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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 $X(t)$ ...

3

In short: $(0.937)^2=0.878$ A bit more detail: \begin{align}(z-re^{j\theta})(z-re^{-j\theta})&=z^2-r\left(e^{j\theta}+e^{-j\theta}\right)z+r^2\\&=z^2-2r\cos(\theta)z+r^2\end{align}

2

Unless your audio data consists of pure sinusoidal note sources, a raw FFT magnitude plot will give you all the harmonics or overtones of pitched musical sounds, as well as any fundamental pitch frequency spectrum (if any). So, with most common polyphonic music recordings, pitch detection/estimation, if possible, usually involves a significant amount of ...

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

Since the bandwidth stays constant at 120 kHz, the symbol rate needs to be reduced: $$R_s = W_m/(1+\alpha) = 120000/2 = 60000 \text{ symbols per second.}$$ Then the bit rate is $R_b = 6R_s=360,000 \text{ bits per second.}$ The transmission duration is then $$T = \frac{48 \times 10^6 \text{ b}}{360 \times 10^3 \text{ b/s}} = 133.3 \text {s}.$$

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 $y(1-4y)=x$, 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.

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 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 ~~~,~~~ y = x \\ y &= 0 ~~~,~~~ y = x \\ y &= 0 ~~~,~~~ y = x \\ y &= 0 ~~~,~~~ y = x \\ \... 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^*}) $$which implies that for every zero z_0 of H(z) there will be three more zeros at z_0^*, 1/z_0, and 1/z_0^*; at conjugate, reciprocal, and conjugate-reciprocal locations respectively. Note that the reciprocal ... 1 don't need the solution. A hint would be good to start. This is bog-standard cyclic-prefix OFDM, and that's really well-covered in literature. 1 HINT: Just express y[n] and x[n] in terms of v[n-k], k=0,1,\ldots, and show the equality. 1 HINTS: \cos(2)-\cos(2)=0 What remains is y(t)=\cos(t-1)-\cos(t). Can this really be the response of a causal system to an input signal that starts at t=0? Maybe you forgot some important detail in your answer. The correct answer must consist of 3 different expressions for the time intervals t<2, 2<t<3, and t>3. Your answer ... 1 Remember that there are four types of linear-phase FIR filters. The filter in your example is a type III filter: odd filter length and odd symmetry. The frequency response of such a system has the form$$H(e^{j\omega})=A(\omega)je^{-j\omega(N-1)/2}\tag{1}$$where N is the filter length (number of taps), and A(\omega) is real-valued and odd, i.e., A(\... 1 You did make small mistakes, the inverse z transfom for\frac 1{1+az^{-1}} is {(-a)}^nu(n) and in partial fractions you have fallow the rule that numerator degree should be less than the denomination degree. so your question becomes \frac {z^{-2}} {1-0.5z^{-2}}=-2+\frac 2 {1-0.5z^{-2}} = -2+\frac 1 {1-\frac1 {√2}z^{-1}}+\frac 1 {1+\frac1 {√2}z^{-1}}.so ... 1 HINT: Write H(z) as$$H(z)=G(z^2)\tag{1}$$with$$G(z)=\frac{z^{-1}}{1-0.5z^{-1}}\tag{2}$$Determine the inverse transform g[n] of G(z). Figure out what replacing z by z^2 means in the time domain. From this, the inverse transform of H(z) is easily obtained from g[n]. EDIT: Your partial fraction expansion lacks the factor z^{-2}, and the ... 1 The poles in the equation are Z=1.5,0.8;hence to be two sided the roc is 0.8<=z<=1.5 so the 0.8 term is right sided and 1.5 term is left sided let us do partial fractions now  X(Z)=1+\frac{1.5-0.85z^{-1}} {(1-0.8z^{-1})(1-1.5z^{-1})}  after doing some maths i got x(n)=\delta(n)-0.5(0.8)^nu(n)-2(1.5)^nu(-n-1) so x(-2)=-0.88888,x(-1)=-1.3333,x(0)=0.... 1 From your physics knowledge you get that sine waves are usually equated, in the general case, as$$y(t) = A * \sin(\omega t)$$where \omega = 2 \pi f and the periodicity is T=1/f Comparing the general equation with your particular case, you have:$$2*\pi*f = \pi / 4$$meaning that$$f=1/8 Hz$$or$$T=1/f=8s The amplitude is $A=3$ If you want to ...

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Assuming you have this board, there are 512kByte of SRAM on the board. That should easily accommodate a 7k delay buffer.

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As Robert B.J. has already indicated, a bare bones FFT analysis is not the recommended method for a professional audio harmonic inspection. Nevertheless it can be very useful in certain cases, one of which is, I think, this one. Be aslo warned that, as hotpaw2 indicated, with this simplistic approach, false positives might be detected. From your provided ...

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x(n)*[d(n)-d(n-1)]=x(n)*d(n)-x(n)*d(n-1) = x(n)-x(n-1)

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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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