Phase and amplitude spectrum

Thread Starter

xxxyyyba

Joined Aug 7, 2012
289
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?
 

t_n_k

Joined Mar 6, 2009
5,455
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?
 

Thread Starter

xxxyyyba

Joined Aug 7, 2012
289
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:

t_n_k

Joined Mar 6, 2009
5,455
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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