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4

HINT: $$\frac{\cos(n\pi /6)}{(n+3)\pi}=\frac{\cos[(n+3)\pi/6 -\pi/2]}{(n+3)\pi}=\frac{\sin[(n+3)\pi/6]}{(n+3)\pi}$$


3

There's nothing wrong about this – it's absolutely what you should be seeing. The sine wave is intact, if you applied a reconstruction filter – there's no amplitude fluctuation of the wave, as soon as you limit its bandwidth to the nyquist bandwidth, the analog signal "has" to follow the full sines.


3

The MATLAB / OCTAVE function freqz(b,a) is typically used for displaying the magnitude and the phase of a Frequency Response function $H(e^{j\omega})$ associated with a discrete-time, linear time-invariant (LTI), system with rational transfer function $H(z)$ (having a corresponding LCCDE representation) given by $$H(z) = \frac{ \sum\limits_{k=0}^{M} b_k z^{-...


2

First of all, the correct equation for an RRC pulse is given here. You are correct that you need to define the sampling frequency Fs and the symbol interval Ts. The number of samples per symbol interval is then Ts*Fs, which I assume to be an integer. You also need to define beta, and the pulse duration D. For simplicity, let's assume D is given as a multiple ...


2

The integration in your question is equivalent to convolution with the unit step function: $$y(t)=\int_{-\infty}^tx(\tau)d\tau=(x\star u)(t)\tag{1}$$ This means that in the Fourier domain we have $$Y(j\omega)=X(j\omega)U(j\omega)\tag{2}$$ With $$U(j\omega)=\frac{1}{j\omega}+\pi\delta(\omega)\tag{3}$$ Eq. $(2)$ becomes $$Y(j\omega)=\frac{X(j\omega)}{j\...


2

I would not teach integrals this way. Issues might arise form the definition of terms. In standard calculus, a (note: "a", not "the) primitive function (also called antiderivative or indefinite integral) of a continuous function $f$ is a kind of converse of the concept of derivation. SO: If $F$ is "a" differentiable function whose derivative is equal ...


2

Every periodic signal can be described as sum of "basis" functions. These basis functions are the sin and cosines at frequencies of integer multiples of a fundamental freqeuency. This fundamental freqeuency is a consequence of the periodicity of the signal. So if a signal is periodic with time $T$, then the fundamental frqeuency is $F_o = \frac{1}{T}$ Hz. ...


1

No, you can't do it much simpler or faster than that. In the old days, subscripts used to cost time, not sure that is true any more. #include <math.h> #include <stdio.h> //========================================================= int main( int argCount, char *argValues[] ) { double x = 1.2; double C = cos( x ); double ...


1

Using the shifting and scaling properties of the Fourier transform is a rather complicated way of computing the Fourier transform of $\cos(at+b)$. A more straightforward way is to realize that $$\cos(at+b)=\frac12\left[e^{jat}e^{jb}+e^{-jat}e^{-jb}\right]\tag{1}$$ From the Fourier transform pair $$e^{jat}\Longleftrightarrow 2\pi\delta(\omega-a)\tag{2}$$ ...


1

Answer : What you are considering as $\Omega_{N_x}$ is equal to $\frac{\Omega_N}{2}$ according to question. So, what you are saying is same as what answer mentions given we are considering Baseband Samping of $y_a(t)$. I think you are confused because of the terms Nyquist rate and Nyquist Frequency. Nyquist rate and Nyquist frequency are two different ...


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Firstly, your book uses the term Nyquist frqeuency as $\Omega_N + 2\Omega_o$, this is incorrect, this is the Nyquist rate (minimum sampling rate) if we consider the signal to be baseband. The maximum frequency content is then $\Omega_o + \frac{\Omega_N}{2}$, since you have defined maximum frequency of $x(t)$ as $\Omega_{N_x}$ this is nothing but $\frac{\...


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Update: I feel there is no need to bring concept of Bandpass sampling (as mentioned in comments below) because this is specifically regarding a simple misunderstanding that OP has. The text referenced asks about Nyquist Rate (which wrongly mentions it as Nyquist Frequency) but OP asks about highest Frequency in $y(t)$. For the component $\frac{1}{2}X(\Omega-...


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