P7-3 Consider the periodic function f(t) with period T,...

P7-3 Consider the periodic function $f(t)$ with period $T$, which is depicted below and $\infty$ can be expressed as $f(t) = \sum_{k=-\infty} c_k e^{jk\omega_0 t}$ where $\omega_0 = \frac{2\pi}{T}$ and $c_k = |c_k|e^{j\phi_k}$. Based on Parseval theory, the power of $f(t)$ is defined as $P = \frac{1}{T} \int_T f^2(t)dt = \sum_{k=-\infty}^{\infty} |c_k|^2$. (A) Calculate $P$. $N$ (B) Determine the maximal value of $N$ which satisfies $\sum_{k=-N}^N |c_k|^2 \le 0.98P.$

Transcript

00:03 We're giving x and x of t and y of t both as functions of time.

00:12 So for part a, we need to sketch a graph that indicates x of t and y to 2.

01:04 So it looks something like this.

01:07 It goes through 1, 2, all the way to and y sub t.

01:49 Then we can calculate then we can calculate y sub t given our plots so x times 1 minus 0 0 .5 t is equal to x times negative 0 .5 times t minus 2 again again we have a graph of x times 1 minus 0 .5 t you have a of 1 comma 2 and negative 2 to negative 6.

03:42 And for the second one, you have y times 2t plus 1 is equal to y times 2 t plus 0 .5.

04:21 The y axis would be y times 2t plus 1.

04:31 The x axis is just t and 0 .5 to 1 .5, now, we could calculate i by taking the integral of x times 1 minus 0 .5, t times y times 2t plus 1, s times t minus 1.

05:46 And our limits are from 0 to 4.

06:04 Just simplifying this integral, you have x times 0 .5, y of 3 times t minus 1 times t minus 1d and this interval is equal to 2 so for part b if we say x of t we're given the x of t is equal to 2 minus 1 .8 t plus 3 sine 2 .7 plus cosine 1 .8 plus cosine 1 .8t so this function you can indicate that yes it is it has a periodic signal because y -sub 1 is equal to 1 .8 and if we divide both sides by 10, so y -sub 1 is 1 .8 which is equal to 18 over 10 and the y -sub 2 is 2 .7 over 10 so that's 2 .7 that's 27 i should say and y -sop 3 is equal to 1 .8, which is 18 over 10.

09:15 Now, why not, to calculate why not, you have 8cf times 1 .8, 2 .7, and 1 .8, so the hcf over the lcm, 18, 27, and 18 divided by the lcm.

10:22 So this gives us 9 over 10, which is 0 .9 rating.

10:42 And then c sub d is equal to over w alpha...

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