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I want to find phase and amplitude spectrum of this signal:

enter image description here

Here is what I have done:

$$f(t)=\sum_{-\infty}^{+\infty}Fne^{jnwt}, Fn=\frac{1}{T}\int_{0}^{T}f(t)e^{-jnwt}dt=\frac{1}{T}\int_{t1}^{t1+\tau}Ee^{-jnwt}dt=...=\frac{E}{T}\frac{1}{jnw}e^{-jnwt1}(1-e^{-jnw\tau})=\frac{E}{T}\frac{1}{jnw}e^{-jnwt1}e^{jnw\frac{\tau}{2}}(\frac{e^{jnw\frac{\tau}{2}}-e^{-jnw\frac{\tau}{2}}}{2})*2=\frac{2E}{Tjnw}e^{-jnw(t1+\frac{\tau}{2})}\sin{(nw\frac{\tau}{2})}$$

I stuck there :( What should I do next?

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The amplitude of the spectrum of this signal is related, as all pulsed signal, with the \$sinc(t)\$ function. You must find a trigonometric relationship between the coefficient associated with the sine function

$$ \dfrac{2E}{T\,j\,n\omega} $$

and the argument of the function.

$$ n\,\omega\dfrac{\tau}{2} $$

Check out the link, and see that there is a relationship between the pulse width and frequency of the \$sinc(t)\$ function.

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  • \$\begingroup\$ Here is solution from my book: $$|Fn|=E\frac{\tau}{T}|\frac{\sin{nw\frac{\tau}{2}}}{nw\frac{\tau}{2}}|$$ I'm trying to get that from my expression in first post but without success :( \$\endgroup\$ – etf Sep 30 '14 at 8:34
  • \$\begingroup\$ Look at the Euler's expression of the sine in terms of exponentials. The denominator must be \$2j\$. \$\endgroup\$ – Martin Petrei Sep 30 '14 at 11:05
  • \$\begingroup\$ Thanks!!! I used 2 in denominator instead of 2j. I don't know how I made such a mistake :) \$\endgroup\$ – etf Sep 30 '14 at 13:47

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