Tag Info

10

For those who still cannot chalk the difference here is an simple example Take long transmission line with simple quasi-sinusoidal signal with an amplitude envelope, $a(t)$, at its input $$x(t) = a(t) \cdot \sin(\omega t)$$ If you measure this signal at the transmission line end, $y(t)$, it might come somewhere like this: \begin{align} y(t) &= a(... 8 The book's formula is right. LetH(w) = 1 - r e^{j(\theta - w)} = [1-r \cos(\theta - w)] + j [-r \sin(\theta - w)]$$Since the group delay \tau is the negative of the derivative of the phase of H(w), we first define the phase as:$$\phi(w) = \tan^{-1}\left( \frac{-r \sin(\theta - w)}{1-r \cos(\theta - w)} \right)$$Using the derivative rule for the ... 7 The group delay of a filter is defined as minus the change in the phase response with respect to frequency. If the phase response of a filter is \Phi(\omega), the corresponding group delay \tau_g is given by: $$\tau_g = -\frac{d\Phi(\omega)}{d\omega}$$ In Matlab code, the group delay of a 4th order Butterworth filter can be ... 6 This is a slightly tedious but nevertheless straightforward exercise in computing the derivative of a function:$$\begin{align}\tau(\omega)&=-\frac{d\phi(\omega)}{d\omega}=-\frac{d}{d\omega}\arctan(f(\omega))\tag{1}\end{align}$$with$$f(\omega)=\frac{r\sin(\omega-\theta)}{1-r\cos(\omega-\theta)}\tag{2}$$From (1) we have$$\tau(\omega)=-\frac{f'(\...

6

There are a couple of interesting aspects of "reconstruction to unity". First, there are two ways of combining two filters: parallel and in series. For a parallel topology it is ALWAYS possible to find a complimentary filter so that the pairs add to unity. It's easy enough, actually. Simply do $\tilde{H}(\omega) = 1-H(\omega)$. In the time domain that means ...

6

I know this is a pretty old question, but I've been looking for a derivation of the expressions for group delay and phase delay on the internet. Not many such derivations exist on the net so I thought I'd share what I found. Also, note that this answer is more of a mathematical description than an intuitive one. For intuitive descriptions, please refer to ...

6

Answer : No, any causal LTI system with frequency response $H(f)$ cannot produce the output $y(t)$ in advance. And, the answer lies in the causality of input signal $x(t)$ being applied to $h(t)$. Any causal input $x(t)$ which has an identifiable beginning cannot truly be Narrow-Band or Band-Limited. It will have non-zero frequency content at all frequencies....

5

I only need the total group delay, not spectrum of group delay. Group delay is a spectrum, so this doesn't make sense. The group delay is the derivative of the phase response of the filter, so in Python it can be calculated as from scipy import signal from numpy import pi, diff, unwrap, angle w, h = signal.freqs(b, a) group_delay = -diff(unwrap(angle(h))...

5

Note that a constant group delay is not sufficient for a band-limited signal to exhibit no dispersion. It is the phase delay that needs to be constant. If the phase is affine, i.e., if we have $$\phi(\omega)=a+b\omega,\qquad \omega>0\tag{1}$$ the group delay is constant $$\tau_g(\omega)=-\frac{d\phi(\omega)}{d\omega}=-b\tag{2}$$ but the phase delay is ...

5

As pointed out by Peter K., it is true that many well-known techniques for designing FIR filters actually only design linear phase filters. However, FIR filters are very well suited for delay equalization, simply because the design process is much simpler than for IIR filters. The reason for this is the fact that the design problem can be formulated in such ...

5

If you are looking for a frequency-independent delay applied to any given input signal by the filter (apart from amplifying and attenuating certain frequency components), then you won't be able to find it because there is no such delay. As you can see in your plots, group delay and phase delay are generally frequency dependent. Furthermore, for general input ...

4

For the Karplus-Strong algorithm, your primary concern is for phase delay, $$\tau_\phi(\omega) \ = \ - \frac{\phi(\omega)}{\omega}$$ rather than group delay. $$\tau_g(\omega) \ = \ - \frac{d \phi(\omega)}{d \omega}$$ where $$\phi(\omega) \ \triangleq \ \arg \left\{ H(e^{j \omega}) \right\}$$ and $H(z)$ is the transfer function of your filter, and $$... 4 This is actually pretty simple. Take a pure-tone sine wave, say$$s(t) = \cos\left(\omega t + \phi \right).$$Now, fix a time t_0 and find the phase \psi(t_0) of s(t_0). We get$$\psi(t_0) = \omega t_0 + \phi$$Now let's vary the frequency \omega and keep the time fixed. The above expression becomes a linear function in \omega! Even though we ... 4 The problem is that the stft function is splitting the signal up into different windows. That means that the signal from time n to n+N_{w}-1 is multiplied by$$ n, n+1, n+2, \ldots, n+N_w-1$$instead of$$ 0, 1, 2, \ldots, N_w-1 $$which is causing the scaling problem. If I apply the group delay calculation from this derivation, I get: where the top ... 4 It's often easier to design FIR filters for compensating group delay. At the same time they could also compensate the magnitude if necessary. The easiest method is to use a complex least squares method, which boils down to solving a system of linear equations for the filter coefficients. The difficult part is to choose an appropriate bulk delay in the ... 4 A real-valued system that doesn't distort the shape of the input signal must have the following input-output relation:$$y(t)=Ax(t-t_0)\tag{1}$$with arbitrary real-valued constants A>0 and t_0. In the frequency domain, Eq. (1) corresponds to$$Y(\omega)=Ae^{-j\omega t_0}X(\omega)\tag{2}$$Consequently, the corresponding system is an LTI system ... 3 Using the logarithmic derivative of the transfer function, as detailed in Julius O. Smith's Numerical Computation of Group Delay, the following computations seem to involve a little less of derivatives (and less risks of mistakes), which could be useful for more complicated frequency responses and related group delays (like rational fractions). And you can (... 3 I agree with Maximilian MatthÃ©'s answer, but I'd like to show you another route to the solution, which might be a bit more straightforward, and which avoids the explicit application of the Hilbert transform. First of all, note that the inverse Fourier transform of the real part of the frequency response corresponds to the even part of the impulse response: ... 3 You have two kinds information in the question: The system is causal The real part of the Frequency response is given. Now, (repeating the steps from Wikipedia Entry on Causal Filter) h(t) is causal, i.e. h(t)=0, t<0. Let g(t)=\frac{1}{2} (h(t)+h^*(-t)) which is non-causal, but has hermitian symmetry, hence its Fourier Transform is real. We ... 3 Any kind of digital filter will cause the the output signal to be delayed by some amount of samples. From what I gather, you are trying to run a signal through a high pass filter (is it an FIR or IIR?) and correct the group delay by "filtering the first time, inverting the response in time...". I personally have never been taught or have read of such an ... 3 The phase delay of any filter is the amount of time delay each frequency component suffers in going through the filters (If a signal consists of several frequencies.) The group delay is the average time delay of the composite signal suffered at each component of frequency. 3 Your integrator is a perfect accumulator, which is an unstable (marginally stable) system with a pole at z=1:$$H(z)=\frac{1}{1-z^{-1}}\tag{1}$$This means that the value of the frequency response at \omega=0 (corresponding to the pole at z=1) is undefined. Consequently, the value of the group delay at \omega=0 is also undefined. Note that you can ... 3 Consider an D-tap FIR filter with liner phase, the group delay (measured in samples) is$$g=\frac{D-1}{2}\tag{1}$$and therefore, if it is measured in seconds it will be$$g=T_s\frac{D-1}{2}\tag{2}$$where T_s=1/F_s. The CIC filter which is also denoted as recursive running sum filter is indeed a special implementation of a moving-average filter. The ... 3 As I mentioned in my comment, you're right that the unit of group delay of a discrete-time system is samples. And \Omega is indeed the normalized frequency in radians. The problem is that your final formula for the group delay of a first-order allpass filter is wrong (I didn't check the original formula). If you have an allpass system$$H(z)=\frac{a+z^{-...

3

Lets put forward the intuition behind the concept of the group delay before further discussing how to find the delay of FIR filters. Consider an input signal $x[n]$ of length $L_x$ which is nonzero between $n=0$ and $n=L_x-1$. And let a simplistic filter impulse response to be $h[n] = \delta[n-d]$. The output is immediately shown to be $y[n] = x[n-d]$ which ...

3

Here is my simplest explanation: The group delay, as the negative derivative of phase, predicts the time delay of the amplitude envelope of a pulse, as shown in the hand-drawn graphic below. The upper part of the sketch shows a sinusoidal waveform varied in amplitude by its envelope. The lower one is showing this same envelope before and after a system ...

3

Here is a actual example with negative group delay that will provide further insight: Below is a plot of the output and input of a pulse through a realizable filter that has negative group delay: It seems like a complete violation of causality, but it is just a clever DSP magic trick. Let's explore further: The filter above that did this had the following ...

3

If there are phase discontinuities after all $2\pi$-jumps have been removed, then these discontinuities are usually caused by zeros of the frequency response. The phase jumps by $\pi$ at these frequencies, and the group delay doesn't exist, or, if one prefers, is non-finite. Note that the group delay is meaningless anyway at frequencies where the frequency ...

2

Most FIR filters are linear phase as their coefficients are (anti-)symmetric. So, most FIR filter design techniques are targeted at linear phase designs. That means FIR filters are not much good at equalizing group delay -- linear phase FIR filters all have constant group delay. IIR filters, on the other hand, generally have non-linear phase. That means ...

2

"Group delay" isn't the delay between the change on the input and the first effect; it's the delay that a packet of oscillations of different frequencies experience. In the case of a linear phase filter (and your moving average, its impulse response being symmetric, is linear phase) that is the "center" of the effects of a single impulse. So, it's the ...

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