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I must do this task:

figure1

Short translate: Picture presents modified sine signal. I must calculate F(k) using Fourier Series, next draw amplitude spectrum and phase spectrum.

I have 2 problems:

  1. Is this proper way to calculate F(k)? (T-period (2pi/w), w-angular frequency)

  2. Can anyone explain how should look amplitude and phase spectrum?

Thanks for any help/

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  • \$\begingroup\$ +1 for a pretty clear question. However you could improve it by using the math notation available here, and also possibly explain your effort in regard to point 1. \$\endgroup\$ – Dzarda Aug 25 '14 at 17:20
  • \$\begingroup\$ What is a bit weird is that the signal appears to have a (smallest) period of \$T/2\$, not \$T\$. \$\endgroup\$ – Matt L. Aug 25 '14 at 17:31
  • \$\begingroup\$ @Dzarda I have problem with integral. First I thought that integral should be from 0 to pi/2w (T/4), but my friend thinks that should be from 0 to pi/4w (T/8). We are trying find somebody who can check our solutions. \$\endgroup\$ – bigben Aug 25 '14 at 17:35
  • \$\begingroup\$ can you show any work you have done? \$\endgroup\$ – cjferes Aug 25 '14 at 17:37
  • \$\begingroup\$ @MattL. I've swapped sine to Euler's formula. \$\endgroup\$ – bigben Aug 25 '14 at 17:38
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This is too long for a comment. I also needed the extra space. Feel free to comment anything.

The thing you are really asking is if the calculation is made with T/8 or T/4 or T/2. The thing is, the signal has a period of T/2, so that's the value you need to use. Now, we also need to take into account the time interval where the signal is 0. To do that, we need to define the signal properly.

The first thing you should do is to describe the signal, that is: $$ f(t)=\left\{\begin{array}{ccc}\Biggl|A\sin\Bigl(2\pi\frac{4}{T}t\Bigr)\Biggr|&&t\in\biggl(0+k\frac{T}{2},\frac{T}{4}+k\frac{T}{2}\biggr),\,k\in\mathbf{Z}\\0&&t\in\biggl(\frac{T}{4}+k\frac{T}{2},\frac{T}{2}+k\frac{T}{2}\biggr),\,k\in\mathbf{Z} \end{array}\right.$$

Note the sinusoid has amplitude A and period T/4, but the second semi-period is positive (so we use the absolute value).

The expression for f(t) is the same as: $$ f(t)=\left\{\begin{array}{ccc}A\sin\Bigl(2\pi\frac{4}{T}t\Bigr)&&t\in\biggl(0+k\frac{T}{2},\frac{T}{8}+k\frac{T}{2}\biggr),\,k\in\mathbf{Z}\\-A\sin\Bigl(2\pi\frac{4}{T}t\Bigr)&&t\in\biggl(\frac{T}{8}+k\frac{T}{2},\frac{T}{4}+k\frac{T}{2}\biggr),\,k\in\mathbf{Z}\\0&&t\in\biggl(\frac{T}{4}+k\frac{T}{2},\frac{T}{2}+k\frac{T}{2}\biggr),\,k\in\mathbf{Z} \end{array}\right.$$ Remember that $$c_n=\frac{1}{P}\int_{t_0}^{P+t_0}f(t)e^{-2\pi j \frac{n}{P}t}\,dt$$

In our case, P is the period (P=T/2), j is the imaginary unit, and t_0 is the initial instant (say t_0=0). Then, $$\begin{array}{rcl} c_n=\frac{2}{T}\int_{0}^{\frac{T}{2}}f(t)e^{-2\pi j \frac{2n}{T}t}\,dt&=&I_1+I_2+I_3 \end{array}$$ where I'll only calculate I_1 to get the idea going: $$\begin{array}{rcl} I_1&=&\frac{2}{T}\int_0^{\frac{T}{8}}A\sin\Bigl(\frac{8\pi}{T}t\Bigr)e^{-2\pi j\frac{2n}{T}t}\,dt\\ &=&\frac{2A}{T}\int_0^{\frac{T}{8}}\frac{e^{j\frac{8\pi}{T}t}-e^{-j\frac{8\pi}{T}t}}{2j}\cdot e^{- j\frac{4\pi n}{T}t}\,dt\\ &=&\frac{A}{jT}\int_0^{\frac{T}{8}}e^{j\frac{8\pi}{T}t}e^{-j\frac{4\pi n}{T}t}\,dt-\frac{A}{2j}\int_0^{\frac{T}{8}}e^{-j\frac{8\pi}{T}t}e^{-j\frac{4\pi n}{T}t}\,dt\\ &=&\frac{A}{jT}\Biggl( \frac{T}{8\pi-4\pi n}e^{j2\pi\frac{2-n}{T}t}\Biggr|_{0}^{\frac{T}{8}}-\frac{T}{-8\pi-4\pi n}e^{-j2\pi\frac{2+n}{T}t}\Biggr|_{0}^{\frac{T}{8}}\Biggr)\\ &=&\frac{A}{jT}\Biggl( \frac{T}{8\pi-4\pi n}(e^{j2\pi\frac{2-n}{T}\frac{T}{8}}-1)+\frac{T}{8\pi+4\pi n}(e^{-j2\pi\frac{2+n}{T}\frac{T}{8}}-1)\Biggr)\\ &=&\frac{A}{j}\Biggl( \frac{1}{8\pi-4\pi n}(e^{j\pi\frac{2-n}{4}}-1)+\frac{1}{8\pi+4\pi n}(e^{-j\pi\frac{2+n}{4}}-1)\Biggr)\\ &&\\ I_2&=&\frac{2}{T}\int_{\frac{T}{8}}^{\frac{T}{4}}-A\sin\Bigl(\frac{8\pi}{T}t\Bigr)e^{-j\frac{4\pi n}{T}t}\,dt\\ &=&...\\ &&\\ I_3&=&\frac{2}{T}\int_{\frac{T}{4}}^{\frac{T}{2}}0e^{2\pi j\frac{2n}{T}t}\,dt\\ &=&0 \end{array}$$

I believe this is all you need to get your calculations correct now... Good luck!

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  • \$\begingroup\$ From 0 to pi/4 this looks awfully like sin^2() function. \$\endgroup\$ – Mike Aug 25 '14 at 21:23
  • \$\begingroup\$ It's not the same, because you're only considering a sinusoid with a different period. Note the period of the sine wave is T/4, but it has an absolute value that makes the behauvior change in the second semi-period. If it were a sin^2, you would need to do a lot of different steps, or use other Fourier transform properties \$\endgroup\$ – cjferes Aug 25 '14 at 21:29
  • \$\begingroup\$ @cjferes Thanks for help. Is it proper way to calculate I1 and next use theory about shifted signal to get I2? \$\endgroup\$ – bigben Aug 26 '14 at 13:12
  • \$\begingroup\$ @bigben I think that's a useful approach, but I haven't tried it. Note that the integrand of I2 is the same as I1 with a minus sign, but the integration limits differ. So I think that things will add up instead of cancel out. Just try to be sure in each calculation step, as these things might mess up if you're not rigorous. Try to do it! \$\endgroup\$ – cjferes Aug 26 '14 at 13:45

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