pi \left(\frac FF_s\right)$$ Then compute $F_s$ ...

4

It appears that your understanding of harmonics is not entirely correct yet. Your example of two sinusoids with frequencies $f_1$ and $f_2$ can actually be explained in terms of harmonics IF $$f_1=n_1f_0\quad\text{and}\quad f_2=n_2f_0,\qquad n_1,n_2\in\mathbb{Z^+}\tag{1}$$ is satisfied for some positive $f_0$. Note that this means that the ratio $f_1/f_2$ is ...

4

so I am wondering where can I go to form an understanding of DSP from the ground up. There is a great list of resources in this answer What Resources Are Recommended for an Introduction to Signal Processing (DSP)? You should spend half an hour with each one and then decide which fits your own learning style the best. My skills in calculus are also quite ...

4

Your result is correct, even if it seems a bit counterintuitive. Note that you don't compute the Fourier transform but the discrete Fourier transform (DFT). The system you want to invert has the impulse response $$g[n]=\frac12\delta[n-19]\tag{1}$$ Obviously, the inverse system must advance the incoming signal by $19$ samples: $$g_{inv}[n]=2\delta[n+19]\tag{2}... 3 The general approach would be to find the zeros of the filter's transfer function. If you do things right it will turn out that there are two complex conjugate zeros on the unit circle at \pm j, i.e., at half the Nyquist frequency. Now you need to choose the sampling frequency such that 80 Hertz corresponds to half the Nyquist frequency. 3 The \omega in the frequency response of a discrete-time system H(e^{j\omega}) is indeed unitless. The frequency response H(e^{j\omega}) is periodic with period 2\pi. If the discrete signal is obtained via sampling with sampling rate f_s, then the relation between \omega, the actual frequency f, and the sampling frequency f_s is$$\omega=2\pi\...

3

The DC gain is simply the sum of filter taps or coefficients. This is the value of the frequency response at DC (i.e. $0\ \rm Hz$), or equivalently $$H(0) = \sum_{n = 0}^{N - 1}h[n]\tag{1}$$ Because, for a digital FIR filter of length $N$ with impulse response given by Equation $(2)$ $$\big\{h[n]\big\}, \quad\text{with}\quad 0\le n\le N -1\tag{2}$$ the ...

3

Proving that the H1 estimator ignores noise on the output: If we assume that the output is the input processed by the transfer function plus some output noise, we have $Y = H(f)X(f) + N$, where everything is in the freq domain. So, $S_{xy}(f) = E[X(f)^*Y(f)] = E[X(f)^*(H(f)X(f) + N(f))]$. where E[] denotes expectation and * denotes complex conjugate. If you ...

3

At first, I thought harmonics come from the signal being periodic. That's correct. However, we know a sine wave is also periodic but contains a single frequency and no harmonics. This is the only periodic signal that doesn't have harmonics. Or to be precise the amplitudes of all harmonics are zero. No, I'm wondering what's the real root-cause for ...

3

My skills in calculus are also quite bad. This is going to be a barrier. It may be the whole barrier. Basically, DSP is a branch of analysis -- it's a collection of mathematical techniques that you can use to easily solve a certain class of differential and difference equations. It may not look like that, but that's because for the differential equations ...

3

The resonant frequency is related to the "significant frequency" (which is the shelf midpoint frequency) by a factor of $\frac{1}{\sqrt{A}}$ for the lowShelf and the reciprocal of that for the highShelf.

3

For an example analysis, I’ve picked up the low-shelf filter in Robert Bristow-Johnson’s Audio EQ Cookbook. In the book, the transfer function is given as; $$H(s) = A\frac{s^2 + \frac{\sqrt{A}}{Q}s + A}{As^2 + \frac{\sqrt{A}}{Q}s + 1}$$ Since the analysis is going to be done by hand, the asymptotic approximation method of Bode plot analysis can be followed. ...

3

Observe the magnitude of frequency response (rescaled to 0 to 1): where \begin{align} H(\omega) &= e^{-j 0\omega} - e^{-j 1\omega} + e^{-j 2\omega} \\ &= 1 - e^{-j 1\omega} + e^{-j 2\omega} \tag{1} \end{align} following the time-shift property: $$x(t - t_0) \Leftrightarrow e^{-j t_0 \omega} X(\omega) \tag{2}$$ Though it isn't particularly ...

2

In general if you have a causal odd length $N$ FIR filter $h[n]$, defined for indices $n\in[0,N-1]$, moving the center bin to index $n=0$ means shifting the impulse response to the left by $(N-1)/2$ samples: $$\tilde{h}[n]=h\left[n+\frac{N-1}{2}\right]\tag{1}$$ In the frequency domain this is equivalent to \tilde{H}(e^{j\omega})=H(e^{j\omega})e^{j\omega (N-...

2

Correct. The magnitude response would be the same and the phase response would be summed with $\omega$ (omega).

2

This statement can only be understood given its context. Right before the sentence you quoted, we have this statement: It is no surprise that the condition for existence of the frequency response is the same as the condition for dominance of the steady-state solution. Indeed, a complex exponential that exists for all $n$ can be thought of as one that is ...

2

Is there an analogous procedure I can do to correct arbitrary phase responses in my signals? Arbitrary all-pass design is tricky since there are some extra constraints to be taken into account. FIR filters cannot be "ideal" all pass filters, since all pass filters have poles and zeroes which inverse of each others. FIR filters have all the poles ...

2

i want to know if taking the cross correlation of these signals would give me the impulse response? No. Let's look at it in the frequency domain. Let's call the input $x$, the output $y$ and the impulse response $h$. Uppercase letters are spectra, lower case time domain signals. For the transfer function we simply have $Y = H\cdot X, H = Y/X$. The spectra ...

2

Let's define the true transfer function $H_0=P_{xy}/P_{xx}=P_{yy}/P_{yx}$. $H_1$: The transfer function is computed as the ratio of the cross spectrum between the input and output signals, to the input autospectrum: $P_{xy}/P_{xx}$. The $H_1$ estimator assumes that there is no noise on the input and consequently that all the input measurements are accurate. ...

2

The significance is a constant time delay for all frequencies. Time delay is the derivative of phase with respect to frequency, so given a linear phase as shown, the time delay is constant. Notice that the abrupt steps in phase actually only occur when the magnitude goes through zero so does not effect the nature of it being a "linear phase" filter....

2

The gain is completely arbitrary and you can scale it as desired for the overall receiver or transmitter design. Where special attention must be paid is with fixed point design where the best practice is to let the filters grow the signal- do not scale the coefficients or the input as that only introduces more quantization noise and degrades SNR. Let the ...

2

The basic idea is that a periodic sound can have a missing fundamental. e.g. If you mix the right multiples of 110 Hz, a human will hear 110 Hz, whether or not there is any non-zero 110 Hz sinusoidal component in the mix (via FT or DFT). So all those multiples are still called the same thing: harmonics. Where do they come from: lots of physical objects/...

2

The question is about measurements in real world, not about simulations, am I right? If this is the case, there is an established technique and instrumentation for doing such measurements. The generic name of an instrument commonly used in laboratories and field testing for this purpose is Vector Network Analyzer. The article describes the operation ...

2

Typically you design a lowpass filter to certain specifications: pass band ripple, stop band attenuation, transition band width, phase distortion, latency, transient preservation, causality, computational complexity, etc. When you compare two filter designs, you can evaluate them individually against all these criteria and compare the results. Often, one ...

2

Looks like clock drift. That happens when you have two independent clocks in the system. Even if they are nominally the same (40 kHz). There will be slightly different and there will be relative drift between them. Typical audio clocks tend to be within 10ppm of their nominal target. So a drift in the order of 1 sample/second is perfectly normal and expected....

1

The group delay is the negative derivative of the phase response as the OP has stated, and specifically for the delay of one clock sample, the phase will go linearly negative to $2\pi$ radians as the frequency goes from 0 to $f_s$ From the picture we see the phase is going approximately 800 degrees at a frequency of $.17\pi$ rad/sample (where $2\pi$ rad/...

1

It looks like artifacts due to the derivative. I used this code in Octave: fs=300; t=[0:1/fs:2]; c=chirp(t, 20, 2, 100); s=c.*(1+0.5*cos(2*pi*10*t)); h=hilbert(s); m=abs(h); a=diff(unwrap(arg(h)))/2/pi*fs; plot(m,"",a) and this is what comes out: I also tested the equivalent of this in LTspice and, without any form of unwrapping, this is the ...

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