# A little help understanding how you get from one step to the next

1. Oct 8, 2011

### SMOF

Hello,

I hope I am posting this is in the right section. It is not a home work question, it is from a solution to a tutorial, and I just cannot see how the lecturer has gotten from one step to the next. If I don't get the solution here, I will ask him, but that will be later next week, and I would like to understand it now.

Anyway. It is a question relating to a digital signal processing class, and we were asked to show analytically that

w(t) = r cos(100t+∅).

In the question, we are told that x(t) = cos(100t) and y(t) = cos(100t + $\pi$) where w(t) = x(t) + y(t).

The part I am having an issue with is from this line

2w(t) = ej100t + e-j100t + ej100t + pi/3 + e -j100t - pi/3

To this line

2w(t) = ej100t(ejpi/3+1) + e-j100t(e-jpi/3+1)

I am sure it is a pretty basic step, but I just cannot see it, and I have no issues with other harder sections, but this is just a mental block for me.

Seán

2. Oct 8, 2011

### mathman

In your next to the last line you left out j multiplying pi/3 in both expressions. The exponent should be j(110t + pi/3), although I am puzzled where the /3 came from, since the original y doesn't have it.

3. Oct 8, 2011

### SMOF

Oh, thats me been silly, I was too worried about getting all the main equations right with the html tags and what have you ...it is y(x) = cos(100t +pi/3).

Sorry.

If by the second to last line you mean 2w(t) = ej100t + e-j100t + ej100t + pi/3 + e -j100t - pi/3 ...that's how it is in the solution.

Seán

4. Oct 8, 2011

### D H

Staff Emeritus
On the first line, rewrite ej100t as ej100t*1, ej100t + pi/3 as ej100t*epi/3. Now collect the terms of ej100t, yielding ej100t + ej100t + pi/3 = ej100t(1+epi/3). Do the same for the other two terms and you get the second line.

There is an easier way to address this if you remember your half angle formulae and some identities based on these. One such identity is

$$\cos a + \cos(a+b) = 2\cos\frac b 2 \cos(a+\frac b 2)$$

It is a good one to keep this identity in your head, at least in structure. You don't need to know the identity by heart, but it is good to know the general nature of the identity as it does come up quite a bit.

5. Oct 9, 2011

### SMOF

That's great, thanks a lot!

Seán

6. Oct 9, 2011

### mathman

It looks to me that a parenthesis pair is missing from the exponent in each term. It is corrected in the last line.