Derivation of the Fourier series of a real signal

[Granger]

Homework Statement

Consider the Fourier series of a signal given by

$$x(t)=\sum_{k=-\infty}^{\infty} a_ke^{jk\omega_0t}$$

Let's consider an approaches to this series given by the truncated series.

$$x_N(t)=\sum_{k=-N}^{N} a_ke^{jk\omega_0t}$$

a- Show that if $x(t)$ is real then the series can be rewritten as

$$x_N(t)=\sum_{k=-N}^{N} A_k\cos(k\omega_0t +\phi_k)$$

b- How can we calculate $A_k$ and $\phi_k$ knowing $a_k$?

Homework Equations

3. The Attempt at a Solution [/B]

Here goes my attempt:

So I first thought of putting together the k term and the (-k term). I have a doubt about how to get the expression of the series by doing this but I will clear it up after my proof.

We will have then:

$$a_ke^{jk\omega_0t} + a_{-k}e^{-jk\omega_0t}$$

Because x is real then $a_{-k}=a_k*$ where $a_k*$ is the conjugate of $a_k$.

We will have then

$$a_ke^{jk\omega_0t} + a_k*e^{-jk\omega_0t}$$
$$a_ke^{jk\omega_0t} + (a_k e^{jk\omega_0t})*$$
$$2 \operatorname{Re} (a_ke^{jk\omega_0t})$$

If we make $a_k = A_k e^{\phi_k}$ we will have:

$$2 \operatorname{Re} (A_ke^{j(k\omega_0t + \phi_k)})$$
$$2 A_k \cos(k\omega_0t + \phi_k)$$

Now my main doubt here is that factor of 2 and how it will affect the way I will express my calculations.

My first thought was: I reached this expression because I summed up two terms of the series. That means one term will be half what I calculates. Therefore, the series will be:

$$x_N(t)=\sum_{k=-N}^{N} A_k\cos(k\omega_0t +\phi_k)$$

Therefore, to answer question b I will just have to remember that $$a_k = A_k e^{\phi_k}$$ and so I will get to:

$$A_k = |a_k|$$
$$\phi_k = \arg(a_k)$$

But I'm not sure if this last few steps are correct, if my elimination of the 2 factor is correct. Or should I do:

$$x_N(t)=\sum_{k=-N}^{N} 2 A_k\cos(k\omega_0t +\phi_k)$$
$$x_N(t)=\sum_{k=-N}^{N} A'_k\cos(k\omega_0t +\phi_k)$$

and then
$$A'_k = |a_k|/2$$
$$\phi_k = \arg(a_k)$$

Can someone clarify this to me please? Which process is correct and why? Thanks!

 

[scottdave]

It doesn't say that a(k) and a (-k) are necessarily equal as a condition of proving it. Yes if they are equal, then it will be real valued, but is the converse always true? Perhaps you'd be better starting with the end result and working back towards what you start with.

 

[Granger]

It doesn't say that a(k) and a (-k) are necessarily equal as a condition of proving it. Yes if they are equal, then it will be real valued, but is the converse always true? Perhaps you'd be better starting with the end result and working back towards what you start with.

But I never said that they were equal. I said one is equal to the conjugate of the other which is always true if the signal is real

 

[vela]

You might find it easier to first rewrite the series as
$$\sum_{k=-N}^{k=N} a_k e^{j k\omega_0 t} = a_0 + \sum_{k=1}^{k=N} (a_k e^{j k\omega_0 t} + a_{-k} e^{-j k \omega_0 t})$$ to make dealing with the factor of 2 less confusing to you. Also, you can just pull the factor of 2 in right from the start: $2a_k = A_k e^{j \phi_k}$.

 

[Granger]

You might find it easier to first rewrite the series as
$$\sum_{k=-N}^{k=N} a_k e^{j k\omega_0 t} = a_0 + \sum_{k=1}^{k=N} (a_k e^{j k\omega_0 t} + a_{-k} e^{-j k \omega_0 t})$$ to make dealing with the factor of 2 less confusing to you. Also, you can just pull the factor of 2 in right from the start: $2a_k = A_k e^{j \phi_k}$.

Hi! Thanks for your reply!

Ok rewriting the series like that makes things easier.
However I didn't quite understand your last suggestion, can you explain it again please?

What I thought to do was, using the rewritten series, I would get to

$$x(t) = a_0 + \sum_{k=1}^{k=N} 2A_k \cos (j k\omega_0 t + \phi_k)$$

Then, using the symmetry of the series between the negative and positive terms I would get to:

$$x(t) = \sum_{k=-N}^{k=N} A_k \cos (j k\omega_0 t + \phi_k)$$

And therefore I would use $a_k = A_k e^{j \phi_k}$ to calculate the values of $A_k$ and $\phi_k$. Is this correct?

 

[vela]

Oh, I didn't notice the limits on the final summation. Is it really from -N instead of from 0 or 1? Usually when you express it as a cosine series, you just have non-negative values for $k$.

 

[Granger]

Oh, I didn't notice the limits on the final summation. Is it really from -N instead of from 0 or 1? Usually when you express it as a cosine series, you just have non-negative values for $k$.

Oh you're right it is from zero to N! I just realized that now. Now I realize what you meant on incorporating the 2 factor when defining $A_k$. Then I can simply write the series as:

$$x(t) = a_0 + \sum_{k=1}^{k=N} A_k \cos ( k\omega_0 t + \phi_k)$$

$$x(t) = \sum_{k=0}^{k=N} A_k \cos (k\omega_0 t + \phi_k)$$

And therefore I would use $2a_k = A_k e^{j \phi_k}$ to calculate the values of $A_k$ and $\phi_k$.

Is this correct now?

 

[vela]

Well, you don't want $j$ in the argument of the cosine. Also, you should probably think about $k=0$ separately.

 

[Granger]

Well, you don't want $j$ in the argument of the cosine. Also, you should probably think about $k=0$ separately.

Yes sorry that was a typo.
But is it incorrect to put k=0 inside the series?
Because that was what my professor did when he gave us the exercise.

 

[Granger]

Yes sorry that was a typo.
But is it incorrect to put k=0 inside the series?
Because that was what my professor did when he gave us the exercise.

Or do you think I should simply make A_0 = a_0 ?

 

[vela]

Or do you think I should simply make A_0 = a_0 ?

Yes, that's probably simplest. Include the term in the summation but recognize that $A_0$ and $a_0$ aren't related in quite the same way as $A_k$ and $a_k$ are for $k\ne 0$.

 

[Granger]

Ok, thank you very much for your help!