Phase and amplitude spectrum

Discussion in 'Homework Help' started by xxxyyyba, Sep 29, 2014.

  1. xxxyyyba

    Thread Starter Member

    Aug 7, 2012
    249
    2
    Hi!
    I want to find phase and amplitude spectrum of this signal:

    postavka1.jpg

    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?
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Is the signal actually periodic in T? Does the picture indicate a single cycle of a periodic pulse train or do you require the Fourier transform of just the single pulse?
     
  3. xxxyyyba

    Thread Starter Member

    Aug 7, 2012
    249
    2
    It's periodic...
    postavka1.jpg

    Here is amplitude spectrum:
    |Fn|=E\frac{\tau}{T}|\frac{\sin{nw\frac{\tau}{2}}}{nw\frac{\tau}{2}}|
    I have no idea how to derive it from my expression in first post :(
     
    Last edited: Sep 30, 2014
  4. xxxyyyba

    Thread Starter Member

    Aug 7, 2012
    249
    2
    Forgot to write that tau=T/3.5
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    If f(t) is given as you have it

    f(t)=\frac{2E}{jn \omega T}\sin(n \omega \frac{\tau}{2})e^{-jn \omega \( t_1+\frac{\tau}{2}\)}

    To obtain the magnitude we can recall the magnitude of the exponential part is always unity - it simply imposes a phase shift of \text{-n\omega (t_1 +\frac{\tau}{2}) } radians. Also recall the complex operator in the function denominator also has no effect on the magnitude - it simply imposes a negative 90 degree value to the overall phase shift.

    So the magnitude then boils down to

    \|f(t) \|= \| \frac{2E}{n \omega T}sin(n \omega \frac{\tau}{2}) \|

    If we multiply both numerator and denominator of the fractional part by
    \text{\frac{\tau}{2}}

    We obtain ...

    \|f(t) \|= \| \frac{2E\frac{\tau}{2}}{n \omega T \frac{\tau}{2}}sin(n \omega \frac{\tau}{2}) \|

    Which can be re-arranged to the form you posted earlier.
     
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