3. Parseval's relation a. Consider the signal x1(t) =...

3. Parseval's relation a. Consider the signal $x_1(t) = 4e^{5t}u(-t)$. Use the Parseval's relation to show that the total energy in time domain is the same as computed in frequency domain as follows: $$E_{x(t)} = \lim_{T \to \infty} \int_{-T}^{T} |x(t)|^2 dt = \int_{-\infty}^{\infty} |x(t)|^2 dt = \frac{1}{2\pi} \int_{-\infty}^{\infty} |X(j\omega)|^2 d\omega$$ b. Consider the signal $x_2(t) = 32 \text{sinc}(8t)$ where $\text{sinc}(t) = \frac{\sin(\pi t)}{\pi t}$. Determine the bandwidth B of the signal $x_2(t)$. Determine the energy of the signal $x_2(t)$ using the Parseval's relation in frequency domain.

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00:01 Here we are given an angle modulating signal which is represented by the time domain that is s of t is equal to 10 cos of 2 pi 10 raised to the power 6 of t plus 3 sin of 2 pi 10 raised to the power 3 of t.

00:17 So this is the here.

00:19 So in the first part we have to assume that the given frequency signal is modulated.

00:23 So we have to calculate the frequency deviation modulation index here.

00:26 So here in this first part we are given a fm signal.

00:31 So here general equation for the fm s divided by g is s of t that is equal to ac cos of omega of ct plus k of f integral from minus infinity to plus infinity m of g of dz.

00:49 So we can compare in the term from here.

00:51 So delta of f become equal to kf multiplied by the m of t.

00:54 So delta of f from here is equal to 1 divided by the 2 pi multiplied by the d of i that is divided by d of t.

01:01 So the value of delta of f from here become equal to 1 divided by 2 pi d divided by the dt of 3 sin of 2 of pi multiplied by the 10 raised to the power 3 of t.

01:12 Solving the term from here, we get the value of delta f that become equal to 3 sin of 2 pi multiplied by the 10 raised to the power 3 of t.

01:20 This is in kilohertz.

01:22 So this is the value of the frequency deviation.

01:25 Now we have to calculate the modulating index of the fm that is represented by beta that is equal to delta of f maximum divided by f of m that is equal to q of f multiplied by the a of m that is divided by f of m where we are having the value of delta of f maximum that is equal to 3 kilohertz and that of fm is equal to 1 kilohertz...

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