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?


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.

| improve this answer | |
  • \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.