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In order for H(z) to be a linear phase filter, it must zeros both on the inside of the unit circle and at the complementary locations (1/z) which are outside the unit circle. Therefore a linear phase circle has no stable causal inverse (since this would necessitate poles outside of the unit circle.) Note that a linear phase filter can be decomposed into ...


3

A continuous-time white noise process $\{X(t)\colon -\infty < t < \infty\}$ is a hypothetical construct that we can treat (in the simplified versions that we use on dsp.SE) as a zero-mean wide-sense stationary process with autocorrelation function $K\delta(\tau)$ where $\delta(\cdot)$ is the Dirac delta. More strongly, all the random variables $X(t)$ ...


2

That's actually a fairly a tricky one. Two ways you can go about it Method 1 Magnitude is that of a ideal lowpass filter. Inverse of that is sinc function Phase is that of an hilbert transformer (times -1). Impulse response is (roughly) $1/\pi t$ convolve the two Method 2 Write out the equation for inverse Fourier transform Pop in values from the graph ...


2

Working with your definitions: $$ v \left( \left( n + 1 \right) {T}_{s} \right) - v \left( n {T}_{s} \right) = \int_{0}^{ \left( n + 1 \right) {T}_{s} } g(u) du - \int_{0}^{ n {T}_{s} } g(u) du = \int_{ n {T}_{s} }^{ \left( n + 1 \right) {T}_{s} } g(u) du $$ So basically we have integration (Which is a Low Pass Filter) of White Noise over a Time Interval ...


2

Your equation does not really define a system, as we don't know what the input/output pair is (is $x$ the input, or $y$?), and where the values of $x$ or $y$ dwell. It is not even well-defined: at $n=0$, we have the equation $y[0](1-4y[0])=x[0]$, which may have two, one or zero solutions. However, let us suppose that $x$ is the input, and $y$ the output. ...


2

Using the definition of linear system as one which can be put in the form: $y_0 + a_1 y_1 + \dots + a_n y_n = b_0 x_0 + \dots + b_m x_m$ because of the multiplicative term $y(n)y(2n)$ we can conclude that the system is nonlinear.


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Inverting a channel can only be done when the channel is a minimum phase system (trailing echos only). A minimum phase system is characterized as having all zeros in the left half plane (for the s plane, or equivalently in a sampled system and the z plane all zeros inside the unit circle). Inverting such a channel results in poles where every zero exists, ...


2

Your understanding is right. With the usual Fourier transform pair notation : $$x(t) \longleftrightarrow X(\omega)$$ and assuming bandwidth of $x(t)$ is $W$, then the four cases will be 1-) $x^2(t) \longleftrightarrow X(\omega) \star X(\omega) \implies Bandwidth = 2W$ 2-) $x(2t) \longleftrightarrow \frac{1}{2}X(\omega/2) ~~~~~ ~\implies Bandwidth = 2W$ ...


1

I would follow a signal block diagram based solution for this problem. First as suggested in the comments, it's very helpful to investigate a few values of $y[n]$ and $x[n]$ for some $n$ : $$ \begin{align} y[0] &= 0 ~~~,~~~ y[1] = x[1] \\ y[2] &= 0 ~~~,~~~ y[3] = x[2] \\ y[4] &= 0 ~~~,~~~ y[5] = x[3] \\ y[6] &= 0 ~~~,~~~ y[7] = x[4] \\ \...


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don't need the solution. A hint would be good to start. This is bog-standard cyclic-prefix OFDM, and that's really well-covered in literature.


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(iii) seems be to correct. Spectrum of two frequencies. The other choices can be verified, (i) Fourier transform of $f(t=0)=1$ is 1. So it cannot be graph (iii) (ii) Signal (iii) is sine wave and there is no spectrum associated with that. (iv) It's Fourier transform of sine wave (iii) which is $$f(t)=0.5\cdot \sin(2\pi Ft)$$ and it's transform is $$\...


1

If you are referring to a buffer as in a logic gate, this question may be more appropriate on the electronics stack exchange site but here are some suggestions. If it is a logic gate, you must be referring to a fixed frequency clock, and shifting the phase for this is quite trivial. Here are some approaches to doing this: For 180° simply invert with a not ...


1

Carry the units and it becomes understandable. T = 0.03125 seconds per sample Thus your sampling rate is 1/T = 32 samples per second Each bin corresponds to the cycles per frame. Your frame has four samples. Suppose the bin index is k. k (cycles per frame) * 32 (samples per second) / 4 (samples per frame) = 8k cycles per second = 8k Hz So the ...


1

There is no real shortcut for computing the impulse response. Partial fractions is the standard way to do it. However, in the case of a second-order transfer function as in your example, the result can be found in tables of $\mathcal{Z}$-transform pairs. E.g., if you use the last two correspondence of this table, you can pretty quickly write down the result. ...


1

HINT: The (zero-state) step response is just the cumulative sum of the impulse response $h[n]$: $$y_{ZS}[n]=u[n]\sum_{k=0}^nh[k]$$ This follows in a straightforward manner from the convolution of a unit step $u[n]$ with the impulse response: $$y_{ZS}[n]=(h\star u)[n]$$


1

You're overcomplicating things here. There's no need for sines and cosines and squares. Note that $\omega=0$ corresponds to $z=1$, and $\omega=\pi$ corresponds to $z=-1$. From the definition of the $\mathcal{Z}$-transform you should be able to figure out that the DC term of the transfer function $H(z)$ is given by $$H(1)=h_1+h_2+h_3\tag{1}$$ and the value ...


1

A real coefficient, minimum phase, FIR filter will have the following property : $$ H(z) = H^*(z^*) = H(\frac{1}{z}) = H^*(\frac{1}{z^*}) $$ which implies that for every zero $z_0$ of $H(z)$ there will be three more zeros at $z_0^*$, $1/z_0$, and $1/z_0^*$; at conjugate, reciprocal, and conjugate-reciprocal locations respectively. Note that the reciprocal ...


1

HINTS: $\cos(2)-\cos(2)=0$ What remains is $y(t)=\cos(t-1)-\cos(t)$. Can this really be the response of a causal system to an input signal that starts at $t=0$? Maybe you forgot some important detail in your answer. The correct answer must consist of $3$ different expressions for the time intervals $t<2$, $2<t<3$, and $t>3$. Your answer ...


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Remember that there are four types of linear-phase FIR filters. The filter in your example is a type III filter: odd filter length and odd symmetry. The frequency response of such a system has the form $$H(e^{j\omega})=A(\omega)je^{-j\omega(N-1)/2}\tag{1}$$ where $N$ is the filter length (number of taps), and $A(\omega)$ is real-valued and odd, i.e., $A(\...


1

HINT: Just express $y[n]$ and $x[n]$ in terms of $v[n-k]$, $k=0,1,\ldots$, and show the equality.


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If you check graph of h(t) which is underdamped oscillations presrnt in 1st and 4th quadrant only. And integration of this h(t) from limits (-infinity to + infinity ) Is finite value. therefore, system is stable


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