Understanding Square Wave Amplitude in Fourier Transforms

[alextsipkis]

Friends,

While studying Fourier Transforms, i saw that a square wave with amplitude +A and -A can be replaced with a sqaure wave with amplitude 2A.

Just wondering how it was done? I'm new to Signals and Systems .

thanks.

 

[Integral]

It is just a coordinate system shift. Translate by -A.

 

[envytea]

If you look at a square wave with amplitude of +/- A on a graph, you will see that it covers a range of 2A (A- -A=2A). This wave has a amplitude that fluctuates between these two values. However, an alternate wave with amplitude 2A has the same range (2A), yet it fluctuates between 0 and 2A. The two are identical in their form, they only differ how values are assigned.

 

[alextsipkis]

So that means that an X shift can be done in the same way. For e.g -5 to +5(on the x-axis) can be from 0 to 10 on the positive side.

 

[NoTime]

I think you meant y-axis (the normal up/down designation), but yes.

 

[alextsipkis]

NO, i meant X-axis . I assumed that when the coordinate shift can be done on Y-axis, the same can be done on X-axis.

Also, when x(t) has amplitude of (-A to +A), the book says that we can write it as x1(t) - A, where x1(t) has an amplitude of 2A. Any explanations.

thanks.

 

[NoTime]

Your statement explicitly says x(t) is an amplitude.
So t is the location on the x-axis and x(t) is a sample point value on the y-axis.
Adding a constant to x(t) just shifts the waveform up or down on the y-axis.

If you want to shift the wave on the x-axis then you would write something like x(t-n).
This has no overall effect on the y-axis amplitude excursions.

 

[Redbelly98]

Your statement explicitly says x(t) is an amplitude.

I don't see that anywhere.

 

[NoTime]

I don't see that anywhere.

Do you think the following says something else?

when x(t) has amplitude of (-A to +A),

 

[Redbelly98]

Missed that, sorry.

At any rate, if this is just a math exercise in shifting coordinates, a signal can be shifted horizontally by one's choice of zero time. Looking back at message #4, we can say either the signal goes from -5 sec to +5 sec, or 0 to 10 sec.

This assumes I'm interpreting alextsipkis's statements correctly.

 

[NoTime]

Missed that, sorry.

At any rate, if this is just a math exercise in shifting coordinates, a signal can be shifted horizontally by one's choice of zero time. Looking back at message #4, we can say either the signal goes from -5 sec to +5 sec, or 0 to 10 sec.

This assumes I'm interpreting alextsipkis's statements correctly.

I agree with you about the coordinate shifting part, you can do both.
Seems to me that the OP defined x to be a square wave generator of amplitude (+A, -A).
That gives A the definition of amplitude.
Unless A got redefined somewhere, I don't see how you could get a time shift by adding a voltage value to the square wave.
Did I miss a redefinition somewhere?
If so, I'm not seeing it, but I've been known to be blind sometimes.

 

[Redbelly98]

Seems to me that the OP defined x to be a square wave generator of amplitude (+A, -A).
That gives A the definition of amplitude.
Unless A got redefined somewhere, I don't see how you could get a time shift by adding a voltage value to the square wave.
Did I miss a redefinition somewhere?
If so, I'm not seeing it, but I've been known to be blind sometimes.

No, adding a voltage can't produce a time shift. It's simply a matter of defining where "zero" is. Whether it's voltage or time, 0V or 0 sec is an arbitrary reference.

If I have a graph that covers a 10 sec range, from 0 to 10 sec, you might ask "when does t=0". And I might answer that t is zero at 2:00 p.m. today. But we could just as easily say t is zero at 5 seconds past 2:00 p.m., in which case the graph would run from -5 to +5 sec.

 

[Phrak]

Offsetting the signal on the amplitude axis introduces a 0 Hz component in the frequency domain.

Shifting the signal on the time axis will introduce a phase change of frequency amplidudes. If you are transforming into sines and cosines the relative amplitudes for at any given frequency will vary inversily. (actually as a^2+b^2=constant, but that may be too much detail)

 

[NoTime]

Offsetting the signal on the amplitude axis introduces a 0 Hz component in the frequency domain.

I admit to being a little rusty on this stuff.
It has after all been a good 20 years since I've been paid to do any of this.

Out of curiosity is a 0 Hz component just a fancy way of saying DC offset or is there a deeper implication.

Rereading this thread it looks like I read "same way" in the following statement to be -5v to +5v, but now notice there is no units specification at all.
I don't know if the OP is confused or not.

So that means that an X shift can be done in the same way. For e.g -5 to +5(on the x-axis) can be from 0 to 10 on the positive side.

 

[Redbelly98]

Out of curiosity is a 0 Hz component just a fancy way of saying DC offset

Yes.