6

This has absolutely nothing to do with causality. The frequency response of a real-valued filter (i.e., one with a real-valued impulse response) is (conjugate) symmetric, i.e., the negative frequencies are redundant. That's why it is sufficient to show the frequency response at non-negative frequencies only. You can easily see that symmetry as follows. The ...


4

Let's first consider periodic signals. Under rather mild conditions, a periodic signal with period $T$ can be represented by its Fourier series: $$f(t)=\sum_{k=-\infty}^{\infty}c_ke^{j2\pi kt/T}\tag{1}$$ Eq. $(1)$ shows that the periodic function $f(t)$ is represented by a sum of weighted complex exponentials with frequencies $\omega_k=2\pi k/T$, which are ...


4

Your first suggestion is correct, the derivatives of the local polynomials are being sampled. From the horse's mouth (Savitzky & Golay 1964): Again, if we restrict ourselves to evaluating the function only at the center point of a set of equally spaced observations, then there esist sets of convoluting integers for the first derivative as well. (...


4

Wouldn't it be much easier to just compare the left-hand side and the right-hand side of $$e^{-j2\pi ft_0}=|H(f)|e^{j\phi(f)}\tag{1}$$ to see that $|H(f)|=1$ and $\phi(f)=-2\pi ft_0$?


3

The sequences $e^{j\omega_0n}$ are eigensequences of discrete-time LTI systems, i.e., the response to such a sequence is the same sequence scaled by a complex constant (the eigenvalue). This can be shown as follows: $$\begin{align}y[n]&=(h\star x)[n]\\&=\sum_{k=-\infty}^{\infty}h[k]x[n-k]\\&=\sum_{k=-\infty}^{\infty}h[k]e^{j\omega_0(n-k)}\\&=...


3

The system is not just a wire, it is in fact time-varying, so it has no frequency response in the conventional sense. The sequence $w[n]$ is of course just a low-pass filtered version of the input signal. The filtered signal is band-limited to half the Nyquist frequency. Now just write the output $y[n]$ in the time domain as the sum of $w[n]$ and the ...


3

No problem at all. Inverse Fourier Transform is totally blind to phase wrapping. You can add any multiple integers of two pi to any phase and you will still get the exact same impulse response


3

I understand that "to have frequencies" can be misleading. This is a shorthand for "having non-zero energy or amplitude at two (specific) frequencies". Rephrased, for the cosine: when a cosine signal is represented in the frequency domain, it has two non-zero components at frequencies $\pm \omega$, and the other values anywhere on the frequency axis have ...


3

HINT: Write $H(e^{j\omega})=e^{-j\omega}G(e^{j\omega})$ and use $(e^{j\omega}+e^{-j\omega})=2\cos(\omega)$. $G(e^{j\omega})$ should turn out to be real-valued.


3

The frequency response can exist, even if there are poles in the right half-plane, as long as there are no poles on the $j\omega$-axis. You cannot use the unilateral Laplace transform, instead you must use the bilateral Laplace transform (with the lower integration limit at $-\infty$), or, simply, the Fourier transform. The corresponding time domain ...


2

To answer your question specifically, no you do not need two poles. A single pole anywhere on the frequency axis will cause a real or complex frequency input to go to infinity (peak). For a digital system, the frequency axis is the unit circle. As Fat32 very well summarized, if your impulse response (which the poles and zeros are the z-transform or Laplace ...


2

If your system has a real impulse response, then pole & zeros will be in complex-conjugate pairs; so a pole at $\omega = \pi/3$ implies another pole at $\omega = -\pi/3$ . Further, by the definition of poles, as the frequency $\omega$ get closer to the pole frequency, the frequency response magnitude increases (so as to say peaks!). So for every peak ...


2

Polynomials are not integrable, hence their Fourier interpretation is complicated. However, locally, there are a few works. Under least-squares, polynomial interpolation's linearity works well with frequency analysis. The class of Savitsky-Golay filters performs least-squares polynomial interpolation, and yields different derivative orders. As they seemed ...


2

TL/DR: For a 2nd order transfer function of a real digital filter given as: $$H(z) = \frac{1}{1-2A_1z^{-1}-A_2z^{-2}}$$ The filter coefficient $A_1$ and $A_2$ are determined from the gain and resonant frequency to be: $$A_1 = Kcos(2\pi f_r/f_s) = Kcos(\omega_r)$$ $$A_2 = -(K^2)$$ Where: $f_r$: Resonant frequency in Hz $f_s$: Sampling frequency in Hz ...


2

Define a frequency response $$\tilde{H}(e^{j\omega})=\frac{e^{-j\omega}-|\alpha|}{1-|\alpha|e^{-j\omega}}\tag{1}$$ and note that the group delay $\tilde{\tau}(\omega)$ corresponding to $(1)$ is related to the group delay $\tau(\omega)$ of the original frequency response $H(e^{j\omega})$ by $$\tau(\omega)=\tilde{\tau}(\omega-\theta)\tag{2}$$ because $(1)$ ...


2

This is easier to see graphically, on an s-plane, knowing that the phase is from the frequency response of the system and the frequency response is determined when we restrict s to be the $j\omega$ axis. Consider the simple ratio of a single pole and a single zero expressed as $$H(s) = \frac{s-z_1}{s-p_1}$$ Given s as any point on the s-plane, and $z_1$ ...


2

ok let me give you a practical (discrete-time) example. Consider a system with input-output relation of the form $$ y[n] = x[n] + 0.5 x[n-1] + 0.25 x[n-2] $$ This input-output relation satisfies linearity and time-invariance properties and hence defines an LTI (linear time invariant) system and therefore it has an impulse response $h[n]$ found to be : $$...


2

If $T$ is a positive real number such that $$x(t) = x(t+T) ~\text{for all}~t \in \mathbb R\tag{1}$$ for continuous-time signals, or if $T$ is a positive integer and $$x[n] = x[n+T]~\text{for all}~n \in \mathbb Z \tag{2}$$ for discrete-time signals, then $x$ is said to be periodic with period $T$ or to have period $T$. A signal $x$ that has period $T$ also ...


2

You cannot solve this problem using the Laplace transform. The reason is that the Laplace transform of the input signal doesn't exist. You could use the Fourier transform, but in this case there's an even simpler way to determine the output signal. You need to know one important property of linear time-invariant (LTI) systems: their response to a sinusoidal ...


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

Your understanding is correct. There is no difference between a phase of $\pi$ and a phase of $-\pi$. You can always add or subtract integer multiples of $2\pi$ to the phase without changing anything because $e^{j2\pi n}=1$ for $n\in\mathbb{Z}$. Clearly you have $e^{j\pi}=e^{-j\pi}=-1$. The probable reason why they used opposite signs in the phase plot for ...


1

Your suspicion is correct. You can show this as follows: $$X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\tag{1}$$ $$X^*(\omega)=\int_{-\infty}^{\infty}x^*(t)e^{j\omega t}dt\tag{2}$$ Now if $x(t)$ is purely imaginary we have $x^*(t)=-x(t)$, and, consequently, $$X^*(\omega)=-\int_{-\infty}^{\infty}x(t)e^{j\omega t}dt=-X(-\omega)\tag{3}$$


1

This is the classical problem of filter design: You've got a frequency response that you want, how to implement it using a FIR? I'm not going to lay out all the theory here, because it can easily be found by looking for Fourier approximation methods for FIR design or windowing methods, but the idea is easy enough: Take your frequency response. Sample ...


1

Just apply the definitions. $$\begin{align} H(e^{j\omega}) &= \Re\Big\{H(e^{j\omega})\Big\} + j \Im\Big\{H(e^{j\omega})\Big\} \\ \\ &= \Big| H(e^{j\omega}) \Big| e^{j \arg\{H(e^{j\omega})\}} \\ \\ &= \Big| H(e^{j\omega}) \Big| e^{j \phi(\omega)} \\ \end{align}$$ Where $$ \phi(\omega)\triangleq \arg\Big\{ H(e^{j\omega}) \Big\} $$ and $$\...


1

I would suggest the frequency sampling design method, which results in a linear phase FIR filter. You sample the desired magnitude on an equidistant grid, add a linear phase term, and then you compute the impulse response by taking an inverse DFT. The resulting impulse response is usually windowed. I've explained the details in this answer, which also ...


1

So, when you remember how the step response, impulse response and frequency response are related, you'll notice that if you, instead of an actual impulse integral (i.e. a step) use something that is wider, then the frequency response simply gets windowed to lower frequencies. In other words, and to put it as an information-gathering problem: if the ...


1

For finite differences, as suggested by Andy, you can use MATLAB/Octave's freqz, but with some tricks to get the phase frequency response right. I'm using Wikipedia's definitions of higher-order finite differences. In Octave: graphics_toolkit("gnuplot"); n = 2; #finite differences derivative order s = 0; #-1 for forward, 0 for central, 1 for backward b = [1]...


1

Regarding the phase: if you set all frequency components to have zero phase, then the signal becomes essentially a sum of sinusoids that tends to an impulse as the frequency increases. In other words, $$\delta(t) = \int_{-\infty}^\infty \cos(2\pi ft) df.$$ In your case, the maximum frequency is limited, which means that the signal will actually become a ...


1

To your second definition it should be added that you only consider causal transfer functions, because it is not difficult to find a smaller phase lag with a non-causal system: A minimum-phase system is a causal and stable system with a phase lag that is smaller than the phase lag of any other causal and stable system with the same magnitude response. ...


1

Yes you are right. Let's practically implement it with Matlab / Octave. N = 8; h = [1,2,3,4,4,3,2,1]; % just an impulse response (FIR) H = fft(h,N); % DFT H[k] of h[n] as samples of H(w) at w = 2*pi*k/N Hm = abs(H); % extract the magnitude of H[k] Hp = angle(H); % extract the phase angle of H[k] Hr = Hm.*exp(j*Hp); ...


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